physicsforyou june2016

7/25/2019 Physicsforyou June2016

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PHYSICS FOR YOU

|

JUNE 16 7

Earth

oon

P

b

r

s

Physics Musing Problem Set 35 8

PMT Practice Paper 11

Core Concept 15

Physics Musing Solution Set 34 19

NEET Phase II Practice Paper 22

NEET Phase I Solved Paper 2016 31

Exam Prep 2016 41

Brain Map 46

Olympiad Problems 52

Ace Your Way - CBSE XII 55

Karnataka CET Solved Paper 2016 63

Kerala PET Solved Paper 2016 72

Live Physics 83

You Ask We Answer 84

Crossword 85

Managing Editor

Mahabir Singh

Editor

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Volume 24 No. 6 June 2016

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8 PHYSICS FOR YOU

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JUNE 16

1. wo blocks o masses 3 kg and 6 kg respectively

are placed on a smooth horizontal surace. Teyare connected by a light spring o orce constant

k= 200 N m1. Initially the spring is unstretched

and the indicated velocities are imparted to the

blocks. Find the maximum extension o the spring.

2. A bullet o mass m strikes a block o mass M

connected to a light spring o stiffness k, with a

speed v0. I the bullet gets embedded in the block

then nd the maximum compression in the spring.

3. A uniorm rod ABo length l

is released rom rest with

AB inclined at angle

with horizontal. It collides

elastically with smooth

horizontal surace afer alling

through a height h.

ake = 60. Find the height Hafer rebound sothat the rod is horizontal rst time when its centreo mass is at the maximum height. (Assume Bdoesnot strike the ground)

4. Assuming a particle to have the orm o a sphere

and to absorb all incident light, nd the radius o a

particle or which its gravitational attraction to the

Sun is counterbalanced by the orce that light exerts

on it. Te power o light radiated by the Sun equals

P= 4 1026W and the density o the particle is

= 1.0 g cm

3

.[Use G =

20

310 11 2 2N m kg and mass o the

Sun = 2 1030kg ]

5. A ray o light is alling on a glass sphere o = 3

such that the incident ray and the emergent ray,

when produced, intersect at a point on the surace

o the sphere. Find the value o angle o incidence.

6. In a YDSE, two thin transparent sheets are used

in ront o the slits S1and S2o reractive indices

1= 1.6 and 2= 1.4 . I both sheets have thickness

t, the central maximum is observed at a distance o5 mm rom centre O. Now the sheets are replaced

by two sheets o same material o reractive index

=1 2

2but having thickness t1 and t2 such

that tt t

=

+1 2

2.Now central maximum is observed

at distance o 8 mm rom centre O on the same

side as beore. Find the thicknesses t1 and t2.

[Given, d= 1 mm , D= 1 m ]

7. A particle is projected rom

point A, that is at a distance

4R rom the centre o the

earth, with speed v1 in a

direction making 30 with

the line joining the centre

o the earth and point A, as

shown. Find the speed v1 i

particle passes grazing the

surace o the earth.

Physics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augmentthe chances of bright students preparing for JEE (Main and Advanced) / NEET / AIIMS / Other PETs with additional study material.In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / NEET / AIIMS / various PETs.The detailed solutions of these problems will be published in next issue of Physics For You.The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those whosend atleast five correct solutions will be published in the next issue.We hope that our readers will enrich their problem solving skills through Physics Musing and stand in better stead while facing the competitive exams.

By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Senior Professor Physics, RAO IIT ACADEMY, Mumbai.


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10 PHYSICS FOR YOU

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JUNE 16

Consider gravitational interaction only betweenthese two.

[UseGM

R=

6 4 107 2 2. m s )

8. A parallel beam of light falls normally on the rstface of a prism of small angle. At the second faceit is partly transmitted and partly reected, thereected beam striking at the rst face again andemerging from it in a direction making an angle6 30with the reversed direction of the incidentbeam. Te refracted beam is found to haveundergone a deviation of 115 from the originaldirection. Find the refractive index of the glass andthe angle of the prism.

9. Some gas (CP/C

V = = 1.25)

follows the cycle ABCDAas shownin the gure. Determine the ratioof the energy given out by the gas

to its surroundings during the isochoric section ofthe cycle to the expansion work done during theisobaric section of the cycle.

10.A system is shown in gure. All contact surfacesare smooth and string is tight and inextensible.Wedge A moves towards right with speed10 m s1 and velocity of B relative to A is indownward direction along the incline havingmagnitude 5 m s1. Find the horizontal and verticalcomponent of velocity of block C.


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PHYSICS FOR YOU

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1. A ball o mass 400 g is dropped rom a height o5 m. A boy on the ground hits the ball vertically

upwards with a bat with an average orce o 100 N.So, that it attains a vertical height o 20 m. Te time

or which the ball remains in contact with the bat is(g= 10 m s2)

(a) 0.12 s (b) 0.08 s (c) 0.04 s (d) 12 s

2. Te moment o inertia o a rod (length L, mass m)about an axis perpendicular to the length o the rodand passing through a point equidistant rom itsmid point and one end is

(a)mL2

12 (b)

7

48

2mL

(c)13

48

2mL (d)19

48

2mL

3. On heating the cathode 1.8 1017 electrons areemitted per second. On using the anode (positivepole) at 400 V, whole the electrons are collectedat positive pole. I the charge o electron is1.6 1019 C, then the maximum positive polecurrent will be(a) 29 mA (b) 2.7 mA(c) 72 A (d) 29 A

4. Which o the ollowing statements about the Bohrsmodel o the hydrogen atom is alse?

(a) Acceleration o electron in n= 2 orbit is less

than that in n= 1 orbit.

(b) Angular momentum o electron in n= 2 orbitis more than that in n= 1 orbit.

(c) Kinetic energy o electron in n = 2 orbit is lessthan that in n= 1 orbit.

(d) Potential energy o electron inn= 2 orbit is lessthan that in n= 1 orbit.

5. A mass m= 100 g is attached at the end o a lightspring which oscillates on a rictionless horizontal

table with an amplitude equal to 0.16 m and timeperiod equal to 2 s. Initially the mass is released

rom rest at t= 0, and displacement x= 0.16 m.Te expression or the displacement o mass at anytime (t) is(a) x= 0.16 cos (t)(b) x= 0.16 sin (t)(c) x= 0.16 cos (t+ )(d) x= 0.16 cos (t+ x)

6. A particle executing simple harmonic motion hasa time period o 4 s. Afer how much interval o

time rom t= 0 will its displacement be hal o itsamplitude?

(a)1

3s (b)

1

2s (c)

2

3s (d)

1

6s

7. wo spheresAand Bo radii aand b, respectivelyare at same electric potential. Te ratio o the suracecharge densities oAand Bis

(a)a

b (b)

b

a (c)

a

b

2

2 (d)

b

a

2

2

8. For a metallic wire, the ratio V/i (V = appliedpotential difference and i= current owing)

(a) is independent o temperature(b) increases as the temperature rises(c) decreases as the temperature rises(d) increases or decreases as temperature rises

depending upon the metal

9. An aircraf executes horizontal loop with a speed o150 m s1when its wings are banked at an angle o12. Te radius o the loop is(a) 10.6 km (b) 5.3 km(c) 7.5 km (d) 8.3 km

*A renowned physics expert, KP Institute o Physics, Chandigarh, 09872662552

*K P Singh

PRACTICE PAPER


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12 PHYSICS FOR YOU

|JUNE 16

10. A carnot engine used rst an ideal monoatomic gas.

If the source and sink temperature on 411 C and

69 C respectively and the engine extracts 1000 J

of heat from the source in each cycle. Ten area

enclosed by the

(a) PVdiagram is 30 J

(b) PVdiagram is 700 J

(c) PVdiagram is 500 J

(d) none of these

11. A square coil 20 cm 20 cm has 100 turns and

carries a current of 1 A. It is placed in a uniform

magnetic eld B = 0.5 with the direction of

magnetic eld parallel to the plane of the coil. Te

magnitude of the torque required to hold this coil,

in this position is

(a) zero (b) 200 N m

(c) 2 N m (d) 10 N m

12. A galvanometer has resistance of 400and deects

full scale for current of 0.2 mA through it. Te shunt

resistance required to convert it into 3 A ammeter is

(a) 0.027 (b) 0.054

(c) 0.0135 (d) 0.27

13. Te eccentricity of earths orbit is 0.0167. Te ratio

of its maximum speed in its orbit to its minimum

speed is

(a) 2.507 (b) 1.0339(c) 8.324 (d) 1.000

14. Aand Bare two wires. Te radius ofAis twice that

of B. Tese are stretched by the same force. Ten,

the stress on Bis

(a) equal to that onA

(b) four times that onA

(c) two times that onA

(d) half that onA

15. Te time period of oscillation of magnet in a

vibration magnetometer is 1.5 s. Te time period of

oscillation of another magnet similar in size, shapeand mass but having one-fourth magnetic moment

that of the rst magnet oscillating at the same place,

will be

(a) 0.75 s (b) 1.5 s

(c) 3.0 s (d) 6.0 s

16. Te number of turns and radius of cross-section of

the coil of a tangent galvanometer are doubled. Te

reduction factor Kwill be

(a) K (b) 2 K (c) 4 K (d) K/4

17. An energy of 484 J is spent in increasing the speed

of ywheel from 60 rpm to 360 rpm. Te moment

of inertia of wheel is

(a) 0.2 kg m2 (b) 0.7 kg m2

(c) 2 kg m2 (d) 3 kg m2

18. Te ratio of radii of two spheres of same material is

1 : 4. Ten the ratio of their heat capacity will be

(a)1

64 (b)

1

32 (c)

1

2 (d)

1

4

19. A convex lens of focal length 12 cm is made up of

a glass of refractive index 3/2. When it is immersed

in a liquid of refractive index 5/4, its focal length

will be

(a) 15 cm (b) 6 cm (c) 30 cm (d) 24 cm

20. For a particle performing SHM, the acceleration ofparticle is plotted against displacement. Te curve

will be a

(a) straight line with positive slope

(b) straight line with negative slope

(c) curve whose nature cant be predicted

(d) parabola

21. Find the ratio of Youngs

modulus of wireAto wire B

(a) 1:1

(b) 1 :1

(c) 1 :3(d) 1: 4

22. Te plane faces of two identical plano-convex

lenses each having focal length of 40 cm, are placed

against each other to form a common convex lens.

Te distance from this lens at which an object must

be placed to obtain a real, inverted image with

magnication equal to unity is

(a) 80 cm (b) 40 cm (c) 20 cm (d) 160 cm

23. wo capillary tubes of same diameter are put

vertically one each in two liquids whose relative

densities are 0.8 and 0.6 and surface tensions are60 and 50 dyn cm1respectively. Ratio of height of

liquid in the two tubes h1/h2is

(a)10

9 (b)

3

10 (c)

10

3 (d)

9

1024. Te temperature of the hydrogen at which the rms

speed of its molecules is equal to that of oxygen

molecules at a temperature of 31 C is

(a) 216 C (b) 235 C

(c) 254 C (d) 264 C


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PHYSICS FOR YOU

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25. When the electron in the hydrogen atom jumpsrom second orbit to rst orbit, the wavelength oradiation emitted is . When the electrons jumprom third orbit to rst orbit, then wavelength oemitted radiation would be

(a)27

32 (b)

32

27 (c)

2

3 (d)

3

2

26. 16 g sample o a radioactive element is taken romMumbai to Delhi in 2 h and it was ound that 1 g othe element remained. Hal-lie o the element is

(a) 2 h (b) 1 h (c)1

2h (d)

1

4h

27. A 5 m aluminium wire (Y = 7 1010 N m2) odiameter 3 mm supports a 40 kg mass. In order

to have the same elongation in a copper wire(Y= 12 1010N m2) o the same length under thesame weight, the diameter should be in mm(a) 1.75 (b) 2.0 (c) 2.3 (d) 5.0

28. A vessel containing 5 L o gas at 0.8 mm pressure, isconnected to an evacuated vessel o volume 3 L. Teresultant pressure inside will be

(a)4

3mm (b) 0.5 mm

(c) 2.0 mm (d)3

4mm

29.

A particle perorming SHM along x-axis, withx= 0 as the mean position is released rom rest atx= 2 cm o t= 0. Te time taken by the particle incrossing the position x= 1.6 cm or the second timeis (ake amplitude o SHM as 2 cm and its periodas 1 s)

(a)1

12s (b)

11

12s

(c)323

360s (d) Inormation insufficient.

30. A man can swim with a speed o 4 km h1 in stillwater. How long does he take to cross a river 1 kmwide, i the river ows steadily 3 km h1 and hemakes his strokes normal to the river current. Howar down the river does he go when he reaches theother bank?(a) 850 m (b) 750 m(c) 650 m (d) None

31. On sounding tuning ork A with another tuningork B o requency 384 Hz, 6 beats are producedper second. Afer loading the prongs oAwith somewax and then sounding it again with B, 4 beats are

produced per second. What is the requency o thetuning orkA?(a) 388 Hz (b) 380 Hz(c) 378 Hz (d) 390 Hz

32. An air capacitor o capacity C= 10 F is connectedto a constant voltage battery o 12 V. Now, the spacebetween the plates is lled with a liquid o dielectricconstant 5. Te charges that ows now rom batteryto the capacitor is(a) 120 C (b) 600 C(c) 480 C (d) 24 C

33. Compressed air in the tube o a wheel o a cycle atnormal temperature suddenly starts coming outrom a puncture. Te air inside(a) starts becoming hotter

(b) remains at the same temperature(c) starts becoming cooler(d) may become hotter or cooler depending upon

the amount o water vapour present

34. Te amplitude o SHMy= 2 (sin 5t + 3 cos 5t) is

(a) 2 (b) 2 2 (c) 4 (d) 2 3

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35. wo perectly elastic particles P and Q o equalmass travelling along the line joining them withvelocities 15 m s1and 10 m s1respectively collide.Teir velocities afer the collision will be (in m s1)

P Q

(a) 0 25(b) 5 20(c) 10 15(d) 20 5

36. A solid sphere is rotating about a diameter at anangular velocity . I it cools so that its radiusreduces to 1/n o its original value, its angularvelocity becomes

(a)

n (b)

n2

(c) n (d) n2

37. Te current through choke coil increases rom 0 to 6 Ain 0.3 s and an induced em o 30 V is produced.Te inductance o the coil o choke is(a) 2.5 H (b) 5 H (c) 1.5 H (d) 2 H

38. A resistance R, inductance L and capacitor C areconnected in series to an oscillator o requency .I resonant requency is rthen current will lag thevoltage when(a) = 0 (b) < r(c) = r (d) > r

39. In relation X = 3YZ, X and Z represents the

dimensions o charge and magnetic eld respectively.Te dimensions o Yin MKS system is(a) [ML21A] (b) [M1L03A2](c) [M2L21A] (d) [ML1A2]

40. Dopplers shif in requency does not depend upon(a) the requency o wave produced(b) the velocity o the source(c) the velocity o the observer(d) distance rom the source to the observer

41. A resistance o 4 and a wire o length 5 m andresistance 5 are joined in series and connected

to a cell o em o 10 V and internal resistance1 . A parallel combination o two identical cellsis balanced across 300 cm o the wire, then em ounknown cell is

(a) 1.5 V (b) 3.0 V (c) 0.67 V (d) 1.33 V

42. A long solenoid is ormed by winding 20 turns/cm

the current necessary to produce a magnetic eld

o 20 m inside the solenoid will be approximately

(

0 7 1

410= m A )

(a) 8.0 A (b) 4.0 A (c) 2.0 A (d) 1.0 A

43. A moving coil galvanometer converted into an

ammeter reads upto 0.03 A by connecting a

shunt o resistance 4 r across it and converted

into an ammeter reads upto 0.06 A, when a shunt

o resistance r is connected across it. What is the

maximum current which can be sent through this

galvanometer i no shunt is used?

(a) 0.01 A (b) 0.02 A (c) 0.03 A (d) 0.04 A

44. In a photoelectric experiment, the stoppingpotential VS is plotted against the requency o

incident light. Te resulting curve is a straight line

which makes an angle with the x-axis. Ten, tan

will be equal to (= work unction o surace)

(a)h

e (b)

e

h (c)

e (d)

eh

45. In a sample o radioactive substance, what

percentage decays in one mean lie time?

(a) 69.3 % (b) 63 % (c) 50 % (d) 36 %

ANSWER KEYS

1. (a) 2. (b) 3. (a) 4. (d) 5. (c)

6. (a) 7. (b) 8. (b) 9. (a) 10. (c)

11. (c) 12. (a) 13. (b) 14. (b) 15. (c)

16. (a) 17. (b) 18. (d) 19. (c) 20. (b)

21. (c) 22. (b) 23. (d) 24. (c) 25. (a)

26. (c) 27. (c) 28. (b) 29. (c) 30. (b)

31. (d) 32. (b) 33. (c) 34. (c) 35. (c)

36. (d) 37. (c) 38. (d) 39. (b) 40. (d)

41. (b) 42. (a) 43. (b) 44. (a) 45. (b)

Solution Senders of Physics Musing

SET-34

1. Asit Srivastava, Meerut (Uttar Pradesh)

2. G. Geeth Nischal, Visakhapatnam (Andhra Pradesh)

3. B. Akhil, Hyderabad (Telangana)

4. Samarth Hawaldar, Indore (Madhya Pradesh)

5. Sudipta Ganguly, Paraganas (West Bengal)


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PHYSICS FOR YOU

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Electric ux has two denitions, one conceptual andthe other mathematical.Conceptually it is defined in t wo different waysdepending on the type of surface we are considering.For open surface : It means the count of eld lines thatintersect the given surface.

For example, the ux through two surfaces S1and S2in front of the point charge qare identical since there

cannot be a single eld line which passes through S1but not through S2, as below:

If the eld lines are parallel to a surface, they are saidto be grazing field lines and such field lines are not

considered in ux since they are not intersecting.For closed surface : It means the net count of eld linesthat either enter or leave a surface.Conventionally, field lines that leave the surface areconsidered to be positive ux whereas eld lines thatenter the surface are negative ux.Let us see through examples what these denitions mean.We consider a hemispherical bowl kept inverted on atable and few charges placed in its surroundings.

+1

4

+3

2

Which of the charges create a non-zero ux throughthe surface?+q1is kept outside the hemisphere, hence if there is anyeld line which enters it at one point will necessarilyleave at some other point since it is placed at thediametric plane as shown :

Hence ux due to +q1= 0.

q4 is also kept outside but there are some eld lineswhich would intersect the surface only at one place

hence would create a non-zero ux. Similarly, for q2

and +q3as well.

Lines marked (1) intersect surface only once hence

they contribute in non-zero ux.

Line marked (2) intersects twice hence zero ux.

Line marked (3) does not intersect even once hence

has a zero ux contribution.

From these discussions, one thing is obvious that

for a closed surface there cannot be any net ux of a

charge placed outside it, a charge placed inside or on

the surface will only have a non-zero ux.

Now let us see the mathematical denition.

Mathematically, ux through any surface is dened

as the integral of dot product of elemental area vector

with the electric eld strength.

A small element of the entire surface of area dA is

chosen where the strength of electric eld is

E. Te

direction of area vector is perpendicular to the surface,

hence we chose an elemental area since the normal's

direction would keep on changing for a curved surface

Contributed By:Bishwajit Barnwal, Aakash Institute, Kolkata


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16 PHYSICS FOR YOU

|JUNE 16

and moreover the strength of electric ux will remain

constant even in variable eld for this small element.

Te electric ux through this small elemental area

d E dA =

otal ux through surface

= = E dA EdAcosLet us try to decode

the expression.

Let an elemental area dA

be inclined at an angle with

E.

dAcoswhich is the shaded portion, is the projectionof the area dAperpendicular to the electric field.

Hence, the ux through the actual surface (inclined

at with electric eld) is equal to the ux through

the shaded surface.For example, consider three surfaces - a sphere, a

cylinder and a cone each of radius R placed in a uniform

electric eld as shown.

If we want to calculate the ux entering or leaving

the surfaces of each, we can take the projection

perpendicular to the uniform eld as below:

sphere= E(R2)

cone =

E R R

1

22 2( )( )

cylinder= E(2R)(2R)

With these basic understanding

of flux clear we can move

towards next section : Gauss's law.

GAUSS'S LAW

It states that the net electric ux through a closed

surface is proportional to the charge enclosed within

the surface, with a proportionality constant of 10

.

Te law is very analytical. Since the count of number

of field lines emanating from a positive charge or

terminating on a negative charge is proportional to the

magnitude of the charge. Terefore, if positive charge

enclosed with a surface is greater than negative charge,

more eld lines will emerge out of the surface than

come inside. And obviously for the outside charges

there would not be any ux as already discussed.

Let us try to understand +1+2 5

3

+4

Gaussian surface

t h i s b e t t e r w i t h a nexample. In the diagram

shown, we have arbitrarily

chosen an irregular

shaped closed surface

which encloses charges

+q2and q3.

According to Gauss's law, ux through the Gaussiansurface drawn,

=+ q q2 3

0

Generalising we can say, the ux through any closedsurface

= =

E dA qen

0

where E dA is the ux through an elemental open

surface of the given closed surface, where obviously

the eld lines of all non-touching charges will create

ux. HenceE is the vector sum of the eld due to all

inside as well as outside charges.

Now, let us see some beautiful applications of Gauss's

law where even though we would not be knowing the

strength of electric eld at all places of the surface but

using symmetry we would come to know the amount

of ux through the surface desired.

Ex. A point charge +Qis kept at(i) body centre of a cube

(ii) face centre of a cube (iii) edge centre of a cube (iv) a vertex of a cube

Find the ux through faces of the cube in each

case.


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PHYSICS FOR YOU

|JUNE 16 17

Soln.: (i) Since the charge's location from each ofthe 6 faces of cube identical, we conclude that theeld lines will equally be distributed amongst allfaces. Te entire cube encloses charge +Q.

=

+=

cube each face

Q

0

6

each face =

1

6 0

Q

(ii) Te charge is placed at

centre of the face in xyplane. With respect tothis charge, four faces areidentically placed - the one

which are perpendicular tothe xy plane. Hence, letux through each of them be .

Te face on the xy plane contains the charge henceeld lines over this face are grazing. Hence will haveno ux through it.

Te other face which is parallel to xy plane will havea non-zero ux and different from the perpendicularfaces, let this value be ||. If we construct anothercube just above the face on which the charge isplaced. Tis will enclose the charge and the ux willequally be divided amongst the two as below:

=

+2

0

cubeQ

= = +

cube

1

24

0

Q

Individually ux values and ||cannot be calculatedfurther.

(iii) o nd the ux through cube, 3 more cubesneeds to be symmetrically placed to enclose thecharge.

=

+

40

cube

Q

=

cube1

4 0

Q

Tis ux is divided amongst 6 faces - (a) 2 touching faces (= 0)

(b) 2 faces with shading and has identical

ux because of identical placement. Let thisux be 1each.

(c) 2 faces with shading and has

identical ux for same reason. Let this ux be

2each.

= = +

cube

1

42 2

01 2

Q

(iv)o enclose the charge, 7 more cubes would berequired.

=+

80

cubeQ

=

cube

1

8 0

Q

Here, 3 faces are touching for which = 0 but

for remaining 3 non-touching faces ux will beidentical.

3

1

8 0

non-touching cube= =

Q

=

non-touching

1

24 0

Q

Once you have understood this, we can move ahead,else visit the article again.

Concept of solid angle in flux calculation

Unlike planar angle, solid angle

cannot be measured using aprotactor. It has to be calculatedas below:Te shaded portion is said to be

cap of sphere and it is said tosubtend a solid angle at the

centre where =A

R2.

Tis value is found to be 2(1 cos).

= = A

R22 1 ( cos )

It is measured in steradian.Note that when = 180, entire sphere is formed forwhich = 4steradian.Field lines from a point charge are uniformly distributedin three dimensional space. Terefore if the solid anglesubtended by a surface on a point charge is known, wecan apply unitary method to calculate the ux throughthe surface as below:

Flux per unit solid angle =Q /

0

4


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18 PHYSICS FOR YOU

|JUNE 16

Flux through the surface subtending solid angle

= = Q Q/ /

( cos )0 04 4

2 1

= Q2

10

( cos )

Ex. wo point charges +Q1 and +Q2 are placed on

opposite sides of a circular disc as shown in the

gure. Find the ux through the disc.

Soln.: Here, tan/

, 1 13

3 60= = = R

R

tan , 2 2

3

1

330= = =

R

R

Te eld lines from both point charges cross from

opposite direction. Hence the difference of the ux

would have to be taken.

= Q Q1

01

2

022

12

1( cos ) ( cos )

=

Q Q1

0

2

021

1

2 21

3

2

Many a times, we see questions in which position

dependent eld is given which is uniform in direction

and then ux for a surface is to be calculated.

Let us see one such

example :

Ex. A cubical surface

of side length a is

placed in a region

where electric eldis given by

E E x

ai= +

0 1

^

where xis the x-coordinate of the point.

Find the charge enclosed within the cube.

Soln.: Te direction of

E suggests that only two

surfaces will have non-zero ux, i.e. the ones parallel

toyz plane while for remaining four faces the eld

lines are grazing.

For surface (1) and

(2), the eld are

different but over

their respective

surfaces they are

constant since

x-coordinate does

not change.

net= exit enter = E2a2 E1a

2

=

=E x x

aa E a0

2 1 20

2( )

= = netenq E a0

02 =q E aen 0 0

2( )

Now, suppose the eld was

E E

y

a i= +

0 1

^

, howwould the answer change?

Again, the ux would be through those two surfaces

only since the direction of the eld has not changed.

But now the eld isn't dependent on x-coordinate

and for the two surfaces (1) and (2), only x-co-

ordinates is different. Terefore ux would be

identical for them and through one of them eld

line enters while through other it leaves.

Hence net= 0. i.e., qen= 0 But if entering or exiting ux was supposed to be

calculated, it could have been found out as below;

d EdA E y

aady = = +

0 1 ( )

= + E a y dy a0

0

( ) = +

E a a

02 2

2 =E

a0

23

2

Albert Einstein

Everybody is a genius.

But if you judge a fish by

its ability to climb a tree

it will live its whole life

believing that it is stupid


14/80

PHYSICS FOR YOU

|JUNE 16 19

SOLUTION SET-34

1. (d) : Taking upward direction as positive.

Initial velocity can be obtained by second kinematic

equation, i.e, s= ut+ (1/2) at2

considering motion from CtoA

14 0 81

210 0 82= u . .

u = 43

2

1m s

Let velocity magnitude at pointA= v

so, v= ugt

v= = 43

2 10 0 827

2 1. m s

Hence time taken from A to B to A i.e. till same level

= =2

2 7v

g. s

Hence the time instant at which the particle comes to

the same level = 0.8 + 2.7 = 3.5 s

2. (b) : vcos = u v= usec dv

dtu

d

dt= sec tan

... (i)

tan =b

y

sec coscos

22 2

2

d

dt

b

y

dy

dt

d

dt

b

y

u= = +

or,d

dt

u

b

=

costan2 ... (ii)

From equations (i) and (ii), we get

=dv

dt

u

b

23tan

3. (a) :

..

.

Fext = 0, using conservation of linear momentum

along the string mucos 30 = mv+ mv

23

2

3

4v u v u= =

4. (c) : Let vbe the speed of particle at B, just when it is

about to loose contact.From Newton's second law, for the particle normal tothe spherical surface,

mv

rmg

2

= sin ... (i)

Applying conservation of energy as the block movesfromAto B

1

22mv mg r r = ( cos sin ) ... (ii)

Solving equations (i) and (ii), we get

3 sin = 2 cos

5. (b) : (max)S= 510 nm, (max)NS= 350 nm

By Weins law, (max)S TS = (max)NSTNS

T

TS

NS

= = =350

510

35

510 6. 9

6. (d) : v N

Nrms =

+ + + +1 2 32 2 2 2....

=+ +

v N N N

Nrms

( )( )1 2 1

6

or v N Nrms =+ +( )( )1 2 1

6 ...(i)

v N

Nav =

+ + + +1 2 3 ....

or v N N

N

Nav =

+=

+( ) ( )1

2

1

2 ...(ii)

From equations (i) and (ii)

v

v

N

Nrms

av

=+

+

22 1

6 1

( )

( )


15/80

20 PHYSICS FOR YOU

|JUNE 16

7. (d) : W P V=1

20 0 , TA=T0, TB= 2T0, TC= 4T0

Heat supplied = QAB+ QBC = CVT0+ CP2T0

= =132

1320 0 0

RT P V

Efficiency of the cyclic process

=

1

213

2

1000 0

0 0

P V

P V

= =1

13100 7 7. %

8. (c) : Te rate of heat ow through the layer of the ice,

iy KA

KA

y

dQ

dtH =

( )

= =0 ( )

/

...(i)

Also, rate of heat gain during fusion of ice,

dQ

dtL

dm

dtL A

dy

dt= = .

...(ii)From equations (i) and (ii), we get

KA

yAL

dy

dt

=

ydy K

Ldt=

0

3600

2

4

y K

Lt

2

2

4

03600

2

=

[ ]

1

216 4

4 10 3600 0

0 9 80

3

=

[ ]

( )

.

=

12

12 0 9 804 3600 10 3

. = 30C

Te atmospheric temperature = = 30C9. (c) : Let particle make an angle with the

horizontal.

tan =

=9 1

4 02

y u t a ty y= +

1

2

2

Now, 1 = usin(1) 1

21 2g( ) or usin = 4

and sin = =2

5

2 5u m s1

x u= = =cos 1 2 51

52 m

10. (c) : dW= dQ dU

dW= nCdT nCVdT

W CdT C dT

a

TdT C T

V

V

=

=

=

a

T

T

T T Rln

0

0

0 0

1

( )

W a RT =

ln( )

( )

1

1 0

Winners (May 2016)

1. Debasrija Mondal, Kharagpur (West Bengal)

2. Vishwajeet Patel, Kota (Rajasthan)

3. K. Srishyam, Mumbai (Maharashtra)

Solution Senders (April 2016)

1. Debasrija Mondal, Kharagpur (West Bengal)

2. Devjit Acharjee, Kolkata (West Bengal)

SOLUTION OF MAY 2016 CROSSWORD


16/80


17/80

22 PHYSICS FOR YOU

|JUNE 16

1. In the circuit shown in thegure, each battery is of5 V and has an internal

resistance of 0.2 . Tereading in the idealvoltmeterVis(a) zero (b) 5 V (c) 7.5 V (d) 10 V

2. A conducting loop, carrying

a current I, is placed in auniform magnetic eldpointing into the plane ofthe paper as shown in thegure. Te loop will have atendency to(a) contract (b) expand

(c) move towards +x-axis(d) move towards x-axis.

3. A spherical black body with a radius of 12 cmradiates 450 W power at 500 K. If the radius ishalved and the temperature is doubled, the powerradiated in watt would be(a) 225 (b) 450 (c) 900 (d) 1800

4. An electromagnetic wave of frequency 3 MHzpasses from vacuum into a dielectric medium withpermittivity r= 4.(a) Te wavelength and frequency both remain

unchanged.(b) Te wavelength is doubled and the frequencyremains unchanged.

(c) Te wavelength is doubled and the frequencybecomes half.

(d) Te wavelength is halved and the frequencyremains unchanged.

5. Te Earths magnetic eld at a given point is0.5 105Wb m2. Tis eld is to be annulled bymagnetic induction at the centre of a circular loopof radius 5 cm. Te current required to be own in

the loop is nearly(a) 0.2 A (b) 0.4 A (c) 4 A (d) 40 A

6. Te molar heat capacity of oxygen gas at SP is

nearly 2.5 R. As the temperature is increased, itgradually increases and approaches 3.5 R. Te mostappropriate reason for this behaviour is that at hightemperatures(a) oxygen does not behave as an ideal gas(b) oxygen molecules dissociate in atoms(c) the molecules collide more frequently(d) molecular vibrations gradually become

effective.

7. wo cells of emfs approximately 5 V and 10 V areto be accurately compared using a potentiometer oflength 400 cm.

(a) Te battery that runs the potentiometer shouldhave voltage of 8 V.

(b) Te battery of potentiometer can have a voltageof 15 V and R adjusted so that the potentialdrop across the wire slightly exceeds 10 V.

(c) Te rst portion of 50 cm of wire itself shouldhave a potential drop of 10 V.

(d) Potentiometer is usually used for comparingresistances and not voltages.

8. In the circuit below, Aand B represent two inputsand C represents the output. Te circuit represents

(a) OR gate (b) NOR gate(c) AND gate (d) NAND gate

9. A metal rod moves at a constant velocity in adirection perpendicular to its length. A constantuniform magnetic eld exists in space in a directionperpendicular to the rod as well as its velocity. Selectthe correct statement from the following.


18/80

PHYSICS FOR YOU

|JUNE 16 23

(a) Te entire rod is at same electric potential.

(b) Tere is an electric eld in the rod.

(c) Te electric potential is highest at the centre of

the rod and decreases towards its ends.

(d) Te electric potential is lowest at the centre of

the rod and increases towards its ends.

10. Te Marina trench is located in the Pacic Ocean

and at one place it is nearly 11 km beneath the

surface of water. Te water pressure at the bottom

of the trench is about 1.01 108Pa. A steel ball of

initial volume 0.32 m3 is dropped into the ocean

and falls to the bottom of the trench. What is the

change in the volume of the ball when it reaches to

the bottom? (Bulk modulus of steel = 1.6 1011Pa)

(a) 1.01 104m3 (b) 2.02 104m3

(c) 3.03 105m3 (d) 4.04 103m3

11. A large open tank has two holes in the wall. One

is a square hole of side Lat a depthyfrom the top

and the other is a circular hole of radius Rat a depth

4yfrom the top. When the tank is completely lled

with water, the quantities of water owing out per

second from both holes are the same. Ten, R is

equal to

(a)L

2 (b) 2L (c) L (d)

L

2

12. An ideal gas is lled in a closed rigid and thermallyinsulated container. A coil of 100 resistor carrying

current 1 A supplies heat for 5 minute to the gas.

Te change in internal energy of gas is

(a) 10 kJ (b) 30 kJ (c) 20 kJ (d) 0 kJ

13. A nucleus of 84210 Po originally at rest emits -particle

with speed v. What will be the recoil speed of the

daughter nucleus?

(a)4

206

v (b)

4

214

v (c)

v

206 (d)

v

214

14.

A circular current loop of magnetic moment M isin an arbitrary orientation in an external magnetic

eld

B. Te work done to rotate the loop by 30

about an axis perpendicular to its plane is

(a) MB (b) 32

MB (c)

MB

2(d) zero

15. A vessel containing water is given a constant

acceleration a towards the right along a straight

horizontal path. Which of the following diagrams

in the gure represent the surface of the liquid?

(a) (b)

(c) (d) none of these

16. A parallel plate capacitor of capacitance C is

connected to a battery and is charged to a potential

difference V. Another capacitor of capacitance 2C

is similarly charged to a potential difference 2V.

Te charging battery is now disconnected and the

capacitors are connected in parallel to each other

in such a way that the positive terminal of one is

connected to the negative terminal of the other. Te

nal energy of the conguration is

(a) zero (b)3

2

2CV (c)

25

6

2CV (d)

9

2

2CV

17. A composite rod made of copper (= 1.8 105K1)

and steel (= 1.2 105K1) is heated. Ten

(a) it bends with steel on concave side

(b) it bends with copper on concave side

(c) it does not expand

(d) data is insufficient

18. Te logic circuit given in the gure performs the

logic operation

(a) ABC (b) A B C (c) A B C (d) A B C

19. Te coordinates of a moving particle at any time

are given by x= t3andy= t3. Te speed of the

particle at time tis given by

(a) 3 2 2t + (b) 3 2 2 2t +

(c) t2 2 2 + (d) 2 2+

20. Tree innitely long

charge sheets are placed

as shown in gure. Te

electric eld at point Pis

(a)2

0

k

(b)4

0

k

(c) 2

0

k (d)

4

0

k


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24 PHYSICS FOR YOU

|JUNE 16

21. A thin wire of length Landuniform linear massdensity is bent into acircular loop with centre

at Oas shown in the gure.Te moment of inertia of the loop about the axisXXis

(a)

L3

28

(b)

L3

216

(c)5

16

3

2

L (d)

3

8

3

2

L

22. wo rods of lengths d1 and d2and coefficients ofthermal conductivities K1and K2are kept in contactwith each other end to end. Te equivalent thermalconductivity is(a) K1d1+ K2d2 (b) K1+ K2

(c)K d K d

d d1 1 2 2

1 2

+

+

(d)d d

d

K

d

K

1 2

1

1

2

2

+

+

23. Monochromatic radiation of wavelengthis incident

on a hydrogen sample in ground state. Hydrogenatoms absorb a fraction of light and subsequentlyemit radiation of six different wavelengths. Find thevalue of .(a) 120 nm (b) 130 nm(c) 97.5 nm (d) 107.5 nm

24. A liquid drop of diameter 4 mm breaks into 1000droplets of equal size. Calculate the resultant changein surface energy, the surface tension of the liquid is

0.07 N m1

.(a) 1.5 103J (b) 2.5 105J(c) 2.1 103J (d) 3.2 105J

25. A thin circular ring of massMand radiusris rotatingabout its axis with a constant angular velocity .wo objects each of mass m, are attached gentlyto the opposite ends of a diameter of the ring. Tewheel now rotates with an angular velocity

(a)M

M m( )+ (b)

( )

( )

M m

M m

+

2

2

(c)M

M m( )+ 2 (d)

( )M m

M

+ 2

26. For a particle executing SHM, the displacement xis given by x= Acost. Identify the graph whichrepresents the variation of potential energy (PE) as

a function of time tand displacement x.

(a) I, III (b) II, IV (c) II, III (d) I, IV

27. An ideal gas goes from the state ito the statefas shown in the gure.Te work done by the gas duringthe process(a) is positive (b) is negative(c) is zero(d) cannot be obtained from this information.

28. A disc is rolling without slippingwith angular velocity . P andQ are two points equidistantfrom the centre C. Te order ofmagnitude of velocity is(a) vQ> vC> vP (b) vP> vC> vQ(c) vP= vC, vQ= vC/2 (d) vP< vC> vQ

29. wo vibrating strings of the same material but

lengths L and 2Lhave radii 2r and r respectively.Tey are stretched under the same tension. Boththe strings vibrate in their fundamental modes, oneof length Lwith frequency 1and the other withfrequency 2. Te ratio 1/2is given by(a) 2 (b) 4 (c) 8 (d) 1

30. A pendulum suspended from the roof of an elevatorat rest has a time period T1. When the elevatormoves up with an acceleration a its time periodbecomes T2, when the elevator moves down withan acceleration a, its time period becomes T3, then

(a) T T T1 2 3= (b) T T T1 22

32

= +

(c) T T T

T T1

2 3

22

32

2=

+

(d) T T T

T T1

2 3

22

32

=

+

31. An equilateral prism is

placed on a horizontalsurface. A ray PQ isincident onto it. Forminimum deviation(a) PQis horizontal (b) QRis horizontal(c) RS is horizontal (d) any one will be horizontal.

32. Ifgis the acceleration due to gravity on the earths

surface, the gain in the potential energy of an objectof mass mraised from the surface of the earth to aheight equal to the radius Rof the earth, is

(a)1

2mgR (b) 2mgR (c) mgR (d)

1

4mgR

33. A vessel contains 1 mole of O2gas (molar mass 32)at a temperature T. Te pressure of the gas is P.An identical vessel containing one mole of He gas(relative molar mass 4) at a temperature 2 Thas apressure of(a) P/8 (b) P (c) 2P (d) 8P


20/80

PHYSICS FOR YOU

|JUNE 16 25

34. A body o mass 4mat rest explodes into three pieces.wo o the pieces each o massmmove with a speedv each in mutually perpendicular directions. Tetotal kinetic energy released is

(a)1

2

2mv (b) mv2 (c)3

22mv (d)

5

2

2mv

35. emperature o oxygen kept in a vessel is raised

by 1C at constant volume. Heat supplied to the

gas may be taken partly as translational and partly

rotational kinetic energies. Teir respective shares are(a) 60%, 40% (b) 50%, 50%(c) 100%, zero (d) 40%, 60%

36. Tree blocks o masses 2 kg, 3 kg and 5 kg areconnected to each other with light string and arethen placed on a rictionless surace as shown in thegure. Te system is pulled by a orce F= 10 N, thentension T1is

(a) 1 N (b) 5 N (c) 8 N (d) 10 N

37. A reshly prepared radioactive source o hal-lie2 h emits radiation o intensity which is 64 timesthe permissible sae level. Te minimum time aferwhich it would be possible to work saely with this

source is(a) 6 h (b) 12 h (c) 24 h (d) 128 h

38. A ball, whose kinetic energy is E, is projected atan angle o 45 to the horizontal. What will be thekinetic energy o the ball at the highest point o itsight?(a) E (b) 2E (c) E/2 (d) E/3

39. wo beams o light having intensities I and 4Iinterere to produce a ringe pattern on a screen. Tephase difference between the beams is /2 at pointAand at point B. Ten the difference between theresultant intensities atAand Bis(a) I (b) 4I (c) 5I (d) 7I

40. A heat engine operates between a cold reservoirat temperature T2= 300 K and a hot reservoir attemperature T1. It takes 200 J o heat rom thehot reservoir and delivers 120 J o heat to the coldreservoir in a cycle. What could be the minimumtemperature o the hot reservoir?

(a) 350 K (b) 400 K (c) 300 K (d) 500 K

41. An engine o mass 6.5 metric ton is going up onincline o 5 in 13 at the rate o 9 km h 1. Calculate

the power o the engine i =1

12andg= 9.8 m s2.

(a) 29.5 kW (b) 73.5 kW

(c) 51.5 kW (d) 105.3 kW42. A ray o light rom a denser medium strikes a rarer

medium at an angle o incidence i(see gure). Tereected and reracted rays make an angle o 90with each other. Te angles o reection and rerac-tion are randr. Te critical angle is

(a) sin1(tanr) (b) sin1(tani)

(c) sin1(tanr) (d) tan1(sini)

43. Te dimensions o1

220E

is

(0: permittivity o ree space, E : electric eld)(a) [ML1] (b) [ML22](c) [ML12] (d) [ML21]

44. Te size o the image o an object, which is at innity,as ormed by a convex lens o ocal length 30 cm is2 cm. I a concave lens o ocal length 20 cm isplaced between the convex lens and the image ata distance o 26 cm rom the convex lens, calculate

the new size o the image.(a) 1.25 cm (b) 2.5 cm(c) 1.05 cm (d) 2 cm

45. In a photoelectric experiment anode potential isplotted against plate current.

(a) Aand Bwill have same intensities while Band

Cwill have different requencies(b) B and C will have different intensities whileAandBwill have different requencies

(c) A and B will have different intensities whileBand Cwill have equal requencies

(d) Band C will have equal intensities whileAandBwill have same requencies.

SOLUTIONS

1. (a) : Current in the circuit, I=

=

8 5

8 0 225

VA

. Reading o voltmeter, V= Ir= 5 25 0.2 = 0


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26 PHYSICS FOR YOU

|JUNE 16

2. (b) : According to Flemingslef hand rule, it is clearthat orce acting on eachpart o the loop is radially

outwards. Te loop willhave a tendency to expandunder the orce acting alongoutward drawn normal.

3. (d) : According to Steans law,Energy radiated, E= eAT4(t)

E

te r T

= 4

2 4

or Power (P) = (4e)r2T4

P

P

r

r

T

T

2

1

2

1

2

2

1

4

=

or =P r

r

T

T

2 1

1

2

1

1

4

450

2 2/

or = 4501

4

16

1= 450 42P

or P2= 1800 W

4. (d) : Frequency remains unchanged with change omedium.

n c

v r r( )reractive index = = =

1

10 0

Since ris very close to 1, n r= = = 4 2

Tus,

medium = =n 2

5. (b) : Magnetic induction at the centre o a currentcarrying circular loop

B I

RI

BR=

=

102

2 10

77

=

( . )( )

( . ).

0 5 10 5 10

2 3 14 100 4

5 2

7A

6. (d)

7. (b) : Te balance point can be obtained only i thepotential difference across the wire is greater than

the ems to be compared (or measured).8. (a) : Te given circuit represents an OR gate. When

eitherAor Bor both inputs are high, the outputCis high.

9. (b) : Let the metal rodABmove parallel to x-axis.Let the uniorm magneticeld point in positivez-direction. Te reeelectrons o the metal rodwill experience magneticorce towards B.

Tere will be excess o electrons on B-end andshortage o electrons onA-end.Te endAbecomes positively charged and the endB becomes negatively charged. An electric eld

is thereore set up in the metal rod. Option (b) iscorrect.

10. (b) : Herep= 1.01 108 Pa, V= 0.32 m3

For steel, = 1.6 1011PaAs =

p

V V /

= =

V pV

1 01 10 0 32

1 60 10

8

11

. .

.= 2.02 104m3

11. (a) : Velocity o efflux, v gh= 2

where hdenotes the depth o the hole.Te quantities o water owing out per second rom

both holes are given to be the same.Applying equation o continuity, we have

a1v1= a2v2 or 2 = 2 (4 )2 2

L gy R g y

or L2 = 2R2 or =2

R L

12. (b) : Internal energy = Heat supplied, as W= 0. U= I2Rt

= (1)2 (100) (5 60) = 30,000 J = 30 kJ.

13. (a) : 84210

82206

24Po X + He

Using linear momentum conservation principle,mv= m

1v

1+ m

2v

2

210 0 = 206 v1+ 4v or v v

14

206=

Recoil speed o daughter nucleus=4

206

v

14. (d) : No work is done to rotate the loop about anaxis perpendicular to its plane as

Mis directedalong the axis. Work is done only when the plane othe loop rotates.

15. (c) : When a vessel containingwater is given a constantacceleration a towards the

right, a orce acts on watertowards right.As an action the water exerts a orce on the wall ovessel towards right. As a reaction the wall pushesthe water towards lef. Te slope o water is as shownin gure. At equilibrium, consider P is a particle owater. Fis the resultant o mgand ma(pseudo orce).Te surace o water should set itsel at 90 to F.Hence (c) represents the surace o liquid.

16. (b) : Cand 2Care in parallel to each other. Resultant capacity, CR= (2C+ C) = 3C


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PHYSICS FOR YOU

|JUNE 16 27

Net potential, VR= 2V V= V+

+

2 , 2

Final energy =

1

2CRVR( )

2

= =1

2 32 2

3

2( )( )C V CV

17. (a) : As copper> steelCopper expands more than steel. So rod bends withcopper on convex side and steel on concave side.

18. (c) : Te output of the circuit is

Y A B C Y A B C = + + = + +Applying De Morgans theorem,

Y A B C Y A B C = = or

19. (b) : As x= t3

= =

v

dx

dt tx 32

Also,y= t3

v dy

dtty = = 3

2

Hence, v v v t t x y= + = +2 2 2 2 2 23 3( ) ( )

= +32 2 2t

20. (c) : All the three charge sheets will produce electriceld at P. Te eld will be along negative Z-axis.

Hence

E Ek k=

2+

2

2+

2( ) or =

2

0 0 00

21. (d) : Let I denote moment of inertia of the loopabout the axisXX.

+I

mRmR mR= =

22 2

2

3

2 Now m= mass of loopor m= L

Again 2R= L or R L

=2

( )

I L

LI

L= or =

3

2 2

3

8

2 3

2

22. (d) : Same amount of heat ows through the two rodsin series combination. Let T0be the temperature ofthe junction.

Q A T T t

d d

K

A T T t

d K

A T T t

d K=

+

=

=( ) ( )

( / )

( )

( / )1 2

1 2

1 0

1 1

0 2

2 2

Also, (T1 T2) = (T1 T0) + (T0 T2)

+

= +

d d

K

d

K

d

K1 2 1

1

2

2 =

+

+

K d d

d

K

d

K

1 2

1

1

2

223. (c) : As the hydrogen atoms emit radiation of six

different wavelengths, some of them must have

been excited to n= 4. Te energy in n = 4 state is

E

E4

124

13 6

160 85= = =

.. .

eVeV

Te energy needed to take a hydrogen atom from its

ground state to n= 4 is 13.6 eV 0.85 eV = 12.75 eV.Te photons of the incident radiation should have12.75 eV of energy. So

hc

=12 75. eV

oreV

=hc

12 75.=

1242

12 7597 5

eV nm

eVnm

..

24. (d) : Volume of 1000 droplets = Volume of larger drop

1000

4

3

4

3

3 3 = r R

r

R= =

=

10

2 10

10

2 103

4m m

Surface area of larger drop = 4R2= 4(2 103)2= 16 106 m2Surface area of 1000 droplets = 4r21000 = 4(2 104)21000 = 16 105m2 Increase in surface area = 16106(10 1) = 144 106m2Te resultant increase in surface energy= Surface tension increase in surface area

= = 0 07 14422

710 3168 106 8. J 3.2 10 J5

25. (c) : Since the two objects are attached gently to the

ring, no external torque is applied to the system.

Terefore angular momentum of the system

remains constant.

I11= I22

+ +

21 1

2

2

2 22 2

= = =I

I

Mr

Mr mr

M

M m( ) ( )

26. (a) : Te rst graph illustrates variation of potential

energy with time.

Te equation x=Acos tdescribes SHM when theparticle starts swinging from the extreme position.

PE is maximum at the extreme position. Hence at

t= 0, the potential energy is maximum. Te curveIrepresents PE versus tgraph.

Te second graph illustrates the variation of PE

with x, PE =1

2

2 2m x

27. (c) : We know, PV= nRT

From given gure, P Tand nand Rare constant, V= constant i.e. process is isochoric.So, work done by the gas during the isochoric

process is zero.


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28 PHYSICS FOR YOU

|JUNE 16

28. (b) : Given: CP= CQ= xwhere Cis the centre of therolling disc. P and Q aretwo points on disc. In purerolling, the instantaneousvelocity is zero at the lowermost point.Tis point is denoted as Owhich is the instantaneouspoint of rotation. Linear velocity v= r, where risthe distance of the point from O.From gure, OP> OC> OQ or rP> rC> rQ vP> vC> vQ

29. (d) : For a vibrating string, in fundamental mode,

=1

2=

Mass

lengthl

T

=volume density

length= =

22r l

lr

=1

2=

1

22l

T

r lr

T

1

2

2

1

2

1

as , are same.=l

l

r

rT

For 1, l1= Land r1 = 2rFor 2, l2= 2Land r2 = r.

1

2

1

2

=2

2= 1

L

L

r

r

30. (c) :T l

g

l

Tg1

2

12

24

= = or

T

l

g a

l

Tg a2

2

22

24

=

+

= +

or

T l

g a

l

Tg a3

2

32

24

=

=

or

+ = =4 4

2 242

22

2

32

2

12

l

T

l

Tg

l

T

or T T TT T

12 3

22

32

2=+

.

31. (b) : In equilateral prism, the refracted ray QRrunsparallel to base. Base is horizontal. Hence QR ishorizontal.

32. (a) : At surface of earth, U GmM

R1 =

At height hfrom surface, U GmM

R h2 =

+

If h= R, U GmM

R2 =

2

U= U2 U1or =

2+U

GmM

R

GmM

R

=2

GmM

R

mgRg

GM

R= =

2 2

Gain in of object raised up =

2PE

mgR

33. (c) : We know, PV= nRT

P

P

T

T

V n1

2

1

2

= ( and are same )

P

P

T

TP P

22

22= or =

34. (c) : otal mass = 4mMass of each of the two small pieces = mMass of third piece = 4m 2m= 2mBy conservation of momentum along horizontaldirection,

2mv= mvcos45 + mvcos45

= 2

1

2mv

=v

v

2otal kinetic energy released

= + +

1

2

1

2

1

22

2

2 22

mv mv m v

= + + =1

2

1

2

1

2

3

22 2 2 2mv mv mv mv

35. (a) : A diatomic oxygen molecule has 3 degrees offreedom due to translatory motion and 2 degrees offreedom due to rotatory motion. Teir associatedkinetic energies will be in the ratio 3 : 2 or 60%and 40%.

36. (c) : Acceleration of the blocks

a =+ +

= 10

2 3 51 2m s

FBD for 2 kg mass

Using Newtons second law of motion,

10 T1= 2a,T1= 10 2a= 10 2 1 = 8 N


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30 PHYSICS FOR YOU

|JUNE 16

37. (b)

38. (c) : Suppose the ball is projected with velocity u.Ten

E mu=1

22

At the highest point, the velocity of the ball will be v= horizontal component of the velocity of projection

= =uu

cos452

Te kinetic energy of the ball at the highest point,

E mv m

u= =

1

2

1

2 2

2= =

1

2

1

2

1

2

2mu E

39. (b) : Resultant intensity of interference pattern

I I I I I ( ) = + + 2 cos1 2 1 2 Here I1= I, I2= 4I

At pointAwhen = /2, IA= I+ 4I= 5I Again at point Bwhen = ,

I I I I I I I I B B= + 4 + 2 4 cos or = 5 4

or IB= I IA IB= 5I Ior IA IB= 4I

40. (d) : Te work done by the engine in a cycle is W= 200 J 120 J = 80 JTe efficiency of the engine is

= = =

W

Q

80

2000 40

J

J.

From Carnots theorem, no engine can have anefficiency greater than that of a Carnot engine.

Tus, 0 40 1 13002

1 1

. = TT T

K

orK300

1 0 40 0 601T

=. .

orK

or KT T1 1300

0 60500

.Te minimum temperature of the hot reservoir hasto be 500 K.

41. (b) : Herem= 6.5 metric ton = 6500 kg,g= 9.8 m s2

v= 9 km h1 = = =9

5

182 5

5

13

1. , sinm s

cos sin = = =1 125

169

12

132

otal force required against which the engine needsto workF= mgsin+f= mg sin+ mgcos= mg(sin+ cos)

= +

=6500 9 85

13

1

12

12

1329400. N

Power of the engine = Fv= 29400 2.5 = 73500 W = 73.5 kW

42. (a) : r+ r + 90 = 180 r = 90 rAlso, i = r

Apply Snells law, D sini= R sinror D sinr = R sin(90 r) = Rcosr

R

D

r= tan

=

=

CR

D

rsin sin (tan )1 1

43. (c) :1

20

2 E = Energy per unit volume

= = =

1

2

Energy

Volume

ML

L M0

22 2

3 E [ LL 1 2 ]

44. (b) : Convex lens forms the image at I1. I1is at thesecond focus of convex lens. Size of I1= 2 cm.I1acts as virtual object for concave lens. Concavelens forms the image ofI1at I2.

Lens formula :1

1

=1

v fu

For concave lens,

1 14

= 120

or1 = 120

+ 14

= 420

= 15v v

or v= 5 cm = Distance of I2from concave lens.

=Magnification = =5

4

v

u

size of image

size of object

or2

= 1.25size of image

or size of image due to concave lens = 2.5 cm

45. (d) : At stopping potential, photoelectric current iszero. It is same forAand B.

Aand Bwill have equal frequencies.Saturation current is proportional to intensity.Band Cwill have equal intensity.Option (d) represents correct answer.


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PHYSICS FOR YOU

|JUNE 16 31

1. A siren emitting a sound of frequency 800 Hz movesaway from an observer towards a cliff at a speed of15 m s1. Ten, the frequency of sound that the

observer hears in the echo reected from the cliff is(ake velocity of sound in air = 330 m s1)(a) 838 Hz (b) 885 Hz (c) 765 Hz (d) 800 Hz

2. Out of the following options which one can be usedto produce a propagating electromagnetic wave?(a) A chargeless particle(b) An accelerating charge(c) A charge moving at constant velocity(d) A stationary charge

3. An inductor 20 mH, a capacitor 50 F and a resistor40 are connected in series across a source of emf

V= 10 sin 340t. Te power loss in A.C. circuit is(a) 0.76 W (b) 0.89 W (c) 0.51 W (d) 0.67 W

4. Match the corresponding entries of column 1withcolumn 2. [Where mis the magnication producedby the mirror] Column 1 Column 2

(A) m= 2 (p) Convex mirror

(B) m =1

2 (q) Concave mirror

(C) m= + 2 (r) Real image

(D)m = +

1

2 (s) Virtual image(a) Ap and s; Bq and r; Cq and s; Dq and r(b) Ar and s; Bq and s; Cq and r; Dp and s(c) Aq and r; Bq and r; Cq and s; Dp and s(d) Ap and r; Bp and s; Cp and q; Dr and s

5. Coefficient of linear expansion of brass and steelrods are 1and 2. Lengths of brass and steel rodsare l1 and l2 respectively. If (l2 l1) is maintainedsame at all temperatures, which one of the followingrelations holds good?

(a) 12l2= 22l1 (b) 1l1= 2l2(c) 1l2= 2l1 (d) 1l22= 2l12

6. At what height from the surface of earth thegravitation potential and the value of g are5.4 107 J kg2and 6.0 m s2 respectively? akethe radius of earth as 6400 km.(a) 1400 km (b) 2000 km(c) 2600 km (d) 1600 km

7. A piece of ice falls from a height hso that it meltscompletely. Only one-quarter of the heat producedis absorbed by the ice and all energy of ice getsconverted into heat during its fall. Te value of his[Latent heat of ice is 3.4 105J/ kg andg= 10 N/kg](a) 136 km (b) 68 km

(c) 34 km (d) 544 km8. In a diffraction pattern due to a single slit of width

a, the rst minimum is observed at an angle 30when light of wavelength 5000 is incident on theslit. Te rst secondary maximum is observed at anangle of

(a) sin

1 1

2 (b) sin

1 3

4

(c) sin

1 1

4 (d) sin

1 2

3

9. A potentiometer wire is 100 cm long and a constantpotential difference is maintained across it. wocells are connected in series rst to support oneanother and then in opposite direction. Te balancepoints are obtained at 50 cm and 10 cm from thepositive end of the wire in the two cases. Te ratioof emfs is(a) 3 : 4 (b) 3 : 2 (c) 5 : 1 (d) 5 : 4

10. A particle of mass 10 g moves along a circle of radius6.4 cm with a constant tangential acceleration.What is the magnitude of this acceleration if the


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32 PHYSICS FOR YOU

|JUNE 16

kinetic energy o the particle becomes equal to8 104J by the end o the second revolution aferthe beginning o the motion?(a) 0.18 m/s2 (b) 0.2 m/s2(c) 0.1 m/s2 (d) 0.15 m/s2

11. An air column, closed at one end and open atthe other, resonates with a tuning ork when thesmallest length o the column is 50 cm. Te nextlarger length o the column resonating with thesame tuning ork is(a) 150 cm (b) 200 cm(c) 66.7 cm (d) 100 cm

12. o get output 1 or the ollowing circuit, the correctchoice or the input is

(a) A= 1, B= 1, C= 0 (b) A= 1, B= 0, C= 1(c) A= 0, B= 1, C= 0 (d) A= 1, B= 0, C= 0

13. A gas is compressed isothermally to hal its initialvolume. Te same gas is compressed separatelythrough an adiabatic process until its volume isagain reduced to hal. Ten(a) Compressing the gas isothermally or

adiabatically will require the same amount owork.

(b) Which o the case (whether compressionthrough isothermal or through adiabaticprocess) requires more work will depend uponthe atomicity o the gas.

(c) Compressing the gas isothermally will requiremore work to be done.

(d) Compressing the gas through adiabatic processwill require more work to be done.

14. Te intensity at the maximum in a Youngs doubleslit experiment is I0. Distance between two slits isd= 5, where is the wavelength o light used inthe experiment. What will be the intensity in ront

o one o the slits on the screen placed at a distanceD= 10d?

(a)3

40I (b)

I02

(c) I0 (d)I04

15. A car is negotiating a curved road o radius R. Teroad is banked at an angle . Te coefficient oriction between the tyres o the car and the road iss. Te maximum sae velocity on this road is

(a)g

R

s

s

+

tan

tan1 (b)

g

R

s

s2 1

+

tan

tan

(c) gR s

s

2

1

+

tan

tan (d) gR

s

s

+

tan

tan1

16.

An electron o mass m and a photon have sameenergy E. Te ratio o de-Broglie wavelengthsassociated with them is

(a) c mE( )2

1

2 (b)1 2

1

2

c

m

E

(c)1

2

1

2

c

E

m

(d) E

m2

1

2

(cbeing velocity o light)

17. A black body is at a temperature o 5760 K.Te energy o radiation emitted by the body at

wavelength 250 nm is U1, at wavelength 500 nmis U2 and that at 1000 nm is U3. Wiens constant,b = 2.88 106 nmK. Which o the ollowing iscorrect?(a) U1> U2 (b) U2> U1(c) U1= 0 (d) U3= 0

18. Given the value o Rydberg constant is 107m1, thewave number o the last line o the Balmer series inhydrogen spectrum will be(a) 0.25 107m1 (b) 2.5 107m1

(c) 0.025 104m1 (d) 0.5 107m1

19.

A npntransistor is connected in common emitterconguration in a given amplier. A load resistanceo 800 is connected in the collector circuit andthe voltage drop across it is 0.8 V. I the currentamplication actor is 0.96 and the input resistance

o the circuit is 192 , the voltage gain and thepower gain o the amplier will respectively be(a) 4, 4 (b) 4, 3.69(c) 4, 3.84 (d) 3.69, 3.84

20. wo non-mixing liquids o densities and n(n> 1) are put in a container. Te height o eachliquid is h. A solid cylinder o length Land density

dis put in this container. Te cylinder oats withits axis vertical and lengthpL(p< 1) in the denserliquid. Te density dis equal to

(a) {2 + (n 1)p} (b) {1 + (n 1)p}(c) {1 + (n+ 1)p} (d) {2 + (n+ 1)p}

21. I the velocity o a particle is v = At+ Bt2, whereAand Bare constants, then the distance travelled by itbetween 1 sand 2 s is

(a)3

2

7

3A B+ (b)

A B

2 3+


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PHYSICS FOR YOU

|JUNE 16 33

(c)3

24A B+ (d) 3A+ 7B

22. A astronomical telescope has objective and eyepiece

of focal lengths 40 cm and 4 cm respectively. oview an object 200 cm away from the objective, thelenses must be separated by a distance(a) 50.0 cm (b) 54.0 cm(c) 37.3 cm (d) 46.0 cm

23. Te ratio of escape velocity at earth (ve) to theescape velocity at a planet (vp) whose radius and

mean density are twice as that of earth is

(a) 1 : 4 (b) 1 2: (c) 1 : 2 (d) 1 2 2:

24. A long straight wire of radius a carries a steadycurrent I. Te current is uniformly distributed over

its cross-section. Te ratio of the magnetic elds Band B, at radial distances

a

2 and 2a respectively,

from the axis of the wire is

(a) 1 (b) 4 (c)1

4 (d)

1

225. A capacitor of 2 F is charged as shown in the

diagram. When the switch Sis turned to position 2,the percentage of its stored energy dissipated is

(a) 75% (b) 80% (c) 0% (d) 20%

26. When a metallic surface is illuminated withradiation of wavelength , the stopping potential isV. If the same surface is illuminated with radiation

of wavelength 2, the stopping potential isV

4. Te

threshold wavelength for the metallic surface is

(a)5

2 (b) 3 (c) 4 (d) 5

27. If the magnitude of sum of two vectors is equal to

the magnitude of difference of the two vectors, theangle between these vectors is(a) 45 (b) 180 (c) 0 (d) 90

28. A body of mass 1 kg begins to move under the

action of a time dependent force

F t i t j= +( ) ,^ ^

2 3 2 N

where andi j^ ^

are unit vectors along x and y axis.What power will be developed by the force at thetime t?(a) (2t3+ 3t4) W (b) (2t3+ 3t5) W(c) (2t2+ 3t3) W (d) (2t2+ 4t4) W

29. Te angle of incidence for a ray of light at arefracting surface of a prism is 45. Te angle ofprism is 60. If the ray suffers minimum deviationthrough the prism, the angle of minimum deviationand refractive index of the material of the prismrespectively, are

(a) 45 2; (b) 301

2;

(c) 451

2; (d) 30 2;

30. A particle moves so that its position vector is given

by

r t x t y = +cos sin^ ^ , where is a constant.Which of the following is true?(a) Velocity of perpendicular to

r and accelerationis directed towards the origin.

(b) Velocity is perpendicular to

r and accelerationis directed away from the origin.

(c) Velocity and acceleration both are perpendicularto

r.

(d) Velocity and acceleration both are parallel to

r.

31. Consider the junction diode as ideal. Te value ofcurrent owing throughABis

(a) 101A (b) 103A (c) 0 A (d) 102A

32.wo identical charged spheres suspended from acommon point by two massless strings of lengthsl, are initially at a distance d(d < < l) apart becauseof their mutual repulsion. Te charges begin to leakfrom both the spheres at a constant rate. As a result,the spheres approach each other with a velocity v.Ten vvaries as a function of the distance xbetweenthe spheres, as(a) vx1/2 (b) vx1

(c) vx1/2 (d) vx

33. A small signal voltage V(t) = V0 sint is appliedacross an ideal capacitor C

(a) Current I(t) is in phase with voltage V(t).(b) Current I(t) leads voltage V(t) by 180.(c) Current I(t), lags voltage V(t) by 90.(d) Over a full cycle the capacitor C does not

consume any energy from the voltage source.

34. Te magnetic susceptibility is negative for(a) ferromagnetic material only(b) paramagnetic and ferromagnetic materials(c) diamagnetic material only(d) paramagnetic material only


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34 PHYSICS FOR YOU

|JUNE 16

35. A square loop ABCD carryinga current i, is placed near andcoplanar with a long straightconductor XY carrying acurrent I, the net force on theloop will be

(a)2

30IiL

(b)

0

2

IiL (c)

2

30

Ii (d)

0

2

Ii

36. A uniform rope of length L and mass m1 hangsvertically from a rigid support. A block of mass m2is attached to the free end of the rope. A transversepulse of wavelength 1is produced at the lower endof the rope. Te wavelength of the pulse when itreaches the top of the rope is 2. Te ratio 2/1is

(a)

m

m2

1 (b)

m m

m

1 2

1

+

(c)m

m1

2

(d)m m

m

1 2

2

+

37. When an -particle of mass m moving withvelocity vbombards on a heavy nucleus of chargeZe, its distance of closest approach from the nucleusdepends on mas

(a)1

2m (b) m (c)

1

m (d)

1

m

38. A disk and a sphere of same radius but different

masses roll off on two inclined planes of the samealtitude and length. Which one of the two objectsgets to the bottom of the plane rst?(a) Both reach at the same time(b) Depends on their masses(c) Disk (d) Sphere

39. From a disc of radius R and mass M, a circularhole of diameter R, whose rim passes through thecentre is cut. What is the moment of inertia of theremaining part of the disc about a perpendicularaxis, passing through the centre?(a) 11MR2/32 (b) 9MR2/32

(c) 15MR2/32 (d) 13MR2/3240. A long solenoid has 1000 turns. When a current of

4 A ows through it, the magnetic ux linked witheach turn of the solenoid is 4 103Wb. Te self-inductance of the solenoid is(a) 2 H (b) 1 H (c) 4 H (d) 3 H

41. What is the minimum velocity with which a bodyof mass mmust enter a vertical loop of radius Rsothat it can complete the loop?

(a) 3gR (b) 5gR (c) gR (d) 2gR

42. Te molecules of a given mass of a gas have r.m.s.velocity of 200 m s1at 27C and 1.0 105N m2pressure. When the temperature and pressure of thegas are respectively, 127C and 0.05 105N m2,the r.m.s. velocity of its molecules in m s1is

(a)100 2

3 (b)

100

3 (c) 100 2 (d)

400

3

43. Te charge owing through a resistance R varieswith time tasQ = at bt2, where aand bare positiveconstants. Te total heat produced in Ris

(a) a R

b

3

2 (b)

a R

b

3 (c)

a R

b

3

6 (d)

a R

b

3

3

44. A refrigerator works between 4C and 30C. Itis required to remove 600 calories of heat everysecond in order to keep the temperature of therefrigerated space constant. Te power required is(ake 1 cal = 4.2 Joules)(a) 236.5 W (b) 2365 W(c) 2.365 W (d) 23.65 W

45. A uniform circular disc of radius 50 cm at rest is freeto turn about an axis which is perpendicular to itsplane and passes through its centre. It is subjectedto a torque which produces a constant angularacceleration of 2.0 rad s2. Its net acceleration in ms2at the end of 2.0 s is approximately

(a) 6.0 (b) 3.0 (c) 8.0 (d) 7.0SOLUTIONS

1. (a) : Here, frequency of sound emitted by siren,0 = 800 HzSpeed of source, vs= 15 m s

1

Speed of sound in air, v= 330 m s1

Apparent frequency of sound at the cliff = frequencyheard by observer = Using Dopplers effect of sound

=

=

v

v vs0

330

330 15800

= =330315

800 838 09. Hz 838 Hz

2. (b) : An accelerating charge is used to produceoscillating electric and magnetic elds, hence theelectromagnetic wave.

3. (c) : Here, L= 20 mH = 20 103H,C= 50 F = 50 106FR= 40 , V= 10 sin 340t = V0 sint= 340 rad s1, V0= 10 VXL= L= 340 20 103 = 6.8


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PHYSICS FOR YOU

|JUNE 16 35

XC

C= =

=

=1 1

340 50 10

10

34 558 82

6

4

.

Z R X X C L

= + 2 2( ) = + ( ) ( . . )40 58 82 6 8

2 2

= + =( ) ( . ) .40 52 02 65 622 2

Te peak current in the circuit is

IV

Z00 10

65 62= =

.A ,cos =

R

Z

40

65 62.

Power loss in A.C. circuit,

= =V I V I rms rms cos cos

1

20 0

= =1

210

10

65 62

40

65 620 46

. .. W

4. (c) : Magnication in the mirror, m

v

u=

m= 2 v= 2uAs vand uhave same signs so the mirror is concave and

image formed is real.

m vu

= = 1

2 2Concave mirror and real image.

m = + 2 v= 2uAs vand uhave different signs but magnication is 2 so

the mirror is concave and image formed is virtual.

m vu

= + = 1

2 2

As vand uhave different signs with magnication1

2

so the mirror is convex and image formed is virtual.

5. (b) : Linear expansion of brass = 1Linear expansion of steel = 2Length of brass rod = l1

Length of steel rod = l2On increasing the temperature of the rods by T, newlengths would be

= +l l T1 1 11( ) ...(i) = +l l T2 2 21( ) ...(ii)Subtracting eqn. (i) from eqn. (ii), we get

= + l l l l l l T 2 1 2 1 2 2 1 1( ) ( ) According to question,

= l l l l 2 1 2 1 (for all temperatures) l22 l11= 0 or l11= l226. (c) : Gravitation potential at a height h from the

surface of earth, Vh= 5.4 107J kg 2

At the same point acceleration due to gravity,

gh= 6 m s2

R= 6400 km = 6.4 106m

We know, VGM

R hh =

+( ), g

GM

R h

V

R hh

h=

+

= +( )2

+ = R hV

g

h

h

= hV

gRh

h

=

( . )

.5 4 10

66 4 10

76

= 9 106 6.4 106= 2600 km

7. (a) : Gravitational potential energy of a piece of ice

at a height (h) = mgh

Heat absorbed by the ice to melt completely

Q mgh=1

4 ...(i)

Also, Q= mL ...(ii)

From eqns. (i) and (ii), mL mgh=1

4 or, h

L

g=

4

Here L= 3.4 105J kg1,g= 10 N kg1

=

= =h4 3 4 10

104 34 10 136

53. km

8. (b) : For rst minimum, the path difference

between extreme waves, asin=

Here = 30 =sin1

2 a= 2 ...(i)

For rst secondary maximum, the path difference

between extreme waves

a sin =3

2

or ( )sin23

2

= [Using eqn (i)]

or sin =3

4 =

sin 13

4

9. (b) : Suppose two cells have emfs 1 and 2 (also1> 2).Potential difference per unit length of the potentiometer

wire = k (say)

When 1 and 2 are in series and support each otherthen

1+ 2= 50 k ...(i)When 1and 2are in opposite direction

1 2= 10

k ...(ii)On adding eqn. (i) and eqn. (ii)

21= 60k1= 30k and2= 50k 30k= 20k

= =

1

2

30

20

3

2

k

k

10. (c) : Here, m= 10 g = 102kg,

R= 6.4 cm = 6.4 102 m, Kf= 8 104J

Ki= 0, at= ?

Using work energy theorem,

Work done by all the forces = Change in KE


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36 PHYSICS FOR YOU

|JUNE 16

Wtangential force+ Wcentripetal force= Kf Ki Ft s+0 = Kf 0 mat (2 2R) = Kf

=

a

K

Rmtf

4=

8 10

422

76 4 10 10

4

2 2.

= 0.099 0.1 m s2

11. (a) : From gure,

First harmonic is

obtained at l= =4

50cm

Tird harmonic is

obtained for resonance,

= = =l3

43 50 150

cm.

12. (b) :

Output of the circuit, Y= (A+ B)C

Y= 1 if C= 1 andA= 0, B= 1 orA= 1,

B= 0 orA= B= 1

13. (d) : V1= V

V2= V/2

On P-Vdiagram,

Area under adiabatic curve

> Area under isothermal curve.

So compressing the gas through adiabatic process will

require more work to be done.

14. (b) : Here, d= 5, D= 10d, y d=

2.

Resultant Intensity at y d

Iy= =2

, ?

Te path difference between two waves at y d=

2

x d d

y

D

d d

d= = =

tan 210

= = =d

20

5

20 4

Corresponding phase difference, = =2

2x .

Now, maximum intensity in Youngs double slit

experiment,

Imax= I1+ I2+ 2I1I2 I0= 4I ( ) I I I1 1= =

=I I0

4.

Required intensity Iy= I1+ I2+ 2I1I2 cos2

= 2I=I02

15. (d) :

For vertical equilibrium on the road,Ncos= mg+fsinmg = Ncosfsin ...(i)Centripetal force for safe turning,

N f mv

Rsin cos + =

2 ...(ii)

From eqns. (i) and (ii), we get

v

Rg

N f

N f

2

=+

sin cos

cos sin

=+

v

Rg

N N

N Ns

s

2max sin cos

cos sin

v Rg s

smax

tan

tan=

+

1

16. (c) : For electron of energy E,

de-Broglie wavelength, eh

p

h

mE= =

2

For photon of energy, E h hc

p

= =

=p hcE ...(ii)

= =

e

p

h

mE

E

hc c

E

m2

1

2

1 2/

17. (b) : According to Weins displacement law

mb

T= =

2 88 10

5760

6. nm K

K= 500 nm

Clearly from graph, U1< U2> U3

18. (a) : Here, R= 107m1

Te wave number of the last line of the Balmer series inhydrogen spectrum is given by

1 1

2

12 2

=

R = = = R

4

10

40 25 10

77 1. m


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PHYSICS FOR YOU

|JUNE 16 37

19. (c) : Here, R0 = 800 , Ri = 192 , current gain, = 0.96Voltage gain = Current gain Resistance gain

= =0 96800

192

4.

Power gain = [Current gain] [Voltage gain] = 0.96 4 = 3.84

20. (b) :

d= density of cylinderA= area of cross-section of cylinderUsing law of oatation,

Weight of cylinder = Upthrust by two liquidsL A d g= n(pL A)g+ (LpL)Agd= np+ (1 p) = (np+ 1 p)d= {1 + (n 1)p}

21. (a) : Velocity of the particle is v=At+ Bt2

ds

dtAt Bt= + 2 , ds At Bt dt = +( )2

= + +s At

Bt

C2 3

2 3

s t A B

C( )= = + +12 3

s , s t A B C ( )= = + +2 28

3s

Required distance = s(t= 2 s) s(t= 1 s)

= 28

3 2 3A B C

A BC+ +

+ +

=

3

2

7

3A B+

22. (b) : Herefo= 40 cm,fe= 4 cmube length(l) = Distance between lenses = vo+feFor objective lens, uo= 200 cm, vo= ?

1 1 1

v u fo o o = or

1 1

200

1

40vo

=

or1 1

40

1

200

4

200vo= = vo= 50 cm

l= 50 + 4 = 54 cm

23. (d) : As escape velocity, v=2GM

R

=2 4

3

8

3

3G

R

RR

G =

v

v

R

Re

p

e

p

e

p

=

=1

2

1

2

1

2 2 = (Rp= 2Reand p= 2e)

24. (a) : Magnetic eld at a point inside the wire at

distance r a=

2from the axis of wire is

B= 0

20

20

2 2 2 4I

ar I

aa I

a= =

Magnetic eld at a point outside the wire at distancer(= 2a) from the axis of wire is

B=

0 0 0

2 2

1

2 4

I

r

I

a

I

a= =

B

B= 1

25. (b) : Initially, the energy stored in 2 F capacitor is

Ui=1

2CV2=

1

2(2 106)V2= V2 106J

Initially, the charge stored in 2 F capacitor isQi = CV= (2 10

6)V = 2V 106coulomb. When

switch S is turned to position 2, the charge owsand both the capacitors share charges till a commonpotential VCis reached.

VC=total charge

total capacitancevolt=

+

=

2 10

2 8 10 5

6

6

V V

( )Finally, the energy stored in both the capacitors

Uf=1

2[(2 + 8) 106]

V V

5 5

2 2

= 106J

% loss of energy, U=U U

U

i f

i

100%

=

( / )V V

V

2 2 6

2 6

5 10

10 100

% = 80%

26. (b) : According to Einsteins photoelectric equation,

eVs=hc hc

0 As per question, eV=

hc hc

0

...(i)

eV hc hc

4 2 0=

...(ii)

From equations (i) and (ii), we get

hc hc hc hc

2 4 40 0 =

hc hc

4

3

4 0 = or 0= 3

27. (d) : Let the two vectors be

A and

B .

Ten, magnitude of sum of

A and

B ,

A B A B AB+ = + +2 2 2 cos

and magnitude of difference of

A and

B ,

A B A B AB = + 2 2 2 cos

A B A B+ = (given)

or A B AB A B AB2 2 2 22 2+ + = + cos cos


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38 PHYSICS FOR YOU

|JUNE 16

4ABcos = 0 4AB0, cos = 0 or = 90

28. (b) : Here,

F ti t j= +( ) ,2 3 2 N m=1 kg

Acceleration of the body,

a F

m

ti t j= =

+( )2 3

1

2 N

kgVelocity of the body at time t,

v adt ti t j dt t i t j= = +( ) = + 2 3 2 2 3 1m s Power developed by the force at time t, P=

F v ti t j t i t j = + +( ) ( )2 3 2 2 3 W = (2t3+ 3t5) W

29. (d) : Given, i= 45,A= 60Since the ray undergoes minimum deviation,therefore, angle of emergence from second face,

e= i= 45 m= i+ eA= 45 + 45 60 = 30

=sin

s

physicsforyou june2016

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