xtraedge may 2010

XtraEdge for IIT-JEE 1 MAY 2010

Dear Students, Find a mentor who can be your role model and your friend ! A mentor is someone you admire and under whom you can study. Throughout history, the mentor-protege relationship has proven quite fruituful. Socrates was one of the early mentors. Plato and Aristotle studied under him and later emerged as great philosophers in their own right. Some basic rules to know mentors : • The best mentors are successful people in their own field. Their behaviors

are directly translatable to your life and will have more meaning to you. • Be suspicious of any mentors who seek to make you dependent on them.

It is better to have them teach you how to fish than to have them catch the fish for you. That way, you will remain in control.

• Turn your mentors into role models by examining their positive traits. Write down their virtues. without identifying to whom they belong. When you are with these mentors, look for even more behavior that reflect their success. Use these virtues as guidelines for achieving excellence in your field.

Be cautious while searching for a mentor : • Select people to be your mentors who have the highest ethical standards

and a genuine willingness to help others. • Choose mentors who have and will share superb personal development

habits with you and will encourage you to follow suit. • Incorporate activities into your mentor relationship that will enable your

mentor to introduce you to people of influence or helpfulness. • Insist that your mentor be diligent about monitoring your progress with

accountability functions. • Encourage your mentor to make you an independent, competent, fully

functioning, productive individual. (In other words, give them full permission to be brutally honest about what you need to change.)

Getting benefited from a role-mode : Acquiring good habits from others will accelerate you towards achieving your goals. Ask yourself these questions to get the most out of your role model/mentors : • What would they do in my situation? • What do they do every day to encourage growth and to move closer to a

goal ? • How do they think in general ? in specific situations ? • Do they have other facts of life in balance ? What effect does that have on

their well-being ? • How do their traits apply to me ? • Which traits are worth working on first ? Later ? A final word : Under the right circumstances mentors make excellent role models. The one-to-one setting is highly conducive to learning as well as to friendship. But the same cautions hold true here as for any role model. It is better to adapt their philosophies to your life than to adopt them. Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Editor : Pramod Maheshwari

No success is possible unless you believe that you can succeed.

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Volume-5 Issue-11 May, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Key Concepts & Problem Solving strategy for IIT-JEE.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics, Chemistry & Maths

Much more IIT-JEE News.

Xtra Edge Test Series for JEE – 2011 & 2012

AIEEE-2010 Examination Paper

S

Success Tips for the Months

• "All of us are born for a reason, but all of us don't discover why. Success in life has nothing to do with what you gain in life or accomplish for yourself. It's what you do for others."

• "Don't confuse fame with success. Madonna is one; Helen Keller is the other."

• "Success is not the result of spontaneous combustion. You must first set yourself on fire."

• "Success does not consist in never making mistakes but in never making the same one a second time."

• "A strong, positive self-image is the best possible preparation for success."

• "Failure is success if we learn from it."

• "The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself."

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 IIT-Develops technology to produce stealth aircraft Urine-processing technologies yield rich cash flow potential

IITian ON THE PATH OF SUCCESS 8 Mr. Sujal Patel

KNOW IIT-JEE 10 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 49

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper

IIT - 2010 Examination Paper with Solution 66

Regulars ..........

DYNAMIC PHYSICS 17

8-Challenging Problems [Set# 1] Students’ Forum Physics Fundamentals Electrostatics-I 1- D Motion, Projectile Motion

CATALYSE CHEMISTRY 33

Key Concept Gaseous State & Real Gases General organic Chemistry Understanding : Physical Chemistry

DICEY MATHS 41

Mathematical Challenges Students’ Forum Key Concept Complex Number Matrices & Determinants

Study Time........

Test Time ..........


IIT develops technology

to produce stealth

aircraft

Materials scientists at the Indian

Institute of Technology in

Roorkee (IIT-R) have developed

microwave absorbing nanocom-

posite coatings that could make

aircraft almost invisible to radar.

The technology for building

invisible, or stealth aircraft, is a

closely guarded secret of

developed countries and a handful

of laboratories in India are doing

research in this area.

Radars that emit pulses of

microwave radiation identify flying

aircraft by detecting the radiation

reflected by the aircraft’s metallic

body. The nanocomposite coatings

developed by Rahul Sharma, R.C.

Agarwala and Vijaya Agarwala at

IIT-R absorb most of the incident

radiation and reflect very little.

Sharma, who revealed his team’s

work at an international

nanomaterials conference held

recently at the Indian Institute of

Science in Bangalore, believes

their nano-product is a significant

step in developing a technology to

enable aircraft escape radar

surveillance and protect its

equipment from electronic

“jamming”.Nanoparticles - so

called because of their very small

size - are known to exhibit unique

physical and chemical properties.

The IIT team found that crystals of

“barium hexaferrite” with particle

size of 10-15 nanometres have the

ability to absorb microwaves.

(Human hair, for comparison, is

100,000 nanometres thick). They

developed special processes for

synthesizing the nanopowder and

formulating it as a coating.

Sharma said that the

nanocomposite coating on the

aluminium sheet absorbed 89

percent of incident microwaves at

15 giga hertz - the frequency

normally used by radars —

reflecting only 11 percent. A

stealth aircraft should ideally

absorb all the incident radiation

and reflect nothing.

Urine-processing

technologies yield rich

cash flow potential

The stink is out of urine, literally

and metaphorically, with a growing

number of researchers spotting

commercial and ecological value in

a liquid most people consider

waste.

The Indian Institute of Technology

(IIT) Delhi, for instance, is working

to harvest this human waste and

convert it into fertiliser. The Delhi

government is willing to consider

a revenue-share commercial

venture selling the phosphates and

nitrates in urine.

On the outskirts of Delhi, a little-

known non-government organisa-

tion, Fountain for Development

Research and Action, is laying the

ground for the first urine bank. It

has diverted urine from two

schools, where it has installed

odour-free urinals, into a tank and

transferred the run-off to a village

nearby for use as fertiliser.

Director Madhab Nayak says the

foundation is working towards

making farmers aware of its

potential as replacement for

expensive urea.

"There is no such thing as waste,"

says Vijayaraghavan M Chariar,

assistant professor at the Centre

for Rural Development and

Technology at IIT. "Urine consists

of a lot of inorganic salts, which

produce gases only when mixed

with water. It is, in fact, pure

fertiliser," he added.

IIT has come up with a cheap,

odour-free, urinal which it has

successfully tested on campus. The

odour-free urinal combines

technology with simple science to


translate into a significant water-

saving initiative (urine smells only

when mixed with water, which

this technology eliminates).

Urine is collected through a tank

placed underground, harvested

and used as liquid fertiliser two to

three metres below the ground on

a five-acre field on campus, said

Chariar, who can talk animatedly

about this human waste and how

its poor treatment alone has led

to sanitation problems.

The public urinal at IIT uses a

simple technology, called Zerodor,

developed by Chariar, that fits

into the waste coupler in the pan

and diverts the urine through a

drain where it is collected and

harvested. The idea is not to

allow it to mix with water at any

stage.

Chariar has already transferred

this technology to Good Yield

Environmental Technologies, a

Kolkata firm, and filed for a patent.

Chariar claims that Zerodor is a

low cost product and would need

replacement in only about two

years.

Meanwhile, the Delhi government,

which has already installed 200

such odourless urinals in different

parts of the city, uses a different

and perhaps more expensive

technology. Amiya Chandra,

deputy commissioner of the city’s

municipal corporation, says,

"Other than problems of

vandalism, these urinals are

working perfectly."

In preparation for the

Commonwealth Games, the Delhi

government is planning to install

1,000 such urinals at a nominal

cost of Rs 3 lakh.

Chariar is already working on the

second phase of his project, which

was initiated by Unicef and

Stockholm Environment Institute,

for setting up a small reactor to

extract nitrates and phosphates

from urine. "This could become a

micro-enterprise from the urinal,"

says Chariar.

The Delhi government is also

looking at installing Chariar’s

technology at a few parks in the

city, while harvesting urine in

those places.

Chariar has even designed similar

urinals for women. "We have filed

for trademark registration and we

are in discussion with companies

for marketing it," he says. With a

little more investment, he says, a

hydrophobic coating on pans

could make it water resistant and

completely drain the urine, leaving

no room for any oxidisation,

which can also cause odour.

In the developed world,

communities have been quick to

realise the huge economic

potential of urine. "Communities

in Germany are exporting urine to

neighbouring countriesthat are

using it on their farms, says

Chariar, explaining how it could

be diverted for use as a nutrient

by a simple plumbing.

The urine tank could deliver the

liquid nutrient directly to plants

about two to three metres below

the soil, he says.

The Centre for Banana Research

in Trichy is already using it for

banana plantations and the

University of agriculture Sciences,

Bangalore, too is looking at its

varied uses.

IIT student produces

electricity from waste

water

Kolkata: Waste water

management is a big issue world

wide and specially in India where

there is acute shortage of the

precious resource in many places

but a 23-year-old student of IIT

Kharagpur claims he has found a

solution.

Apart from finding solutions

management of waste water he

has also demonstrated producing

electricity from it, which could go

a long way in protecting the

earth's resources.

A look at reservoirs used

for water supply

Manoj Mandelia, who is pursuing

integrated MTech at IIT

Kharagpur, there was no policy in

the country which examined


waste as part of a cycle of

production-consumption-ecovery.

"Waste management still

constituted a linear system of

collection and disposal which

creates health and environmental

hazards," he said.

"I developed a product which uses

the concept of microbial fuel cell

(MFC is a bio-electrochemical

system that drives a current by

mimicking bacterial interactions

found in nature), which could not

only treat waste water but also

produce electricity in the

process," explains Mandelia who

heads a team of five people in the

project.

The Water Diviner

The project, named LOCUS which

stands for Localised Operation of

Bio-cells Using Sewage, can

achieve chemical oxygen demand

(COD) reduction levels in waste

water to about 60-80 per cent.

IIT Kharagpur Calls for

Nominations for the

Nina Saxena Excellence

in Technology Award IIT Kharagpur announces the

fourth edition of Nina Saxena

Excellence in Technology Award

to invite entries of this year in

areas of technical innovations.

Nominations for entries to the

award are open until April 30,

2010. Entries can be submitted

either by post to the Director, IIT

Kharagpur, West Bengal,PIN -

721302, India or by email at

[email protected]. The

nomination form is also available

on the official award website

The objective behind the award is

to commemorate the spirit and

drive of Dr. Nina Saxena, who

personified technical excellence.

The fourth edition will continue

the tradition of rewarding

pioneering innovations for

betterment of society.

A distinguished committee is being

formed by IIT Kharagpur to

adjudge the nominations for this

award. The award committee is

chaired by Director of IIT

Kharagpur and is comprised of

Deans and selected faculty

members of IIT Kharagpur and

well known alumnus, based in

India and in the US.

IIT’s New On-Campus

Wind Turbine to

Support Green Jobs,

Research, and Education

The consortium members will

research the wind energy

challenges identified in the U.S.

Department of Energy's "20%

Wind Energy by 2030" report,

including wind technology, grid

system integration, and workforce

challenges. The consortium's plan

relies on IIT experts in electrical

and computer engineering;

mechanical, materials, and

aerospace engineering;

architecture; business; and

members of the Wanger Institute

for Sustainable Energy Research to

tackle these challenges.

Many of the university's

departments and research centers

will also work together to offer

wind energy courses addressing

the technical, operational, social,

and environmental aspects of wind

energy in consultation with

industry. To ensure student

involvement in the project,

fellowships will be offered annually

to undergraduate and graduate

students in wind energy

engineering fields of study. Faculty

and students from international

university consortium members

will also be invited to IIT to attend

workshops and to share ideas with

their American counterparts.

The wind energy consortium will

work with small wind turbine

manufacturer Viryd Technologies

to procure and install an 8KW

Viryd wind turbine on IIT's Main

Campus, and to deliver a second

turbine to one of IIT's engineering

laboratories to perform turbine

reliability studies. The consortium

will also work with wind energy

developer Invenergy to install a

1.5MW GE wind turbine adjacent

to a wind farm in Marseilles, Ill.

The close proximity of IIT's


Marseilles turbine to an existing

wind farm provides an ideal

opportunity to study turbine-to-

turbine wake interaction, wind

farm interaction, and wind energy

efficiencies in addition to turbine

reliability studies.

Hyderabad boy tops in

GATE 2010

HYDERABAD : Malladi

Harikrishna, a final year computer

science engineering student from

the city, has topped the national

level Graduate Aptitude Test in

Engineering (GATE-2010). He

achieved the feat in his first

attempt, scoring 99.99 percentile

by scoring 83.55 per cent in

GATE.

The results were announced on

March 15 by IIT-Guwahati, which

conducted

GATE this

year. Mallad,

who topped

the national-

level

Graduate Aptitude Test in

Engineering (GATE-2010), said,

“My aim is to join ME at the Indian

Institute of Science, Bengaluru. I

want to become a scientist,”

Harikrishna, 21, said.

Students who clear GATE are

eligible for admission to masters’

degree courses in engineering,

technology, architecture,

pharmacy, science in premier

institutes like IITs and NITs.

Many students from the state

made it to the top-100.

Srujana (JNTU, Kukatpally) ranked

22, Karthik Nagarjuna 44,

Mufaquam Ali 51 and Pavan

Kishore got the 102nd rank in

ECE stream.

Srinivas Reddy got the 68th rank

in EEE, and V. Suryanarayana 31 in

Mechanical.

GATE 2010 score is valid for two

years from the date of

announcement of the results,

according to the details published

on GATE website

Terminated IIT students

seek intervention of

Prez,PM

Terminated for "bad

performance", 38 students of IIT

Kanpur have taken the issue to

President Pratibha Patil and Prime

Minister Manmohan Singh with a

plea that the institute reconsider

the decision.

The students have written to

Prime Minister and the PMO has

forwarded the matter to the HRD

Ministry for "appropriate action".

The Ministry has again forwarded

it to the institute for action, a

ministry official said.

The 38 students, including 24 from

under-graduate and 14 post-

graduate levels, were denied

admission into fresh semester in

January this year for "bad

academic record"

.Prof V N Pal, who is an alumni of

the institute, has taken the

students' issue to President Patil,

who is the Visitor of the institute.

Prof Pal met the President on

April one at Rashtrapati Bhawan

and discussed the issue of

termination of students of IIT

Kanpur at length.

He explained the socio-economic

condition of these students.

Vidya Balan to address

IIT

She has been invited by one of the

Indian Institutes of

Technology to be a keynote

speaker at an upcoming seminar.

In an interview to a leading daily,

she confirmed her invitation from

one of the IITs. She said, “I have

been approached by one of the

IITs for being a keynote speaker at

a seminar they are holding. The

topic is ‘The changing face of the

Indian heroine’. I am excited about

it, but am figuring out dates and

prior commitments at the

moment. But, I hope this comes

through.”

A MA in sociology, Vidya Balan

believes her education has

groomed her and given her the

confidence to address seminars at

these prestigious universities.


To his competitors, Sujal Patel is now a name to reckon

with. His company Isilon Systems, in the clustered storage

space, has not only earned its position as the fastest

growing technology company in North America, but in

the five years of selling its products, Patel has transformed

the company from zero sales, into a company with a $100

million run rate, $80 million in cash, and no debt.

“Slightly better is not a good term,” says Patel, the CEO

of Isilon Systems and a pioneer in the clustered storage

space. “For a technology to be adopted in an existing

market, you really need to have a technology that is

substantially better — 10 (times) better than what is in

the marketplace today,” adds Patel. “It has to be so much

better that it is overwhelming for people who buy that

product and service.” When Patel says so, he is not being

merely theoretical. The unsatiated desire to bring out the

best led this 35 year old entrepreneur to steer his

company successfully, in a market space, which was

crowded by biggies like EMC and NetApp; not to mention

the 250 odd startups in the storage space.

Patel founded Isilon at the age of 26. Prior to this, Patel

spent his initial career days at Real Networks. As an

engineer, Patel used to solve some of Real Networks’

most complex back-end operational challenges. That

experience gave him the insight for a new type of solution,

a type of virtualized storage optimized for media. His

experience gave him the insight to a real customer need,

and his deep technical knowledge gave him the ability to

spot a solution.

I did not want to wait 10-15 years, treading cautiously at

every step before taking the plunge. So when I got the

chance to found Isilon, I jumped at the opportunity,”

beams Patel.

That’s also a reason which led venture capitalists to take

this 26 year old lad seriously. A few months after Patel

founded Isilon in 2001, the NASDAQ came down

crashing, bringing the ‘Dot Com Burst’. The venture

capital market was in disarray. With the existing

companies dropping their revenues, there was not much

hope for new companies to find potential investment. To

make matters worse, there were about 250 other

startups in the storage space. Patel was undeterred. Sure

about his ideas, he approached close to 40 venture

capitalists, and with perseverance, eventually he managed

to gain their confidence. Five months after the company’s

beginning, Patel had managed to raise $8.4 million venture

MR. SUJAL PATEL

Success StoryThis article contains story of a person who get succeed after graduation from different IIT's


capital in what was one of the toughest markets to raise

money.

But all was not rosy yet. The road ahead proved to be

more challenging than raising the funds. The next three

years were spent in building the products. Developing the

company’s IP not only took a tremendous amount of

money, but also ate in to the time to get into the market.

By the time the product was ready to debut in

the markets in 2003, the company’s debt was near

$20 million.

Cruising over Obstacles

Not only confident and unshakable, Patel is also a man of

clear vision, with no illusions about his capabilities. At 26,

as the Founder of the company, he served as the CEO for

the initial three years. But he knew that, if he wanted to

make Isilon the next big thing in storage space, he needed

someone who knew the dynamics of running a large

organization. He promptly stepped down and appointed

as a new CEO, who knew what it takes to grow into the

larger league. “I floated the company but knew my

limitations in the business front. We had three products

ready to debut in the market and if we wanted then to

succeed, we needed an expert who knew the right strings

to pull,” says Patel. His focus and timely decisions were

fruitful in the subsequent years when Isilon was on a

dream run, literally growing at 200 percent year-on-year.

By the time the company went public, it was a $60 million

company.

“My goal was to see Isilon become a $100 million

company by 2007 and become a player to be reckoned in

the $4 billion global storage market for its technology.

And we were still $40 million short. I knew that despite

the economic instability, the company I had founded had

great potential if one could maximize it,” says Patel.

To start with, he re-structured the entire organization,

replacing every executive in the management, including

CTO, CFO, Head of engineering, Head of operations and

others. Such an action is quite unheard of, especially

immediately after a company had gone public.

Next, Patel began revamping the business strategy by

reaching out to the broader enterprise segments. It was

not easy. The segment he wanted to target comprised of

Fortune 50 companies who did not have much

expectation from a startup like Isilon. With the

organizational re-structuring, he completely overhauled

the company’s services and products to meet the

expectations of the large enterprise customers.

Eventually, more and more Fortune 50 companies began

to adopt Isilon’s products.

Finally he decided to increase the company’s focus on

R&D. Innovation has always been an essential part of his

life. A very innovative and inquisitive person, Patel is

known to get at least six ideas every day. Even as a

kid he was known for asking around 500 questions

about everything under the Sun. Thus, Patel diverted

about 25 percent of the company’s revenue

towards R&D and fostered innovation within the

organization.

So what drives this confident and zealous man? It is the

self-belief, passion and the creative way of looking at

problems and coming up with solutions, says Patel. Even

during his college days, while working on the Internet, he

looked for opportunities to think about new ways, solve

problems, or to bring innovative techniques/technologies

to the market place.

To run a business successfully, it is also important that

one should be honest with oneself, the team and the

stakeholders. The foremost thing that Patel did after

taking over the company was to communicate with

customers and partners and update them about the

happenings within the organization, reinstating their faith

in him and the company. “I went and talked to each of our

investors and customers, telling what we planned

and how much earnings we expected through the

quarters. I believe that apart from your technology

offerings, one reason companies want to do business

with you is the goodwill you develop in tough times,”

says Patel.


PHYSICS

1. In Searle's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 100 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm). Find maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free from error.

[IIT-2004] Sol. Maximum percentage error in Y is given by

Y =

4DW

2π×

XL

.maxY

Y

∆ = 2

∆

DD +

xx∆ +

LL∆

= 2

+

+

1101.0

125.0001.0

05.0001.0 = 0.0489

So maximum percentage error = 4.89%.

2. Particles P and Q of mass 20 gm and 40 gm respectively are simultaneously projected from points A and B on the ground. The initial velocities of P and Q make 45º and 135º angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49 m/s. The separation AB is 245 m. [IIT-1982]

A

P Q

135º

B45º

Both particle travel in the same vertical plane and

undergo a collision After the collision, P retraces its path, Determine the position of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground? Take g = 9.8 m/s2.

Sol. mp = 20 g mQ = 40 g The horizontal velocities of both the particles are

same and since both are projected simultaneously, these particle will meet exactly in the middle of AB (horizontally).

To find the vertical velocity at the time of collision let us consider the motion of P in vertical and horizontal directions.

A

P Q

135º

B45º

49m/s

49m/s

s/m2

49

s/m2/49 s/m2/49245m

s/m2

49

Horizontal direction Sx = 122.5 Tx = ?

vx = 249

∴ velocity =time

ntdisplaceme

⇒ 2

49 =xt

5.122

∴ tx = 492)5.122(

Vertical Direction vy = ?

uy =2

49

ty = 492)5.122(

ay = – 9.8 m/s2

∴ vy = uy + aty =2

49 – 9.8 ×49

2)5.122( = 0

mpVp mQVQ – +

Before collision

mpVp mQVQ′

After collision Also, v2 – u2 = 2as

∴ –2

349

= 2 (– 9.8) × s

⇒ s = 61.25

KNOW IIT-JEEBy Previous Exam Questions


⇒ The collision takes place at the maximum height where the velocities of both the particles will be in the horizontal direction.

Applying conservation of linear momentum in the horizontal direction with the information that P retraces its path therefore its momentum will be same in magnitude but different in direction.

Momentum of system after collision ∴ mpmp – mQvQ = – mPvP + mQv′Q

∴ v′Q =Q

QQPP

mvmvm2 −

=040.0

)2/49(040.0)2/49(02.02 −××

=

−1

040.004.0

249 =

249 × 0 = 0

New Path of Q after Collision Considering vertical Motion of Q uy = 0 sy = – 61.25 ay = – 9.8 ty = ?

S = ut +21 at2 =

21 × (– 9.8) × t2 = (–61.25)

∴ t = 3.53 sec Considering Horizontal motion of Q : Since the '

QV = 0, therefore the particle Q falls down vertically so it falls down on the mid point of AB.

3. Three particles, each of mass m, are situated at the

vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. [IIT-1988]

Sol. The radius of the circle

r =4

aa32 2

2 −

=3

a

m

m m

a/ 3

v

F FR

F a

Let v be the velocity given. The centripetal force is provided by the resultant gravitational attraction of the two masses

FR = º60cosF2FF 222 ++

= 3 F = 3 G 2amm×

∴ 3 G 2

2

am =

rmv2

⇒ v2 =3a

maG32 ×

⇒ v =a

Gm

Time period of circular motion

T =v

r2π =

aGm

3a2π = 2πGm3a3

4. Two fixed charges –2Q and Q are located at the points with coordinates (–3a, 0) and (+3a, 0) respectively in the x-y plane. [IIT-1991]

(a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre.

(b) Give the expression V(x) at a general point on the x-axis and sketch the function V(x) on the whole x-axis.

(c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantative argument that the particle eventually crosses the circle. Find its speed when it does so.

Sol. (a) Let P be a point in the X-Y plane with co-ordinates (x, y) at which the potential due to charges –2Q and +Q placed at A and B respectively be zero.

P(x,y)

XOX′

Ax (–3a,0)

B

y

Y

+Q C (5a,0)

(3a-x) (–3a,0)

Y′(3a+x)

∴ 22 y)xa3(

)Q2(K

++=

22 y)xa3(

)Q(K

+−

+

⇒ 2 22 y)xa3( +− = 22 y)xa3( ++ ⇒ 4[(3a – x)2 + y2] = [(3a + x)2 + y2] ⇒ 4[6a2 + x2 – 12ax + y2] = [6a2 + x2 + 12ax + y2] ⇒ 3x2 + 3y2 – 30ax + 27a2 = 0 ⇒ x2 + y2 – 10ax + 9a2 = 0 ⇒ (x – 5a)2 + (y – 0)2 = (4a)2


This is the equation of a circle with centre at (5a, 0) and radius 4a. Thus c (5a, 0) is the centre of the circle.

(b) For x > 3a To find V(x) at any point on X-axis let us consider a

point (arbitrary) M at a distance x from the origin. +Q

M

–2Q (3a,x) (–3a,0) O x

The potential at M will be

V(x) =a3x

)Q2(K+− +

)a3x()Q(K

−+ where k =

041πε

∴ V(x) = KQ

+−

− a3x2

a3x1 For |x| > 3a

Similarly, for

0 < |x| < 3a V(x) = KQ

+−

− xa32

xa31

Since circle of zero potential cuts the x-axis at (a, 0) and (9a, 0) hence V(x) = 0 at x = a at x = 9a

• From the above expressions V(x) → ∞ at x → 3a and V(x) → – ∞ and x → – 3a. • V(x) → 0 as x → + ∞

• V(x) varies at x1 in general.

X

V

–3aa 3a

(c) Applying Energy Conservation (K.E. + P.E.)centre = (KE. + P.E.)circumference

0 + K

−

a8Qq2

a2Qq =

21 mv2 + K

−

a12Qq2

a6Qq

⇒ 21 mv-2 =

a4KQq ⇒ v =

ma2KQq =

πε ma2Qq

41

0

5. A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12Ω are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [IIT-2002]

X

B A C D

12Ω

(a) Are there positive and negative terminals on the

galvanometer ? (b) Copy the figure in your answer book and show the

battery and the galvanometer (with jockey) connected at appropriate points.

(c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance of X.

Sol. (a) No. There are no positive and negative terminals on the galvanometer.

(b) & (c) Q Bridge is balanced JB

AJ

RR

=ρρ

4.06.0 =

X12Ω

⇒ X = 8 Ω where ρ is the resistance per unit length.

J A

C D

X

G

12Ω

CHEMISTRY

6. The oxides of sodium and potassium contained in a 0.5 g sample of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.1180 g. Subsequent treatment of the chlorides with silver nitrate gave 0.2451 g of silver chloride. What is the percentage of Na2O and K2O in the mixture ? [IIT-1979]

Sol. Mass of sample of feldspar containing Na2O and K2O = 0.5 g.

According to the question, Na2O + 2HCl → 2NaCl + H2O ..(1) 2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g K2O + 2HCl → 2KCl + H2O ...(2) 2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g Let, Mass of NaCl = x g ∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate, NaCl + AgNO3 → AgCl + NaNO3 ...(3) 23 + 35.5 = 58.5g 108 + 35.5 = 143.5g KCl + AgNO3 → AgCl + KNO3 ...(4) 39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3)


58.5 g of NaCl yields = 143.5 g AgCl

∴ x g of NaCl yields = 5.585.143 x g AgCl

And from eq. (4), 74.5 g of KCl yields = 143.5 g of AgCl ∴ (0.1180 – x)g of KCl yields

= 5.745.143 (0.1180 – x)g AgCl

Total mass of AgCl

5.585.143 x +

5.745.143 (0.1180 – x) = 0.2451

which gives, x = 0.0342 Hence, Mass of NaCl = x = 0.0342 g And Mass of KCl = 0.1180 – 0.0342 = 0.0838g Step 2. From eq.(1), 117 g of NaCl is obtained from = 62 g Na2O ∴ 0.0342 g NaCl is obtained from

= 11762 × 0.032 = 0.018 g Na2O

From eq. (2), 149 g of KCl is obtained from = 94 g K2O ∴ 0.0838 g of KCl is obtained from

= 14994 × 0.0838 = 0.053 g K2O

Step 3. % of Na2O in feldspar = 5.0

018.0 × 100 = 3.6%

% of K2O in feldspar =5.0

053.0 × 100 = 10.6 %

7. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ..... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ? [IIT-1996]

Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in fig. Three such cells form one hcp unit.

For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside, hence

Number of atoms per unit cell = 88 + 1 = 2

N b

a

O

60º

Area of the base = b × ON = b × a sin 60º = 23 a2

( Q b = a) Volume of the hexagonal cell

= Area of the base × height = 23 a2. c

But c = 322 a

c b α β

a γ

∴ Volume of the hexagonal cell

= 23 a2 .

322 a = a3 2

and radius of the atom, r = a/2 Hence, fraction of total volume of atomic packing

factor = cellhexagonaltheofVolume

atoms2ofVolume

= 2a

r342

3

3π× =

2a2a

342

3

3

π×

= 23

π

= 0.74 = 74% ∴ The percentage of void space = 100 – 74 = 26%

8. A basic nitrogen compound gave a foul smelling gas when treated with CHCl3 and alc. KOH. A 0.295 g sample of the substance, dissolved in aq. HCl and treated with NaNO2 solution at 0ºC liberated a colourless, odourless gas whose volume corresponds to 112 ml at STP. After the evolution of gas was completed, the aq. solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and I2 gave a yellow precipitate. Identify the original sustance. Assume that it contains one N atom per molecule. [IIT-1993]

Sol. As the compound on heating with CHCl3 and alc. KOH gives foul smelling gas, it should be any primary amine.

RNH2 + CHCl3 + 3KOH →∆ RN C(Alkyl isocyanide (foul smelling gas)

+ 3KCl + 3H2O

Since the compound on treating with NaNO2 and HCl at 0ºC produces a colourless gas, the compound must be a p-aliphatic amine, because if it was aromatic diazonium salt might have been produced


NaNO2 + HCl → NaCl + HNO2 RNH2 + HNO2 → ROH + N2 + H2O Thus, the gas liberated is N2.

Amount of gas produced = ml400,22

ml112 = 2001 ml

From the above equation, it is obvious that amount of

compound RNH2 = 2001 mol.

If M is the molar mass of RNH2, then

)M(massMolar

g295.0 = 2001 mol

∴ M = 0.295 × 200 g mol–1 = 59 g mol–1 Thus, the molar mass of alkyl group (R—) will be, 59 – 16 = 43 g mol–1. Hence, R = C3H7 —, i.e., CH3CH2CH2 –

or CH3

CH3 CH–

The original compound may be either CH3CH2CH2NH2 or CH3 – CH – CH3

NH2

From equation, it is clear that the liquid obtained after distillation is ROH. Since it gives yellow ppt.

with NaOH and I2, it must have CH3 — C —

OH

group.

Hence, it is concluded that ROH is CH3 – CH – CH3

OH

.

Thus, the original compound is CH3 – CH – CH3

NH2

.

The different equations are :

CH3

Isopropyl amine

CH– NH2 + CHCl3 + 3KOH ∆ CH3

CH3

CH3 CH – N C + 3KCl +

Isopropyl isocyanide

CH3

CH3 CH – NH2 + HNO2 →

Isopropyl alcohol

CH3

CH3 CH – OH + N2 + H2O

CH3 – CH – CH3 + I2 + 2NaOH → OH

O Acetone

CH3 – C – CH3 + 2NaI + 2H2O

CH3 – C – CH3 + 3I2 + 3NaOH → O

O

CI3 – C – CH3 + 3NaI + 3H2O

CI3 – C – CH3 + NaOH → CH3COONa + CHI3

O Yellow ppt.

∆

9. An organic compound (A) C8H6, on treatment with

dil. H2SO4 containing HgSO4 gives a compound (B), which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous AlCl3. The compound (B) when treated with iodine in aq. KOH, yields (C) and a yellow compound (D). Identify (A), (B), (C) and (D) with justification. Show, how (B) is formed from (A).

[IIT-1994] Sol. The given reactions may be formulated as follows :

C8H6(A)

Dil H2SO4

∆ (B)

I2 + KOH

(C) + (D)

AlCl3

∆ C6H6 + Acid chlorideHgSO4

The reaction of compound (B) with I2 in KOH is

iodoform reaction. The compound (B) must have a –COCH3 group so as to exhibit iodoform reaction. Since (B) is obtained from benzene by Friedal-Crafts reaction, it is an aromatic ketone (C6H5COCH3). The compound (C) must be potassium salt of an acid.

The compound (A) may be represented as C6H5C2H. Since it gives C6H5COCH3 on treating with dil. H2SO4 and HgSO4, it must contain a triple bond (–C ≡ CH) in the side chain. Here, the given reactions may be formulated as follows :

C≡CH

(A)

dil H2SO4

HgSO4; H2O

C = CH2

OH

COCH3

Acetophenone(B)

Benzene

CH3COCl AlCl3; – HCl


– C – CH3

O

+ 3I2 + 4KOH –3KI;–3H2O

∆

(C) Potassium benzoate

COOK + CHI3

(D)

(B)

Hence. (A) C≡CH

Phenyl acetylene

(B)COCH3

Acetophenone

(C)

COOK

Potassium benzoate

(D) Idoform

3CHI

10. When 20.02 g of a white solid (X) is heated, 4.4 g of

an acid gas (A) and 1.8g of a neutral gas (B) are evolved leaving behind a solid residue (Y) of weight 13.8 g. Gas (A) turns lime water milky and (B) condenses into a liquid which changes anhydrous CuSO4 blue. The aqueous solution of (Y) is alkaline to litmus and gives 19.7 g of a white precipitate (Z) with BaCl2 solution. (Z) gives CO2 with an acid. Identify (A), (B), (X), (Y) and (Z). [IIT-1989]

Sol. (i) Since acidic gas (A) turns lime water milky hence it is CO2 or SO2, both of which form white insoluble compound with Ca(OH)2

(ii) Neutral gas (B) condenses into a liquid which turns anhydrous CuSO4 (white) into blue (CuSO4.5H2O), hence (B) is H2O.

(iii) (Y) gives alkaline solution and its solution forms white precipitate (Z) with BaCl2 solution. (Z) on heating gives the acid gas CO2, hence (Z) is BaCO3 and therefore (Y) is a metal carbonate.

(iv) Since (Y) and (A) are produced from (X), thus (X) is a metal bicarbonate.

g02.20

)X( → g4.4)A( +

g80.1)B( +

g8.13)Y(

From the above values we may write a general equation for a bicarbonate.

)X(

3MHCO2 →∆ )A(2CO + H2

)B(O +

)Y(32COM

Q 4.4g CO2 is obtained from 20.02 g of MHCO3 ∴ 4g CO2 is obtained from 200.2 g of MHCO3

Molecular weight of MHCO3 = 2

2.200 = 100.1

Atomic weight of M = 39 Thus, the metal M is potassium and then (X) is

KHCO3. The equations are :

)X(

3KHCO2 →∆ )Y(

32COK + )A(2CO +

)B(2OH

)Y(

32COK + BaCl2 → 2KCl + )Z(

3BaCO

)Z(

3BaCO →∆ BaO + )A(2CO ↑

Hence, (A) is CO2 (B) is H2O (X) is KHCO3 (Y) is K2CO3 and (Z) is BaCO3.

MATHEMATICS

11. From a point A common tangents are drawn to the

circle x2 + y2 = a2/2 and parabola y2 = 4ax. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. [IIT-1996]

Sol. Equation of any tangent to the parabola, y2 = 4ax is y = mx + a/m.

This line will touch the circle x2 + y2 = a2/2

A(–a, 0)

C

O

B y

D

L

E

πx =

– a

/2

x = a

x

If 2

ma

=

2a 2

(m2 + 1)

⇒ 2m1 =

21 (m2 + 1) ⇒ 2 = m4 + m2

⇒ m4 + m2 – 2 = 0 ⇒ (m2 – 1)(m2 + 2) = 0 ⇒ m2 – 1 = 0, m2 = – 2 (which is not possible). ⇒ m = ± 1 Therefore, two common tangents are y = x + a and y = –x – a These two intersect at A(–a, 0) The chord of contact of A(–a, 0) for the circle

x2 + y2 = a2/2 is (–a)x + 0.y = a2/2 or x = – a/2 and chord of contact of A(–a, 0) for the parabola

y2 = 4ax is 0.y = 2a(x – a) or x = a Again length of BC = 2BK


= 2 22 OKOB −

= 24

a2

a 22− = 2

4a 2

= a

and we know that DE is the latus rectum of the parabola so its length is 4a.

Thus area of the trapezium

BCDE = 21 (BC + DE) (KL)

= 21 (a + 4a)

2a3 =

4a15 2

12. Let V be the volume of the parallelopiped formed by the vectors

→a = a1 i + a2 j + a3 k ;

→b = b1 i + b2 j + b3 k

→c = c1 i + c2 j + c3 k

If ar, br, cr, where r = 1, 2, 3 are non-negative real

numbers and ∑=

++3

1rrrr )cba( = 3L. Show that

V ≤ L3. [IIT-2002]

Sol. V = |)cb.(a|→→→

× ≤ 23

22

21 aaa ++

23

22

21 bbb ++ 2

322

21 ccc ++ ...(1)

Now, L = 3

)ccc()bbb()aaa( 321321321 ++++++++

[(a1 + a2 + a3) (b1 + b2 + b3) (c1 + c2 + c3)]1/3 ∴ L3 ≥ [(a1 + a2 + a3)(b1 + b2 + b3)(c1 + c2 + c3)] ..(2) Now, (a1 + a2 + a3)2 = 2

1a + 22a + 2

3a + 2a1a2 + 2a1a3 + 2a2a3 ≥ 21a + 2

2a + 23a

⇒ (a1 + a2 + a3) ≥ 23

22

21 aaa ++

Similarly, (b1 + b2 + b3) ≥ 23

22

21 bbb ++

and (c1 + c2 + c3) ≥ 23

22

21 ccc ++

∴ from (1) and (2)

L3 ≥ [( 21a + 2

2a + 23a )( 2

322

21 bbb ++ )( 2

322

21 ccc ++ )]1/3 ≥ V

13. T is a prallelopiped in which A, B, C and D are vertices of one face and the just above it has corresponding vertices A´, B´, C´, D´, T is now compressed to S with face ABCD remaining same and A´, B´, C´, D´ shifted to A´´, B´´, C´´, D´´ in S. The volume of parallelopiped S is reduced to 90% of T. Prove that locus of A´´ is a plane. [IIT-2004]

Sol. Let the equation of the plane ABCD be ax + by + cz + d = 0, the point A´´ be (α, β, γ) and

the height of the parallelopiped ABCD be h.

⇒ 222 cba

|dcba|

++

+γ+β+α = 90%. h

⇒ aα + bβ + cγ + d = ± 0.9h 222 cba ++

⇒ locus is, ax + by + cz + d = ±0.9h 222 cba ++

⇒ locus of A´ is a plane parallel to the plane ABCD

14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001]

Sol. Let us define at onto function F from A : [r1, r2 ... rn] to B : [1, 2, 3] where r1r2 .... rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws.

Number of such functions, M = N – [n(1) – n(2) + n(3)] where N = total number of functions and n(t) = number of function having exactly t elements

in the range. Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0 ⇒ M = 3n – 3.2n + 3 Hence the total number of favourable cases = (3n – 3.2n + 3). 6C3

⇒ required probability = n3

6nn

6C)32.33( ×+−

15. A straight line L through the origin meets the line

x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q two straight lines L1 and L2 are drawn, parallel to 2x – y = 5 and 3x + y = 5 respectively. Lines L1 and L2 intersect at R, shown that the locus of R as L varies, is a straight line.

[IIT-2002] Sol. Let the equation of straight line L be y = mx

P ≡

++ 1mm,

1m1 ; Q ≡

++ 1mm3,

1m3

Now equation of L1 : y – 2x = 1m2m

+− ...(1)

equation of L2 : y + 3x = 1m9m3

++ ...(2)

By eliminating 'm' from equation (1) and (2), we get locus of R as x – 3y + 5 = 0, which represents a straight line.


1. Two capacitors C1 and C2, can be charged to a

potential V/2 each by having

C1 C2

O R R

S2 S1 V

(A) S1 closed and S2 open

(B) S1 open and S2 closed

(C) S1 and S2 both closed

(D) cannot be charged at V/2

2. Energy liberated in the de-excitation of hydrogen

atom from 3rd level to 1st level falls on a photo-

cathode. Later when the same photo-cathode is

exposed to a spectrum of some unknown

hydrogen like gas, excited to 2nd energy level, it is

found that the de-Broglie wavelength of the

fastest photoelectrons, now ejected has decreased

by a factor of 3. For this new gas, difference of

energies of 2nd Lyman line and 1st Balmer line if

found to be 3 times the ionization potential of the

hydrogen atom. Select the correct statement(s)

(A) The gas is lithium

(B) The gas is helium

(C) The work function of photo-cathode is 8.5eV

(D) The work function of photo-cathode is 5.5eV

3. In the figure shown there exists a uniform time

varying magnetic field B = [(4T/s) t + 0.3T] in a

cylindrical region of radius 4m. An equilateral

triangular conducting loop is placed in the

magnetic field with its centroide on the axis of the

field and its plane perpendicular to the field.

++++

+++++++ +

+ + + + + + + +

+ + + + +

+ + +

+ C B

A

(A) e.m.f. induced in any one rod is 16V

(B) e.m.f. induced in the complete ABC∆ is

V348

(C) e.m.f. induced in the complete ABC∆ is 48V

(D) e.m.f. induced in any one rod is V316

4. 6 parallel plates are arranged as shown. Each

plate has an area A and Distance between them is

as shown. Plate 1-4 and plates 3-6 are connected

equivalent capacitance across 2 and 5 can be

writted as d

nA 0∈. Find min value of n. (n, d are

natural numbers)

dd

dd

2d

1 2 3

4 5 6

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set # 1


5. Match the following Column – I Column – II

(A) A light conducting (P) Magnetic field B circular flexible is doubled. loop of wire of radius r carrying current I is placed in uniform magnetic field B, the tension in the loop is doubled if

(B) Magnetic field at a (Q) Inductance is point due to a long increased by four straight current times. carrying wire at a point near the wire is doubled if

(C) The energy stored (R) Current I is in the inductor will doubled become four times

(D) The force acting on a (S) Radius r is moving charge, doubled moving in a constant magnetic field will be doubled if (T) Velocity v is Doubled

Passage # (Q. No. 6 to Q. No. 8 )

A solid, insulating ball of radius ‘a’ is surrounded

by a conducting spherical shell of inner radius ‘b’

and outer radius ‘c’ as shown in the figure. The

inner ball has a charge Q which is uniformly

distribute throughout is volume. The conducting

spherical shell has a charge –Q.

Answer the following questions.

–Q

Q a

c b

6. Assuming the potential at infinity to be zero, the

potential at a point located at a distance a/2 from

the centre of the sphere will be

(A)

−

πε b1

a2

4Q

0 (B)

−

πε b1

a811

4Q

0

(C)

−

πε b1

a1

4Q

0 (D) None of these

7. Work done by external agent in taking a charge q

slowly from inner surface of the shell to surface

of the sphericalball will be

(A)

−

c1

a1kQq (B)

−

a1

b1kQq

(C)

−

b1

a1kQq (D)

−

a1

c1kQq

8. Now the outer shell is grounded, i.e., the outer

surface is fixed to be zero. Now the charge on the

inner ball will be

(A) zero (B) Q

(C)

−+

b1

c1

a1

CQ (D)

−+

b1

c1

a1

bQ

Cartoon Law of Physics Any body passing through solid matter will leave a

perforation conforming to its perimeter.

Also called the silhouette of passage, this

phenomenon is the specialty of victims of directed-

pressure explosions and of reckless cowards who

are so eager to escape that they exit directly

through the wall of a house, leaving a cookie-

cutout-perfect hole. The threat of skunks or

matrimony often catalyzes this reaction.


1. A fighter plane flies at a velocity of 300

secm . On

the fighter plane there is a gun which shoots at a rate of 40 rounds per second with a muzzle velocity of

1200

secm . The shots are aimed at another fighter

plane flying at a velocity of 200

secm . Find the rate

at which the projectiles hit the target plane : (a) When the two planes move in the same direction, and

the target plane is in front of the shooting plane. (b) The same as (a), when the target plane is in the rear

of the shooting plane. (c) When the two planes move towards one another. (d) When the two planes move away from one another. Sol. Denote by vs the velocity of the plane from which the

shots are fired, by vt the velocity of the target plane and by L the distance between them at the certain moment of time when the shooting plane starts to shoot. Denote by r the rate of fire of the gun and by v the muzzle velocity.

(a) The time it takes for the first projectile to reach the target plane is:

t1 =ts vvv

L−+

…(1)

After a time of r1 the second projectile is shot, and

the distance between the planes at this time is given by:

L' = L – r

vv ts − …(2)

Thus, the time it takes the second projectile to arrive at the target plane is:

t2 =ts

ts

vvvr

vvL

−+

−−

…(3)

which is

∆t = t2 + r1 – t1

=r1 –

)vvv(rvv

ts

ts

−+−

=)vvv(r

v

ts −+ …(4)

after the first shot. Naturally, the time increment does not depend on the initial distance; thus the rate of hitting is:

r′ =t

1∆

= rv

)vvv( ts −+

= 40 ×12001300 = 43.33

ondsechits … (5)

(b) Using the same reasoning for this case, we obtain:

r′ = rv

vvv ts +−

= 40 ×12001100 = 36.66

ondsechits … (6)

(c) In this case:

r′ = rv

vvv ts ++

= 40 ×12001700 = 56.67

ondsechits … (7)

(d) Here,

r′ = rv

vvv ts −−

= 40 ×1200700 = 23.33

ondsechits … (8)

2. Consider the system described in figure.

m1

m2

(a) Use the equations of energy conservation to find the velocities of masses m1 and m2 after they are released from rest and pass a distance y (assume m2 > m1).

(b) Use the expression obtained in the first section to find the acceleration of the masses.

Sol. (a) We write the change in the potential energy of the masses :

m1 : = ∆Ep = m1gy ...(1) m2 : ∆Ep = – m2gy ...(2) The change in the kinetic energy is

∆Ek = 21 m1v1

2 + 21 m2v2

2 ...(3)

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumPHYSICS


Because | 1vr

| = | 2vr

| ≡ v, we obtain

∆Ek = 21 (m1 + m2)v2 ...(4)

Since there are no external forces except gravity, which is a conservative force, we know, using energy conservation that :

∆Ep + ∆Ek = 0 ...(5) By substituting

(m1 – m2)gy + 21 (m1 + m2)v2 = 0 ...(6)

Hence, v(y) = 2/1

21

12

mm)mm(g2

+

− y ...(7)

(b) By definition,

a = dtdv =

2/1

21

12

mm)mm(g2

+

− .dt

yd

= y2

1mm

)mm(g22/1

21

12

+

− . dtdy ...(8)

and because dtdy = v we substitute Eq. (7) into Eq.

(8) and obtain.

a = 12

12

mmmm

+− g

3. A smooth incline of lift angle α is accelerated at a

rate α. A block of mass m is placed on the incline. At t = 0 the block is released, and begins moving (see figure.) N

m

mg

a

y´

x´α

(a) Write the equations of motion for the block. (b) What is the maximal value of 'a' for which the

block will remain attached to the incline ? (c) How much time is required for the block to slide

a distance L along the incline ? (d) If we accelerate the incline in the opposite

direction, what is the minimal value of a necessary for the block to slide up the incline ?

Sol. (a) D'alembert's force exists in the acclerated system, on the plane of the incline. Therefore,

m ar

= mg x ´ sin α – mg y ´ cos α + N y ´ – m ar

a

r is expressed by the unit vectors of the accelerated

system as : a

r = – a x ´ cos α – a y ´ sin α

and in component form :

≥α+α−=

α+α=

)0N(sinacosgmNa

cosasinga

´y

´x

(b) The maximal value is obtained when N = 0 and ay´ = 0. (The block is still upon the inclined surface, as stipulated.) Therefore,

a = g cot α (c) The equation of motion with a constant

acceleration 'a' is :

x(t) = x0 + v0t + 21 at2

In our case, we have L = ´xa21 t2. Therefore,

t = α+α cosasing

L2

(d) Because 'a' is changed to (–a), the equation of motion in the ´x direction becomes :

´xa = g sin α – a cos α In order for the mass to slide up the incline, the

condition ´xa > 0 must be met. Thus, a > g tan α

4. Two long wires are placed on a smooth horizontal table. Wires have equal but opposite charges. Magnitude of linear charge density on each wire is λ. Calculate (for unit length of wires) work required to increase the separation between the wires from a to 2a.

Sol. Since, wires have opposite charge, therefore, they attract each other. To increase separation between the wires, work is to be done against this force of attraction.

Let at some instant separation between the wires be x as shown in Fig.

To calculate force of attraction between the wires, first electric field due to charge on one wire at position of the second wire is to be calculated.

Therefore, considering a cylindrical surface of radius x and of unit length, co-axial with positively charged wire,

x

+

+

+

+

–

–

–

–

Its area = 2π x × 1 Charge enclosed within it = λ

∴ Flux passing through the cylindrical surface =0ε

λ

Electric field, E = Flux passing per unit area.

=x2

)/( 0

πελ

=x2 0πε

λ

Magnitude of charge on unit length of second wire = λ ∴ Force of attraction per unit length is F = λ E


or F =x2 0

2

πελ

To increase the separation, wires are to be pulled apart by applying an infinitesimally greater force (F + dF).

∴ Work done to increase separation from x to (x + dx), dW = (F + dF) dx ≈ F. dx

or Total work done = ∫=

=

a2x

axdxF = ∫

=

= πελa2x

ax 0

2dx

x2

=0

2

2πελ loge 2

5. Two short electric dipoles having dipole moment p1 and p2 are placed co-axially and uni-directionally, at a distance r apart. Calculate nature and magnitude of force between them.

Sol. Let second dipole having dipole moment p2 consist of charges (+ q2) and (– q2) which are separated by an elemental distance 2dr as shown in Fig.

Then p2 = q2 2dr …(1)

r

p1 →

+– q2 q2

2.dr

Since, dipoles are separated by a distance r, it means distance between their centres is r. Distance of charges (– q2) & (+ q2) from centre of first dipole is (r – dr) & (r + dr), respectively. If electric field strength due to and at distance r from dipole having dipole moment p1 is E, then electric field strength at position of two charges will be (E – dE) & (E + dE), respectively.

Where E =04

1επ 3

1

rp2 (rightward)

∴ dE = – 3 .04

1επ

. 41

rp2 dr

Force on charge (– q2) is F1 = (E – dE) q2 (leftward) and that on charge (+ q2) is F2 = (E + dE) q2 (rightwards)

Hence, net force on second dipole is F = F2 – F1 (rightward)

or F = dE 2q2 = – 304

1επ

. 41

rp2 dr 2q2

But 2q2 dr = p2

∴ Net force = –04

1πε 4

21

rpp6

(–ve) sign indicates that actual direction of force on second dipole is leftward or force between two dipoles is of attraction and its magnitude is

F =04

1πε

. 421

rpp6

1. The typical size of a meteor is about one

cubic centimeter, which is equivalent to the size of a sugar cube.

2. Each day, Earth accumulate 10 to 100 tons of material.

3. There are over 100 billion galaxies in the universe.

4. The largest galaxies contain nearly 400 billion stars.

5. The risk of a falling meteorite striking a human occurs once every 9,300 years.

6. A piece of a neutron star the size of a pin point would way 1 million tons.

7. Europa, Jupiter’s moon, is completely covered in ice.

8. Light reflecting off the moon takes 1.2822 seconds to reach Earth.

9. There has only been one satellite destroyed by a meteor, it was the European Space Agency’s Olympus in 1993.

10. The International Space Station orbits at 248 miles above the Earth.

11. The Earth orbits the Sun at 66,700mph.

12. Venus spins in the opposite direction compared to the Earth and most other planets. This means that the Sun rises in the West and sets in the East.

13. The Moon is moving away from the Earth at about 34cm per year.

14. The Sun, composed mostly of helium and hydrogen, has a surface temperature of 6000 degrees Celsius.

15. A manned rocket reaches the moon in less time than it took a stagecoach to travel the length of England.

16. The nearest known black hole is 1,600 light years (10 quadrillion miles/16 quadrillion kilometers) away.


• Coulomb's Law :

F0 = 221

0 rqq

41πε

(in vacuum)

Vectorially →F = 2

21

0 rqq

41πε

r

In any material medium F = 221

r0 rqq

41

επε

where εr is a constant of the material medium called its relative permittivity, and ε0 is a universal constant, called the permittivity of free space.

ε0 = 8.85 × 10–12 or 04

1πε

= 9 × 109

The unit of ε0 is C2N m–2 or farad per metre.

Also F = 221

r0 rqq

41

επε

Where ε is called the absolute permittivity of the medium. Obviously, εr = F0/F. Remember εr = ∞ for conductors. Conductors and insulators Each body contains enormous amounts of equal and opposite charges. A 'charged' body contains an excess of either positive or negative charge. In a conductor, some of the negative charges are free to move around. In an insulator (also called a dielectric), the charges cannot move. They can only undergto small localized displacements, causing polarization. Induction When a charged body A is brought near another body B, unlike charges are induced on the near surface of B (called bound charges) and like charges appear on the far surface of B (called free charges) If B is a conductor, the free charges can be removed by earthing B, e.g., by touching it. If B is an insulator, separation of like and unlike charges will still occur due to induction. However, the like charges cannot then be removed by earthing B.

• Electric Field And Potential Electric Field An electric field of strength E is said to exist at a point if a test charge ∆q at that point experiences a force given by

→→

∆=∆ FqF or qFE

∆∆

=

→→

The unit of electric field is Newton per coulomb or volt per metre. The electric field strength at a distance r from a point charge q in a medium of permittivity ε is given by

E =πε41

2rq

Vectorially →E =

πε41

2rq r

With reference to any origin

→E =

πε4q

3

rR

rR→→

→→

−

−

Where →R is the position vector of the field point and

→r , the position vector of q.

Due to a number of discrete charges

→E = 3

i

iNi

1i

1

rR

rR4q

→→

→→=

= −

−πε∑

Electric Potential The electric potential at a point is the work done by an external agent in bringing a unit positive charge from infinity up to that point along any arbitrary path.

VP =q

)agentexternalanby(W P

∆∆ →∞ volt(V) or JC–1

The potential difference between two points P and Q is given by

VP – VQ =q

)agentby(W PQ

∆

∆ → volt (V)

The potential at a distance r from a point charge q in a medium of permittivity ε is

ϕ or V =πε41

rq =

πε41

→→− rR

q

with reference to any arbitrary origin. Due to a number of charges

ϕ or V =→→

=

= −πε∑

i

1Ni

1i rR

q41

Electrostatics-I

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


In a conductor, all points have the same potential. If charge q (coulomb) is placed at a point where the potential is V (volt), the potential energy of the system is qV (joule). It follows that if charges q1, q2 are separated by distance r, the mutual potential

energy of the system is r4

qq 21

πε.

• Relation Between Field (E) And Potential (V) The negative of the rate of change of potential along a given direction is equal to the component of the field that direction.

Er = –rV

∂∂ along r

and E⊥ =θ∂

∂r

V perpendicular to r

When two points have different potentials, an electric field will exist between them, directed from the higher to the lower potential.

• Lines of Force A line of force in an electric field is such a curve that the tangent to it at any point gives the direction of the field at that point. Lines of force cannot intersect each other because it is physically impossible for an electric field to have two directions simultaneously.

• Equipotential Surfaces The locus of points of equal potential is called an equipotential surface. Equipotential surfaces lie at right angles to the electric field. Like lines of force, they can never intersect.

Note: For solving problems involving electrostatic units, remember the following conversion factors:

3 × 109 esu of charge = 1 C 1 esu of potential = 300 V • Electric Flux

The electric flux over a surface is the product of its surface area and the normal component of the electric field strength on that surface. Thus,

dϕ = (E cos θ) ds = En ds =→E .

→ds

ds

O

E

N

The total electric flux over a surface is obtained by summing :

ϕE =∑→→

∆ s.E or ∫→→sd.E

Gauss's Theorem The total electric flux across a

closed surface is equal to 0

1ε

times the total charge

inside the surface.

Mathematically ∑→→

∆ s.E = q/ε0

where q is the total charge enclosed by the surface.

Problems in electrostatics can be greatly simplified by the use of Gaussian surfaces. These are imaginary surfaces in which the electric intensity is either parallel to or perpendicular to the surface every-where. There are no restrictions in constructing a Gaussian surface. The following results follow from Gauss's law

1. In a charged conductor, the entire charge resides only on the outer surface. (It must always be remembered that the electric field is zero inside a conductor.)

2. Near a large plane conductor with a charge density σ (i.e., charge per unit area), the electric intensity is

E = σ/ε0 along the normal to the plane 3. Near an infinite plane sheet of charge with a

charge density σ, the electric intensity is E = σ/2ε0 along the normal to the plane 4. The electric intensity at a distance r from the axis

of a long cylinder with λ charge per unit length (called the linear density of charge), is

E =02

1πε r

λ along→r

Problem solving strategy: Coulomb's Law : Step 1 : The relevant concepts : Coulomb's law

comes into play whenever you need to know the electric force acting between charged particles.

Step 2 : The problem using the following steps : Make a drawing showing the locations of the

charged particles and label each particle with its charge. This step is particularly important if more than two charged particles are present.

If three or more charges are present and they do not all lie on the same line, set up an xy-coordinate system.

Often you will need to find the electric force on just one particle. If so, identify that particle.

Step 3 : The solution as follows : For each particle that exerts a force on the particle

of interest, calculate the magnitude of that force

using equation F = 221

0 r|qq|

41πε

Sketch the electric force vectors acting on the particle(s) of interest due to each of the other particles (that is, make a free-body diagram). Remember that the force exerted by particle 1 on particle 2 points from particle 2 toward particle 1 if the two charges have opposite signs, but points from particle 2 directly away from particle 1 if the charges have the same sign.

Calculate the total electric force on the particle(s) of interest. Remember that the electric force, like any force, is a vector. When the forces acting on a charge are caused by two or more other charges, the total force on the charge is the vector sum of


the indivual forces. It's often helpful to use components in an xy-coordinate system. Be sure to use correct vector notation; if a symbol represents a vector quantity, put an arrow over it. If you get sloppy with your notation, you will also get sloppy with your thinking.

As always, using consistent units is essential. With the value of k = 1/4πε0 given above, distances must be in meters, charge in coulombs, and force in newtons. If you are given distance in centimeters, inches, or furlongs, donot forget to convert ! When a charge is given in microcoulombs (µC) or nanocoulombs (nC), remember that 1µC = 10–6C and 1nC = 10–9C.

Some example and problems in this and later chapters involve a continuous distribution of charge along a line or over a surface. In these cases the vector sum described in Step 3 becomes a vector integral, usually carried out by use of components. We divide the total charge distribution into infinitesimal pieces, use Coulomb's law for each piece, and then integrate to find the vector sum. Sometimes this process can be done without explicit use of integration.

In many situations the charge distribution will be symmetrical. For example, you might be asked to find the force on a charge Q in the presence of two other identical charges q, one above and to the left of Q and the other below and to the left of Q. If the distance from Q to each of the other charges are the same, the force on Q from each charge has the same magnitude; if each force vector makes the same angle with the horizontal axis, adding these vectors to find the net force is particularly easy. Whenever possible, exploit any symmetries to simplify the problem-solving process.

Step 4 : your answer : Check whether your numerical results are reasonable, and confirm that the like charges repel opposite charges attract.

Problem solving strategy : Electric-field calculations Step 1: the relevant concepts : Use the principle of

superposition whenever you need to calculate the electric field due to a charge distribution (two or more point charges, a distribution over a line, surface, or volume or a combination of these).

Step 2: The problem using the following steps : Make a drawing that clearly shows the locations

of the charges and your choice of coordinate axes. On your drawing, indicate the position of the field

point (the point at which you want to calculate the electric field E

r). Sometimes the field point will

be at some arbitrary position along a line. For example, you may be asked to find E

r at point on

the x-axis. Step 3 : The solution as follows :

Be sure to use a consistent set of units. Distances must be in meters and charge must be in coulombs. If you are given centimeters or nanocoulombs, do not forget to convert.

When adding up the electric fields caused by different parts of the charge distribution, remember that electric field is a vector, so you must use vector addition. Don't simply add together the magnitude of the individual fields: the directions are important, too.

Take advantage of any symmetries in the charge distribution. For example, if a positive charge and a negative charge of equal magnitude are placed symmetrically with respect to the field point, they produce electric fields of the same magnitude but with mirror-image directions. Exploiting these symmetries will simplify your calculations.

Must often you will use components to compute vector sums. Use proper vector notation; distinguish carefully between scalars, vectors, and components of vectors. Be certain the components are consistent with your choice of coordinate axes.

In working out the directions of Er

vectors, be careful to distinguish between the source point and the field point. The field produced by a point charge always points from source point to field point if the charge is positive; it points in the opposite direction if the charge is negative.

In some situations you will have a continuous distribution of charge along a line, over a surface, or through a volume. Then you must define a small element of charge that can be considered as a point, finds of all charge elements. Usually it is easiest to do this for each component of E

r

separately, and often you will need to evaluate one or more integrals. Make certain the limits on your integrals are correct; especially when the situation has symmetry, make sure you don't count the charge twice.

Step 4 : your answer : Check that the direction of Er

is reason able. If your result for the electric-field magnitude E is a function of position (say, the coordinate x), check your result in any limits for which you know what the magnitude should be. When possible, check your answer by calculating it in a different way.

Problem solving strategy : Gauss's Law Step 1 : Identify the relevant concepts : Gauss's law

is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniform over a plane. In these situations we determine the direction of E

r from the symmetry

of the charge distribution. If we are given the charge distribution. we can use Gauss's law to find the the magnitude of E

r. Alternatively, if we are given the

field, we can use Gauss's law to determine the details


of the charge distribution. In either case, begin your analysis by asking the question, "What is the symmetry ?"

Step 2 : Set up the problem using the following steps Select the surface that you will use with Gauss's

law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface.

The Gaussian surface does not have to be a real physical surface, such as a surface of a solid body. Often the appropriate surface is an imaginary geometric surface; it may be in empty space, embedded in a solid body, or both.

Usually you can evaluate the integral in Gauss's law (without using a computer) only if the Gaussian surface and the charge distribution have some symmetry property. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively.

Step 3 : Execute the solution as follows : Carry out the integral in Eq.

ΦE = ∫ φ dAcosE = ∫ dAE = ∫ Ad.Err

= 0

enclQε

(various forms of Gauss's law) This may look like a daunting task, but the

symmetry of the charge distribution and your careful choice of a Gaussian surface makes it straightforward.

Often you can think of the closed surface as being made up of several separate surfaces, such as the side and ends of a cylinder. The integral ∫ dAE

over the entire closed surface is always equal to the sum of the integrals over all the separate surfaces. Some of these integrals may be zero, as in points 4 and 5 below.

If Er

is perpendicular (normal) at every point to a surface with area A, if points outward from the interior of the surface, and if it equal to EA. If instead E

r is perpendicular and inward, then E⊥ =

– E and ∫ ⊥dAE = – EA.

If Er

is tangent to a surface at every point, then E⊥ = 0 and the integral over that surface is zero.

If Er

= 0 at every point on a surface, the integral is zero.

In the integral ∫ ⊥dAE , E⊥ is always the

perpendicular component of the total electric field at each point on the closed Gaussian surface. In general, this field may be caused partly by charges within the surface and partly by charges outside it. Even when there is no charge within the surface, the field at points on the Gaussian

surface is not necessarily zero. In that case, however, the integral over the Gaussian surface – is always zero.

Once you have evaluated the integral, use eq. to solve for your target variable.

Step 4 : Evaluate your answer : Often your result will be a function that describes how the magnitude of the electric field varies with position. Examine this function with a critical eye to see whether it make sense.

1. Supposing that the earth has a charge surface density

of 1 electron/metre2, calculate (i) earth's potential, (ii) electric field just outside earths surface. The electronic charge is – 1.6 × 10–19 coulomb and earth's radius is 6.4×106 metre (ε0 = 8.9 × 10–12 coul2/nt–m2).

Sol. Let R and σ be the radius and charge surface density of earth respectively. The total charge, q on the earth surface is given by

q = 4 p R2 σ (i) The potential V at a point on earth's surface is same

as if the entire charge q were concentrated at its centre. Thus,

V =Rq.

41

0πε

=04

1πε

.RR4 2σπ =

0

.Rε

σ

Substituting the given values

V =)mnt/coul109.8(

)metre/coul106.1()metre104.6(2212

2196

−××−××

−

−−

= – 0.115coul

mnt − = – 0.115couljoule = – 0.115 volt.

(ii) E =04

1πε 2R

q =04

1πε

. 2

2

RR4 σπ =

0εσ

= 2212

219

mnt/coul109.8metre/coul106.1

−××−

−

−

= – 1.8 × 10–8 nt/coul.

The negative sign shows that E is radially inward. 2. Determine the electric field strength vector if the

potential of this field depends on x, y co-ordinates as (a) V = a(x2 – y2) and (b) V = axy.

Sol. (a) V = a(x2 – y2)

Hence, Ex = –xV

∂∂ = – 2ax, Ey = –

yV

∂∂ = + 2ay

∴ E = – 2axi + 2ayj or E = – 2a(xi – yj) (b) V = a x y

Solved Examples


Hence, Ex = –xV

∂∂ = –ay, Ey = –

yV

∂∂ = – ax

∴ E = – ayi – axj = – a[yi + xj]

3. A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such the surface densities are equal. Find the potential at the common centre.

q

O

R

r

q′

Sol. Let q and q′ be the charges on inner and outer sphere.

Then q + q′ = Q …(1) As the surface densities are equal, hence

2r4qπ

= 2R4'q

π

(∴ Surface density = charge/area) ∴ q R2 = q′ r2 …(2) From eq. (1) q′ = (Q – q), hence q R2 = (Q – q)r2 q(R2 + r2) = Q r2

∴ q = 22

2

rRrQ+

and q′ = Q – q = 22

2

rRRQ+

Now potential at O is given by

V =04

1πε r

q +04

1πε r

'q

=04

1πε r)rR(

rQ22

2

++

041πε r)rR(

rQ22

2

+

=04

Qπε )rR(

)Rr(22 +

+

4. S1 and S2 are two parallel concentric spheres enclosing charges q and 2q respectively as shown in fig.

(a) What is the ratio of electric flux through S1 and S2 ? (b) How will the electric flux through the sphere S1

change, if a medium of dielectric constant 5 is introduced in the space inside S1 in place of air ?

q

2q

S1 S2

Sol. (a) Let Φ1 and Φ2 be the electric flux through spheres S1 and S2 respectively.

Φ1 =0

qε

and Φ2 = 00

q3q2qε

=ε+

∴ 2

1

ΦΦ =

0

0

/q3/q

εε

=31

(b) Let E be the electric field intensity on the surface of sphere S1 due to charge q placed inside the sphere. When dielectric medium of dielectric constant K is introduced inside sphere S1, then electric field intensity E′ is given by

E′ = E/K Now the flux Φ′ through S1 becomes

Φ′ = ∫∫ ε==

0

'

KqdS.E

K1dS.E

∴ Φ′ = 05

qε

5. A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1 cm. It is covered by a concentric, hollow conducting sphere of a radius 5 cm. (a) Find the electric field at a point 2 cm away from the centre. (b) A charge of 6 × 10–8 C is placed on the hollow sphere. Find the surface charge density on the outer surface of the hollow sphere.

Sol. (a) See fig. (a) Let P be a point where we have to calculate the electric field. We draw a Gaussian surface (shown dotted) through point P. The flux through this surface is

2cmP5cm

q = 6 × 10–8 C

Fig. (a) Fig. (b) Φ = ∫∫ −×π== E)102(4dSEdS.E 22

According to Gauss's law, Φ = q/ε0 ∴ 4π × (2 × 10–2)2 E = q/ε0

or E = 220 )102(4

q−××πε

= 4

89

104)104()109(

−

−

××××

= 9 × 105 N/C (b) See fig. (b) We draw a Gaussian surface (shown

dotted) through the material of hollow sphere. We know that the electric field in a conducting material is zero, therefore the flux through this Gaussian surface is zero. Using Gauss's law, the total charge enclosed must be zero. So, the charge on the inner surface of hollow sphere is 6 × 10–8 C. So, the charge on the outer surface will be 10 × 10–8 C.


Kinematics :

Velocity (in a particular direction)

= takenTime

)directionthatin(ntDisplaceme

xAB)V(r

= AxVr

– BxVr

and xAB)V(r

= t

dx

Where dx is the displacement in the x direction in time t.

Swimmer crossing a river

d vscosθ vs θ

vssinθ vr

Time taken to cross the river = θcosV

ds

For minimum time, θ should be zero.

vR vs

vr

x

in this case resultant velocity

VR = 2r

2s vV + and t =

svd .

Also x = vr × t For reaching a point just opposite the horizontal

component of velocity should be zero. v sinθ = Vr

|Average Velocity| = time

|ntDisplaceme|

ar

= t

u–vrr

= t

)u(–vrr

+

⇒ |a| = t

cosuv2–uv 22 θ+

Where θ is the angle between v and u. The direction of acceleration is along the resultant of

vr

and (– ur

).

Graphs During analysis of a graph, the first thing is see the

physical quantities drawn along x-axis and y-axis. If y = mx, the graph is a straight line passing through

the origin with slope = m. [see fig. (a)]

θ

Y

X(i)

m = tanθ θ is acute and m is positive

Y

Xθ

(ii)

m = tanθ θ is abtuse and m is positive

fig.(a) if y = mx + c, the graph is a straight line not passing

through the origin and having an intercept c which may be positive or negative [see fig. (b,) (c)]]

c

Y

X(i)

m is '+' ve

c

Y

X(ii)

c is negativem is '+' ve

Y

X (ii)

c is positive m is negative

fig(b)

fig(c) For y = kx2, where k is a constant, we get parabola

[see fig (d)]

Y

XParabola Fig.(d)

x2 + y2 = r2 is equation of a circle with centre at origin and radius r.

1-D Motion, Projectile Motion PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


For (x – a)2 + (y – b)2 = r2, the motion is in a circular path with centre at (a, b) and radius r

2

2

2

2

by

ax

+ = 1 is equation of an ellipse

x × y = constant give a rectangular hyperbola.

Note : To decide the path of motion of a body, a

relationship between x and y is required. Area under-t graph represents change in velocity.

Calculus method is used for all types of motion (a = 0 or a = constt or a = variable)

a = f(t)

Differentiate w.r.t time

v = f(t) s = f(t)

Differentiate w.r.t time

integrate w.r.t. time

integrate w.r.t. time

S stand for displacement

a = v dsdv =

dtdv = 2

2

dtsd

Also vx = dtdx ⇒ vy =

dtdy

ax = dt

dvx = 2

2

dtxd and ay =

dtdvy = 2

2

dtyd

The same concept can be applied for z-co-ordintae. Projectile motion :

ucosθ

ucosθ

ucosθ θ

usinθ

usinθ

θ u

Q

P

u

ucosθ

ucosθ

F g F

g F g

F g

Projectile motion is a uniformly accelerated motion. For a projectile motion, the horizontal component of

velocity does not change during the path because there is no force in the horizontal direction. The vertical component of velocity goes on decreasing with time from O to P. At he highest point it becomes zero. From P to Q again. the vertical component of velocity increases but in downwards direction. Therefore the minimum velocity is at the topmost

point and it is u cos θ directed in the horizontal direction.

The mechanical energy of a projectile remain constant throughout the path.

the following approach should be adopted for solving problems in two-dimensional motion :

Resolve the 2-D motion in two 1-D motions in two mutually perpendicular directions (x and y direction) Resolve the vector quantitative along these directions. Now use equations of motion separately for x-direction and y-directions.

If you do not resolve a 2-D motions in two 1-D motions in two 1-D motion then use equations of motion in vector form

vr

= ur

+ at ; sr

= ut + 21 a

rt2 ; v

r. vr

– ur

. u = 2 ar

sr

s = 21 ( u

r + v

r)t

When y = f(x) and we are interested to find (a) The values of x for which y is maximum for

minimum (b) The maximum/minimum values of y then we may

use the concept of maxima and minima. Problem solving strategy : Motion with constant Acceleration : Step 1: Identify the relevant concepts : In most

straight-line motion problems, you can use the constant-acceleration equations. Occasionally, however, you will encounter a situation in which the acceleration isn't constant. In such a case, you'll need a different approach

ax = dt

d xυ = dtd

dtdx = 2

2

dtxd

Step 2: Set up the problem using the following steps: You must decide at the beginning of a problem

where the origin of coordinates are usually a matter of convenience. If is often easiest to place the particle at the origin at time t = 0; then x0 = 0. It is always helpful to make a motion diagram showing these choices and some later positions of the particle.

Remember that your choice of the positive axis direction automatically determines the positive directions for velocity and acceleration. If x is positive to the right of the origin, the vx and ax are also positive toward the right.

Restate the problem in words first, and then translate this description into symbols and equations. When does the particle arrive at a certain point (that is, what is the value of t)? where is the particle when its velocity has a specified value (that is, what is the value of x when vx has the specified value)? "where is the motorcyclist when his velocity is 25m/s?"


Translated into symbols, this becomes "What is the value of x when vx = 25 m/s?"

Make a list of quantities such as x, x0,vx,v0x,ax and t. In general, some of the them will be known quantities, and decide which of the unknowns are the target variables. Be on the lookout for implicit information. For example. "A are sits at a stoplight" Usually means v0x = 0.

Step 3 : Execute the solution : Choose an equation from Equation vx = v0x + axt

x = x0 + v0xt + 21 axt2 (constant acceleration only)

2xv = 2

x0v + 2ax(x – x0) (constant accelerations only)

x – x0 =

+

2vv xx0 t (constant acceleration only)

that contains only one of the target variables. Solve this equation for the equation for the target variable, using symbols only. then substitute the known values and compute the value of the target variable. sometimes you will have to solve two simultaneous equations for two unknown quantities.

Step 4 : Evaluate your answer : Take a herd look at your results to see whether they make sense. Are they within the general range of values you expected?

Problem solving strategy :

Projectile Motion : Step 1 : Identify the relevant concepts : The key

concept to remember is the throughout projectile motion, the acceleration is downward and has a constant magnitude g. Be on the lookout for aspects of the problem that do not involve projectile motion. For example, the projectile-motion equations don't apply to throwing a ball, because during the throw the ball is acted on by both the thrower's hand and gravity. These equations come into play only after the ball leaves the thrower's hand.

Step 2 : Set up the problem using the following steps Define your coordinate system and make a sketch

showing axes. Usually it's easiest to place the origin to place the origin at the initial (t = 0) position of the projectile. (If the projectile is a thrown ball or a dart shot from a gun, the thrower's hand or exits the muzzle of the gun.) Also, it's usually best to take the x-axis as being horizontal and the y-axis as being upward. Then the initial position is x0 = 0 and y0 = 0, and the components of the (constant) acceleration are ax = 0, ay = – g.

List the unknown and known quantities, and decide which unknowns are your target variables. In some problems you'll be given the initial

velocity (either in terms of components or in terms of magnitude and direction) and asked to find the coordinates and velocity components as some later time. In other problems you might be given two points on the trajectory and asked to find the initial velocity. In any case, you'll be using equations

x = (v0 cosα0)t (projectile motion) through ...(1)

vy = v0 sin α0 – gt (projectile motion) ...(2) make sure that you have as many equations as

there are target variables to be found. It often helps to state the problem in words and

then translate those words into symbols. For example, when does the particle arrive at a certain point ? (That is at what value of t?) Where is the particle when its velocity has a certain value? (That is, what are the values of x and y when vx or vy has the specified value ?) At the highest point in a trajectory, vy = 0. so the question "When does the particle reach its highest points ?" translates into "When does the projectile return to its initial elevation?" translates into "What is the value of t when y = y0 ?"

Step 3 : Execute the solution use equation (1) & (2) to find the target variables. As you do so, resist the

temptation to break the trajectory into segments and analyze each segment separately. You don't have to start all over, with a new axis and a new time scale, when the projectile reaches its highest point ! It's almost always easier to set up equation (1) & (2)

at the starts and continue to use the same axes and time scale throughout the problem.

Step 4 : Evaluate your answer : As always, look at your results to see whether they make sense and whether the numerical values seem reasonable.

Relative Velocity : Step 1 : Identify the relevant concepts : Whenever

you see the phrase "velocity relative to" or "velocity with respect to", it's likely that the concepts of relative will be helpful.

Step 2 : Set up the problem : Label each frame of reference in the problem. Each moving object has its own frame of reference; in addition, you'll almost always have to include the frame of reference of the earth's surface. (Statements such as "The car is traveling north at 90 km/h" implicitly refer to the car's velocity relative to the surface of the earth.) Use the labels to help identify the target variable. For example, if you want to find the velocity of a car (C) with respect to a bus (B), your target variable is vC/B.

Step 3 : Execute the solution : Solve for the target variable using equation

vP/A = vP/B + vB/A (relative velocity along a line) ...(1)


(If the velocities are not along the same direction, you'll need to use the vector from of this equations, derived later in this section.) It's important to note the order of the double subscripts in equation (1) vA/B always means "velocity of A relative to B." These subscripts obey an interesting kind of algebra, as equation (1) shown. If regard each one as a fraction, then the fraction on the left side is the product of the fractions on the right sides : P/A = (P/B) (B/A). This is a handy rule you can use when applying Equation (1) to any number of frames of reference. For example, if there are three different frames of reference A, B, and C, we can write immediately.

vP/A = vP/C + vC/B + VB/A Step 4 : Evaluate your answer : Be on the lookout for

stray minus signs in your answer. If the target variable is the velocity of a car relative to a bus (vV/B), make sure that you haven't accidentally calculated the velocity of the bus relative of the car (vB/C). If you have made this mistake, you can recover using equation.

vA/B = – vB/A

Solved Examples

1. A small glass ball is pushed with a speed V from A. It moves on a smooth surface and collides with the wall at B. If it loses half of its speed during the collision, find the distance, average speed and velocity of the ball till it reaches at its initial position.

V 0.5V BA

d Sol. The ball moves from A to B with a constant speed V.

Since it loses half of its speed on collision, it returns from B to A with a constant speed V/2.

∴ V1 = V and V2 = V/2

Using the formula, VaV =)V/d()V/d(

dd

2211

21

++

Putting d1 = d2 = d; V1 = V and V2 = V/2

We obtain, VaV =)V5.0/d()dV(

dd 21

++ =

3V2

From the formula,

average velocity =net

net

2

2

1

1

21aV

t|s|

Vs

Vs

|ss|V→→→

→=

+

+=

Since s1 = s2 = d and snet = |ss| 21→→

+ = 0

(As the ball returns to its initial position, the change in position, the change in position vector of the ball, that is the net displacement will be zero).

∴ |V| aV→

= 0.

2. A long belt is moving horizontally with a speed of 4 Km/hour. A child runs on this belt to and fro with a speed of 9 Km/hour (with respect to the belt) between his father and mother located 50 m apart on the moving belt. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt,

(b) speed of the child running opposite to the direction of motion of the belt and

(c) time taken by the child in case (a) and (b) ? Which of the answers change, if motion is viewed by

one of the parents ? Sol. Let us consider positive direction of x-axis from left

to right (a) Here, vB = + 4 Km/hour Speed of child w.r.t. belt, vC = = 9 Km/hour ∴ Speed of child w.r.t. stationary observer, vC′ = vC + vB or vC′ = 9 + 4 = 13 Km/hour (b) Here, vB = + 4 Km/hour, vC = – 9 Km/hour ∴ Speed of child w.r.t. stationary observer, vC′ = vC + vB or vC′ = – 9 + 4 = –5 Km/hour The negative sign shows that the child appears to run

in a direction opposite to the direction of motion of the belt.

(c) Distance between the parents, s = 50 m = 0.05 Km Since parents and child are located on the same belt,

the speed of the child as observe by stationary observer in either direction (either father to mother or from mother to father) will be 9 Km/hour.

Time taken by the child in case (a) and (b),

t =hour/km9km50.0 = 20 sec.

If the motion is observed by one of parents, answer to case (a) case (b) gets altred. It is because the speed of the child w.r.t. either of mother or father is 9 Km/hour.

3. A particle is projected with velocity v0 = 100 m/s at

an angle θ = 30º with the horizontal. Find : (a) velocity of the particle after 2 sec. (b) angle between initial velocity and the velocity after 2 sec. (c) the maximum height reached by the projectile (d) horizontal range of the projectile


Sol. (a) jvivv ytxtt

→→→+=

where i and j are the unit vectors along +ve x and +ve y-axis respectively

→

tv =(ux + axt) i + (uy + ayt) j

Here, ux = v0 cos θ = 50 3 m/s, ax = 0 uy = v0 sin θ = 50 m/s, ay = – g (Q g acts downwards)

→

tv = 50 i3 + (50 – 10 × 2) j

=[50 i3 + 30 j ] m/s

∴ |→

2v | = )vv( 2y

2x + = 22 )30()350( +

(b) →

0v = 50 i3 + 50 j

→

2v = 50 i3 + 30 j

∴ →

0v .→

2v = 7500 + 1500 = 9000

If α is the angle between →

0v and→

2v

Then, cos α =|v||v|

v.v

20

20→→

→→

×=

65.911009000×

α = cos–1 (0.98) = 10.8º (c) vy

2 – u2y = 2ayy

At y = ymax, vy = 0 ∴ 0 – v0

2 sin2 θ = 2 (–g)ymax

∴ ymax = g2sinv 22

0 θ = 125 m

(d) R =g

2sinu 2 θ = 1732 m

4. A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle α with the horizontal. Having fallen the distance 'h', the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time ?

α

α α

Sol. Just before impact magnitude of velocity of the ball,

v = )gh2(

As the ball collides elastically and the inclined plane is fixed, the ball follows the law of reflection.

Now along the incline, velocity component after impact is v sin α and acceleration is g sin α. Perpendicular to the incline, velocity component is vcos α and acceleration (– g cos α). Hence, if we measure x and y-coordinate along the incline and perpendicular to the incline, then

x = (v sin α) t + ½ (g sin α)t2 and y = (v cos α) t – ½ (g cos α)t2 When the ball hits the plane for a second time, y = 0, (v cos α)t – ½(g cos α)t2 or t = (2v/g) Putting this value of t in x,

x =gsinv4 2 α = 8h sin α

5. A batsman hits a ball at a height of 1.22m above the ground so that ball leaves the bat at an angle 45º with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m from the position of the batsman. Will the ball clear the wall if its range is 106.68 m. Take g = 10 m/s2

Sol. R(range) =g

2sinv20 θ

or, 20v =

θ2sinRg = Rg as θ = 45º

1.22mA

v0

45º

B106.68m

or, v0 = )Rg( …(1)

Equation of trajectory

y = x tan 45º –º45cosv2

gx22

0

2

or, y = x –.½Rg2

gx2= x –

Rggx2

Putting x = 97.53, we get

y = 97.53 –1068.106

)53.97(10 2

×× = 8.35 cm

Hence, height of the ball from the ground level is h = 8.35 + 1.22 = 9.577 m As height of the wall is 7.31 m so the ball will clear

the wall.


Real Gases : Deviation from Ideal Behaviour : Real gases do not obey the ideal gas laws exactly

under all conditions of temperature and pressure. Experiments show that at low pressures and moderately high temperatures, gases obey the laws of Boyle, Charles and Avogadro approximately, but as the pressure is increased or the temperature is decreased, a marked departure from ideal behaviour is observed.

V

p

H2

Ideal gas

Plot of p versus V of hydrogen, as compared to that of an ideal gas The curve for the real gas has a tendency to coincide

with that of an ideal gas at low pressures when the volume is large. At higher pressures, however, deviations are observed.

Compressibility Factor : The deviations can be displayed more clearly, by

plotting the ratio of the observed molar volume Vm to the ideal molar volume Vm,ideal (= RT/p) as a function of pressure at constant temperature. This ratio is called the compressibility factor Z and can be expressed as

Z = ideal,m

m

VV

= RTp Vm

Plots of Compressibility Factor versus Pressure : For an ideal gas Z = 1 and is independent of pressure

and temperature. For a real gas, Z = f(T, p), a function of both temperature and pressure.

A graph between Z and p for some gases at 273.15 K, the pressure range in this graph is very large. It can be noted that:

(1) Z is always greater than 1 for H2. (2) For N2, Z < 1 in the lower pressure range and is

greater than 1 at higher pressures. It decreases with increase of pressure in the lower pressure region, passes through a minimum at some pressure and then

increases continuously with pressure in the higher pressure region.

(3) For CO2, there is a large dip in the beginning. In fact, for gases which are easily liquefied, Z dips sharply below the ideal line in the low pressure region.

100Z

ideal gas CH4

N2 H2 t = 0ºC

p/101.325 bar200 300

1.0

0

CO2

Plots of Z versus p of a few gases This graph gives an impression that the nature of the

deviations depend upon the nature of the gas. In fact, it is not so. The determining factor is the temperature relative to the critical temperature of the particular gas; near the critical temperature, the pV curves are like those for CO2, but when far away, the curves are like those for H2 (below fig.)

200Z

ideal gas T1

T1>T2>T3>T4

p/101.325 kPa400 600

1.0

0

T2

T3T4

Plots of Z versus p of a single gas

at various temperatures Provided the pressure is of the order of 1 bar or less,

and the temperature is not too near the point of liquefaction, the observed deviations from the ideal gas laws are not more than a few percent. Under these conditions, therefore, the equation pV = nRT and related expressions may be used.

Van der Waals Equation of state for a Real gas Causes of Deviations from Ideal Behaviour : The ideal gas laws can be derived from the kinetic

theory of gases which is based on the following two important assumptions:

Physical Chemistry

Fundamentals

GASEOUS STATE & REAL GASES

KEY CONCEPT


(i) The volume occupied by the molecules is negligible in comparison to the total volume of the gas.

(ii) The molecules exert no forces of attraction upon one another.

Derivation of van der Waals Equation : Van der Waals was the first to introduce

systematically the correction terms due to the above two invalid assumptions in the ideal gas equation piVi = nRT. His corrections are given below.

Correction for volume : Vi in the ideal gas equation represents an ideal

volume where the molecules can move freely. In real gases, a part of the total volume is, however, occupied by the molecules of the gas. Hence, the free volume Vi is the total volume V minus the volume occupied by the molecules. If b represents the effective volume occupied by the molecules of 1 mole of a gas, then for the amount n of the gas Vi is given by

Vi = V – nb ...(1) Where b is called the excluded volume or co-volume.

The numerical value of b is four times the actual volume occupied by the gas molecules. This can be shown as follows.

If we consider only bimolecular collisions, then the volume occupied by the sphere of radius 2r represents the excluded volume per pair of molecules as shown in below Fig.

excluded volume

2r

Excluded volume per pair of molecules Thus, excluded volume per pair of molecules

= 34

π(2r)3 = 8

π 3r

34

Excluded volume per molecule

=

π 3r

348

21 = 4

π 3r

34

= 4 (volume occupied by a molecule) Since b represents excluded volume per mole of the

gas, it is obvious that

b =

π 3

A r344N

Correction for Forces of Attraction : Consider a molecule A in the bulk of a vessel as

shown in Fig. This molecule is surrounded by other molecules in a symmetrical manner, with the result that this molecule on the whole experiences no net force of attraction.

AB

Arrangement of molecules within and near the surface of a vessel Now, consider a molecule B near the side of the

vessel, which is about to strike one of its sides, thus contributing towards the total pressure of the gas. There are molecules only on one side of the vessel, i.e. towards its centre, with the result that this molecule experiences a net force of attraction towards the centre of the vessel. This results in decreasing the velocity of the molecule, and hence its momentum. Thus, the molecule does not contribute as much force as it would have, had there been no force of attraction. Thus, the pressure of a real gas would be smaller than the corresponding pressure of an ideal gas, i.e.

pi = p + correction term ...(2) This correction term depends upon two factors: (i) The number of molecules per unit volume of the

vessel Large this number, larger will be the net force of attraction with which the molecule B is dragged behind. This results in a greater decrease in the velocity of the molecule B and hence a greater decrease in the rate of change of momentum. Consequently, the correction term also has a large value. If n is the amount of the gas present in the volume V of the container, the number of molecules per unit volume of the container is given as

N' = V

nNA or N' ∝ Vn

Thus, the correction term is given as : Correction term ∝ n/V ...( 2a) (ii) The number of molecules striking the side of the

vessel per unit time Larger this number, larger will be the decrease in the rate of change of momentum. Consequently, the correction term also has a larger value,. Now, the number of molecules striking the side of vessel in a unit time also depends upon the number of molecules present in unit volume of the container, and hence in the present case:

Correction term ∝ n / V ...(2b) Taking both these factors together, we have


Correction term ∝

Vn

Vn

or Correction term = a 2

2

Vn ...( 3)

Where a is the proportionality constant and is a measure of the forces of attraction between the molecules. Thus

pi = p + a 2

2

Vn ...(4)

The unit of the term an2/V2 will be the same as that of the pressure. Thus, the SI unit of a will be Pa m6 mol–2. It may be conveniently expressed in kPa dm6 mol–2.

When the expressions as given by Eqs (1) and (4) are substituted in the ideal gas equation piVi = nRT, we get

+ 2

2

Vanp (V – nb) = nRT ...(5)

This equation is applicable to real gases and is known as the van der Waals equation.

Values of van der Waals Constants : The constants a and b in van der Waals equation are

called van der Waals constants and their values depend upon the nature of the gas. They

Van Der Waals Constants

Gas 2–6 moldmkPaa 1–3 moldm

b

H2 He N2

O2 Cl2 NO NO2 H2O CH4 C2H6 C3H8 C4H10(n) C4H10(iso) C5H12(n) CO CO2

21.764 3.457

140.842 137.802 657.903 135.776 535.401 553.639 228.285 556.173 877.880 1466.173 1304.053 1926.188 150.468 363.959

0.026 61 0.023 70 0.039 13 0.031 83 0.056 22 0.027 89 0.044 24 0.030 49 0.042 78 0.063 80 0.084 45 0.122 6 0.114 2 0.146 0 0.039 85 0.042 67

are characteristics of the gas. The values of these constants are determined by the critical constants of the gas. Actually, the so-called constant vary to some extent with temperature and this shows that the van

der Waals equation is not a complete solution of the behaviour of real gases.

Applicability of the Van Der Waals Equation : Since the van der Waals equation is applicable to real

gases, it is worth considering how far this equation can explain the experimental behaviours of real gases. The van der Waals equation for 1 mole of a gas is

+ 2

mVap (Vm – b) = RT ..(i)

At low pressure When pressure is low, the volume is sufficiently large and b can be ignored in comparison to Vm in Eq. (i). Thus, we have

+ 2

mVap Vm = RT or pVm +

mVa =RT

or Z = 1 – RTV

a

m ...(ii)

From the above equation it is clear that in the low pressure region, Z is less than 1. On increasing the pressure in this region, the value of the term (a/VmRT) increase as V is inversely proportional to p. Consequently, Z decreases with increase of p.

At high pressure When p is large , Vm will be small and one cannot ignore b in comparison to Vm. However, the term 2

mV/a may be considered negligible in comparison to p in Eq. (i) Thus,

p(Vm – b) = RT or Z = 1 + RTpb ...(iiii)

Here Z is greater than 1 and increases linearly with pressure. This explains the nature of the graph in the high pressure region.

A high temperature and low pressure If temperature is high, Vm will also be sufficiently large and thus the term 2

mV/a will be negligibly small. At this stage, b may also be negligible in comparison to Vm. Under these conditions, Eq. (i) reduces to an ideal gas equation of state:

pVm = RT Hydrogen and helium The value of a is extremely

small for these gases as they are difficult to liquefy. Thus, we have the equation of state as p(Vm – b) = RT, obtained from the van der Waals equation by ignoring the term 2

mV/a . Hence, Z is always greater than 1 and it increases with increase of p.

The van dar Waals equation is a distinct improvement over the ideal gas law in that it gives qualitative reasons for the deviations from ideal behaviour. However, the generality of the equation is lost as it contains two constants, the values of which depend upon the nature of the gas.


Stability of different types of carbocations in decreasing order :

C ⊕

> CH ⊕

>

⊕

>

(Ph)3

⊕C > (Ph)2

⊕CH > Ph –

⊕C H2 ≥

CH2 = CH – CH2 ≥ R – C – R > R – CH – R

⊕ ⊕ ⊕

R

> R – CH2 > CH2 = CH ⊕ ⊕

A special stability is associated with cycloproyl

methyl cations and this stability increases with every additional cyclopropyl group.

This is undoubtedly because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p-orbital of the cation carbon.

C H

H

H H

H

H Cyclopropyl methyl cation orbital representation conjugation

with the p-like orbital of the ring Nucleophilicity versus basicity : If the nucleophilic atoms are from the same period of

the periodic table, strength as a nucleophile parallels strength as a base. For example :

H2O < NH3 CH3OH ≈ H2O < CH3CO2

Θ < CH3OΘ ≈ OHΘ

Increasing base strength Increasing nucleophile strength ⇒ Nucleophile strength increases down a column of the

periodic table (in solvents that can have hydrogen bond, such as water and alcohols). For example :

ΘOR <

ΘSR

R3N < R3P

ΘF <

ΘCl <

ΘBr <

ΘI

increasing nucleophilic strength decreasing base strength ⇒ Steric bulk decreases nucleophilicity. For

example :

H3C – C – O

CH3

CH3

Θ < HOΘ

weaker nucleophileStronger base

stronger nucleophileweaker base

Leaving Groups : A good leaving groups is the one which becomes a

stable ion after its departure. As most leaving groups leave as a negative ion, the good leaving groups are those ions which stabilize this negative charge most effectively. The weak bases do this best, thus the best groups are weak bases. If a group is a weak base i.e., the conjugate base of a strong acid, it will generally be a good leaving group. In an SN2 reaction the leaving group begins to gain negative charge as the transition state is reached. The more the negative charge is stabilized, the lower is the energy of the transition state; this lowers the energy of activation and thereby increases the rate of reaction.

The acids HCl, HBr, HI and H2SO4 are all strong acids since the anions Cl–, Br–, I– and HSO4

– are stable anions these anions (weak bases) are also good leaving groups in SN2 reactions. Of the halogens, an iodide ion is the best leaving group and the fluoride ion is the poorest :

I– > Br– > Cl– > F– The order of basicity is opposite : F– > Cl– > Br– > I–,

the reason that alkyl fluorides are ineffective substrates in SN2 reactions is related, to the relatively low acidity of HF (pKa = 3). Sulfonic acids, R SO2OH are similar to sulfuric acid in acidity and the sulfonate ion RSO3

– is a very good leaving group. Alky benzenesulfonates, alkyl p-toluenesulfonates are therefore, very good substrates in SN2 reactions.

The triflate ion (CF3SO3–) is one of the best leaving

groups known, it is the anion of CF3SO3H which is a strong acid much stronger than sulfuric acid.

Organic Chemistry

Fundamentals

GENERAL ORGANIC CHEMISTRY

KEY CONCEPT


Kinetic Isotope Effects : The kinetic isotope effect is a change of rate that

occurs upon isotopic substitution and is generally expressed as a ratio of the rate constants, k light/k heavy. A normal isotope effect is one where the ratio of k light to k heavy is greater than 1. In an inverse isotope effect, the ratio is less than 1. A primary isotope effect is one which results from the making or breaking of a bond to an isotopically substituted atom and this must occur in the rate determining step. A secondary isotope effect is attributable to isotopic substitution of an atom not involved in bond making or breaking in the rate determining step. Thus when a hydrogen in a substrate is replaced by deuterium, there is often a change in the rate. Such changes are known as deuterium, isotope effects and are expressed by the ratio kH/kD, the typical value for this ratio is 7. The ground state vibrational energy (the zero-point vibrational energy) of a bond depends on the mass of the atoms and is lower when the reduced mass is higher. Consequently, D – C, D – O, D – N bonds, etc., have lower energies in the ground state than the corresponding H – C, H – O, H – N bonds, etc. Thus, complete dissociation of deuterium bond would require more energy than that for a corresponding hydrogen bond in the same environment. In case a H – C, H –O, or H – N bond is not broken at all in a reaction or is broken in a non-rate-determining step, substitution of deuterium for hydrogen generally does not lead to a change in the rate, however, if the bond is broken in the rate-determining step, the rate must be lowered by the substitution. This helps in determination of mechanism. In the bromination of acetone, the rate determining step is the tautomerization of acetone which involves cleavage of a C–H bond. In case this mechanistic assignment is correct, one should observe a substantial isotop effect on the bromination of deuterated acetone. Indeed kH/kD was found to be around 7.

CH3COCH3 + Br2 → CH3COCH2Br

rate-determining step Bromoacetone

CH3COCH3 CH3C = CH2

OH

Several mechanisms get support from kinetic isotope

effect. Some of these are, oxidation of alcohols with chromic acid and electrophilic aromatic substitution. An example of a secondary isotope effect, where it is sure that the C – H bond does not break at all in the reaction. Secondary isotope effects for kH/kD are generally between 0.6 and 2.0.

(CZ3)2CHBr + H2O → (CZ3)2CHOH + HBr the solvolysis of isopropyl bromide where Z = H or D, kH/kD is

1.34 Secondary isotope effect.

The substitution of tritium for hydrogen gives isotope effects which are numerically larger (kH/kT = 16).

E2 elimination like SN2 process takes place in one step (without the formation of any intermediates). As the attacking base begins to abstract a proton from a carbon next to the leaving group, the C – H bond begins to break, a new carbon-carbon double bond begins to form and leaving group begins to depart. In confirmation with this mechanism, the base induced elimination of HBr from (I) proceeds 7.11 times faster than the elimination of DBr from (II). Thus C–H or C – D bond is broken in the rate determining step. If it was not so there would not have been any rate difference.

– C – CH2Br – CH = CH2Base

Faster reaction

H

H1-Bromo-2-phenylethane (I)

– C – CH2Br – CD = CH2Base

Slower reaction

D

D1-Bromo-2,2-dideuterio-2-phenylethane (II)

No deuterium isotope effect is found in E1 reactions since the rupture of C – H (or C – D) bond occurs after the rate determing step, rather than during it. Thus no rate difference can be measured between a deuterated and a non deuterated substrate.

Mechanism Review : Substitution versus Elimination

SN2 Primary substrate Back-side attack of Nu : with respect to LG Strong/polarizable unhindered nucleophile

Bimolecular in rate-determining step Concerted bond forming/bond breaking Inverse of stereochemistry Favored by polar aprotic solvent.

SN1 and E1

Tertiary substrate Carbocation intermediate Weak nucleophile/base (e.g., solvent)

Unimolecular in rate-determining step Racemization if SN1 Removal of β-hydrogen if E1 Protic solvent assists ionization of LG Low temperature (SN1)/high temperature (E2)

SN2 and E2 Secondary or primary substrate Strong unhindered base/nucleophile leads to SN2 Strong hindered base/nucleophile leads to E2 Low temperatrue (SN2)/high temperature (E2)

E2 Tertiary or secondary substrate Concerted anti-coplanar TS Bimolecular in rate-determining step Strong hindered base High temperature


1. What is the solubility of AgCl in 0.20 M NH3 ? Given : Ksp(AgCl) = 1.7 × 10–10 M2 K1 = [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103 M–1 and K2 = [Ag(NH3)2

+]/[Ag(NH3)+][NH3] = 7.14 × 103 M–1 Sol. If x be the concentration of AgCl in the solution, then [Cl–] = x From the Ksp for AgCl, we derive

[Ag+] = ]Cl[

Ksp−

= x

M107.1 210−×

If we assume that the majority of the dissolved Ag+ goes into solution as Ag(NH3)2

+ then [Ag(NH3)2+] = x

Since two molecules of NH3 are required for every Ag(NH3)2

+ ion formed, we have [NH3] = 0.20 M – 2x Therefore,

Kinst =])NH(Ag[

]NH][Ag[

23

23+

+

= x

)x2M20.0(x

M107.1 2210

−

× −

= 6.0 × 10–8 M2 From which we derive

2

2

x)x2M20.0( − = 210

28

M107.1M100.6

−

−

×× = 3.5 × 102

which gives x = [Ag(NH3)2+] = 9.6 × 10–3 M, which

is the solubility of AgCl in 0.20 M NH3 2. The values of Λ∞ for HCl, NaCl and NaAc (sodium

acetate) are 420, 126 and 91 Ω–1 cm2 mol–1, respectively. The resistance of a conductivity cell is 520 Ω when filled with 0.1 M acetic acid and drops to 122 Ω when enough NaCl is added to make the solution 0.1 M in NaCl as well. Calculate the cell constant and hydrogen-ion concentration of the solution. Given :

∞Λm (HCl) = 420 Ω–1 cm2 mol–1,

∞Λm (NaCl) = 126 Ω–1 cm2 mol–1, and Λm(NaAc) = 91 Ω–1 cm2 mol–1 Sol. Resistance of 0.1 M HAc = 520 Ω Resistance of 0.1 M HAc + 0.1 M NaCl = 122 Ω Conductance due to 0.1 M NaCl,

G = Ω122

1 – Ω520

1 = 0.00627 Ω–1

Conductivity of 0.1 M NaCl solution k = Λmc = (126 Ω–1 cm2 mol–1)(0.1 mol dm–3) = 12.6 Ω–1cm2 dm–3 = 12.6 Ω–1 cm2(10 cm)–3 = 0.0126 Ω–1 cm–1 Cell constant,

K = Gk =

)00627.0()cm0126.0(

1

11

−

−−

ΩΩ = 2.01 cm–1

Conductivity of 0.1 M HAc solution

k = RK =

Ω

−

520cm01.2 1

Molar conductivity of 0.1 M HAc solution

Λm(HAc) = ck =

)dmmol1.0(cm)520/01.2(3

11

−

−−Ω

= 0.038 65 Ω–1 cm–1 dm3 mol–1 = 38.65 Ω–1 cm2 mol–1 According to Kohlrausch law, Λ∞(HAc) is given by ∞Λm (HAc) = ∞Λm (HCl) + ∞Λm (NaAc) – ∞Λm (NaCl) = (420 + 91 – 126) Ω–1 cm2 mol–1 = 385 Ω–1 cm2 mol–1 Therefore, the degree of dissociation of acetic acid is

given as

α = ∞Λ

Λ

m

m = )molcm385(

)molcm65.38(121

121

−−

−−

ΩΩ ≈ 0.1

and the hydrogen-ion concentration of 0.1 M HAc solution is

[H+] = cα = (0.1 M)(0.1) = 0.01 M Thus, its pH is pH = – log[H+]/M = – log(0.01) = 2 3. Potassium alum is KA1(SO4)2.12H2O. As a strong

electrolyte, it is considered to be 100% dissociated into K+, Al3+, and SO4

2–. The solution is acidic because of the hydrolysis of Al3+, but not so acidic as might be expected, because the SO4

2– can sponge up some of the H3O+ by forming HSO4

–. Given a solution made by dissolving 11.4 g of KA1(SO4)2.12H2O in enough water to make 0.10 dm3 of solution, calculate its [H3O+] :

(a) Considering the hydrolysis Al3+ + 2H2O Al(OH)2+ + H3O+ with Kh = 1.4 × 10–5 M (b) Allowing also for the equilibrium HSO4

– + H2O H3O+ + SO42–

with K2 = 1.26 × 10–2 M

Sol. (a) Amount of alum = 1molg38.474g4.11

− = 0.024 mol

Molarity of the prepared solution = 3dm1.0mol024.0

= 0.24 M Hydrolysis of Al3+ is

UNDERSTANDINGPhysical Chemistry


Al3+ + 2H2O Al(OH)2+ + H3O+

Kh = ]Al[

]OH][)OH(Al[3

32

+

++

If x is the concentration of Al3+ that has hydrolyzed, we have

Kh = xM24.0

)x)(x(−

= 1.4 × 10–5 M

Solving for x, we get [H3O+] = x = 1.82 × 10–3 M (b) We will have to consider the following equilibria. Al3+ + 2H2O Al(OH)2+ + H3O+ H3O+ + SO4

2– HSO4– + H2O

Let z be the concentration of SO42– that combines

with H3O+ and y be the net concentration of H3O+ that is present in the solution. Since the concentration z of SO4

2– combines with the concentration z of H3O+, it is obvious that the net concentration of H3O+ produced in the hydrolysis reaction of Al3+ is (y + z). Thus, the concentration (y + z) of Al3+ out of 0.24 M hydrolyzes in the solution. With these, the concentrations of various species in the solution are

zyM24.0

3Al−−

+ + 2H2O zy

2)OH(Al+

+ + y

3OH +

y

3OH + + zM48.0

24SO

−

− z

4HSO− + H2O

Thus, Kh = )zyM24.0(

)y)(zy(−−

+ = 1.4 × 10–5 M ...(i)

K2 = )zM48.0(y

z−

= M1026.1

12−×

...(ii)

From Eq. (ii), we get

z = y)M1026.1(

y)M48.0(2 +× −

Substituting this in Eq. (i), we get

+×−−

+×+

−

−

y)M1026.1(y)M48.0(y24.0

yy)M1026.1(

y)M48.0(y

2

2

= 1.4 × 10–5

Making an assumption that y <<1.26 × 10–2 M, and then solving for y, we get

[H3O+] = y = 2.932 × 10–4 M

4. The critical temperature and pressure for NO are 177 K and 6.485 MPa, respectively, and for CCl4 these are 550 K and 4.56 MPa, respectively. Which gas (i) has smaller value for the van der Walls constant b; (ii) has smaller value of constant a; (iii) has larger critical volume; and (iv) is most nearly ideal in behaviour at 300 K and 1.013 MPa.

Sol. We have Tc(NO) = 177 K Tc(CCl4) = 550 K pc(NO) = 6.485 MPa pc(CCl4) = 4.56 MPa

(i) Since c

c

Tp

= Rb27/a8b27/a 2

= b8

R therefore, b = c

c

p8RT

Thus,

b(NO) = )MPa485.6)(8(

)molKcmMPa314.8)(K177( 11–3 −

= 28.36 cm3 mol–1 and

b(CCl4) = )MPa56.4)(8(

)molKcmMPa314.8)(K550 113 −−

= 125.35 cm3 mol–1 Hence b(NO) < b(CCl4) (ii) Since a = 27pcb2 therefore a(NO) = (27) (6.485 MPa) (28.36 cm3 mol–1)2 = 140827 MPa cm6mol–2 ≡ 140.827 kPa dm6 mol–2 a(CCl4) = (27) (4.56 MPa) (125.35 cm3 mol–1)2 = 1934538 MPa cm6 mol–2 ≡ 1934.538 KPa dm6mol–2 Hence a(NO) < a(CCl4) (iii) Since Vc = 3b therefore, Vc(NO) = 3 × (28.36 cm3 mol–1) = 85.08 cm3 mol–1 Vc(CCl4) = 3 × (125.35 cm3 mol–1) = 376.05 cm3 mol–1 Hence Vc(NO) < Vc(CCl4) (iv) NO is more ideal in behaviour at 300 K and

1.013 MPa, because its critical temperature is less than 300 K, whereas for CCl4 the corresponding critical temperature is greater than 300 K.

5. At 298 K, the emf of the cell Hg

22

-3

ClHg with .satKCldm mol 01.0 1 mol dm–3 KNO3

HgO with .satKOHdm mol 01.0 -3 Hg

is found to be 0.1634 V and the temperature coefficient of the emf to be 0.000837 VK–1. Calculate the enthalpy and entropy changes of the reaction. What is the reaction that occurs in the cell ?

Sol. For the given cell, we have Electrode Reduction reaction Right HgO(s) + H2O(1) + 2e– = Hg(1) + 2OH–(aq) ...(i) Left Hg2Cl2(s) + 2e– = 2Hg(1) + 2Cl–(aq) ...(ii) Subtracting Eq. (ii) from Eq. (i), we get HgO(s) + H2O(1) + Hg(1) + 2Cl–(aq) = Hg2Cl2(s) + 2OH–(aq) The number of electrons involved in the electrode

reactions is 2. Thus ∆G = –nFE = –2(96500 C mol–1)(0.1634 V) = 31536.2 J mol–1

∆H = – nF

∂∂

−pT

ETE

= –2(96500 C mol–1)[(0.1634V) – (298 K × 0.000837VK–1)] = 16603 J mol–1

∆S = nFpT

E

∂∂ = 2(96500 C mol–1)(0.000837 V K–1)

= 161.54 JK–1mol–1


Passage :

A bag contains ‘n’ cards marked 1, 2, 3, ......, n. ‘X’ draws a card from the bag and the card is put back into the bag. Then ‘Y’ draws a card. The probability that ‘X’ draws.

1. The same card as ‘Y’ is –

(A) n1 (B)

n21

(C) 2n1 (D)

n2

2. a higher card than ‘Y’ is –

(A) n

1n − (B) n21n −

(C) 2n1n − (D) 2n2

1n −

3. a lower card than ‘Y’ is –

(A) n

1n − (B) n21n −

(C) 2n1n − (d) 2n2

1n −

4. Evaluate : 228∫

∫−

−10

4924

10

5025

dx)x1(x

dx)x1(x = ?

5. Find the minimum value of

(x1 – x2)2 + 2

22

21 )13x)(x17(

20x

−−−

where x1 ∈ R+ and x2 ∈ (13, 17).

6. Let f(x) = a1 tan x + a2 tan 2x + a3 tan

3x + ...... + an

tannx , where a1, a2, a3, ... an ∈ R and n ∈ N. If | f(x) |

≤ | tan x | for ∀ x ∈

ππ−

2,

2, Prove that ∑

=

n

1i

i

ia ≤ 1

7. Let az2 + bz + c be a polynomial with complex coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region.

| z | ≤ + ab +

bc

8. Find the fifth degree polynomial which leaves remainder 1 when divided by (x – 1)3 and remainder –1 when divided by (x + 1)3.

9. A quadrilateral ABCD is inscribed in a circle of radius R such that AB2 + CD2 = 4R2. Using vector method prove that its diagonals are at right angle.

10. Through a focus of an ellipse two chords are drawn and a conic is described to pass through their extremities, and also through the centre of the ellipse. Prove that it cuts the major axis in another fixed point.

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions wil l be published in next issue

1Set

High Speed Avalanches

Although an avalanche can mean the fall of any material e.g. snow, soil and even rocks, in common usage it generally refers to a falling mass of ice and snow which breaks away from the side of a mountain or cliff and surges down at great speed.


1. An ellipse of eccentricity 2/3 is inscribed in an ellipse

of equal eccentricity and area equals to 9 square units in such a way that both the ellipses touch each other at one end of their common major axis. If length of major axis of smaller ellipse is equal to length of minor axis of bigger ellipse, find the area of the bigger ellipse outside the smaller ellipse.

Sol. The required figure will be drawn as follows

x

y

and we can redraw the figure for our purpose (i.e.

keeping the area out side the smaller ellipse and inside the bigger ellipse same) as

x(a, 0) (b, 0)

(b, 0) (c, 0)

(–a, 0) (–b, 0)

(–c, 0) (–b, 0)

y

Therefore, we can let the ellipses be 2

2

ax + 2

2

by = 1

and 2

2

bx + 2

2

cy = 1

Required area = π ab – π bc = π b (a – c) Now b2 = a2 (1 – e2) and c2 = b2 (1 – e2) (1 – e2)2 ⇒ c = a (1 – e2) ⇒ a – c = ae2 Thus required area = πb (ae2) = πabe2

= 9 × 2

32

= 4 sq. units.

2. Given a point P on the circumference of the circle |z| = 1, and vertices A1, A2, ......, An of an inscribed regular polygon of n sides. Prove using complex numbers that

(PA1)2 + (PA2)2 + ......... + (PAn)2 is a constant. Sol. Without loss of generality we can take P as

1 + 0i. i.e., P ≡ C is 0

A2

A1

A3

P

θ2 θ1

θn

An

Let Ar ≡ C is θr, r = 1, 2, ......, n. PAr = |Cis θr – Cis 0| = |(cosθr – 1) + i(sinθr)| PAr

2 = (cos θr – 1)2 + (sinθr)2 = 2 – 2cos θr

⇒ ∑=

n

1r

2r )PA( = 2n – 2∑

=

θn

1rrcos

Now, ∑=

θn

1rrcos = Re

θ∑

=

n

1rrCis

= Re ]e.......ee[ n21 iii θθθ +++

= Re

−

−

π

πθ

n2i

n

n2ii

e1

e1e 1

Q θ2 – θ1 = θ3 – θ2 = ..... = θn – θn–1 = n

2π

= Re

−

−π

θ

n2i

i

e1

)11(e 1

= 0

Hence, ∑=

n

1r

2r )PA( = 2n = constant.

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' Forum

MATHS


3. In a class of 20 students, the probability that exactly x students pass the examination is directly proportional to x2 (0 ≤ x ≤ 20). Find out the probability that a student selected at random has passed the examination. If a selected students has been found to pass the examination find out the probability that he/she is only student to have passed the examination.

Sol. Let Ex : event that exactly x out of 20 students pass the examination and A : event that a particular student passes the examination ⇒ P(Ex) = kx2 (k is the proportionality constant) Now, E0, E2, ....., E20 are mutually exclusive and

exhaustive events. ⇒ P(E0) + P(E1) + P(E2) + ... + P(E20) = 1 ⇒ 0 + k(1)2 + k(2)2 + .... + k(20)2 = 1

⇒ k

++

6)140)(120)(20( = 1

⇒ k = 2870

1

Now, P(A) = ∑=

20

0xxx )E/A(P).E(P

= ∑=

20

0x

2kx . 20x = ∑

=

20

0x

3x20k

= 2

2)120(20

2870201

+

× =

8263

and P(E1/A) = )A(P

)E/A(P).E(P 11

=

8263

201.)1(

28701 2

= 44100

1

4. Find the set of values of ‘a’ for which minimum value of x3 – 6ax2 + 9a2x + 7, x ∈ [–1, 2] is 3.

Sol. Let f(x) = x3 – 6ax2 + 9a2x + 7 a ≠ 0, otherwise f(x) = x3 + 7, which is always

increasing and hence min f = f(–1) = 6 ≠ 3. Now f´(x) = 3x2 – 12ax + 9a2 = 0 for stationary points ⇒ x = a, 3a CaseI : a > 0 ⇒ –1 is always in the left of a. Case I. (a) : 2 ≤ a, then 3 = min f = f(–1) = –1 – 6a – 9a2 + 7 ⇒ 3a2 + 2a – 1 = 0, no admissible value of a is

obtained. (a,f(a))

(3a,f(3a))

Case I. (b) : –1 < a < 2 < 3a

i.e., 32 < a < 2, then

3 = min f = minf(–1), f(2) = min –1 –6a – 9a2 + 7, 8 – 24 a + 18a2 + 7 = –1 – 6a – 9a2 + 7 as – 1 – 6a – 9a2 + 7 < 8 – 24a + 18a2 + 7 i.e., 3a2 – 2a + 1 > 0, which is true Hence 3 = –1 – 6a – 9a2 + 7

⇒ a = –1 or 31 , none of which is possible.

Case I(c) : 3a ≤ 2 ⇒ 3 = min f = minf(–1), f(3a) = –1 – 6a –9a2 + 7, 18a3 + 7 = –1 – 6a – 9a2 + 7, as 18a3 + 77 – 1 – 6a – 9a2 + 7 i.e., 18a3 + 9a2 + 6a + 170

which is true as a > 0. Hence a = – 1 or 31 ,

in which a = 31 is permissible.

Case II : a < 0 ⇒ 2 is always in the right of a Case II (a) a ≤ –1 ⇒ 3 = min f = f(–1)

⇒ a = –1, as a = 31

Hence a = –1 is one possibility

(a,f(a))

(3a,f(3a))

Case II (b) 3a ≤ –1 < a

⇒ –1, as a = 31

⇒ 3 = min f = f(a) = a3 – 6a3 + 9a3 + 7 ⇒ 4a3 = – 4 ⇒ a = – 1, not possible Case II(c) –1 < 3a ⇒ 3 = min f = min f(–1), f(a) = min –1 –6a – 9a2 + 7, 4a3 + 7 = 4a3 + 7, as 4a3 + 7 < –1 – 6a – 9a2 + 7 as (a + 1)2 (4a + 1) < 0. Hence a = –1, not possible Hence a = –1 or a = 1/3

5. A man standing at a distance 5m in front of the base of a building 10m high on which a flagstaff is mounted observes that the top of the building and the top of a mountain behind the building are along the same straight line. When he recedes by a distance of 48 m he observes that now the top of the flagstaff and the top of the mountain are along the same straight line. If at both the locations, the flagstaff subtends the same angle at the man’s eye, find the height of mountain.


Sol. CD : Flagstaf DE : Building KF : Mountain (height = h say) The figure illustrates the situation. Since, ∠CBD = ∠CAD = α say, points A, B, C and

D are concyclic. ⇒ ∠ABD = ∠ACD = 90º – (α + β) ⇒ ∠ABC = 90º – (α + β) + α = 90º – β = ∠KCH

K

H

G

FE A 548B

90º – β

α β

α 10

D

C 90º – β

β

Now, h = KH + HF = (CH) tan (90º – β) + (BE) tan(90º – β) (Q HF = CE) = [DG + (BA + AE) cot β = [KG cot β + (48 + 5)] cot β ⇒ h = [(h – 10)cotβ + 53] cot β (Q KG = KF – GF)

Putting cot β = 105 =

21 , we get

h =

+

− 53210h

21

⇒ 4h = h – 10 + 106 ⇒ 3h = 96

⇒ h = 32 m

6. If a, b, c and n are positive integers such that a + b + c = n, show that

(aabbcc)1/n + (abbcca)1/n + (acbacb)1/n ≤ n. Sol. Since a, b, c are integers, from A.M. – G.M.

inequality we can write

cba)timesc...cc()timesb...bb()timesa....aa(

++++++++++

≥ [(a.a....a times)(b.b....b times)(c.c...c times)]1/a +b+c

⇒

cbac.cb.ba.a

++++ ≥ ( ) cba

1cba cba ++

Similarly, bac

c.bb.aa.c++++ ≥ ( ) cba

1cac cba ++

and acb

c.ab.ca.b++++

≥ ( ) acb1

acb cba ++

Adding these three inequalities, we get

cba

ca2bc2ab2cba 222

+++++++ ≥ (aabbcc)1/n

+ (acbacb)1/n + (abbcca)1/n

where LHS = cba)cba( 2

++++ = a + b + c Hence proved

How do Satellites Stay Up?

Satellites orbit the earth because of the force of gravity. To understand why this happens and why the satellite does not get pulled in and fall, we have to understand what forces do. A force will change the motion of an object; it might speed it up, slow it down or change its direction. For example, if you are running and someone pushes you from behind, you speed up (the force is in the direction of your motion). But if someone pushes you in the chest when you are running, you slow down (the force is in the opposite direction to your motion). If you are running and someone pushes you from the side, you move away from them, changing your direction. (the force is at right angles to the motion). This idea is called Newton’s First Law. To make something move in a circle it must be moving and have a force that is always at right angles to the motion so that it constantly changes direction. This force is called the centripetal force. Imagine swinging a rock on a string around your head. The tension in the string pulls the rock round in a circle (this is the force at right angles to the motion). So the tension is the centripetal force. If we cut the string, the rock will continue in a straight line because there is no longer a force to change its direction. For a satellite, the centripetal force is the gravitational force, the pull of the earth. If we could switch gravity off, we would lose all our satellites as they move off in straight lines! Going back to the rock example, we need to put energy into keeping the rock moving because the rock is moving through air and is losing energy constantly because of air resistance. We don’t need to do this with satellites because they are moving through space where there is no air, so no air resistance acts on the satellites and they don’t slow down. Many people think there is a centrifugal force acting which pulls the satellite (or rock) outwards. This is not the case; there is no such thing as a centrifugal force. Imagine riding in the back seat of a car as it turns the corner. Let us assume the back seat is very slippery and you don’t have a seatbelt on. As the car turns the corner, you slide from the inside to the outside. You could argue that a force is pushing you outwards. In fact the reason you move outwards is that there is no force keeping you moving in the same circle as the car. What you really do is continue in a straight line. Eventually you will hit the far door and the door then pushes you providing the centripetal force to keep you moving in the same circle as the car. If the car door was open you would not have a centripetal force acting and you would continue in a straight line out of the car! (Don’t try this at home!)


1− is denoted by ‘i’ and is pronounced as ‘iota’.

i = 1− ⇒ i2 = –1, i3 = –i, i4 = 1.

If a, b ∈ R and i = 1− then a + ib is called a complex number. The complex number a + ib is also denoted by the ordered pair (a, b)

If z = a + ib is a complex number, then : (i) a is called the real part of z and we write Re (z) = a. (ii) b is called the imaginary part of z and we write Im (z) = b

Two complex numbers z1 and z2 are said to be equal complex numbers if Re (z1) = Re (z2) and Im (z1) = Im (z2).

If z = x + iy is a non zero complex number, then 1/z is called the multiplicative inverse of z.

If x + iy is a complex number, then the complex number x – iy is called the conjugate of the complex number x + iy and we write iyx + = x – iy.

Algebra of Complex Numbers (i) Addition : (a + ib) + (c + id) = (a + c) + i(b + d) (ii) Subtraction : (a + ib) – (c + id) = (a – c) + i(b – d) (iii) Multiplication : (a + ib) + (c + id) = (ac – bd) + i(ab + bc) (iv) Division by a non-zero complex number :

idciba

++ = 22 dc

bdac++ + i 22 dc

adbc+− , (c + id) ≠ 0

Properties : If z1, z2 are complex numbers, then

(i) )z( 1 = z1

(ii) z + z = 2 Re (z) (iii) z – z = 2i Im (z) (iv) z = z iff z is purely real (v) z = z iff z is purely imaginary

(vi) 2121 zzzz +=+

(vii) 2121 z–zz–z =

(viii) 2121 z.zz.z =

(ix) 2

1

2

1zz

zz

=

provided z2 ≠ 0

If x + iy is a complex number, then the non-negative

ral number 22 yx + is called the modulus of the complex number x + iy and write

| x + iy| = 22 yx +

Properties : If z1, z2 are complex numbers, then (i) | z1 | = 0 iff z1 = 0 (ii) | z1 | = | z 1 | = | – z1 | (iii) – | z1 | ≤ Re (z1) ≤ | z1 | (iv) – | z1 | ≤ Im (z1) ≤ | z1 | (v) | z1 z 1 | = | z1 |2 (vi) | z1 + z2 | ≤ | z1 | + | z2 | (vii) | z1 – z2 | ≥ | z1 | – | z2 | (viii) | z1 z2 | = | z1 | | z2 |

(ix) 2

1zz =

|z||z|

2

1 , provided z2 ≠ 0

(x) | z1 + z2 |2 = | z1 |2 + | z2 |2 + 2 Re (z1 2z )

(xi) | z1 – z2 |2 = | z1 |2 + | z2 |2 – 2 Re (z1 2z )

(xi) | z1 + z2 |2 + | z1 – z2 |2 = 2 [| z1 |2 + | z2 |2 ]. De Moivre’s Theorem

(i) If n is any integer (positive or negative), then (cos θ + i sin θ)n = cos nθ + i sin nθ (ii) If n is a rational number, then the value or one of

the values of (cos θ + i sin θ)n is cos nθ + i sin nθ Euler’s Formula

eiθ = cos θ + i sin θ and e–iθ = cos θ – i sin θ

Square root of complex number Square root of z = a + ib are given by

±

−

+

+

2a|z|i

2a|z| for b > 0 and

−

+

±2

a|z|i–2

a|z| for b < 0.

COMPLEX NUMBER Mathematics Fundamentals M

AT

HS


If ω = 2

3i1+− , then the cube roots of unity are 1, ω

and ω2. We have: (i) 1 + ω + ω2 = 0 (ii) ω3 = 1

Let z = x + iy be any complex number. Let z = r (cos θ + i sin θ) where r > 0. ∴ x = r cos θ and y = r sin θ ∴ x2 + y2 = r2

⇒ r = 22 yx + (Q r > 0)

∴ cos θ = 22 yx

x

+ and sin θ =

22 yx

y

+

The value of θ is found by solving these equations. θ is called the argument (or amplitude) of z.

If – p < θ ≤ π, then θ is called the principal argument of z.

Identification of θ –

x y arg(z) Interval of θ

+ + θ

π

<θ<2

0

+ – –θ

<θ<

π 02

–

– + (π – θ)

π<θ<

π2

– – –(π – θ)

π

<θ<π2

––

If z1 and z2 are two complex numbers then

(i) | z1 – z2 | is the distance between the points with affixes z1 and z2.

(ii) nmnzmz 12

++ is the affix of the point dividing the

line joining the points with affixes z1 and z2 in the ratio m : n internally.

(iii) n–mnz–mz 12 is the affix of the point dividing the

line joining the points with affixes z1 and z2 in the ratio m : n externally where m ≠ n.

(iv) If z1, z2, z3 are the affixes of the vertices of a

triangle then the affix of its centroid is 3

zzz 321 ++.

(v) z = tz1 + (1 – t)z2 is the equation of the line joining points with affixes z1 and z2. Here ‘t’ is a parameter.

(vi) 12

1

12

1zz

zzzz

zz−

−=

−− is the equation of the line

joining points with affixes z1 and z2. Three points with affixes z1, z2, z3 are collinear if

1zz1zz1zz

33

22

11 = 0.

The general equation of a straight line is 0bzaza =++ , where b is any real number.

(i) | z – z1 | < r represents the circle with centre z1 and radius r.

(ii) | z – z1 | < r represents the interior of the circle with centre z1 and radius r.

1

1zzzz

−− = k represents a circle line which is the

perpendicular bisector of the line segment joining points with affixes z1 and z2.

(z – z1) )zz( 2− + )zz( 1− + (z – z2) = 0 represents the circle with line joining points with affixes z1 and z2 as a diameter.

| z – z1 | + | z – z2 | = 2k, k ∈ R+ represents the ellipse with foci at points with affixes z1 and z2.

If z1, z2, z3 be the affixes of the points A, B, C respectively, then the angle between AB and AC is

given by arg

−−

12

13zzzz

.

If z1, z2, z3, z4 are the affixes of the points A, B, C, D respectively, then the angle between AB and CD is

given by arg

−−

34

12zzzz .

nth roots of a complex number Let z = r (cos θ + i sin θ), r > 0 be any complex

number. nth root o z = z1/n

= r1/n

θ+π

+θ+π

nk2sini

nk2cos ,

where k = 0, 1, 2, ………, n – 1. There are n distinct values and sum of all these

values is 0. Logarithm of a complex number

Let z = reiθ be any complex number. Then log z = log reiθ = log r + log eiθ = log r + iθ log e = log r + iθ. ∴ log z = log | z | + i amp (z).


Matrices :

An m × n matrix is a rectangular array of mn numbers (real or complex) arranged in an ordered set of m horizontal lines called rows and n vertical lines called columns enclosed in parentheses. An m × n matrix A is usually written as :

A =

mnmj2m1m

inij2i1i

n2j22221

n1j11211

a...a...aa

a...a...aa

a...a...aaa...a...aa

MM

MM

Where 1 ≤ i ≤ m and 1 ≤ j ≤ n and is written in compact form as A = [aij]m× n

A matrix A = [aij]m × n is called (i) a rectangular matrix if m ≠ n (ii) a square matrix if m = n (iii) a row matrix or row vector if m = 1 (iv) a column matrix or column vector if n = 1 (v) a null matrix if aij = 0 for all i, j and is denoted by

Om× n (vi) a diagonal matrix if aij = 0 for i ≠ j (vii) a scalar matrix if aij = 0 for i ≠ j and all diagonal

elements aii are equal Two matrices can be added only when thye are of same

order. If A = [aij]m × n and B = [bij]m × n, then sum of A and B is denoted by A + B and is a matrix [aij + bij]m × n

The product of two matrices A and B, written as AB, is defined in this very order of matrices if number of columns of A (pre factor) is equal to the number of rows of B (post factor). If AB is defined , we say that A and B are conformable for multiplication in the order AB.

If A = [aij]m × n and B = [bij]n × p, then their product AB is a matrix C = [cij]m × p where

Cij = sum of the products of elements of ith row of A with the corresponding elements of jth column of B.

Types of matrices : (i) Idempotent if A2 = A (ii) Periodic if Ak+1 = A for some positive integer k.

The least value of k is called the period of A.

(iii) Nilpotent if Ak = O when k is a positive integer. Least value of k is called the index of the nilpotent matrix.

(iv) Involutary if A2 = I. The matrix obtained from a matrix A = [aij]m × n by

changing its rows into columns and columns of A into rows is called the transpose of A and is denoted by A′.

A square matrix a = [aij]n × n is said to be (i) Symmetric if aij = aji for all i and j i.e. if A′ = A. (ii) Skew-symmetric if aij = – aji for all i and j i.e., if A′ = –A.

Every square matrix A can be uniquely written as sum of a symmetric and a skew-symmetric matrix.

A = 21 (A + A′) +

21 (A – A′) where

21 (A + A′) is

symmetric and 21 (A – A′) is skew-symmetric.

Let A = [aij]m × n be a given matrix. Then the matrix obtained from A by replacing all the elements by their conjugate complex is called the conjugate of the matrix A and is denoted by ]a[A ij= .

Properties :

(i) ( )A = A

(ii) )BA( + = A + B

(iii) )A(λ = λ A , where λ is a scalar

(iv) BA)BA( = . Determinant : Consider the set of linear equations a1x + b1y = 0 and

a2x + b2y = 0, where on eliminating x and y we get the eliminant a1b2 – a2b1 = 0; or symbolically, we write in the determinant notation

22

11

baba

≡ a1b2 – a2b1 = 0

Here the scalar a1b2 – a2b1 is said to be the expansion

of the 2 × 2 order determinant 22

11

baba

having 2

rows and 2 columns. Similarly, a determinant of 3 × 3 order can be

expanded as :

MATRICES & DETERMINANTS

Mathematics Fundamentals MA

TH

S


333

222

111

cbacbacba

= a133

22

cbcb

– b133

22

caca

+ c133

22

baba

= a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) = a1(b2c3 – b3c2) – a2(b1c3 – b3c1) + a3(b1c2 – b2c1) = Σ(± aibjck)

To every square matrix A = [aij]m × n is associated a number of function called the determinant of A and is denoted by | A | or det A.

Thus, | A | =

nn2n1n

n22221

n11211

a...aa

a...aaa...aa

MMM

If A = [aij]n × n, then the matrix obtained from A after deleting ith row and jth column is called a submatrix of A. The determinant of this submatrix is called a minor or aij.

Sum of products of elements of a row (or column) in a det with their corresponding cofactors is equal to the value of the determinant.

i.e., ∑=

n

1iija Cij = | A | and ∑

=

n

1jija Cij = | A |.

(i) If all the elements of any two rows or two columns of a determinant ate either identical or proportional, then the determinant is zero.

(ii) If A is a square matrix of order n, then | kA | = kn | A |. (iii) If ∆ is determinant of order n and ∆′ is the

determinant obtained from ∆ by replacing the elements by the corresponding cofactors, then

∆′ = ∆n–1 (iv) Determinant of a skew-symmetric matrix of odd

order is always zero. The determinant of a square matrix can be evaluated

by expanding from any row or column. If A = [aij]n × n is a square matrix and Cij is the

cofactor of aij in A, then the transpose of the matrix obtained from A after replacing each element by the corresponding cofactor is called the adjoint of A and is denoted by adj. A.

Thus, adj. A = [Cij]′. Properties of adjoint of a square matrix (i) If A is a square matrix of order n, then A . (adj. A) = (adj . A) A = | A | In. (ii) If | A | = 0, then A (adj. A) = (adj. A) A = O. (iii) | adj . A | = | A |n –1 if | A | ≠ 0 (iv) adj. (AB) = (adj. B) (adj. A). (v) adj. (adj. A) = | A |n – 2 A.

Let A be a square matrix of order n. Then the inverse of

A is given by A–1 = |A|

1 adj. A.

Reversal law : If A, B, C are invertible matrices of same order, then

(i) (AB)–1 = B–1 A–1 (ii) (ABC)–1 = C–1 B–1 A–1

Criterion of consistency of a system of linear equations (i) The non-homogeneous system AX = B, B ≠ 0 has

unique solution if | A | ≠ 0 and the unique solution is given by X = A–1B.

(ii) Cramer’s Rule : If | A | ≠ 0 and X = (x1, x2,..., xn)′

then for each i =1, 2, 3, …, n ; xi = |A||A| i where

Ai is the matrix obtained from A by replacing the ith column with B.

(iii) If | A | = 0 and (adj. A) B = O, then the system AX = B is consistent and has infinitely many solutions.

(iv) If | A | = 0 and (adj. A) B ≠ O, then the system AX = B is inconsistent.

(v) If | A | ≠ 0 then the homogeneous system AX = O has only null solution or trivial solution

(i.e., x1 = 0, x2 = 0, …. xn = 0) (vi) If | A | = 0, then the system AX = O has non-null

solution. (i) Area of a triangle having vertices at (x1, y1), (x2, y2)

and (x3, y3) is given by 1yx1yx1yx

21

33

22

11

(ii) Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear iff area of ∆ABC = 0.

A square matrix A is called an orthogonal matrix if AA′ = AA′ = I.

A square matrix A is called unitary if AAθ = AθA = I (i) The determinant of a unitary matrix is of modulus

unity.

(ii) If A is a unitary matrix then A′, A , Aθ, A–1 are unitary.

(iii) Product of two unitary matrices is unitary. Differentiation of Determinants :

Let A = | C1 C2 C3 | is a determinant then

dxdA = | C′1 C2 C3 | + | C1 C′2 C3 | + | C1 C2 C′3 |

Same process we have for row. Thus, to differentiate a determinant, we differentiate one

column (or row) at a time, keeping others unchanged.

XtraEdge for IIT-JEE MAY 2010 49

a

PHYSICS

Question 1 to 8 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. If a1 and a2 be the retardations offered to the bullet by wood and iron plates respectively then -

(A) a1 = 2a2 (B) a2 = 2a1 (C) a1 = a2 (D) Data Insufficient

2. The cone falling with a speed v0 strikes and penetrates the block of packing material. The acceleration of the cone after impact is a = g – cx2, where c is a positive constant and x is the penetration distance. If the maximum penetration depth is xm. Then c equals -

x

v0

(A) 2m

20m

xvgx2 +

(B) 2m

20m

xvgx2 −

(C) 3m

20m

x2v3gx6 −

(D) 3m

20m

x2v3gx6 +

3. A dipole consists of two particles, one of charge Q, mass m and the other of charge –Q and mass 2m separated by a distance L. For small oscillations about the equilibrium position the time after which the dipole will align itself in the direction of the uniform field E is -

m

E→

2m

Q

–Q

(A) 2πQE3mL2 (B) π

QE3mL2

(C) 2π

QE3mL2 (D)

4π

QE3mL2

4. In the circuit shown, a potential difference of 60 V is applied across AB. The potential difference between the points P and Q is -

IIT-JEE 2011

XtraEdge Test Series # 1

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current, Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, Chemical Kinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle, Function, Limit, Continiuty, Differentiation, Application of Differentiation (Tangent & Normal, Monotonicity, Maxima & Minima)

Instructions : Section - I • Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for

correct answer and -1 mark for wrong answer. • Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks will be awarded for

correct answer and -1 mark for wrong answer. • Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer

and -1 mark for wrong answer. Section - II • Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly

matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.


S

R

A

B

P

Q

C 90V C

2C

2C

(A) 15 V (B) 30 V (C) 45 V (D) 60 V

5. A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic

vector field →B at a point having coordinates (x, y)

in the z = 0 plane is -

(A)

+−

πµ

220

yxjxiy

2I (B)

++

πµ

220

yxjyix

2I

(C)

+−

πµ

220

yxiyjx

2I (D)

+−

πµ

220

yxjyix

2I

6. Two coils, X and Y, are linked such that emf E is induced in Y when the current in X is changing at

the rate

=

dtdII

•. If a current I0 is now made to

flow through Y, the flux linked with X will be -

(A) EI0•I (B)

•I

E I0

(C) )IE(•

I0 (D) E

II•

0

7. In the circuit shown, A is joined to B for a long time, and then A is joined to C. The total heat produced in R is -

R

E

A 2L

+

BC

–

2R

(A) 2

2

RLE (B) 2

2

R2LE

(C) 2

2

R4LE (D) 2

2

R8LE

8. In a transformer, NP and NS are 1000 and 3000 respectively. If the primary is connected across 80 V A.C., the potential difference across each turn of the secondary will be -

(A) 240 V (B) 0.24 V (C) 0.9 V (D) 0.08 V

Questions 9 to 12 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

9. Two objects have life times given by t1 and t2. if t is the life time of an object lying midway between these two times on the logarithmic scale then -

(A) log10(t) = 21 [log10(t1) + log10(t2)]

(B) t =2

tt 21 +

(C) t = 21tt

(D) t1 =

21

+

21 t1

t1

10. The position of a particle traveling along x-axis is given by xt = t3 – 9t2 + 6t where xt is in cm and t is in second Then -

(A) the body comes to rest firstly at (3 – 7 ) s and

then at (3 + 7 ) s (B) the total displacement of the particle in

traveling from the first zero of velocity to the second zero of velocity is zero

(C) the total displacement of the particle in traveling from the first zero of velocity to the second zero of velocity is – 74 cm.

(D) the particle reverses its velocity at (3 – 7 ) s

and then at (3 + 7 ) s and has a negative

velocity for (3 – 7 ) < t < (3 + 7 )

11. A projectile is fired upward with velocity v0 at an angle θ and strikes a point P(x, y) on the roof of the building (as shown). Then,

y

α x

Roof

v0

θ

P(x,y) h

(A) the projectile hits the roof in minimum time if

θ + α =2π .

(B) the projectile hits the roof in minimum time if

θ + α =4π .

(C) the minimum time taken by the projectile to hit

the roof is tmin = α

α−−

cosgcosgh2vv 22

00 .

(D) the projectile never reaches the roof for v0 < αcosgh2


12. In the given circuit the point A is 9 V higher than point B -

A

6V 1Ω

R

B C D

15V 2Ω

24V 1Ω

(A) R = 1 Ω (B) R = 7 Ω (C) Potential difference between B and D is 30 V (D) Potential difference between B and C is 15 V

This section contains 2 paragraphs, each has 3 multiple choice questions. (Question 13 to 18) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 13 to 15) In the diagram (given below), the broken lines represent the paths followed by particles W, X, Y and Z respectively through the constant field E. The numbers below the field represent meters.

X

E

Z

WY

0 1 2 3 4

13. If all particles started from rest, and all are positively charged, which particles must have been acted upon by a force other than that produced by the electric field –

(A) W and Y (B) X and Z (C) X, Y and Z (D) W, X, Y and Z

14. If the particles are positively charged, which particles increased their electric potential energy –

(A) X and Z (B) Y and Z (C) W, X, Y and Z

(D) Since the electric field is constant, none of the particles increased their electric potential energy

15. Suppose that the field strength E is 10 N/C and particle Y has a charge of –10 C. When particle Y is released from rest, it follows the path as shown and accelerates to a velocity of 10 m/s. What is the mass of particle Y –

(A) 1 kg (B) 2 kg (C) 3 kg (D) 4 kg

Passage : II (No. 16 to 18) A set of experiments in the physics lab is designed

to develop understanding of simple electrical circuit principles for direct current circuits . The student is given a variety of batteries, resistors, and DC meters ; and is directed to wire series and parallel combinations of resistors and batteries making measurements of the currents and voltage drops using the ammeters and voltmeters. The student calculates expected current and voltage values using Ohm’s law and Kirchoff’s circuit rules and then checks the results with the meters.

16. Resistors of 4 ohms and 8 ohms are connected in series. A battery of 6 volts is connected across the series combination . How much power, in watts, is consumed in the 8-ohm resistor ?

(A) 0.67 W (B) 2 W (C) 12 W (D) 24 W

17. Two 4-ohm resistors are connected in series and this pair is connected in parallel with an 8-ohm resistor. A 12 volt battery is connected across the ends of this parallel set. What power, in watts, is consumed in the 8-ohm resistor in this case ?

(A) 0.9 W (B) 2.0 W (C) 4.4 W (D) 18 W

18. A 6-volt battery is connected across a 2- ohm resistor. What is the heat energy dissipated in the resistor in 5 minutes ?

(A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.


19. A circular current carrying loop is placed in x-y plane as shown. A uniform magnetic field B = B0 k is present in the region. Match the following :

x

y

Column-I Column-II (A) Magnetic moment of loop (P) Zero (B) Torque on the loop (Q) Maximum (C) Potential energy of loop (R) Along +ve z-axis (D) Equilibrium of loop (S) Stable (T) Minimum

20. An L-C circuit consists of an inductor with L = 0.09 H and a capacitor of C = 4 × 10–4 F. The initial charge on the capacitor is 5 µC, and the initial current in the inductor is zero. Match the following:

Column-I Column-II (A) Maximum voltage across (P) 8.33 × 10–4 capacitor (S.I. unit) (B) Maximum current in the (Q) 3.125 × 10–8 inductor (S.I. unit) (C) Maximum energy stored (R) 4.33 × 10–6 In the inductor (S.I. unit) (D) Charge on the capacitor (S) 1.25 × 10–2

when current in the inductor (S.I. unit) has half its maximum value (T) None

CHEMISTRY


1. A reaction follows the given concentration (C) vs time graph. The rate for this reaction at 20 seconds will be –

0.50.4

0.30.20.1

0 20 40 60 80 100Time/second

(A) 4 × 10–3 Ms–1 (B) 8 × 10–2 Ms–1 (C) 2 × 10–2 Ms–1 (D) 7 × 10–3 Ms–1

2. The potential of the Daniell cell,

Zn)M1(

ZnSO4

)M1(CuSO4 Cu was reported by Buckbee,

Surdzial, and Metz as Eº = 1.1028 – 0.641 × 10–3 T + 0.72 × 10–5 T2,

where T is the celcius temperature. Calculate ∆Sº for the cell reaction at 25º C -

(A) – 45.32 (B) – 34.52 (C) – 25.43 (D) – 54.23

3. In a hypothetical solid C atoms form CCP lattice with A atoms occupying all the Tetrahedral voids and B atoms occupying all the octahedral voids. A and B atoms are of the appropriate size such that there is no distortion in the CCP lattice. Now if a plane is cut (as shown) then the cross section would like –

Plane

CCP unit cell

(A)

C C

C C

B

BBBA

A

B

(B)

C C

C

BBB

C C

CC

(C)

C

C

BBB

C C

CC

A

A A

A

(D)

C C

C

BBB

C C

CC

4. The melting point of RbBr is 682ºC while that of NaF is 988ºC. The principal reason that the melting point of NaF is much higher than that of RbBr is that :

(A) the molar mass of NaF is smaller than that of RbBr

(B) the bond in RbBr has more covalent character than the bond in NaF

(C) the difference in electronegativity between Rb and Br is smaller than the difference between Na and F

(D) the internuclear distance, rc + ra is greater for RbBr than for NaF

5. Following is the graph between (a – x)–1 and time t for a second order reaction

θ = tan–1(0.5) OA = 2L mol–1

Hence, rate at the start of the reaction is –


O time t

A

(a–x

)–1

θ

(A) 1.25 mol–1 min–1

(B) 0.5 mol L–1 min–1 (C) 0.125 mol L–1 min–1 (D) 1.25 mol L–1 min–1

6. A solution contains Na2CO3 and NaHCO3. 10 ml of the solution requires 2.5 ml of 0.1 M H2SO4 for neutralization using phenolphthalein as the indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M H2SO4 was required. The amount of Na2CO3 in the g/litre is

(A) 5.3 (B) 4.2 (C) 10.6 (D) 8.4

7. 1 mole mixture of CH4 & air (containing 80% N2, 20% O2 by volume) of a composition such that when underwent combustion gave maximum heat (assume combustion of only CH4). Then which of the statements are correct, regarding composition of initial mixture:

(A) 118X,

112X,

111X

224 NOCH ===

(B) 21X,

81X,

83X

224 NOCH ===

(C) 32X,

61X,

61X

224 NOCH ===

(D) Data insufficient

8. Consider the following Ist order competing reactions :

X → 1k A + B and Y → 2k C + D if 50% of the reaction of X was completed when

96% of the reaction of Y was completed, the ratio of their rate constants (k2 / k1) is :

(A) 4.06 (B) 0.215 (C) 1.1 (D) 4.65


9. Select the correct statements from the following regarding sols –

(A) Viscosity of lyophilic sols (emulsoid) is much higher than that of solvent

(B) Surface tension of lyophobic sols (suspensoid) is usually low

(C) The particles of lyophilic sols always carry a characteristics charge either positive or negative

(D) Hydrophobic sols can easily be coagulated by addition of electrolytes

10. Which of the following is/are correct ? (A) α-rays are more penetrating than β-rays (B) α-rays have greater ionizing power than β-rays (C) β-particles are not present in the nucleus, yet

they are emitted from the nucleus (D) γ-rays are not emitted simultaneously with α

and β-rays.

11. Choose the correct statement(s) - (A) At the anode, the species having minimum

reduction potential is formed from the oxidation of corresponding oxidizable species

(B) In highly alkaline medium, the anodic process during the electrolytic process is

4OH– → O2 + 2H2O + 4e– (C) The standard potential of Cl– | AgCl | Ag half–

cell is related to that of Ag+ | Ag through the expression

ºAg/Ag

E + = ºAg|AgCl|Cl–E +

FRT ln Ksp (AgCl)

(D) Compounds of active metals (Zn, Na, Mg) are reducible by H2 whereas those of noble metals (Cu, Ag, Au) are not reducible.

12. The co-ordination number of FCC structure for metals is 12, since -

(A) each atom touches 4 others in same layer, 3 in layer above and 3 in layer below

(B) each atom touches 4 others in same layer, 4 in layer above and 4 in layer below

(C) each atom touches 6 others in same layer, 3 in layer above and 3 in layer below

(D) each atom touches 3 others in same layer, 6 in layer above and 6 in layer below


Passage : I (No. 13 to 15) The cell potential for the unbalanced chemical reaction : Hg2

2+ + NO3– + H3O+ → Hg2+ +HNO2 + H2O

is measured under standard state conditions in the electrochemical cell shown in the accompanying diagram. The cell voltage is positive: EºCell = 0.02V Eº = 0.02 V

Dish A anode Dish B cathode Given NO3

– + 3H3O + 2e– → HNO2 + 4H2O Eº = 0.94 V


13. Which of the following statements must be true of the solutions in order for the cell to operate with the voltage indicated ?

(A) The solution in Dish A must be acidic (B) The solution in Dish B must be acidic (C) The solutions in both Dish A and Dish B must

be acidic (D) No acid may be in either Dish A or Dish B

14. At what pH will the cell potential be zero if the activity of other components are equal to one ?

(A) 059.02

02.0×

(B) –059.002.0

(C) 059.004.0 (D)

059.002.0 ×

32

15. How many moles of electrons pass through the circuit when 0.6 mole of Hg2+ and 0.30 mole of HNO2 are produced in the cell that contains 0.5 mole of Hg2

2+ and 0.40 mole of NO3– at the

beginning of the reaction ? (A) 0.6 mole (B) 0.8 mole (C) 0.3 mole (D) 1 mole

Passage : II (No. 16 to 18) At any fixed temperature, the vapour phase is

always richer in more volatile component as compared to the solution phase. In other words, mole fraction of more volatile component is always greater in vapour phase than in solution phase. We can also say that vapour phase is relatively richer in the component whose addition to liquid mixture results in an increase in total vapour pressure.

16. If 2 moles of A and 3 moles of B are mixed to form an ideal solution, vapour pressures of A and B are 120 and 180 mm of Hg respectively, the total vapour pressure of solution will be

(A) 48 mm Hg (B) 108 mm Hg (C) 156 mm Hg (D) 15.6 mm Hg

17. From the statement in question 75, the composition of A and B in the vapour phase when the first traces of vapours are formed is :

(A) A = 0.407, B = 0.592 (B) A = 0.8, B = 0.1 (C) A = 0.109, B = 0.791 (D) A = 0.307, B = 0.692

18. From the statement in question 75, at what pressure will the last traces of liquid disappear ?

(A) 100 mm Hg (B) 130 mm Hg (C) 140 mm Hg (D) 150 mm Hg

This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the following : Column-I Column-II (A) Intermolecular (P) Ne H-bonding (B) Intramolecular (Q) NaCl H-bonding (C) Vander Waal's forces (R) H2O

(D) Strongest bonding (S)

OHCHO

(T) Chloroal hydrate 20. Match the following : Column-I Column-II (A) PH3 (P) ≈ 90º or = 90º (B) H2O (Q) 100º < B.A.<109º28' (C) PF5 (R) 120º (D) IF7 (S) 72º

(T) 180º

MATHEMATICS


1.

π

+8

cos1

π

+8

3cos1

π

+8

5cos1

π

+8

7cos1

is equal to - (A) 1/2 (B) cos π/8

(C) 1/8 (D) 2221+


2. If a, b, c, d are the sides of a quadrilateral, then

the minimum value of 2

222

dcba ++ is

(A) 1 (B) 1/2 (C)1/3 (D) 1/4

3. If the function f(x) = cos |x| – 2ax + b increases along the entire number scale, the range of values of a is given by

(A) a ≤ b (B) a = b/2 (C) a ≤ – 1/2 (D) a ≥ – 3/2

4. If φ(x) = 3f

3x2

+ f(3– x2) ∀ x ∈(–3, 4)

where f"(x) > 0 ∀ x ∈ (–3, 4), then φ(x) is

(A) increasing in

4,

23

(B) decreasing in

−−

23,3

(C) increasing in

− 0,

23

(D) decreasing in

23,0

5. The point in the interval [0, 2π] where f(x) = ex sin x has maximum slope is

(A) 4π (B)

2π

(C) π (D) None of these

Q.6 If in ∆ABC, R9

cbaAsincCsinbBsinaCcoscBcosbAcosa ++

=++++

then the triangle ABC is (A) isosceles (B) equilateral (C) right angled (D) None of these

Q.7 The slope of the normal at the point with abscissa x = – 2 of the graph of the function f(x) = |x2 – x| is of form p/q (where p & q are coprime) then -

(A) p = 4 (B) q = 2 (C) p + q = 6 (D) p – q = – 2

Q.8 In a ∆ABC, b2 + c2 = 1999a2, then

=+

AcotCcotBcot

(A) 999

1 (B) 1999

1

(C) 999 (D) 1999

Questions 9 to 12 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 9. For what triplets of real numbers (a, b, c)

with a ≠ 0, the function : f(x) =

++

≤

otherwise;cbxax

1x;x2

is continous for all real x : (A) (a, 1 – 2a ,a)/a ∈ R, a ≠ 0 (B) (2a, 1 – 2a ,0)/a ∈ R, a ≠ 0 (C) (a, b, c)/a, b, c ∈ R, a + b + c = 1, a ≠ 0 (D) (a, 1 – 2a,c)/ a, c ∈ R, a ≠ 0

10. Let f(x) =

≥<

0x2–x0x2–

and g(x) = |f(x)| then : (A) g(x) is continuous for all values of x. (B) g(x) is differentiable for all value of x (C) g(x) is not differentibale at x = 0 (D) g(x) is not differentiable at x = 2

11. If 5f(x) + 3f

x1 = x + 2 and y = xf(x) then

1xdxdy

=

is equal to :

(A) 14 (B) 7/8 (C) 1 (D) None of these

12. If f(x) and g(x) are differentiable function in [1, 5] and φ(x) = max f(x), g(x), f(x) – g(x) = 0 has exactly one root α in [1, 5] then :

(A) φ(x) continuous and differentiable at all points in [1, 5]

(B) φ(x) differentiable in [1, 5] (C) φ(x) necessarily differentiable in [1, 5]– α (D) φ(x) is not differentiable at x = α This section contains 2 paragraphs, each has 3 multiple choice questions. (Question 13 to 18) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 13 to 15)

Let f

+

nyx =

n)2–n(b)y(f)x(f ++ and

f '(0) = a ; n ∈ N but n ≠ 2. Let g(x) = f(x) ; x ≥ 0 = 3x + sin 2x ; x < 0 If f(x) & g(x) both are differentiable function, then -


13. Range of sgn g(|x|) includes : (A) 2 (B) 1 (C) –1 (D) all of the above

14. Which one of the following statements doesn't hold good :

(A) g(x) is many one function (B) g(x) is invertible (C) f(x) is invertible (D) g(|x|) is non differentiable at x = 0

15. Let h(x) = |g(x) –5x|, then wrong statement is : (A) h(|x|) is a single valued function (B) |h(|x|)| is always positive (C) |h(|x|)| is differentiable every where (D) h(x) is differentiable every where

Passage : II (No. 16 to 18) Let z denotes the set of integers. Let p be a

prime number and let z1 ≡ 0, 1. Let f : z → z and g : z → z1 are two functions defined as follows:

f(n) = pn ; if n ∈ z and g(n) = 1 ; if n is a perfect square = 0 ; otherwise

16. g(f(x)) is : (A) Manyone into (B) Manyone onto (C) one-one onto (D) one-one into 17. f(g(x)) = p has : (A) no real root (B) at most one real root (C) infinitely many roots (D) exactly two real root

18. Wrong statement about g(f((x)) is : (A) it is non periodic function (B) it is neither even nor odd function (C) it is even function (D) it is many one function This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the following : Column –I Column –II (A) f : R → R is defined as (P) 2

f(x) =

<+≥++

0xfor3kx20xfor3kxx 2

if f(x) is injective then 'k' can be equal to

(B) If 2x

9)x(flim2x −

−→

= 3 then (Q) 5

)x(flim2x→

is

(C) If 6787lim x5

kx

x ++

∞→ does not (R) 9

exists then 'k' can be (D) Let f(x) & g(x) satisfy the (S) 12 following properties f(3) = 2, g(3) = 4, g(0) = 3, f ´ (3) = –1, g´(3) = 0, g´(0) = 2, If T(x) = f(g(x)) and U(x) = ln(f(x)) then |T´(0) + 6U´3)| can be equal (T) –1 20. Match the items of column-I with column-II Column –I Column –II

(A) Function f :

π

3,0 → [0, 1] (P) one-one

defined by f(x) = xsin is function (B) Function f :(1, ∞) → (1, ∞) (Q) many-one

defined by f(x) = 1x3x

−+ is function

(C) Function f:

ππ−

34,

2→ (R) into

[–1, 1] defined by function f(x) = sin x is (D) Function f : (2, ∞)→[8, ∞) (S) onto

defined by f(x) = 2x

x 2

−is function

(T) odd Function


PHYSICS


1. If E, m, L and G denote energy, mass, angular momentum and universal gravitation constant

respectively then dimensions of 25

2

GmEL will be that

of - (A) angle (B) length (C) mass (D) time

2. Three vectors →P ,

→Q and

→R are such that |

→P | = |

→Q |,

|→R | = 2 |

→P | and

→P +

→Q +

→R =

→0 . The angles

between →P &

→Q ,

→Q &

→R and

→P &

→R are respectively

(in degrees) - (A) 90, 135, 135 (B) 90, 45, 45 (C) 45, 90, 90 (D) 45, 135, 135

3. The velocity of a boat in still water in η times less than the velocity of flow of the river (η > 1). The angle with the stream direction at which the boat must move to minimize drifting is -

(A) sin–1

η1 (B) cot–1

η1

(C) 2π + sin–1

η1 (D)

2π + cot–1

η1

4. The position of a particle is given by →r = a cos(ωt)

i + a sin (ωt) j + bt k where ω =T2π and T is time

period for one revolution of the particle following a helical path. The distance moved by the particle in one full turn of the helix is -

(A) ωπ4 222 ba ω+ (B)

ωπ2 222 ba +ω

(C) ωπ2 222 ba ω+ (D)

ωπ4 222 ba +ω

5. If the velocity v of a particle moving along a straight line decreases linearly with its displacement s from 20 ms–1 to a value approaching zero at s = 30 m, then acceleration of the particle at s = 15 m is -

O

20

30s (in m)

v(in

ms–1

)

(A) 32 ms–2 (B) –

32 ms–2

(C) 3

20 ms–2 (D) –3

20 ms–2

Time : 3 Hours Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle

Instructions : Section - I • Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer

and -1 mark for wrong answer. Section - II • Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly

matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

IIT-JEE 2012

XtraEdge Test Series # 1

Based on New Pattern


6. A self-propelled vehicle of mass M whose engine delivers constant power P has an acceleration

a =MvP . To increase the velocity of the vehicle

from v1 to v2, the distance traveled by it (assuming no friction) is -

(A) s =MP3 ( 2

2v – 21v ) (B) s =

P3M ( 2

2v – 21v )

(C) s =P3

M ( 32v – 3

1v ) (D) s =MP3 ( 3

2v – 31v )

7. A particle starts from rest and traverses a distance l with uniform acceleration, then moves uniformly over a further distance 2l and finally comes to rest after moving a further distance 3l under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its way is -

(A) 51 (B)

52 (C)

53 (D)

54

8. An express elevator can accelerate or decelerate with values whose magnitudes are limited to 0.4g. The elevator attains a maximum vertical speed of 400 metre per minute. The minimum time required by the elevator to start from rest from the 10th floor and to stop at the 30th floor, a distance 100 m apart is -

(A) 1.67 s (B) 16.7 s (C) 167 s (D) 1670 s


9. A physical quantity is measured by using an instrument having a least count l. Then -

(A) error in the measurement of the physical quantity can exceed the least count l

(B) error in the measurement of the physical quantity equals the least count l

(C) error in the measurement of the physical quantity can be less than the least count l

(D) all (A), (B) and (C) are incorrect

10. A bead is free to slide down a smooth wire tightly stretched between the points P1 and P2 on a vertical circle of radius R. If the bead starts from rest from P1, the highest point on the circle and P2 lies anywhere on the circumference of the circle. Then,

P1

R P2

g θ

(A) time taken by bead to go from P1 to P2 is dependent on position of P2 and equals

2gR cos θ

(B) time taken by bead to go from P1 to P2 is

independent of position of P2 and equals 2gR

(C) acceleration of bead along the wire is g cos θ (D) velocity of bead when it arrives at P2 is

gR2 cos θ

11. A body is moving along a straight line. Its distance xt from a point on its path at a time t after passing that point is given by xt = 8t2 – 3t3, where xt is in metre and t in second.

(A) Average speed during the interval t = 0 s to t = 4 s is 20.21 ms–1

(B) Average velocity during the interval t = 0 s to t = 4 s is – 16 ms–1

(C) The body starts from rest and at t =9

16 s it

reverses its direction of motion at xt = 8.43 m from the start

(D) It has an acceleration of –56 ms–2 at t = 4s

12. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that -

(A) its velocity is constant (B) its acceleration is constant (C) its kinetic energy is constant (D) it moves in a circular path


Passage : I (No. 13 to 15) A student performs an experiment to determine

how the range of a ball depends on the velocity with which it is released. The “range” is the distance between where the ball lands and where it was released, assuming it lands at the same height from which it was released.

In each trial, the student uses the same baseball, and launches it at the same angle. Table I shows the experimental results

Table 1 Trial Launch speed (m/s) Range (m)

1 2 3 4

10 20 30 40

8.0 31.8 70.7 122.5


Based on this data, the student then hypothesizes that the range, R, depends on the initial speed, v0, according to the following equation : R = Cv0

n, where C is a constant, and n is another constant.

13. Based on this data, the best guess for the value of n is –

(A) 1/2 (B) 1 (C) 2 (D) 3

14. The student speculates that the constant C depends on :

I. The angle at which the ball was launched II. The ball’s mass III. The ball’s diameter If we neglect air resistance, then C actually depends

on – (A) I only (B) I and II (C) I and III (D) I, II and III

15. The student performs another trial in which the ball is launched at speed 5.0 m/s. Its range is approximately –

(A) 1.0 meters (B) 2.0 meters (C) 3.0 meters (D) 4.0 meters

Passage : II (No. 16 to 18)

When an airplane flies, its total velocity with respect to the ground is :

vtotal = vplane + vwind Where vplane denotes the plane's velocity through

motionless air, and vwind denotes the wind's velocity. Crucially, all the quantities in this equations are vectors. The magnitude of a velocity vector is often called the "speed".

Consider an airplane whose speed through motionless air is 100m/s. To reach its destination, the plane must fly east.

The "heading" of a plane of the direction in which the nose of the plane points. So, it is the direction in which the engines propel the plane.

16. If the plane has an eastward heading, and a 20 m/s wind blows towards the southwest, then the plane's speed is :

(A) 80 m/s (B) more than 80 m/s but less than 100 m/s (C) 100 m/s (D) more than 100 m/s

17. The pilot maintains an eastward heading while a 20 m/s wind blows northward. The plane's velocity is deflected from due east by what angle ?

(A) sin–1(1/5) (B) cos–1(1/5) (C) tan–1(1/5) (D) None of these

18. Because the 20 m/s northward wind persists, the pilot adjusts the heading so that the plane's total velocity is eastward. By what angle does the new heading differ form due east ?

(A) sin–11/5 (B) cos–11/5 (C) tan–11/5 (D) None of these

This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Column-I Column-II

(A) →→b.a (P) Area of the

Parallelogram formed (B) L (Q) Area below y-x graph within the limits given

(C) →→

× ba (R) Projection of

→b on

→a

(D) ∫f

t

x

x

ydx (S) Aerial velocity

(T) Zero for →a ⊥

→b

20. Column-I Column-II (A) Horizontal range is (P) π/2 maximum, when the projection is vertical (B) Vertical height is (Q) π maximum, when the angle of projection is (C) x and y coordinates (R) π/4 changes in projectile motion with (D) The magnitude of (S) Time velocity with the time in projection (T) First decrease then increase


CHEMISTRY


1. A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as CaC2O4. After filtering and washing the precipitate, it requires 40 ml of 0.250 N KMnO4, solution acidified with H2SO4 to titrate it as,

MnO4– + H+ + C2O4

–2 → Mn+2 + CO2 + 2H2O The percentage of CaO in the sample is (A) 54.0% (B) 27.1% (C) 42% (D) 84%

2. CO2 is a gas but SiO2 is a solid, because – (A) CO2 has smaller molecular size but SiO2 has

greater molecular size (B) In CO2 pπ – pπ bond exists between C and O (C) In SiO2, pπ – pπ bond exists between Si and O (D) CO2 is a discrete molecule where weak

molecular force of attraction exist, while SiO2 is a large polymeric molecule

3. In BrF3 molecule, the lone pairs occupy equatorial position to minimize –

(A) lone pair-bond pair repulsion only (B) bond pair-bond pair repulsion only (C) lone pair-lone pair repulsion and lone pair-bond

pair repulsion (D) lone pair-lone pair repulsion only

4. Which is/are correct statements ? (A) A solute will dissolve in water if hydration

energy is greater than lattice energy (B) If the anion is large compared to the cation, the

lattice energy will remain almost constant (C) solubility of II A hydroxide is in order Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 (D) none is correct

5. Sulphur trioxide is prepared by the following two reactions.

S8(s) + 8O2(g) → 8SO2(g) 2SO2(g) + O2(g) → 2SO3(g) How many grams of SO3 are produced from 1 mol

of S8 ? (A) 1280.0 (B) 640.0 (C) 960.0 (D) 320.0 6. Which of the following statements is correct ? (A) The magnitude of the second electron affinity

of Sulphur is greater than that of Oxygen (B) The magnitude of the second electron affinity

of Sulphur is less than that of Oxygen (C) The first electron affinities of Bromine and

Iodine are approximately the same (D) The first electron affinity of Fluorine is greater

than that of Chlorine

7. Iron forms two oxides, in first oxide 56 gram. Iron is found to be combined with 16 gram oxygen and in second oxide 112 gram. Iron is found to be combined with 48 gram oxygen. This data satisfy the law of -

(A) Conservation of mass (B) Reciprocal proportion (C) Multiple proportion (D) Combining volume

8. Lattice energy of BeCO3 (I) , MgCO3 (II) and CaCO3 (III) are in the order -

(A) I > II > III (B) I < II < III (C) I < III < II (D) II < I < III Questions 9 to 12 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

9. Which of the following statements is/are correct ? (A) Group 12(IIB) elements do not show

characteristic properties of transition metals (B) Among transition elements, tungsten has the

highest melting point (C) Among transition elements, group 3 (IIIB)

elements have lowest densities (D) Transition metals are more electropositive than

alkaline earth metals.

10. 11.2 g of mixture of MCl (volatile) and NaCl gave 28.7 g of white ppt with excess of AgNO3 solution. 11.2 g of same mixture on heating gave a gas that on passing into AgNO3 solution gave 14.35 g of white ppt. Hence ?

(A) Ionic mass of M+ is 18 (B) Mixture has equal mol fraction of MCl and

NaCl (C) MCl and NaCl are in 1 : 2 molar ratio (D) Ionic mass of M+ is 10

11. Specify the coordination geometry around the hybridization of N and B atoms in a 1 : 1 complex of BF3 and NH3 -

(A) N : tetrahedral, sp3 ;B : tetrahedral, sp3

(B) N : pyramidal, sp3 ; B : pyramidal, sp3 (C) N : pyramidal, sp3 ; B : planar, sp2 (D) N : pyramidal, sp3 ; B : tetrahedral, sp3

12. The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio 1 : 4. The energy difference between them may be :

(A) Either 12.09 eV or 3.4 eV (B) Either 2.55 eV or 10.2 eV (C) Either 13.6 eV or 3.4 eV (D) Either 3.4 eV or 0.85 eV



Passage : I (No. 13 to 15) The radius of the nucleus of an atom can be

approximately determined as, rnu = (1.4×10–13)A1/3

where A is mass number of the atoms and M is the charge of the electron = 4.8 × 10–10 esu.

The mass of α-particle = 4 × mass of H-atom

mass of hydrogen atom = 6

10 ×10–24 gm.

Consider during collision kinetic energy of α-particle just equal to coulombic force of repulsion.

The mass number of Au = 197 The mass number of He = 4 The atomic number of Au = 79 Given : (4)1/3 = 1.59 and (197)1/3 = 5.82

374.10

)8.4(792 2×× = 351

351.3 × = 3.245 Plank’s constant, h = 6.625 × 10–34JS

13. What is the distance between the α-particle and Au nucleus during the collision –

(A) 10.374 × 10–13 cm (B) 10.374 Å (C) 10.374 × 10–10 cm (D) 10.374 nm 14. What should be the minimum velocity of the

α-particle to strike the nucleus of 79Au197 ? (A) 3.245 × 108 m/s (B) 3.245 × 109 m/s (C) 3.245 × 105 m/s (D) 3.245 × 107 m/s 15. What is the de-broglie’s wave length associated

with a α-particle while it is moving to colloide with the Au nucleus ?

(A) 245.34

6625.6×

× ×1025m (B)245.3

6625.6 × ×10–15m

(C) 245.34

6625.6×

× ×10–15m (D)245.35

6625.6×

× ×10–15m

Passage : II (No. 16 to 18) When we use the concept that one mole of one

substance contains the same number of elementary entities as one mole of any other substance we don't actually need to know what that number is. Sometimes however we will need to work with the actual number of elementary entities in a mole of substance. This number is called Avogadro's number.

NA = 6.022137 × 1023 mol–1 The unit mol–1 'which we read as' per mole signifies

that a collection of NA molecular level entities is equivalent to one mole at the macroscopic level. For example a mole of carbon contains 6.02 × 1023 atoms of C. A mole of oxygen contains 6.02 × 1023 molecules of O2.

16. The number of atoms present in 8 g of ozone is (A) NA (B) 3NA

(C) 6

NA (D) 2

NA

17. Which of the following is a reasonable value for the number of atoms in 1.00 g of Helium ?

(A) 0.25 (B) 4.0 (C) 4.1 × 10–23 (D) 1.5 × 1023

18. How many years it would take to spend Avogadro's number of rupees at the rate of 10 Lac rupees per second ?

(A) 1.9 × 1010 years (B) 1.6 × 1010 years (C) 5 × 106 years (D) 6.4 × 105 years This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the following : Column-I Column-II (A) ICl2

– (P) Linear (B) BrF2

+ (Q) Pyramidal (C) ClF4

– (R) Tetrahedral (D) AlCl4

– (S) Square planar (T) Angular 20. Match the following : Column-I Column-II (A) 0.5 mole of SO2(g) (P) Occupy 11.2L at STP (B) 1g of H2(g) (Q) Weighs 24g (C) 0.5 mole of O3(g) (R) Total no. of atoms = 1.5NA (D) 1g molecule of O2(g) (S) Weighs 32g (T) Total no. of atoms ≈ 9 × 1023


MATHEMATICS Question 1 to 8 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. 3 cosec 20° – sec 20° is equal to : (A) 1 (B) 2 (C) 4 (D) none of these 2. If cos 20° – sin 20° = p then cos 40° is equal to-

(A) 2p2p −− (B) 2p2p −

(C) 2p2p −+ (D) none of these 3. The expression

απ+

α

π )–3(sin–2

3sin3 44

απ+

α+

π )–5(sin2

sin2– 66

is equal to -

(A) 0 (B) 1 (C) 2 (D) 5

4. If α + β = 2π and β + γ = α, then

γ+βαtan2tan

tan =

(A) 0 (B) 1 (C) 2 (D) 3

5. If 1< x < 2 , then number of solutions of the equation

tan–1(x – 1) + tan–1 x + tan–1(x + 1) = tan–1 3x, is/are

(A) 0 (B) 1 (C) 2 (D) 3 6. In a triangle ABC if BC =1 and AC = 2. Then the

maximum possible value of angle A is- (A) π/6 (B) π/4 (C) π/3 (D) π/2

7. If α ∈

π−

π− ,

23 , then the value of

tan–1(cot α) – cot–1(tan α) + sin–1(sin α) + cos–1(cos α) is equal to -

(A) 2π + α (B) π + α (C) 0 (D) π – α

8. Solution of the equation

3 sin–1

+ 2x1x2 – 4 cos–1

+

−2

2

x1x1 +

2 tan–1

− 2x1x2 =

3π is -

(A) x = 3 (B) x =3

1

(C) x = 1 (D) x = 0

Questions 9 to 12 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 9. If sin θ + sin φ = a and cos θ + cos φ = b , then -

(A) )ba(21

2–cos 22 +±=

φθ

(B) )b–a(21

2–cos 22±=

φθ

(C)

+±=

φθ

22

22

bab–a–4

2–tan

(D) cos (θ – φ) = 2

2–ba 22 +

10. Let y = sin x . sin (60º + x) . sin(60º - x). Then for

all real x - (A) the minimum value of y is 2 (B) the maximum value of y is 1 (C) y ≤ 1/4

(D) y ≥ –1/4 11. If cosec–1 x = sin–1(1/x), then x may be- (A) 1 (B) –1/2 (C) 3/2 (D) – 3/2 12. If in a triangle ABC, θ is the angle determined by

cos θ = (a – b)/c, then

(A) ab2sin)ba( θ+ = cos

2BA −

(B) ab2sin)ba( θ+ = cos

2BA +

(C) ab2

sinc θ = cos 2

BA −

(D) ab2

sinc θ = cos 2

BA +



Passage : I (No. 13 to 15) f(x) = sin cot–1 (x + 1) – cos (tan–1 x) a = cos tan–1 sin cot–1 x b = cos (2 cos–1 x + sin–1 x)

13. The value of x for which f(x) = 0 is (A) – 1/2 (B) 0 (C) 1/2 (D) 1

14. If f(x) = 0 then a2 is equal to (A) 1/2 (B) 2/3 (C) 5/9 (D) 9/5

15. If a2 = 26/51, then b2 is equal to (A) 1/25 (B) 24/25 (C) 25/26 (D) 50/51

Passage : II (No. 16 to 18)

In a ∆ABC, if cos A. cos B. cos C = 8

13 −

sin A. sin B. sin C = 8

33 + , then

16. Value of tan A + tan B + tan C is

(A) 1333

−

+ (B) 1343

−

+

(C) 1336

−

− (D) None of these

17. Value of Σ tan A .tan B = (A) 5 – 4 3 (B) 5 + 4 3

(C) 6 + 4 3 (D) 6 – 4 3

18. Value of tan A, tan B and tan C are (A) 1, 3 , 2 (B) 1, 3 , 2

(C) 1, 2, 3 (D) None of these This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct

matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the items of Column I with the items of Column II. The principal value of

Column-I Column-II

(A)

π−

65sinsin 1 is (P)

2017π

(B)

π

−−

67sincos 1 is (Q)

21

(C)

π−

415tantancos 1 is (R)

3π

(D)

π

−π−

109sin

109cos

21cos 1 is (S)

6π

(T) 2π

20. Column-I Column-II

(A) If in a ∆ABC the angles are (P) 125π

in AP and b : c = 3 : 2 then ∠A =

(B) If the length of the side of a (Q) 3

2π

triangle are 3, 5, 7 then largest angle of the triangle is

(C) If the sides of a triangle are (R) 6π

in ratio 2 : 6 : ( 3 + 1) then largest angle of the triangle is

(D) If in a triangle ABC, b = 3 , (S) π c = 1 and B – C = 90º then ∠A = (T) 3π/5


CHEMISTRY

SECTION – I Single Correct Choice Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The bond energy (in kcal mol–1) of a C–C single

bond is approximately (A) 1 (B) 10

(C) 100 (D) 1000 Ans. [C] Sol. Value is 82.6 kcal/mol. 2. The species which by definition has ZERO

standard molar enthalpy of formation at 298 K is (A) Br2(g) (B) Cl2 (g)

(C) H2O (g) (D) CH4 (g) Ans. [B] Sol. Because standard state of Cl2 is gas. 3. The ionization isomer of [Cr(H2O)4 Cl(NO2)] Cl

is (A) [Cr(H2O)4 (O2N)] Cl2

(B) [Cr(H2O)4 Cl2](NO2) (C) [Cr(H2O)4 Cl(ONO)]Cl (D)

[Cr(H2O)4 Cl2(NO2)]. H2O Ans. [B]

Sol. Ionization isomer have different ions in solution

4. The correct structure of ethylenediaminetetraacetic

acid (EDTA) is -

(A)

HOOC–CH2 CH2 –COOH

N–CH=CH–N

HOOC–CH2 CH2–COOH

(B)

HOOC COOH

N–CH2–CH2–N

HOOC COOH

(C)


N–CH2–CH2–N


(D)

HOOC–CH2 H N–CH–CH–N

H CH2–COOH

CH2 COOH

CH2 HOOC

Ans. [C] Sol. Structure is


N–CH2–CH2–N


5. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

(A) BrCH2CH2CH2CH2CH3 and CH3CH2C≡CH (B) BrCH2CH2CH3 and CH3CH2 CH2 C≡CH (C) BrCH2CH2CH2 CH2CH3 and CH3C≡CH (D) BrCH2CH2CH2CH3 and CH3CH2 C ≡CH

IIT-JEE 2010 PAPER-I (PAPER & SOLUTION)

Time : 3 Hours Total Marks : 243

Instructions : • The question paper consists of 3 parts (chemistry, Mathematics and Physics). Each part consists of four sections. • For each question in Section I, you will be awarded 3 marks if you have darkened only the bubble corresponding to

the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.

• For each question in Section II, you will be awarded 3 marks if you darkem only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. Partial marks will be awarded for partially correct answers. No negative marks will be awarded in this section.

• For each question in Section III, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.

• For each question in Section IV, you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for in this section.


Ans. [D]

Sol. (i) CH3–CH2–C≡CH NaNH2 CH3CH2 C≡ +NaC–

(ii) CH3–CH2–CH2–CH2–Br + CH3CH2–C≡ +NaC–

––→ CH3–CH2–CH2–CH2–C ≡C–CH2–CH3

6. The correct statement about the following disaccharide is

H

OCH2CH2O H

H

OH OH

H

CH2OH

O

OH OH

H

OH

H

H

CH2OH

(b) (a)

CH2OH

OH

(A) Ring (a) is pyranose with α-glycosidic link

(B) Ring (a) is furanose with α-glycosidic link (C) Ring (b) is furanose with α-glycosidic link

(D) Ring (b) is Pyranose with β-glycosidic link Ans. [A] Sol. Ring (a) is pyranose with α-glycosidic link

7. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is -

(A)

T

k (B)

T

k

(C)

T

k (D)

T

k

Ans. [A]

Sol. k = Ae–Ea/RT ; k ∝ RT/E– ae

T

k

8. In the reaction OCH3 HBr the

products are

(A) OCH3 and H2 Br

(B) Br and CH3 Br

(C) Br and CH3OH

(D) OH and CH3Br

Ans. [D] Sol.

OCH3HBr OH + CH3Br

SECTION – II Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

9. In the reaction

NaOH(aq)/Br2

OH

the

intermediate(s) is (are)

(A)

OBr

Br (B)

O

BrBr

(C)

O

Br (D)

O

Br

Ans. [A, B, C]

Sol.

NaOH(aq)/Br2

OH OH Br

Br

Br

Bromination take place at ortho & para position due to activation of benzene ring by –OH group.

10. The reagent(s) used for softening the temporary

hardness of water is (are) - (A) Ca3(PO4)2 (B) Ca(OH)2

(C) Na2CO3 (D) NaOCl

Ans. [B, C, D]

11. Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair(s) of solutions which from a buffer upon mixing is(are)


(A) HNO3 and CH3COOH (B) KOH and CH3COONa

(C) HNO3 and CH3COONa (D) CH3COOH and CH3COONa

Ans. [C, D] Sol. Acidic buffer is made up of weak acid & it's

conjugate ion 12. In the Newman projection for 2,2-dimethylbutane

CH3 H3C

H H

Y

X

X and Y can respectively be

(A) H and H (B) H and C2H5 (C) C2H5 and H (D) CH3 and CH3

Ans. [B, D]

Sol.

CH3–C–CH2–CH3

CH3

CH3

≡

CH3C H3

HH

CH3

CH3

1 2 3 4

along C2–C3 Bond Option [D]

X = CH3 Y = CH3

CH3–C–CH2–CH3

CH3

CH3

≡

CH3C H3

HH

H

C2H5

1 2 3 4

along C1–C2

Option [B] X = H Y = C2H5 13. Among the following, the intensive property is

(properties are) (A) molar conductivity

(B) electromotive force (C) resistance (D) heat capacity

Ans. [A, B]

SECTION – III Paragraph Type

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Paragraph for Questions No. 14 to 15 The concentration of potassium ions inside a

biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

M(s) | M+ (aq; 0.05 mloar) || M+ (aq; 1 molar) | M(s)

For the above electrolytic cell the magnitude of the potential |Ecell| = 70 mV.

14. For the above cell (A) Ecell < 0; ∆G > 0 (B) Ecell > 0; ∆G < 0 (C) Ecell < 0; ∆Gº > 0 (D) Ecell > 0; ∆Gº < 0 Ans. [B] Sol. For concentration cell

Ecell = – nk log

105.0 [where k is +ve constant]

= + ve ∴ Ecell > 0 ; ∆G < 0 15. If the 0.05 molar solution of M+ is replaced by a

0.0025 molar M+ solution, then the magnitude of the cell potential would be

(A) 35 mV (B) 70 mV (C) 140 mV (D) 700 mV

Ans. [C]

Sol. Ecell = – nk log

10025.0

= 2

105.0log

nk– = 2 × 70 = 140

Paragraph for Question No. 16 to 18

Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO4. 5H2O), atacamite (Cu2Cl (OH) 3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However, 18% of the world copper production come from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.


16. Partial roasting of chalcopyrite produces (A) Cu2S and FeO (B) Cu2O and FeO (C) CuS and Fe2O3 (D) Cu2O and Fe2O3

Ans. [A] 17. Iron is removed from chalcopyrite as - (A) FeO (B) FeS

(C) Fe2O3 (D) FeSiO3

Ans. [D]

18. In self-reduction, the reducing species is - (A) S (B) O2–

(C) S2– (D) SO2

Ans. [C]

SECTION – IV Integer Type

This section contains. TEN questions. The answer to each questions is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

19. The number of neutrons emitted when U23592

undergoes controlled nuclear fission to SrandXe 90

3814254 is

Ans. [4] Sol. pynxSrXenU 1

110

9038

14254

10

23592 +++→+

235 + 1 = 142 + 90 + x x = 4

20. The total number of basic groups in the following

form of lysine is

H3N–CH2–CH2–CH2–CH2

CH–C

O

OH2N

⊕

Ans. [2] 21. The total number of cyclic isomers possible for a

hydrocarbon with the molecular formula C4H6 is Ans. [5]

Sol. , , , ,

22. In the scheme given below, the total number of

intramolecular aldol condensation products form 'Y' is

1. O3 2. Zn, H2O

Y 1. NaOH(aq) 2. heat

Ans. [1]

Sol.

1. O3 2. Zn, H2O

O

O O

O

OH–

O

OH

Heating (Only one product)

O

23. Amongst the following, the total number of compound soluble in aqueous NaOH is

NCH3

H3CCOOH OCH2CH3

CH2OHOH

NO2 OH

NCH3 H3C

CH2CH3 CH2CH3

COOH

Ans. [4]

Sol.

COOH OH

OH

NCH3H3C

COOH

,

,

24. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is

KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2

FeCl3 K2CO3 NH4NO3 LiCN Ans. [3] Sol. Basic salt are KCN, LiCN, K2CO3

25. Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is

Ans. [0]

26. The value of n in the molecular formula BenAl2Si6O18 is


Ans. [3] Sol. Be3Al2Si6O18

Si6O18–12 ion

27. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value is

Ans. [3]

Sol. 3

0.2525.252.25 ++ = 25.15

Significant figure = 3 Significant figure in the answer can not be more

than least significant figure any given value.

28. The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained :

[R] (molar)

1.0 0.75 0.40 0.10

t (min.) 0.0 0.05 0.12 0.18 The order of the reaction is Ans. [0] Sol. R → P Assume zero order R = [R0] – kt

k = t

]R[]R[ 0 −

k1 = 5.075.01− = 5

k2 = 12.0

4.01− = 5

k3 = 18.0

1.01− = 5

∴ order of reaction should be zero.

MATHEMATICS


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

29. Let f , g and h be real-valued functions defined on

the interval [0, 1] by f(x) = 2x2x ee −+ ,

g(x) =2x2x exe −+ and h(x) =

2x2x2 eex −+ . If a , b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then

(A) a = b and c ≠ b (B) a = c and a ≠ b

(C) a ≠ b and c ≠ b (D) a = b = c

Ans.[D]

Sol. f '(x) = 2x(

− − 22 xx ee

g'(x) = ( )1x2x2e 2x2+−

h'(x) = 2x3ex2

Q all f'(x) , g'(x), h'(x) are positive so all attains absolute maxima at x = 1

So Q f (1) = g(1) = h(1) = e + e–1 = a = b = c

30. Let p and q be real numbers such that p ≠ 0, p3 ≠ q and p3 ≠ – q. If α and β are non zero complex numbers satisfying α + β = – p and α3 + β3 = q,

then a quadratic equation having βα and

αβ as its

roots is -

(A) (p3 + q) x2 – (p3 + 2q) x + (p3 + q) =0

(B) (p3 + q) x2 – (p3 – 2q) x + (p3 + q) =0

(C) (p3 – q) x2 – (5p3 – 2q) x + (p3 – q) =0

(D) (p3 – q) x2 – (5p3 + 2q) x + (p3 – q) =0

Ans.[B]

Sol. α + β = – p ……….(1)

α3 + β3 = q

⇒ (α+ β) ( α2 + β2 – αβ) = q

⇒ (α+ β) (( α + β)2 – 3αβ) = q

⇒ (–p) (p2 – 3αβ) = q

αβ = p3pq 3+ …..(2)

Now S = αβ

+βα =

αβαβ−β+α 2)( 2

(Sum of root) S = qpq2p

3

3

+

− using (1) and (2)


(Product of root) P = αβ

βα . = 1

31. Equation of the plane containing the straight line

4z

3y

2x

== and perpendicular to the plane

containing the straight lines 2z

4y

3x

== and

3z

2y

4x

== is

(A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0

(C) x – 2y + z = 0 (D) 5x + 2y – 4z = 0

Ans.[C]

Sol. Plane passing through origin (0, 0, 0) and normal vector to plane is perpendicular to k2j4i3 ++ ,

k3j2i4 ++ and k4j3i2 ++ i.e. normal vector

to plane is kj2i +− so equation to plane is x – 2y + z = 0.

32. If the angle A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression

A2sinacC2sin

ca

+ is -

(A) 21 (B)

23

(C) 1 (D) 3

Ans.[D]

Sol. A2sinacC2sin

ca

+

AcosAsin2acCcosCsin2

ca

+

= R

AcoscCcosa + = Rb =

RBsinR2 = 3

33. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times If r1, r2 and r3 are the numbers obtained on the die, then the probability that 321 rrr ω+ω+ω = 0 is

(A) 181 (B)

91 (C)

92 (D)

361

Ans.[C]

Sol. n(s) = 63

Similarly total number of elements in events set is 48

= 21648 =

5412 =

92

34. Let P, Q, R and S be the points on the plane with position vectors – ji2 − , i4 , j3i3 + and

j2i3 +− respectively. The quadrilateral PQRS must be a -

(A) Parallelogram, which is neither a rhombus nor a rectangle

(B) Square

(C) Rectangle, but not a square

(D) Rhombus, but not a square

Ans.[A]

Sol. ji6PQ +=

ji6RS +=

j3iRQ −=

j3iSP −=

RQPQ ≠ (∴ not a rhombus or a rectangle)

PQ || RS

r1 r2 r3 1 1

2 3 4 5 6

x 3,6 2,5 x

3,6 2,5

2 1 2 3 4 5 6

3,6 x

1,4 3,6 x

1,4 3 1

2 3 4 5 6

2,5 1,4 x

2,5 1,4 x


RQ || SP

Also PQ . RQ 0≠

∴ PQRS is not a square

⇒ PQRS is a parallelogram

35. The value of 0x

lim→

dt4t

)t1(ntx1 x

043 ∫ +

+l is

(A) 0 (B) 121

(C) 241 (D)

641

Ans.[B]

Sol. Use L'hospital rule in

0x

Lim→ ∫ +

+x

0 43 dt4t

)t1(ntx1 l

= 0x

Lim→ 24 x3)4x(

)x1(nx+

+l

= 0x

Lim→

)4x(3.x

)x1(n2 +

+l

= 4.3

1 = 121

36. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system

=

001

zyx

A has exactly two distinct solution, is -

(A) 0 (B) 29 – 1 (C) 168 (D) 2

Ans.[A]


This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.

37. Let ABC be a triangle such that ∠ ACB =6π and

let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x +1, b = x2 – 1 and c = 2x + 1 is (are)

(A) ( )32 +− (B) 1+ 3

(C) 32 + (D) 4 3

Ans.[B] Sol. As sum of two sides is always greater than third

side, So x > 1 Now

Cos 6π =

ab2c–ba 222 +

23 =

ab2c–ab2)b–a( 22 +

3 – 2 =ab

c–)b–a( 22

3 – 2 = )1–x()1xx(

)1x2(–)2x(22

22

++++

3 – 2 = 1xx

3–2 ++

x2 + x + 1 = 3–2

3

x2 + x + 1 = 3(2 + 3 )

x2 + x – 5 – 3 3 = 0

(x – (1 + 3 ) (x + (2 + 3 )) = 0

x = 1 + 3 , – (2 + 3 )

So x = 3 + 1 38. Let A and B be two distinct points on the

parabola y2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be -

(A) r1

− (B) r1

(C) r2 (D)

r2

−

Ans.[C, D] Sol. Let A ( )1

21 t2,t B ( )2

22 t2,t

Slope = 21

22

12

tt)tt(2

−

− = 21 tt

2

+

Equation of circle will be )tx( 2

1− )tx( 22− + )t2y( 1− )t2y( 2− = 0

22 yx + – )tt(x 22

21 + – )tt(y2 21 + + 2

221 tt + 4t1t2

= 0 As it touches x axis so

22

21 tt + 4t1t2 =

4)tt( 22

221 +

4 22

21 tt + 16 t1t2 = 4

241 tt + + 2 2

221 tt

( )222

21 tt − = 16 t1t2 . . . (1)

AB is diameter so

222

21 )tt( − + 4 (t1– t2)2 = 4r2 . . . (2)


From (1) and (2) 4 t1 t2 + (t1– t2)2 = r2 (t1+ t2)2 = r2 t1+ t2 = ± r

∴ Slope = ±r2

39. The value(s) of ∫ +

−1

02

44

x1

)x1(xdx is (are) -

(A) π−722 (B)

1052

(C) 0 (D) 2

31571 π

−

Ans.[A]

Sol. I = ∫ +−

1

02

44

x1)x1(x

= ∫ +

+−+−1

02

45678dx

x1xx4x6x4x

= ∫ +−+−+−

1

02

2456 dx)x1

44x4x5x4x(

= π−722

40. Let z1 and z2 be two distinct complex numbers let

z = (1–t) z1 + tz2 for some real number t with 0 < t < 1. If Arg (w) denotes the principal argument of a nonzero complex number w, then

(A) | z – z1 | + | z – z2 | = | z1 – z2 | (B) Arg (z – z1) = Arg (z – z2)

(C) 1212

11zzzz

zzzz−−

−−= 0

(D) Arg (z – z1) = Arg (z2 – z1)

Ans.[A,C,D]

Sol. t = 12

1

zzzz

−−

So, 12

1

zzzz

−− = t eio V t ∈ (0, 1)

So Geometrically

z1 z z2

So option A, C, D are true.

41. Let f be a real valued function defined on the

interval (0, ∞ ) by f(x) = nl x + ∫ +x

0

dttsin1 .

Then which of the following statement(s) is (are) true ?

(A) f"(x) exists for all x ∈ (0, ∞ ) (B) f '(x) exists for all x ∈ (0, ∞ ) and f ' is

continuous on (0, ∞ ), but not differentiable on (0, ∞ )

(C) there exists α > 1 such that |f '(x)| < | f (x) | for all x ∈ (α, ∞ )

(D) there exists β > 0 such that |f '(x)| + | f '(x) | ≤ β for all x ∈ (0, ∞ )

Sol.[B,C]

f(x) = ln x + tdtsin1x

0∫ +

f ′(x) = xsin1x1

++ = x1 + |cos

2x + sin

2x |

(A is not correct)

x1 + xsin1+ < ln x + dttsin1

x

0∫ +

Q ln x > x1 for some x = α ∀ α > 1

and xsin1+ < ∫ +x

0

dttsin1

for some x = α ∀ α > 1 so option C is correct


This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


The circle x2 + y2 – 8x = 0 and hyperbola 9

x2–

4y2

= 1 intersect at the points A and B

42. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is -

(A) 2x – 5 y – 20 = 0

(B) 2x – 5 y + 4 = 0 (C) 3x – 4y + 8 = 0

(D) 4x – 3y + 4 = 0 Ans.[B]


Sol. y = m(x – 4) ± 4 2m1+

y = mx ± 4m9 2 −

– 4m ± 4 2m1+ = ± 4m9 2 −

16m2 + 16 + 16m2 m 32 m 2m1+ = 9m2 – 4

m 32m 2m1+ = – 23m2 – 20 1024m2 + 1024 m4 = 529m4 + 400 + 920 m2 495 m4 + 104 m2 – 400 = 0 (5m2 – 4) (99m2 + 100) = 0

∴ m2 = 54 ∴ m = ±

52

So tangent with positive slope

y = 5

2 x ± 5

4

2x – 5 y ± 4 = 0 43. Equation of the circle with AB as its diameter is (A) x2 + y2 – 12 x + 24 = 0

(B) x2 + y2 + 12 x + 24 = 0 (C) x2 + y2 + 24 x – 12 = 0

(D) x2 + y2 – 24x – 12 = 0 Ans.[A] Sol. x2 + y2 – 8x = 0 4x2 – 9y2 = 36

x2 +

−9

36x4 2– 8x = 0

13x2 – 72x – 36 = 0 (x – 6) (13x + 6) = 0

x = 6, 13

6−

x = 6, y = ± 12 ∴ Equation of required circle (x – 6)2 + (y – 12 ) (y + 12 ) = 0 x2 + y2 – 12x + 24 = 0


Let p be an odd prime number and Tp be the following set of 2 × 2 matrices :

−∈

== 1p,......,2,1,0c,b,a:

acba

ATp

44. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is -

(A) (p – 1)2 (B) 2(p – 1) (C) (p –1)2 + 1 (D) 2p –1

Ans.[D]

Sol. |A | = a2 – bc if A is symmetric b = c then | A | = (a + b) (a – b) So a & b can attain 2(p – 1) solution It A is skew symmetric then a = b = c = 0 So total no. of solution = 2p – 2 + 1 = 2p – 1 45. The number of A in Tp such that the trace of A is

not divisible by p but det (A) is divisible by p is [Note : The trace of a matrix is the sum of its

diagonal entries.] (A) (p – 1) (p2 – p + 1) (B) p 3 – (p – 1)2

(C) (p –1)2 (D) (p –1) (p2 – 2) Ans.[C] 46. The number of A in Tp such that det (A) is not

divisible by p is - (A) 2p2 (B) p3 – 5p

(C) p3 –3p (D) p3 – p2

Ans. [D]

SECTION – IV Integer type

This section contains TEN paragraphs. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

47. Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to –

Ans.[9]

Sol. Y – y =dxdy ( X – x)

y – x dxdy = x3

xdxdy – y = – x3

2

1

TT –

xy = – x2

I.F. = x1

xy = ∫ − xdx + C

xy =

2x2− + C

1 = 21

− + C ∴ C = 23

∴ y = 2

)3x(x 2 +−


∴ f(x) = 2

)3x(x 2 +−

f(–3) = 2

)39(3 +−− = 9

48. The number of values of θ in the interval

ππ−

2,

2 such that

5nπ

≠θ for n = 0 , ± 1, ± 2

and tan θ = cot 5 θ as well as sin 2 θ = cos 4 θ is – Ans.[3]

Sol. tan θ = cot 5θ ⇒ θ = (2n + 1) 12π

So θ = ± 12π , ±

4π , ±

125π

∴ Sin 2θ = cos 4θ ⇒ sin 2θ = 21 or – 1

only θ = 12π ,

125π ,

4–π satisfies the given

conditions So total number of solution = 3

49. The maximum value of the expression

θ+θθ+θ 22 cos5cossin3sin1 is

Sol.[2] f (θ) = θ+θθ+θ 22 cos5cossin3sin

1

= θθ+θ+ cossin3cos41

12

= θ+θ++ 2sin

23)2cos1(21

1

= θ+θ+ 2sin

232cos23

1

So f(θ)max =

494–3

1

+

= 2

50. If ar

and br

are vector is space given by

5j2ia −

=r

and 14

k3ji2b

++= , then the value of

( ) ( ) ( )[ ]b2aba.ba2rrrrrr

−××+ is Ans.[5] Sol. 0ba&1|b||a| =⋅==

)]b2–a()ba[()ba2( ××⋅+

= 22 |a|4|b|]a2b[)ba2( +=+⋅+ = 5

51. The line 2x + y = 1 is tangent to the hyperbola

2

2

2

2

by

ax

− =1. If this line passes through the point

of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

Ans.[2] Sol. 1 = 4a2 – b2 ... (1)

ea2 = 1

a =2e ... (2)

also b2 = a2 (e2 – 1) ... (3) (1) & (3) 1 = 4a2 – a2e2 + a2 ⇒ 1 = 5a2 – a2e2

⇒ 1 = 4e–

4e5 42

⇒ e4 – 5e2 + 4 = 0 ⇒ (e2 – 4) (e2 – 1) = 0 ∴ e = 2 52. If the distance between the plane Ax – 2y + z = d

and the plane containing the lines

43z

32y

21x −

=−

=− and

54z

43y

32x −

=−

=− is

6 , then | d | is – Ans.[6]

Sol. 543432kji

= )1(k)2(j)1(i −+−−−

Plane is normal to vector kj2i +− 1(X – 1) – 2 (Y –2) + 1(Z – 3) = 0 X – 2Y + Z = 0

6 = 6|d|⇒ | d | = 6

53. For any real number x, let [x] denote the largest

integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by

−+−

=evenis]x[ifx]x[1oddis]x[if]x[x

)x(f

Then the value of ∫−

π 10

10

2)x(f

10 cos πx dx is

Ans.[4]

Sol. f(x) =

= evenis]x[Vx1oddis]x[Vx

graph of y = f(x) is


–5 –3 –1 1 2 3 –2 –4 0

Q f(x) & cos πx both are even functions

So, I = 10

2π ∫−

10

10

)x(f cos πx dx

= 5

2π∫ π10

0

dx)xcos()x(f

∴ f(x) & cos πx both are periodic then

I = π2 ∫2

0

)x(f cos(πx) dx

=

π−+π−π ∫ ∫

1

0

2

1

2 dx)xcos()1x(dx)x(xcos)x1(

=

π+

π 22 22 = 4

54. Let ω be the complex number cos 3

2sini3

2 π+

π .

Then the number of distinct complex number z

satisfying ω+ω

ω+ωωω+

z11z

1z

2

2

2

= 0 is equal to =

Ans.[1] Sol. On solving the determinant It become z3 = 0 So no. of solutions = 1 55. Let Sk, k = 1, 2, ……, 100, denote the sum of the

infinite geometric series whose first term is k

!k1k − and the common ratio is

k1 . Then the value

of !100

1002+ ( )∑

=+−

100

1kk

2 S1k3k is –

Ans.[4]

Sol. Sk = K

K

∑=

100

1K

|(k2 – 3k + 1)Sk|

= 1 + 1 + 1–k

)1k3–k( 2100

3K

+∑=

= 2 + 1–k

k–2–k

1–k∑

= 2 + 2 – 99

100 = 4

56. The number of all possible values of θ, where 0 < θ < π, for which the system of equations

(y + z) cos 3 θ = (xyz) sin 3θ

x sin 3θ = z

3sin2y

3cos2 θ+

θ

(xyz) sin3θ = (y + 2z) cos 3θ + y sin 3θ) have a solution (x0, y0, z0) with y0z0 ≠ 0, is Ans.[3] Sol. (xyz) sin 3θ + y (– cos 3θ) + z (– cos 3θ) = 0 (xyz) sin 3θ + y (–2 sin 3θ) + z (–2 cos 3θ) = 0 (xyz) sin 3θ + y (– cos 3θ − sin 3θ ) + z (– 2cos

3θ) = 0 For y0z0 ≠ 0 ⇒ Nontrivial solution

θ−θ−θ−θθ−θ−θ

θ−θ−θ

3cos23sin3cos3sin3cos23sin23sin

3cos3cos3sin = 0

sin 3θ cos3θ 23sin3cos123sin2113cos1

θ+θθθ

= 0

sin3θ cos3θ [(4sin 3θ – 2 cos 3θ – 2sin 3θ) – (2cos 3θ – cos 3θ – sin 3θ) + 2 cos 3θ – 2 sin 3θ] = 0

⇒ (sin3θ cos3θ) [2 sin3θ – 2cos 3θ – cos 3θ + sin 3θ + 2cos 3θ – 2sin 3θ] = 0

⇒ (sin3θ cos3θ) (sin3θ – cos 3θ) = 0

⇒ sin 3θ = 0 ⇒ θ = 3π ,

32π

⇒ cos 3θ = 0 ⇒ θ = 6π ,

2π ,

65π ,

67π

These two donot satisfy system of equations

⇒ sin 3θ = cos 3θ ⇒ 3θ = 4π ,

45π ,

49π

⇒ θ = 12π ,

125π ,

43π

= 3

No. of solutions = 3

PHYSICS


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.


57. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is-

4R

4R 3R

P

(A) )524(

R7GM2

− (B) )524(R7

GM2−−

(C) R4

GM (D) )12(R5

GM2−

Ans. [A] Sol. ∆Wext = U2 – U1 for unit +ve mass U1 = V1 and U2 = V2 = 0

V1 = ∫dV = ∫ +− 2/122 )R16r(

Gdm

= ∫ +ππ

− 2/1222 )R16r(R7rdr2GM

Put r2 + 16R2 = t2

V1 = ∫−R24

R52 dt

R7GM2 = )524(

R7GM2

−−

∆Wext = U2 – U1 = )524(R7

GM2−

58. A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is µ and tan θ > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sinθ – µ cosθ) to P2 = mg(sinθ + µ cosθ), the frictional force f versus P graph will look like –

P

θ

(A)

P2

P1 P

f

(B)

P2P1 P

f

(C)

P2

P1

P

f

(D)

P2P1

P

f

Ans. [A] Sol. In the given range block is in equilibrium so P – mg sin θ + f = 0 f = mg sinθ – P Equation of straight line with negative slope.

59. A real gas behaves like an ideal gas if its - (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high Ans. [D] Sol. Reason : PV = nRT holds true in case of low

pressure and high temperature conditions.

60. Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is-

tL

(A) directly proportional to L (B) directly proportional to t

(C) independent of L (D) independent of t

Ans. [C]

Sol. R = ALρ =

LtLρ

R = tρ

R is independent of L

61. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistances is-

(A) 6040100 R

1R1

R1

+= (B) R100 = R40 + R60

(C) R100 > R60 > R40 (D) 4060100 R

1R1

R1

+>

Ans. [D]

Sol. Rated power = R

V2

R ∝ powerRated

1 P1 > P2 > P3

∴ 321 R

1R1

R1

>>

4060100 R

1R1

R1

>>


62. To verify Ohm's law, student is provided with a test resistor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct to carry out the experiment is-

(A)

R2

G1

RT

G2 R1

V

(B)

R1

G1

RT

G2 R2

V

(C)

R1

RT

G2

R2

V

G1

(D)

R2

RT

G2

R1

V

G1

Ans. [C]

Sol. Converted ammeter =

G

R2

Converted ammeter =

G R1

Voltmeter should be connected in parallel to RT and Ammeter should be connected in series with RT.

63. A thin flexible wire of length L is connected to two adjacent fixed points and carries a current i in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into plane of the paper, the wire takes the shape of a circle. The tension in the wire is-

x x x x x x xx x x x x x xx x x x x x xx x x x x x xx x x x x x xx x x x x x x

(A) IBL (B) π

IBL

(C) π2

IBL (D) π4

IBL

Ans. [C] Sol.

2dcosT θ

2dsinT2 θ

2dcosT θ

Fmagnetic = iB×Rdθ

iTT

dθ

T = tension

iB × R dθ = 2

dsinT2 θ

T = iBR 2πR = L

R = π2

L ∴ T = π2

iBL

64. An AC voltage source of variable angular

frequency ω and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased -

(A) the bulb glows dimmer (B) the bulb glows brighter

(C) total impedance of the circuit is unchanged (D) total impedance of the circuit increases

Ans. [B]

Sol.

R

~

C

V0, ω


Z = 2

2

C1R

ω+

ω increased z decreased ∴ current in circuit increase ∴ Bulb glow brighter.


This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

65. A student uses a simple pendulum 1 m length to determine g, the acceleration due to gravity. He uses a stop watch the least count of 1 sec for this records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true ?

(A) Error ∆T in measuring T, the time period, is 0.05 seconds

(B) Error ∆T in measuring T, the time period, is 1 second

(C) Percentage error in the determination of g is 5%

(D) Percentage error in the determination of g is 2.5 %

Ans. [A,C]

Sol. Time period (T) = nsoscillatioof.no

)t(timeTotal

So TT∆ =

tt∆ =

sec40sec1

∆T = 2401

× = 0.05 sec

T = g

2 lπ ; T2 =

g4 2lπ ; g = 2

2

T4 lπ ;

gg∆ =

l

l∆ + TT2 ∆ ;

l

l∆ = 0

∆T = 0.05 sec ; T = 2 sec.

Putting we get l

l∆ = 2

05.02× = 0.05

gg∆ × 100 = 5 %

66. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that

QQ

(A) |Q1| > |Q2|

(B) |Q1| < |Q2| (C) at a finite distance to the left of Q1 the electric

field is zero (D) at a finite distance to the right of Q2 the

electric field is zero Ans. [A,D] Sol. Number of field lines emitting from Q1 is more

than number of field lines reaching at Q2 So | Q1 | > | Q2 | and if so E

r at a point which is right to Q2 will be

zero.

67. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, As shown in figure. Its pressure at A is P0. Choose the correct option(s) from the following

V4V B

A CV

T T (A) Internal energies at A and B are the same (B) Work done by the gas in process AB is P0V0

ln 4 (C) Pressure at C is P0/4

(D) Temperature at C is 4

T0

Ans. [A,B] Sol. From figure AB → isothermal process So TA = TB ⇒ Internal energies will be same.

WAB = nRT0 ln

1

2

VV = P0V0 ln 4

It is not given that line BC passes through origin. So we can't find pressure or temperature at point C.

68. A Point mass 1 kg collides elastically with a

stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are) correct for the system of these two masses ?

(A) Total momentum of the system is 3 kg ms–1 (B) Momentum of 5 kg mass after collision is 4

kg ms–1 (C) Kinetic energy of the centre of mass is 0.75 J (D) Total kinetic energy of the system is 4 J


Ans. [A,C] Sol.

1kg 5kg

v1

Before collision

1kg 5kgv2

After collision

2m/sec

Collision is elastic so v2 + 2 = v1 ......(i) Conservation of momentum, 1 × v1 + 0 = – 2 × 1 + 5 × v2

....(ii) Solving v1 = 3 m/sec v2 = 1 m/sec Total momentum systemp

r = 1 × v1 = 3 kg-m/sec

Momentum of 5 kg = 5 × v2 = 5 kg-m/sec

vCM = 6

0531 ×+× = 0.5 m/sec

kCM = 2CM21 v)MM(

21

+× = 0.75 joule

ksystem = 21 × 1 × 9 = 4.5 joule.

69. A rat OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60º (see figure). If the refractive index of the material of the prism is 3 , which of the following is (are) correct ?

60º

90º 75º

135ºC

B

A D

60º

(A) The ray gets totally internally reflected at face

CD (B) The ray comes out through face AD (C) The angle between the incident ray and the

emergent ray is 90º (D) The angle between the incident ray and the

emergent ray is 120º Ans. [A,B,C]

Sol. Refraction at first surface AB : rsin

º60sin = 13

r = 30º

60º

90ºr1

30º = r135º

60º

E

D

C

B

O

Ar2 45º

60º

it hits at E By geometry angle r1 = 45º, r2 = 45º

We know sinθC = 3

1 and sin 45º = 2

1

2

1 > 3

1

so 45º > θC So total internal reflection occurs. After reflection angle of incidence at AD will be

30º so ray comes out making an angle 60º with the normal at AD.

60º 30º60º

Final Ray Incident Ray


This section contains 2 paragraphs.. Based upon the first paragraph 2 multiple choice question and based upon the second paragraph 3 multiple choice question have to be answered. Each of these questions has four choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.


Electrical resistance of certain materials, knows as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that their critical temperature becomes smaller than TC (0) if they are placed in a magnetic field, i.e. the critical temperature TC(B) is a function of the magnetic field strength B. The dependence of TC (B) on B is shown in the figure.

B

TC(B)

O

TC(0)


70. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1, which of the following graphs shows the correct variation of R with T in these fields ?

(A)

T

R

O

B1 B2

(B)

T

R

O

B1

B2

(C)

T

R

O

B2 B1

(D)

T

R

O

B2

B1

Ans. [A] Sol.

B2 > B1 so TC(B2) < TC(B1) Dashed Solid line line Resistance ∝ Temperature above critical . 71. A superconductor has TC(0) = 100 K. When a

magnetic field of 7.5. Tesla is applied, its TC decreases to 75 K. For this material one can definitely say that when

(A) B = 5 Tesla, TC (B) = 80 K (B) B = 5 Tesla, 75 K < TC (B) < 100 K (C) B = 10 Tesla, 75 K < TC (B) < 100 K (D) B = Tesla, TC(B) = 70 K

Ans. [B] Sol. TC(0) = 100 K, B = 0 B = 7.5 tesla TC (B) = 75 If B = 5 Tesla, TC(B) should be greater than 75 K

Paragraph for Question No. 72 to 74 When a particle of mass m moves on the x-axis in

a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time

period is proportional to km , as can be seen

easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = αx4 (α > 0) for | x | near the origin and becomes a constant equal to V0 for | x | ≥ X0 (see figure)

V0

X0 x

V(x)

72. If the total energy of the particle is E, it will

perform periodic motion only if- (A) E < 0 (B) E > 0

(C) V0 > E > 0 (D) E > V0 Ans. [C] Sol. Energy Total should be less than maximum

potential energy so E < V0 and E > 0.

73. For periodic motion of small amplitude A, the time period T of this particle is proportional to-

(A) αmA (B)

αm

A1

(C) m

A α (D) mA

1 α

Ans [B] Sol. V(x) = α x4

[α] = 4]x[)]x(V[ =

]L[]TML[

4

22 −

= [ML–2T–2]

Time period ∝ (Amplitude)x (α)y (Mass)z [T] = [L]x [ML–2T–2]y [M]2

Solving x = –1, y = 21

− , z = 21

T = A–1 α–1/2 M1/2 = αM

A1

74. The acceleration of this particle for | x | > X0 is - (A) proportional to V0

(B) proportional to 0

0

mXV

(C) proportional to 0

0

mXV

(D) zero


Ans. [D] Sol. for |x| > x0 U = constant

F = – 0dxdU

=

acceleration = zero.

SECTION – IV Integer Type

This section contains TEN questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 75. A stationary source is emitting sound at a fixed

frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the car in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m/s.

Ans. [7]

Sol.

v1 v2

f0 * S

f1 = f0

+v

vv 1

f2 = f1

− 1vvv =

)vv()vv(f

1

10

−+

)vv()vv(f

'f2

202 −

+=

0

22

ff'f − =

)vv()vv(

2

2

−+ –

)vv()vv(

1

1

−+ = 0.012

)vv)(vv(

)vv)(vv()vv)(vv(

12

2112

−−−+−−+ = 0.012

)vv)(vv()vv(vvvvvv)vv(vv

12

21212

21122

−−−−+−−−+ = 0.012

)vv)(vv(

)vv(v2

21

12

−−− = 0.012

)vv(v2

12 − = 0.012

(v2 – v1) = 0.006 × 330 × 5

18

= 7.128

76. The focal length of thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its

image changes from m25 to m50. The ratio 50

25

mm

is ? Ans. [6]

Sol. m = uf

f+

m25 = 2520

20−

= –4 ; m50 = 5020

20−

= –32

50

25

mm

= 6

77. An α-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are λα and λp

respectively. The ratio αλ

λp , to the nearest integer

is ? Ans. [3] Sol. After accelerating through V0 KE of a particle

becomes = qV0 evolts so KEα = 200 eV KEp = 100 eV

λdebroglied = MKE2h

αλ

λP = ppKEM

KEM αα = 10012004

×× = 22

= 2 × 1.414 ~= 3

78. When two identical batteries of internal resistance 1Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2 then the value of R is Ω is ?

Ans. [4] Sol.

R

E E

1Ω 1Ω

i1

R

E

E

1Ω

1Ω

i2

i1 = 2R

E2+

Eeq = E

J1 = R.2R

E2 2

+ req = 0.5Ω


i2 = 5.0R

E+

J2 = R5.0R

E2

+

From J1 = 2.25 J2

R2R

E22

+

= 2.25 R5.0R

E2

+

2R

2+

= 5.0R

5.1+

2R + 1 = 1.5 R + 3 0.5 R = 2

R = 5.0

2 = 4 Ω

79. Two spherical bodies A(radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B ?

Ans. [9] Sol. λA = 500 nm λB = 1500 nm λATA = λBTB

2

1

TT =

B

A

TT = 3 … (i)

B

A

rr =

31

From Stefan's law

B

A

EE =

)T)(gr4(

)T)(r4(42B

41A

2

2

σ

πσ=

2

31

× (3)4 = 9

80. When two progressive waves y1 = 4sin(2x – 6t)

and y2 = 3 sin

π

−−2

t6x2 are superimposed,

the amplitude of the resultant wave is ? Ans. [5] Sol. y1 = 4 sin (2x –6t) y2 = 3 sin (2x –6t –π/2)

φ = 2π

Ares = φ++ cosAA2AA 2122

21

Ares = 043 22 ++ = 5

81. A 0.1 kg mass is suspended from a wire of

negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9 × 10–7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad/s. If the Young's modulus of the material of the wire is n × 109 Nm–2, the value of n is ?

Ans. [4]

Sol.

x

strain = l

x

strainstress = Y

stress = Yx (l = 1 m)

AF = Yx

F = Ayx Ayx = ma

a = m

AYx ; ω = m

AY

140 = 0.1

10n104.9 9–7 ×××

140 = 70 n n = 4

82. A binary star consists of two stars (mass 2.2 MS) and B(mass 11 MS), where MS is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is ?

Ans. [6] Sol.

rA rB d


LA = mAω rA2

LB = mBωrB2

Ratio (K) = B

BA

LLL + =

B

A

LL + 1 = 2

BB

2AA

rmrm + 1

52.2

11mm

rr

A

B

B

A ===

Ratio (K) = 51 × 52 + 1 = 6

83. Gravitational acceleration on the surface of a

planet is 11

6 g, where g is the gravitational

acceleration on the surface of the earth. The

average mass density of the planet is 32 times

that of the earth. If the escape speed on the surface of the earth is taken to be 11 km/s, the escape speed on the surface of the planet in km/s will be ?

Ans. [3]

Sol. c

p

υ

υ=

ee

pp

Rg2Rg2

= e

p

e

p

RR

gg

× …

(1)

⇒ p

3

p

R34

M

π=

32

3e

e

R34M

π⇒

e

P

MM =

32

3e

3p

R

R … (2)

2p

p

R

GM=

116

2e

e

RGM

⇒ c

p

MM

= 11

62e

2p

R

R …(3)

from (2) and (3)

32

3e

3p

R

R=

116

2e

2p

R

R⇒

e

p

RR

= 22

63 … (4)

from (1) and (4)

e

p

υ

υ=

2263

116

× = 24218 =

113

vp = 3 km /sec. 84. A piece of ice (heat capacity = 2100 Jkg–1ºC–1 and

latent heat = 3.36 × 108 J kg–1) of mass m grams is at –5ºC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is ?

Ans. [8] Sol. The amount of heat required to raise the temp

from –5°C to 0°C. Q1 = m × 2100 × 10–3 × 5 = 10.5 m Joule The amount of heat required to melt 1 gm ice = 10–3 × 3.36 × 105 = 336 J 420 = 336 + 10.5 m 10.5 = 84

m = 8 gm.00

ATTITUDE

• Great effort springs naturally from a great

attitude.

• Like success, failure is many things to many people. With Positive Mental Attitude, failure is a learning experience, a rung on the ladder, a plateau at which to get your thoughts in order and prepare to try again.

• Your attitude, not your aptitude, will determine your altitude.

• Develop an attitude of gratitude, and give thanks for everything that happens to you, knowing that every step forward is a step toward achieving something bigger and better than your current situation.

• You can adopt the attitude there is nothing you can do, or you can see the challenge as your call to action.

• "An optimist is a person who sees a green light everywhere, while the pessimist sees only the red stoplight... The truly wise person is colorblind."

• Positive thinking will let you do everything better than negative thinking will.

• You cannot control what happens to you, but you can control your attitude toward what happens to you, and in that, you will be mastering change rather than allowing it to master you.

• You can do it if you believe you can!


CHEMISTRY



1. Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is

(A) 1 and diamagnetic (B) 0 and diamagnetic

(C) 1 and paramagnetic (D) 0 and paramagnetic

Ans. [A] Sol. 2s1σ 2s1

*σ 2s2σ 2s2*σ 2

p2 xπ =

yp2π

B. Ο = 21 (6 – 4) = 1

Diamagnetic

2. The compounds P, Q and S COOH

P HO

OCH3

Q H3C

C

S

|| O

O

were separately subjected to nitration using

HNO3/H2SO4 mixture. The major product formed

in each case respectively, is

(A) COOH

NO2

HO

OCH3

H3C

C||O

O

NO2

O2N

(B)

COOH

NO2 HO

OCH3

H3C

C||O

O

NO2

NO2

IIT-JEE 2010 PAPER-II (PAPER & SOLUTION)

Time : 3 Hours Total Marks : 237

Instructions : • The question paper consists of 3 Parts (Chemistry, Mathematics and Physics). And each part consists of four

Sections. • For each question in Section I: you will be awarded 5 marks if you have darkened only the bubble corresponding

to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus two (–2) mark will be awarded.

• For each question in Section II: you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubbles is darkened. No negative marks will be awarded for incorrect answers in this section.

• For each question in Section III: you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.

• For each question in Section IV: you will be awarded 2 marks for each row in which you have darkened the bubbles(s) corresponding to the correct answer. Thus, each question in this section carries a maximums of 8 marks. There is no negative marks awarded for incorrect answer(s) in this section.


(C) COOH

HO

OCH3

H3C

C

|| O

O

NO2

NO2

NO2

(D) COOH

HO

OCH3

H3C

C

|| O

O

NO2

NO2

NO2

Ans. [C]

Sol.

COOH

HNO3/H2SO4

OH

COOH

OH

NO2

OCH3

HNO3/H2SO4

CH3

OCH3

CH3

NO2

C HNO3/H2SO4

C

|| O

O

|| O

O

NO2

3. The packing efficiency of the two-dimensional square unit cell shown below is

L

(A) 39.27 % (B) 68.02% (C) 74.05% (D) 78.54% Ans. [D] Sol. Area of square = L2

& 4R = L2

R = 22

L

% packing efficiency (η) = 100squareofareacircleofarea

×

= 100L

R22

2×

π×

= 100L

24L2

2

2

××π×

= 78. 5%

4. The species having pyramidal shape is (A) SO3 (B) BrF3

(C) –23SiO (D) OSF2

Ans. [D]

Sol. SO

FF

5. In the reaction

O H3C

(1) NaOH/Br2C

NH2

OC

Cl(2)

T

the structure of the product T is

(A) O H3C C

O– C

O

(B) NHC

OCH3


(C)

NH C

O

H3C

(D) O H3C C

NH–C

O

Ans. [C] Sol.

O CH3

NaOH/Br2 C NH2

NH2

NH2 +

Cl C O

–HCl

NH–C||O

Hoffmann's degradation

CH3

CH3

CH3

6. The complex showing a spin-only magnetic

moment of 2.82 B.M. is (A) Ni(CO)4 (B) [NiCl4]2–

(C) Ni((PPh3)4 (D) [Ni(CN)4]2–

Ans. [B] Sol. [NiCl4]–2 Ni+2 1s2 2s2 2p6 3s2 3d8 4s°

Cl– Cl– Cl– Cl–

hyb = sp3 no. of unpaired electrons = 2

µ = )4(2 = 8 = 2.82 BM

SECTION – II Integer Type

This section contains a group of 5 questions. The answer to each questions is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

7. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. The value of x is

Ans. [7]

Sol. d = Vm ⇒ 10.5 g/cc

Number of atoms of Ag in 1cc ⇒ AN108

5.10×

In 1cm, number of atoms of Ag = 3 AN108

5.10×

In 1cm2, number of atoms of Ag = 3/2

AN108

5.10

×

In 10–12 m2 or 10–8 cm2, number of atoms of

Ag =3/2

AN108

5.10

× 10–8

= 3/223

1081002.65.10

×× × 10–8

= 1.5 × 107 Thus, x = 7

8. Among the following, the number of elements

showing only one non-zero oxidation state is O, Cl, F, N. P, Sn, Tl, Na, TI Ans. [2] Sol. F & Na only show one non zero oxidation state

that are – 1 & + 1 respectively. 9. One mole of an ideal gas is taken from a to b

along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line paths is wd, then the integer closest to the ratio wd\ws is

4.4.3.3.02.2.01.51.00.50.0

0.0 0.5 1.0 1.5 2.0 2.5 4.5 5.53.0 3.5 4.0 5.0 6.0

a

b

P(atm)

V(lit) Ans. [2]

Sol. For solid line path show approxy isothermal process

∴ work done |WS| = 2.303 (PV) log 5.5.5

= 2.303 × 4 × .5 × log 11 ~– 4.79 for dashed line path work done wd = 4 × |2 – .5| + 1 × |3 – 2| + .5 × |5.5 –3| = 6 + 1 + 1.25 = 8.25

∴ s

d

ww

= 8.4

25.8 = 1.71 ~– 2

10. The total number of diprotic acids among the following is

H3PO4 H2SO4 H3PO3 H2CO3 H2S2O7 H3BO3 H3PO2 H2CrO4 H2SO3


Ans. [6] Sol.

HO – S – OH|| O

|| O

HO – P – OH|| O

C||O

OH OH

H–O – S – O– S – O – H || O

|| O

|| O OH

|| O

Cr O

OH

O S

OH

||O

OH

11. Total number of geometrical isomers for the

complex [RhCl(CO)(PPh3)(NH3)] is Ans. [3] Sol.

Cl

CO

PPh3

NH3

Rh

Cl

CO PPh3

NH3

Rh

Cl

NH3 CO

PPh3

Rh

SECTION – III

Paragraph Type

This section contains 2 paragraphs. Based upon each of the paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for questions No. 12 to 14

Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below –

O

H3C H3C OH

O

12. The compounds P and Q respectively are -

(A)

H3C CH

CH3

C

O

H and H3C C

O

H

(B)

H3CCH

CH3

C

O

H and H C

O

H

(C) H3C

CH

CH3

CH2

O

andH3C

C

O

HC H

(D) H3C

CH

CH3

CH2

O

andH

C

O

HC H

Ans. [B] 13. The compound R is

(A) H3CC

C

O

H

H3C

CH2OH

(B) H3CC

C

O

H

H3C

CHOH H3C

(C)

H3CCH C

O

OH

CH3

CH2

CH H

(D)

H3CCH C

O

OH

CH3

CH

CH H

H3C

Ans. [A]


14. The compound S is

(A)

H3CCH C

O

CN

CH3

CH2

CH H

(B)

H3CC

C

O

CNCH2

H H3C

(C)

H3CCH CH

CN

OH

CH3

CH2

CH OH

(D)

H3CC

CH

CN

OH CH2

OH H3C

Ans. [D]

Sol. (12 to 14)

(i)

O

CH3 – CH C

CH3

H →–OH

O

CH3 – CΘ

C

CH3

H

(ii)

O

CH3 – CΘ C

CH3

H O

C H + H

CH3 – C – CH2

CH3

C

OH

OH

[R]

(iii) CH3 – C – CH2 + HCN

CH3

C

O H

OH C – CH2

CH3

H – C – CN

OH

OHCH3

[S]

(iv) CH3 – C – CH2

CH3

OHH – C – CN

OH

∆ →

+OH3 CH3 – C — CH2

CH3

HC OCOHO

Final product

Paragraph for Questions No. 15 to 17

The hydrogen like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

15. The state S1 is – (A) 1s (B) 2s (C) 2p (D) 3s Ans. [B] Sol. Q One radial node

∴ n – l – 1 = 1 or n – l = 2 l = 0 n = 2 Orbital name = 2s

16. Energy of the state S1 in units of the hydrogen

atom ground state energy is – (A) 0.75 (B) 1.50

(C) 2.25 (D) 4.50 Ans. [C]

Sol. S1 = Energy of e of H in ground state × 2

2

23

= 2.25 × energy of e of H in ground state 17. The orbital angular momentum quantum number

of the state S2 is – (A) 0 (B) 1 (C) 2 (D) 3 Sol. [B] For S2 = n – l – 1

n – l = 2 n = 3, l = 1 Orbital = 3p ∴ l = 1

[ S2 = energy of e of H in ground state × 2

2

n3 , n = 3]


SECTION – IV

Matrix Type This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

18. Match the reactions in Column I with appropriate options in Column II.

Column I

(A) N2Cl + OH C0

OH/NaOH 2

° →

N = N OH

(B)

H3C – C – C – CH3

OH

CH3

OH

CH3 → 42SOH

H3C

OCH3C

CCH3CH3

(C)

C CH3

O

+

→OH.2

LiAlH.1

3

4

CHCH3

OH

(D) HS Cl

→Base S

Column II (p) Racemic mixture (q) Addition reaction (r) Substitution reaction

(s) Coupling reaction (t) Carbocation intermediate

Ans. [A → r,s,t; B → t; C → p,q; D → r] 19. All the compounds listed in Column I react with

water. Match the result of the respective reactions with the appropriate options listed in Column II.

Column I Column II (A) (CH3)2SiCl2 (p) Hydrogen halide formation (B) XeF4 (q) Redox reaction (C) Cl2 (r) Reacts with glass (D) VCl5 (s) Polymerization (t) O2 formation

Ans. [A → p,s; B → p,q,r,t; C → p,q,t; D → p]

MATHEMATICS



20. A signal which can be green or red with

probability 54 and

51 respectively, is received by

station A and then transmitted to station B. The probability of each station receiving the signal

correctly is 43 . If the signal received at station B

is green, then the probability that the original signal was green is

(A) 53 (B)

76

(C) 2320 (D)

209

Ans. [C] Sol. Event (1) : original signal OG : Original signal is green OR : Original signal is red Event (2) : Signal received by A. AG : A received green AR : A received Red Event (3) : Signal received by B BG : B received green BR : B received Red

P

BGOG =

+

ORBGP).OR(P

OGBGP).OG(P

OGBGP).OG(P


=

++

+

+

41.

43

43.

41

51

41.

41

43.

43

54

41.

41

43.

43

54

= 2320

21. If the distance of the point P (1, –2, 1) from the

plane x + 2y –2z = α, where α > 0, is 5, then the foot of the perpendicular from P to the plane is

(A)

−

37,

34,

38 (B)

−

31,

34,

34

(C)

310,

32,

31 (D)

−

25,

31,

32

Ans. [A]

Sol. 1

1x − = 2

2y + = 21z

−− = λ 5

3|241|

=α−−−

foot (1 + λ, –2 + 2λ, 1 –2λ) |α + 5| = 15 (1 + λ) + 2(–2 + 2λ) – 2 (1 –2λ) = 10

α = 10 (correct), –20 (wrong) 1 + λ – 4 + 4λ –2 + 4λ = 10 9λ = 15 , ⇒ λ = 5/3

foot =

−

37,

34,

38

22. Two adjacent sides of a parallelogram ABCD are

given by AB= 2 i + 10 j + 11 k and AD = – i +

2 j + 2 k The side AD is rotated by an acute angle α in the plane of the parallelogram so that AD becomes AD′. If AD′ makes a right angle with the side AB, then the cosine of the angle α is given by

(A) 98 (B)

917

(C) 91 (D)

954

Ans. [B]

Sol.

D

D′

C

B A

α

α−

π2

2 i + 10 j + 11 k

– i + 2 j + 2 k

cos

α−

π2

= |AD||AB|

AD.AB = )15(3

40 = 98

sin α = 98 ⇒ cos α =

917

23. Let S = 1, 2, 3, 4. The total number of unordered pairs of disjoint subsets of S is equal to

(A) 25 (B) 34 (C) 42 (D) 41

Ans. [D] Sol. S = 1, 2, 3, 4 Possible subsets No. of elements in Ways Set A Set B 0 0 = 1 1 0 = 4C1 = 4 2 0 = 4C2 = 6 1 1 = 4C2 = 6 3 0 = 4C3 = 4 2 1 = 4C2. 2C1 = 12 4 0 = 4C4 = 1

3 1 =!1!3

!4 = 4

2 2 =!2!2!2

!4 = 3

Total ⇒ 1 + 4 + 6 + 6 + 4 + 12 + 1 + 4 + 3 = 41

24. Let f be a real-valued function defined on the

interval (–1, 1) such that e–x f(x) = 2 +

∫ +x

0

4 dt1t , for all x ∈ (–1, 1) , and let f–1 be the

inverse function of f. Then (f–1)′ (2) is equal to

(A) 1 (B) 31

(C) 21 (D)

e1

Ans. [B]

Sol. f(x) = 2ex + ex ∫ +x

0

4 dt1t

x = 0 ; f(0) = 2

f ′(x) = 2ex + ex ∫ +x

0

4 dt1t + ex 4x1+

f ′(0) = 2 + 1 = 3 (f–1)′ (2) = )0(f

1′

= 31

25. For r = 0, 1, …, 10, let Ar, Br and Cr denote, respectively, the coefficient of xr in the expansions of (1 + x)10, (1 + x)20 and (1+ x)30.

Then ∑=

−10

1rr10r10r )ACBB(A is equal to

(A) B10 – C10 (B) A10 ( )1010210 ACB −

(C) 0 (D) C10– B10

Ans. [D] Sol. Ar = 10Cr, Br = 20Cr , Cr = 30Cr

∑=

10

1rr

10 C (20C10 20Cr – 30C10 10Cr)


= 20C10 ∑=

10

1rr

10 C 20C20–r – 30C10 r

1010

1rr

10 C.C∑=

= 20C10 [30C20 – 10C0 20C20] – 30C10 [20C10 – (10C0)2] = 20C10

30C20 – 20C10 – 30C10 20C10 + 30C10 = 30C10 – 20C10

= C10 – B10

SECTION – II

Integer type

This section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question no. in the ORS is to be bubbled.

26. Let a1, a2, a3, …….., a11 be real numbers satisfying a1 = 15, 27 –2a2 > 0 and ak =2ak–1 – ak–2 for k = 3, 4, …., 11.

If 11

a.....aa 211

22

21 +++ = 90, then the value of

11a.....aa 1121 +++ is equal to

Ans. [0]

Sol. Q ak–1 = 2aa 2kk −+

so 11

a.....aa 211

22

21 +++ = 90

⇒ Σ(a + (r –1) d)2 = 11 × 90 ⇒ Σ(a2 + 2ad (r –1) + (r –1)2d2) = 11 × 90

11a2 + 2ad 2

1110× + 6

211110 ×× d2 = 11 × 90

so on solving d = –3

so 11

a......aa 1121 +++

= 2

11 . 111 . (2 × a1 + (11 –1) (–3))

= 21 (30 – 30) = 0

27. Let f be a function defined on R (the set of all real

numbers) such that f ′(x) = 2010 (x –2009) (x –2010)2 (x –2011)3

(x –2012)4, for all x ∈ R. If g is a function defined on R with values in the

interval (0, ∞) such that f(x) = ln g(x), for all x ∈ R, then the number of

points in R at which g has a local maximum is

Ans. [1] Sol. g(x) = ef(x) g′(x) = ef(x) f ′ (x) g′(x) = 0 ⇒ f ′ (x) = 0 ⇒ x = 2009, 2010, 2011,

2012 Points of local maxima = 2009, ⇒ only one point 28. Let k be a positive real number and let

A =

−−−

−

1k2k2k21k2k2k21k2

and

B =

−−−

−

0k2kk20k21

k1k20

If det (adj A) + det(adj B) = 106, then [k] is equal to

[Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k]

Ans. [4]

Sol. det (A) = 1k2k2k21k2k2k21k2

−−−

−

= (2k –1) [–1 + 4k2] –2 k [–2 k –4k k ]

+ 2 k [4k k + 2 k ] det (A) = (2k –1) (4k2 –1) + 4 k (2k + 1) + 4k (2k +1) = (2k –1) (4k2 –1) + 8k (2k + 1) det (B) = 0 det (adj A) = (det A)2 = 106 det A = 103 8k3 + 1 –2k –4k2 + 16k2 + 8k = 103 8k3 + 12k2 + 6k – 999 = 0 k = 2 → 64 + 48 + 12 – 999 < 0 k = 3 → 8(27) + 109 + 18 – 999 < 0 k = 4 → 8(64) + 12 (16) + 24 – 999 512 + 192 + 24 – 999 < 0 k = 5 → 8(125) + 12 (25) + 6(5) – 999 > 0 so [k] = 4 29. Two parallel chords of a circle of radius 2 are at a

distance 3 + 1 apart. If the chords subtend at the

center, angles of kπ and

k2π , where k > 0, then

the value of [k] is [Note : [k] denotes the largest integer less than or

equal to k] Ans. [3]


Sol. π/k

π/2k

d = 2 cos kπ + 2 cos

k2π

413 + = cos

k43π cos

k4π

⇒ cos k4

π cos k4

3π =2213 + .

21

k4

π = 12π

4k = 12 k = 3 30. Consider a triangle ABC and let a, b and c denote

the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3 . If ∠ACB is obtuse and if r denotes the radius of the in-circle of the triangle, then r2 is equal to

Ans. [3]

Sol.

B 6 C

10

A

∆ =21 ab sin C

15 3 = 21 6(10) sin C ⇒ sin C = 3 /2

⇒ C = 120°

cos C = 6.10.2

c36100 2−+⇒ C2 = 136 + 120° (1/2)

⇒ C2 = 196 ⇒ C =14 s = 15

r = s∆ = 3

r2 = 3


This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f(x) and let t = |s|.

31. The real number s lies in the interval

(A)

− 0,

41 (B)

−−

43,11

(C)

−−

21,

43 (D)

41,0

Ans. [C] Sol. f(x) = 4x3 + 3x2 + 2x + 1 f ′(x) = 12x2 + 6x + 2 is always positive

f(0) = 1, f(–1/2) = 1/4, f(–3/4) = –21

so root ∈

−−

21,

43

Q the equation have only one real root so

s ∈

−−

21,

43 and t ∈

43,

21

32. The area bounded by the curve y = f(x) and the

lines x = 0, y = 0 and x = t, lies in the interval

(A)

3,

43 (B)

1611,

6421

(C) (9, 10) (D)

6421,0

Ans. [A]

Sol. A(t) = ∫t

0

)x(d)x(f = t4 + t3 + t2 + t = t

−−

t1t1 4

A(1/2) = 15/16 & A (3/4) = 3

256175

So A(t) ∈

3,

43

33. The function f ′(x) is

(A) increasing in

−−

41,t and decreasing

in

− t,

41

(B) decreasing in

−−

41,t and increasing

in

− t,

41


(C) increasing in (–t, t) (D) decreasing in (–t, t) Ans. [B] Sol. f ′′(x) = 6 (4x + 1)

Paragraph for questions No. 34 to 36

Tangents are drawn from the point P(3, 4) to the

ellipse 4

y9

x 22+ = 1, touching the ellipse at points

A and B. 34. The coordinates of A and B are (A) (3, 0) and (0, 2)

(B)

−

151612,

58 and

−

58,

59

(C)

−

151612,

58 and (0, 2)

(D) (3, 0) and

−

58,

59

Ans. [D] Sol. Equation of tangent

y = mx ± 4m9 2 + as it passes through (3, 4)

so 4 = 3m ± 4m9 2 +

m = 21 and undefined.

So equation of the tangents will be x –2y + 5 = 0 and x = 3

so point of contacts are (3, 0) and

−

58,

59

35. The orthocentre of the triangle PAB is

(A)

78,5 (B)

825,

57

(C)

58,

511 (D)

57,

258

Ans. [C]

Sol.

P(3,4)

θ

H A(3,0)

−

58,

59B

Equation of two altitudes PH and AQ are 3x – y – 5 = 0 and 2x + y – 6 = 0 respectively

so orthocentre will be

58,

511

36. The equation of the locus of the point whose distances from the point P and the line AB are equal, is

(A) 9x2 + y2 – 6xy –54 x – 62 y + 241 = 0 (B) x2 + 9y2 + 6xy –54x + 62 y –241 = 0 (C) 9x2 + 9y2 –6xy –54 x –62 y – 241 = 0 (D) x2 + y2 –2xy + 27x + 31y – 120 = 0

Ans. [A] Sol. Equation of AB is x + 3y – 3 = 0 so required locus will be

(x – 3)2 + (y –4)2 = 10

)3–y3x( 2+

⇒ 9x2 + y2 – 6xy –54x –62y + 241 = 0

SECTION – IV Matrix Type

This section contains 2 questions. Each question has four statements (A, B, C and D)given in Column I and five statements (p, q, r, s and t) in column-II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column-II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

37. Match the statements in Column-I with those in Column-II.

[Note : Here z takes values in the complex plane and lm z and Re z denote, respectively, the imaginary part and the real part of z]

COLUMN-I COLUMN-II

(A) The set of points (p) an ellipse with

z satisfying eccentricity 54

|z – i|z|| = |z + i|z|| is contained in or e

qual to (q) the set of points z satisfying Im z = 0

(B) The set of points z (r) the set of points z Satisfying satisfying |Im z| ≤ 1 |z + 4| + |z –4| = 10 is contained in or equal to

(C) If |w| = 2, then the set (s) the set of points z of points satisfying |Re z| ≤ 2

z = w – w1 is contained

in or equal to (D) If |w| =1, then the set (t) the set of points z

of points satisfying |z| ≤ 3

z = w + w1 is contained

in or equal to


Ans. A → q, r ; B → p ; C → p, s, t ; D → q, r, s, t Sol. (A) Let |Z| = r ∀ r ∈ R

irZirZ

+− = 1Which is the equation of line of

perpendicular bisector of y = r & y = – r that is y = 0 (B) |Z + 4| + |Z – 4| = 10 it will represent on ellipse having foci (–4, 0), (4, 0)

so its equation will be 9y

25x 22

+ = 1

whose eccentricity is 4/5 (C) Let w = 2eiθ.

z = 23 cos +

25 i sin θ

(D) Let w = eiθ Z = eiθ + e–iθ = 2 cos θ.

38. Match the statements in Column-I with the values in Column-II.

COLUMN-I COLUMN-II (A) A line from the origin (p) –4 meets the lines

1

2x − = 21y

−− =

11z + and

2

38x −

= 13y

−+ =

11z −

at P and Q respectively. If length PQ = d, then d2 is (B) The values of x satisfying (q) 0

tan–1 (x + 3) – tan–1 (x –3)

= sin–1

53 are,

(C) Non-zero vectors ar

, br

(r) 4 and c

r satis1fy a

r. b

r= 0,

( br

– ar

). ( br

+ cr

) = 0 and 2 | b

r+ c

r|= | b

r– a

r|.

If ar

= µ br

+ 4 cr

, then the possible values of µ are

(D) Let f be the function on (s) 5 [–π, π] given by f(0) = 9 and f(x)

= sin

2xsin

2x9 for x ≠ 0

The value of π2

∫π

π−

dx)x(f is

(t) 6

Ans. A → t ; B → p, r ; C → q, s ; D → r

Sol. (A) Let P ≡ (λ + 2, 1 –2 λ, λ – 1)

Q ≡ (2µ + 32 ; – µ –3, µ + 1)

equation line PQ

→r = (λ + 2) i + (1 – 2λ) j + (λ – 1) k

+ α ((2µ – λ +32 ) i + (2λ – µ – 4) j

+ (µ + 2 – λ) k

∴ This line passing through origin so.

λ + 2 + α (2µ − λ + 32

) = 0

1 – 2λ + α(2λ – µ – 4) = 0 λ – 1 + α(µ – λ + 2) = 0

on solving above three µ = 31

& λ = 3

So P ≡ (5, – 5, 2) & Q ≡ (3

10 ,310− ,

34 )

So PQ = 6 ⇒ (PQ)2 = 6

(B) tan–1(x + 3) – tan–1 (x –3) = sin–1 53

tan–1 8x

62 −

= tan–1 43

⇒ x2 – 8 = 8 ⇒ x2 = 16 ⇒ x = ± 4

(C) | br

|2 + br

. cr

= ar

. cr

…. (1)

put ar

= µ br

+ 4 cr

∀ ar

. br

= 0 ⇒ br

. cr

= –4µ | b

r|2

…(2) from (1) and (2)

2

2

cb = 24

16µ+µ−

… (3)

∴ 2| br

+ cr

| = | br

– ar

| and ar

= µ br

+ 4 cr

2

2

cb = 223

12µ+µ−

… (4)

from (3) and (4) m = 0,5

(D) f(x) =

2xsin

2x9sin

= xsinx5sin +

xsinx4sin

I = π2

∫π

π−

dx)x(f


= π4

∫π

0

dx)x(f

=π4

∫π

0xsinx5sin

= π8

∫π 2/

0 xsinx5sin dx

= π8

∫π +2/

0

dxxsin

)x2x3(sin = π8

∫π

+2/

0

dx)x4cos21(

= 4

PHYSICS



39. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to -

F F

(A) 22

0R1

σε

(B) R1 2

0σ

ε

(C) R

1 2

0

σε

(D) 2

2

0 R1 σε

Ans. [A] Sol.

dθ

0

2

2εσ (Electrostatic pressure)

θ

dA = 2πR sin θ × Rdθ

dF = 0

2

2εσ × dA

Component of dF along vertical axis = dF cos θ

Total force = θθπε

σ∫π

d2sinR2

22/

0 0

2 = 2

0

2R

2π×

εσ

40. A block of mass 2 kg is free to move along the x-

axis. It is at rest and from t = 0 onwards it is subjected to a time dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is -

4N

O 3s 4.5s

t

F(t)

(A) 4.50 J (B) 7.50 J

(C) 5.06 J (D) 14.06 J Ans. [C] Sol.

4N

O 3 4.5

t (sec)

F

–2N

m = 34

−

At t = 4.5 sec →F = – 2N

Total Impulse I =

×× 43

21 –

×× 5.12

21

⇒ = 6 – 1.5 = 4.5 SI unit Impulse = change in momentum 4.5 = 2[v – 0]

v = 25.4 = 2.25 m/sec

K.E. = 21 × 2 × (2.25)2 = 5.06 J

41. A tiny spherical oil drop carrying a net charge q is

balanced in still air with a vertical uniform

electric field of strength 5107

81×

π Vm–1. When

the field is switched off, the drop is observed to fall with terminal velocity 2 × 10–3 ms–1. Given g = 9.8 m s–2, viscosity of the air = 1.8 × 10–5 Nsm–2 and the density of oil = 900 kg m–3, the magnitude of q is -

(A) 1.6 × 10–19 C (B) 3.2 × 10–19 C (C) 4.8 × 10–19 C (D) 8.0 × 10–19 C

Ans. [D]


Sol. qE = mg

⇒

×

π 5107

81q = 900 × 34

πr3 × 9.8

q = 5

3

1081378.9r4900

×××××× ...... (1)

vT = 2 × 10–3 m/sec

2 × 10–3 = 92 × 5

2

108.18.9900r

−×××

r2 = 8.9900210108.118 35

××××× −−

= 0.1836 × 10–

10 = 18.36 × 10–12 r = 4.284 × 10–6 m

q = 51024378.93600

××× × 78.62 × 10–18

q = 0.799 × 10–18 ≈ 8 × 10–19 C

42. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string is -

(A) 5 grams (B) 10 grams (C) 20 grams (D) 40 grams

Ans. [B]

Sol. Fundamental frequency of closed pipe = l4

v

⇒ 8.04

320×

= 2.3

320 = 100 Hz

Frequency of 2nd Harmonic of string =l

v = µT1

l

100 = ll /m

501

⇒ 100 = 5.0m

50×

⇒ 100 = m

100

10000 = m

100 ⇒ m = 10–2 kg = 10 gm

43. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is -

(A) virtual and at a distance of 16 cm from the mirror

(B) real and at a distance of 16 cm from the mirror

(C) virtual and at a distance of 20 cm from the mirror

(D) real and at a distance of 20 cm from the mirror

Ans. [B] Sol.

30cm 10cm

10 cm20 cm6 cm

Refraction of reflected light by lens f = + 15 cm u = + 10 cm

v1 –

u1 =

f1 ⇒

101

v1

− = 151

v = 6 cm as incident rays are converging so refracted rays

will converge more and final image is real. 44. A vernier calipers has 1 mm marks on the main

scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is -

(A) 0.02 mm (B) 0.05 mm (C) 0.1 mm (D) 0.2 mm

Ans. [D] Sol. Least Count = M.S. Reading – V.S. Reading

.... (1) 20 V.S. = 16 M.S. or 16 mm

1 V.S. = 2016 M.S. or

2016 mm

In equation (1)

Least Count =

−

20161 mm

= 0.2 mm

SECTION – II Integer Type

This section contains Five questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 45. A large glass slab (µ = 5/3) of thickness 8 cm is

placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R ?

Sol. [6]

sin θcr = 53 ⇒ tan θcr =

43


θcr

R

8 cm

R = 8 tan θcr

= 8 × 43 = 6 cm

46. Image of an object approaching a convex mirror

of radius of curvature 20 m along its optical axis

is observed to move from 325 m to

750 m in 30

seconds. What is the speed of the object in km per hour ?

Ans. [3] Sol. For position of object initially when image

was at 325 m

– 101 =

253

− + u1

253 –

101 =

u1

100

10–12 = u1

u1 = 50

u1

O

For position of object when image is at 7

50 m

– 101 =

507

− + u1

507 –

101 =

u1

u2 = 25

u2

O

Speed of object =

302550 − =

3025 m/sec

= 3025 ×

10003600 = 3 km/hr

47. To determine the half life of a radioactive

element, a student plots a graph of dt

)t(dNnl

versus t. Here dt

)t(dN is the rate of radioactive

decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is -

654321

2 3 4 5 6 7 8Years

dt)t(dNnl

Ans. 8

Sol. From graph slope = 21 = 0.5 year–1

dtdN = Ne–λt

ln

dtdN = ln (N) – λt

so comparing we get λ = 0.5 year–1

t1/2 =

5.0693.0 year

t = 4.16 years

so No. of half lives = 693.016.4 × 0.5 = 3

N0 → 2

N0 → 4

N0 → 8

N0 ⇒ p = 8

48. A diatomic ideal gas is compressed adiabatically

to 321 of its initial volume. In the initial

temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is -

Ans. [4] Sol. for adiabatic process TVγ–1 = const.

⇒ Ti 1

57

V−

= aTi 1

57

32V −

⇒ a = 4

49. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V ?

(Take : ln 5 = 1.6, ln 3 = 1.1]


2MΩ

2MΩ A B

2µF

2µF Ans. [2]

Sol. 2×106

2×106

2×10–6

B

2×10–6

A

106

10V

4×10–6

q = CV0 (1 – e–t/RC) V = V0(1 – e–t/RC) 4 = 10(1 – e–t/4) 3 = 5e–t/4

log3 = log 5 – 4t

1.1 – 1.6 = – 4t ⇒ t = 2 sec


This section contains 2 paragraphs.. Based upon each of paragraph 3 multiple choice question have to be answered. Each of these questions has four choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph for Questions No. 50 to 52

When liquid medicine of density ρ is to be put in the eye, it is done with the help of a dropper. As the bulb on the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

50. If the radius of the opening of the dropper is r, the

vertical force due to the surface tension on the drop of radius R (assuming r << R) is

(A) 2πrT (B) 2πRT

(C) R

Tr2 2π (D) r

TR2 2π

Ans. [C]

Sol. FT

R

r

FT = 2πrT Net vertically upward force

⇒ 2πrT

Rr =

RTr2 2π

51. If r = 5× 10–4 m, ρ = 103 kgm–3, g = 10ms–2, T =

0.11 Nm–1, the radius of the drop when it detaches from the dropper is approximately

(A) 1.4 × 10–3 m (B) 3.3 × 10–3 m (C) 2.0 × 10–3 m (D) 4.1 × 10–3 m

Ans. [A]

Sol. R

Tr2 2π = 3R34

π × ρ × g

⇒ R

11.010252 8– ××× = 34 × R3 × 103 × 10

⇒ R4 = 4

8–

1041011.0350

××××

⇒ R4 = 4.125 × 10–12 ⇒ R = 1.4 × 10–3 m

52. After the drop detaches, its surface energy is (A) 1.4 × 10–6 J (B) 2.7 × 10–6 J

(C) 5.4 × 10–6 J (D) 8.1 × 10–6 J Ans. [B] Sol. Surface energy = T(A) = T × 4πR2 ⇒ 0.11 × 4 × 3.14 × 1.96 × 10–6

⇒ 2.7 × 10–6 J


The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

53. A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n = 0 is not allowed) is

(A)

π I8h

n1

2

2

2 (B)

π I8h

n1

2

2

(C)

π I8hn 2

2 (D)

π I8hn 2

22


Ans. [D] Sol. Bohr quantization principle

L = π2

nh = Iω ⇒ ω = I2

nhπ

Rotational KE = 2I21

ω = 2

I2nhI

21

π=

I8hn2

22

π

54. It is found that the excitation frequency from ground to the first excited state of rotation for the

CO moelcule is close to π4 × 1011 Hz. Then the

moment of inertia of CO molecule about its center of mass is close to (Take h = 2π × 10–34 J s)

(A) 2.76 × 10–46 kg m2 (B) 1.87 × 10–46 kg m2 (C) 4.67 × 10–47 kg m2 (D) 1.17 × 10–47 kg m2 Ans. [B] Sol. ∆E = E2 – E1

= I8

h22

22

π –

I8h12

22

π =

I8h3

2

2

π= hν

When ν = π4 × 1011 Hz

Solving I = 1.87 ×10–46 kg – m2 55. In a CO molecule, the distance between C (mass

= 12 a.m.u.) and O (mass = 16 a.m.u.) where 1

a.m.u. = 35 ×10–27 kg, is close to

(A) 2.4 × 10–10 m (B) 1.9 × 10–10

m (C) 1.3 × 10–10

m (D) 4.4 × 10–11 m

Ans. [C] Sol. I = 2

11rm + 222rm

Where m1 = 12 amu m2 = 16 amu m1r1 = m2r2 r1 + r2 = r where r → distance between C & O. Putting and solving r = 1.279 × 10–10

m

~_ 1.3 × 10–10 m

SECTION – IV

Matrix Type

This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

56. Two transparent media of refractive indices µ1 and µ3 have a solid lens shaped transparent material of refractive index µ2 between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between µ1, µ2 and µ3 are given. Match them to the ray diagrams shown in Column II.

Column I Column II

(A) µ1 < µ2 (p)

µ2 µ1 µ3

(B) µ1 > µ2 (q)

µ2 µ1 µ3

(C) µ2 = µ3 (r) µ2 µ1 µ3

(D) µ2 > µ3 (s) µ2 µ1 µ3

(t) µ2 µ1 µ3

Ans. [A → p,r; B → q,s,t; C → p,r,t D → q,s] Sol. For (p) µ2 > µ1 as light rays bend towards normal at first refraction µ2 = µ3 as no refraction occurs at second refraction Option : (A), (C) For (q) µ2 < µ1 as bend away from normal at first refraction µ3 < µ2 as bends away from normal at second refraction Option (B), (D) For (r)

µ2 > µ1 as bend towards the normal at first refraction µ2 = µ3 as no refraction occurs at second refraction

Option (A), (C) For (s)

µ2 < µ1 as bend away from normal at first refraction µ3 < µ2 as bend away from normal at second refraction

Option (B), (D) For (t)

µ2 < µ1 as bend away from normal at first refraction µ2 = µ3 as no refraction occurs at second refraction

Option (B), (C)


57. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2. (indicated in circuits) are related as shown in Column I. Match the two

Column I Column II

(A) (p) V1

6mH

V2

V

3µF

(B) I ≠ 0, V2 > V1 (q) V1

6mH

V2

V

2Ω

(C) V1 = 0, V2 = V (r) V1

6mH

V2

2Ω

~ V

(D) (s) V1

6mH

V2

3µF~ V

(t) V1

1kΩ

V2

3µF~ V

Ans. [A → r,s,t; B → q,r,s,t; C → p,q; D →

q,r,s,t] Sol. For (p) Insteady state when I = constant VL = 0 = V1 So V2 = V Option (C) For (q) V1 = 0 again as I = constant V2 = V

Also V2 = IR ⇒ Propotional to I. Option (B), (C), (D)

For (r) XL = ωL = (100 π) 6 × 10–3 1.88 Ω

R = 2Ω

V1 = I XL; V2 = IR

So V2 > V1

V2 ∝ I

also V1 ∝ I Option (A), (B), (D)

For (s) V1 = I XL

V2 = I XC where XC = C

1ω

1061 Ω

again V1 ∝ I; V2 ∝ I, I ≠ 0 Option (A), (B) (D)

For (t) V1 = IR when R = 1000 Ω

V2 = I XC when XC 1061 Ω V2 > V1

V1, V2 ∝ I and I ≠ 0 Option (A), (B), (D)

I ≠ 0, V1 is proportional to I

I ≠ 0, V2 is proportional to I

WHICH IS THE HIGHEST WATERFALL IN THE WORLD ?

The highest waterfall in the world is the

Angel Falls in Venezuela. At a towering height of

979m did you know that each drop of water takes 14

seconds to fall from the top to the bottom. The

water flows from the top of a “Tepui” which is a flat

topped mountain with vertical sides.

The waterfall which despite being known to the local

indians for thousands of years was originally called

the “Churun Meru” but for some reason they were

renamed by an American bush pilot called Jimmy

Angel, who noticed them in 1935 whilst flying over

the area looking for gold.


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans B D C C A B C D A,C A,C,D Ques 11 12 13 14 15 16 17 18 Ans A,C,D B,C C B D B D D 19 A → R B → P C → T D → S 20 A → S B → P C → Q D → R

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans D D C D C A A D A,D B,C,D Ques 11 12 13 14 15 16 17 18 Ans A,B B,C C D A C D D 19 A → P,Q,S B → P,Q,R,S C → P D → Q,R,S,T 20 A → P B → Q C → P,R D → S,P

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans C C C A B B C A A,B,C A,C,D Ques 11 12 13 14 15 16 17 18 Ans B C B A B B C C 19 A → P,Q,R,S B → R C → R,S D → Q 20 A → P,R B → P,S C → Q,S D → Q,S

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 Ans A A C B D C C B B B,C,D Ques 11 12 13 14 15 16 17 18 Ans A,B,C,D C,D C A B B C A 19 A → R,T B → S C → P D → Q 20 A → R B → P C → S D → T

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans A D C D B B C A A,B,C A,B Ques 11 12 13 14 15 16 17 18 Ans A B A D C D D A 19 A → P B → T C → S D → R 20 A → P,R,S,T B → P C → P,Q,R,T D → S

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans C B B B A A C B A,C,D A,C Ques 11 12 13 14 15 16 17 18 Ans A,C,D A,D A C B A B D 19 A → S B → R C → Q D → P 20 A → P B → Q C → P D → R

IIT- JEE 2011 (May issue)

IIT- JEE 2012 (May issue)

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