kalyan’s physics challenge

Kalyan’s Physics Challenge

NSEA – 2010

1. As a consequence of increase in the distance between the earth and the moon, there will be

(a) a decrease in the number of total solar eclipses

(b) an increase in the number of total solar eclipses

(c) no change in the number of total solar eclipses

(d) an increase in the duration of the total solar eclipses

Sol A total solar eclipse occurs when the Moon completely covers the sun’s disc as seen from the earth.

When the distance between the Earth and the Moon increases, the apparent size of Moon’s disc decreases

as seen from the earth.

the Moon will no longer be able to cover the Sun’s disc completely, thus decreasing the number of solar

eclipses occurs over a period of time.

2. The lengths of two rods measured using a measuring device of least count 0.01 cm are 4.23 cm and

4.26 cm respectively. If the two rods are joined together end to end, the combined length should be

expressed as

a) (8.49 ± 0.03) cm b) (8.49 ± 0.01) cm c) (8.49 ± 0.02) cm d) (8.49 ± 0.10) cm

Sol Least count of the measuring device: 0.01 cm

Length of the first rod: 𝑙1 = 4.23 ± 0.01 cm

Length of the second rod: 𝑙2 = 4.26 ± 0.01 cm

Length of the combined rod: 𝑙1+ 𝑙2 = (4.23 ± 0.01) + (4.26 ± 0.01) = (8.49 ± 0.02) cm

3. The centre of mass of a uniform semi-circular disc of mass M and radius R is located from the midpoint of

its diameter at a distance of

a) 𝑅

𝜋

b) 𝑅

2𝜋

c) 𝟒𝑹

𝟑𝝅

d) 4𝑅

3𝜋2

Sol Consider an elemental ring of radius y, thickness dy.

Mass of the elemental ring: dm = 𝜎 dA = 2𝑀

𝜋𝑅2𝜋y dy =

2𝑀

𝑅2 y dy

Centre of mass of the ring: 2𝑦

𝜋

Centre of mass of the disc: ycm = 1

𝑀 ∫𝑦 𝑑𝑚

ycm = 1

𝑀 ∫

2𝑦

𝜋

2𝑀

𝑅2 y dy

𝑅

0 =

4

𝜋𝑅2 ∫ 𝑦2

𝑅

0𝑑𝑦 =

4

𝜋𝑅2𝑦3

3

Center of mass of the disc: 𝟒𝑹

𝟑𝝅

4. The world’s first artificial satellite was launched on

a) 21st June, 1956 b) 4th October, 1957 c) 25th July, 1958 d) 31st January 1958

Sol Sputnik 1 was the first artificial Earth satellite. The Soviet Union launched it

into an elliptical low Earth orbit on 4 October 1957, orbiting for three weeks

before its batteries died, then silently for two more months before falling back

into the atmosphere. It was a 58 cm (23 in) diameter polished metal sphere,

with four external radio antennas to broadcast radio pulses. Its radio signal was

easily detectable even by radio amateurs, and the 65° inclination and duration

of its orbit made its flight path cover virtually the entire inhabited Earth.


NSEA – 2010

5. The displacement-time graph for two cyclists A and B going home from school are as shown. Then, their

relative velocity

(a) first increases and then decreases (b) first decreases and then increases

(c) zero (d) is non-zero constant

Sol slope of displacement-time graph: velocity

From the dig: slope of line A is greater than that of B.

From the dig: velocity of A and B are constant.

The relative velocity of A and B: vAB = vA – vB is constant with time.

6. The dimensions of the product of two physical quantities P and Q are [ML2T-2] while those of𝑃

𝑄 are

[MT-2]. Then, quantities P and Q respectively are

(a) force & velocity (b) momentum & displacement

(c) force & displacement (d) work & velocity

Sol Dimensions of the product: PQ: [ML2T-2] ---- (1)

Dimensional of the division: 𝑃

𝑄 = [MT-2] ---- (2)

(1) x (2): P2 = [M2L2T-4] → p = [MLT-2] – force

(1) / (2): Q2 = [L2] → Q = [L] – displacement

7. One ‘astronomical unit’ is

(A) the diameter of the sun (B) the average distance between earth and moon

(C) the average distance between earth and sun (D) the average distance between sun and Saturn

Sol Astronomical Unit: AU: The average distance between Sun and the Earth.

1 AU = 1.5 x 108 km = 9.3 x 107 mi = 4.8 x 10-6 par sec (pc) = 1.6 x 10-5 light years (ly)

8. ‘Chandrasekhar Limit’ is associated with

a) planetary nebulae b) white dwarf c) binary stars d) variable stars

Sol • The Chandrasekhar limit is the maximum mass of a stable white dwarf star. The currently

accepted value of the Chandrasekhar limit is about 1.4 M = 2.765×1030 kg

• White dwarfs resist gravitational collapse primarily through electron degeneracy pressure. The

Chandrasekhar limit is the mass above which electron degeneracy pressure in the star's core is

insufficient to balance the star's own gravitational self-attraction. Consequently, a white dwarf

with a mass greater than the limit is subject to further gravitational collapse, evolving into a

different type of stellar remnant, such as a neutron star or black hole. Those with masses under the

limit remain stable as white dwarfs.

• The limit was named after Subrahmanyam Chandrasekhar, the Indian astrophysicist who

improved upon the accuracy of the calculation in 1930, at the age of 20, in India by calculating the

limit for a polytrope model of a star in hydrostatic equilibrium.


NSEA – 2010

9. An object covers half of the distance left in each successive equal interval of time. The distance-time

graph will be

Sol Given: An object covers half of the distance left in each successive equal interval of time.

[This situation is similar to radioactive decay: N = N0 e-λt = N0 e-0.693t/T → N ∝ e-t]

Let the equal interval of time be T.

Let s0 be the total distance to be covered and s, the remaining distance.

We can write s = s0 e-kt/T → s ∝ e-t

10. The equations of motion of a projectile are given by x = 36t and 2y = 96t – 9.8t2. The angle of projection is

a) sin-1 (4/5) b) sin-1 (3/5) c) sin-1 (2/5) d) sin-1 (3/4)

Sol Displacement in horizontal direction: x = 36t ---- (1)

Displacement in vertical direction: y = 48t – 1

2 9.8t2 ---- (2)

Displacement in horizontal direction: x = u cos θ (t)

Displacement in vertical direction: y = u sin θ (t) −1

2 gt2

u cos θ = 36 and u sin θ = 48

𝑢 sin𝜃

𝑢 cos𝜃= 48

36→ tan θ =

4

3→ sin θ =

𝟒

𝟓

11. Uranus was discovered in the constellation of

a) Taurus b) Capricornus c) Aquarius d) Gemini

Sol When William Herschel discovered Uranus on 13 March 1781 it

was located near η Gem.

Eta Geminorum (η Geminorum, abbreviated Eta Gem, η Gem),

formally named Propus is a triple star system in the constellation of

Gemini.

It is a naked-eye variable star around 380 light years from the Sun.


NSEA – 2010

12. A particle P is sliding down a frictionless hemispherical bowl with center at O. It passes the point A at t =

0 with a speed 10 cm/s. At this instant, a bead Q of the same mass as that of P is ejected from A with a

speed 5 cm/s along the string AB. There is no friction between the bead and the string. Then,

(a) P and Q will reach point B at the same time

(b) P will reach point B earlier than Q

(c) P will reach point B later than Q

(d) from A to B, the magnitude of average velocity of bead Q is higher than that of particle P

Sol

Speed of the particle P, at A: 10 cm/s (along the tangent)

Speed of bead Q at A: 5 cm/s (along AB)

Length of the line AB: 2 x R sin 60 = √3R / Time taken by the bead: tb = 𝐴𝐵

5= √3 R

5 sec

Vertical component of velocity of p decreases from A to C and then increases from C to A.

The average vertical component of velocity of particle p is zero.

Horizontal component of velocity: 10 cos 60 = 5 cm/s [constant value]

So, particle p also takes the same time to go from A to B.

13. The points of intersection of the Moon’s orbit with the path of the Sun are called

a) equinoxes b) nodes c) solstices d) Poles

Sol The lunar nodes are the two points where the Moon's orbital path crosses the ecliptic, the Sun's apparent

yearly path on the celestial sphere.

The ascending (or north) node is where the Moon moves into the northern ecliptic hemisphere, while the

descending (or south) node is where the Moon enters the southern ecliptic hemisphere.

A lunar eclipse can occur only when the full Moon is near (within 11° 38' ecliptic longitude) either lunar

node,

while a solar eclipse can occur only when the new Moon is near (within 17° 25') either lunar node.

14. Distance of the plane 6x – 3y + 2z – 21 = 0 from origin is

a) 7 b) 21 c) 3 d) √70

Sol Perpendicular distance between plane and origin is P =|𝑑|

√𝑎2+𝑏2+𝑐2

Given equation is 6x – 3y + 2z – 21 = 0, compare with standard equation we will get

𝑎 = 6, 𝑏 = −3, 𝑐 = 2, 𝑑 = −21

Hence P = |−21|

√36+9+4=

21

√49=

21

7= 𝟑


NSEA – 2010

15. The planet having the most inclined orbit to the ecliptic is

a) Uranus b) Earth c) Mars d) Mercury

Sol The ecliptic is the mean plane of the apparent path in the Earth's sky that the Sun follows over the course

of one year; it is the basis of the ecliptic coordinate system. This plane of reference is coplanar with Earth's

orbit around the Sun.

S.NO PLANET INCLINATION

1 MERCURY 7.010

2 VENUS 3.390

3 EARTH 00

4 MARS 1.850

5 JUPITER 1.310

6 SATURN 2.490

7 URANUS 0.770

8 NEPTUNE 1.770

16. Which of the following statements is true in case of multiplication of matrices?

(A) when matrices A and B are conformable with each other, AB and BA are both square matrices of the

same order

(B) If A and B are square matrices of the same order, then AB = BA

(C) If A and B are matrices of suitable order, (AB)1 = B1A1

(D) If A and B are matrices of suitable order, then AB = 0 implies A = 0 or B = 0

Sol If A and B are matrices of suitable order, then (AB)1 = B1A1 (property of transpose of matrices)

Other options are not true in general.

17. In a certain region of space, electric and magnetic fields are uniform and parallel to each other. If a proton

is projected at right angles to the two fields, then the path of the proton will be

a) a parabola b) a circle

c) a straight line d) helix with increasing pitch

Sol Uniform electric and magnetic fields are parallel to each other.

Force on a charged particle due to electric field: Fe = qE

Force on a charged particle due to magnetic field:

Fm = qvB sin θ = qvB (θ = 900)

The magnetic field takes the proton into circular path:

centripetal force

The magnetic field takes the proton in a straight line along the

field direction.

The resultant of these two paths: helix with increasing pitch


NSEA – 2010

18. Two persons A and B stand at a distance x away from each other in an open field. Wind blows from A to

B with a velocity w. When A beats a drum, sound travels from A to B in time t1, while when B beats a

drum it travels from B to A in time t2. If the velocity of sound in still air is v, then v equals

a) 2𝑥

𝑡1 + 𝑡2

b) 𝒙

𝟐[𝟏

𝒕𝟏+

𝟏

𝒕𝟐] c) 𝑥

2[1

𝑡1−

1

𝑡2] d) 2𝑥

√𝑡1𝑡2

Sol Speed of sound in still air: v

Given: Speed of wind: w

Time taken by sound to go from A to B: t1 = 𝑥

𝑣 + 𝑤→ v + w =

𝑥

𝑡1 ---- (1)

Time taken by sound to go from B to A: t2 = 𝑥

𝑣 − 𝑤→ v – w =

𝑥

𝑡2 ---- (2)

(1) + (2): 2v = 𝑥

𝑡1 +

𝑥

𝑡2→ v =

𝒙

𝟐[𝟏

𝒕𝟏+

𝟏

𝒕𝟐]

19. Which of the following definite integrals has a value zero?

a) ∫ 𝑠𝑖𝑛 (𝑥

2+

𝜋

3)

2𝜋

0𝑑𝑥 b) ∫ 𝑐𝑜𝑠 (

𝑥

2+

𝜋

4) 𝑑𝑥

9𝜋/4

𝜋/4

c) ∫ 𝐬𝐢𝐧 (𝟏𝟐𝒙 −𝝅

𝟕)𝒅𝒙

𝟕𝝅/𝟏𝟐

𝝅/𝟒d) ∫ cos(5𝑥) 𝑑𝑥

𝜋/5

𝜋/10

Sol Here I = ∫ 𝑠𝑖𝑛 (12𝑥 −𝜋

7) 𝑑𝑥

7𝜋/12

𝜋/4

=−𝑐𝑜𝑠(12𝑥−

𝜋

7)

12|𝜋

4

7𝜋

2

=𝑐𝑜𝑠(12×

𝜋

4−𝜋

7)

12−

𝑐𝑜𝑠(12×7𝜋

12−𝜋

7)

12 =

−𝑐𝑜𝑠(𝜋

7)

12−

−𝑐𝑜𝑠(𝜋

7)

12= 𝟎

Remaining are not equal to zero

20. A square loop of wire with side 2.0 m is perpendicular to a magnetic field with half the area of the loop in

the field as shown. The field varies with time as B = 0.04 – 0.9t where B is in T and t is in seconds. The

loop contains a 2.0 volt cell with negligible resistance. The net emf in the circuit is

(A) 2 volts directed clockwise (B) 3.8 volt directed anticlockwise

(C) 0.2 volt directed clockwise (D) 0.2 volt directed anticlockwise

Sol Magnetic field in the region: B = 0.04 – 0.9t

Magnetic flux through the given region:

φ = BA = 2 (0.04 – 0.9t) = 0.08 – 1.8t

Rate of change of flux: 𝑑φ

𝑑𝑡 = 0 – 1.8 = –1.8

Emf induced in the square loop: e = −𝑑φ

𝑑𝑡 = 1.8 volt

Direction of induced current: anticlockwise

The cell connected in the loop also gives current in the same direction.

So, the effective emf in the circuit: E = 1.8 + 2 = 3.8 volt

21. If ∫ 𝑒𝑥 log 𝑥 (𝑥 log𝑥 + 2

𝑥)𝑑𝑥 = 𝜑(x) + A, then 𝜑(𝑥) is equal to

a) 𝑒𝑥 log x b) 𝑒𝑥 𝑙𝑜𝑔 𝑥 c) 𝑒𝑥

log 𝑥

d) 𝒆𝒙 (log x)2


NSEA – 2010

Sol Let given integral is equal to I

I = ∫ 𝑒𝑥 log 𝑥 (𝑥 log 𝑥 + 2

𝑥) 𝑑𝑥 = 𝜑(x) + A

= ∫𝑒𝑥 [𝑥(𝑙𝑜𝑔𝑥)2+2𝑙𝑜𝑔𝑥

𝑥] 𝑑𝑥

= ∫𝑒𝑥 [(𝑙𝑜𝑔𝑥)2 +2𝑙𝑜𝑔𝑥

𝑥] 𝑑𝑥

= ∫𝑒𝑥 [(𝑙𝑜𝑔𝑥)2 +𝑑

𝑑𝑥(𝑙𝑜𝑔𝑥)2] 𝑑𝑥

= 𝑒𝑥(𝑙𝑜𝑔𝑥)2 + 𝐴 (∵ ∫ 𝑒𝑥[𝑓(𝑥) + 𝑓′(𝑥)]𝑑𝑥 = 𝑒𝑥𝑓(𝑥) + 𝐶 )

Hence 𝜑(x) = 𝒆𝒙(𝒍𝒐𝒈𝒙)𝟐

22. ‘Chandrayaan – I’ was launched in the year

a) 2006 b) 2007 c) 2008 d) 2009

Sol Chandrayaan-1 was India's first lunar probe. It was launched by the Indian Space Research Organization

[ISRO] on 22nd October 2008 and operated until August 2009.

Mass:1,380 kg at launch, 675 kg at lunar orbit and 523 kg after releasing the impactor.

Dimensions: Cuboid in shape of approximately 1.5 m (4.9 ft)

Power: The spacecraft was mainly powered by its solar array, which included one solar panel covering a

total area of 2.15 × 1.8 m generating 750 W of peak power, which was stored in a 36 Ah lithium-ion

battery for use during eclipses.

Propulsion: The spacecraft used a bipropellant integrated propulsion system to reach lunar orbit as well as

orbit and altitude maintenance while orbiting the Moon. The power plant consisted of one 440 N engine

and eight 22 N thrusters. Fuel and oxidizer were stored in two tanks of 390 liters (100 US gal) each.

23. In a plane there are two families of straight lines given by y = x + r and y = r – x where r ∈ [0, 1, 2, 3, 4].

Some of the squares formed by these lines have diagonals of length 2 units. The number of such squares is

a) 8 b) 9 c) 12 d) 16

Sol Two families of straight lines given by 𝑦 = 𝑥 + 𝑟 and 𝑦 = 𝑟 − 𝑥 where 𝑟 ∈ 0,1,2,3,4. Also given that

some of the squares formed by these lines have diagonals of length 2 units.

Family of lines given as

𝑦 = 𝑥 + 𝑟 : 𝑦 = 𝑟 − 𝑥

= 𝑥 + 0, → 𝑙1 : = 0 − 𝑥, → 𝑚1

= 𝑥 + 1, → 𝑙2 : = 1 − 𝑥, → 𝑚2

= 𝑥 + 2, → 𝑙3 : = 2 − 𝑥, → 𝑚3

= 𝑥 + 3, → 𝑙4 : = 3 − 𝑥, → 𝑚4

= 𝑥 + 4, → 𝑙5 : = 4 − 𝑥, → 𝑚5

𝑦 = 𝑚𝑥 + 𝑐 : 𝑦 = 𝑚𝑥 + 𝑐

Here m = 1 : Here m = –1

⇒ 𝑡𝑎𝑛𝜃 = 1 : ⇒ 𝑡𝑎𝑛𝜃 = −1⇒ 𝜃 = 450 : ⇒ 𝜃 = 1350

Lets draw the diagram

For diagonal length 2 square can be formed by the lines from first family (𝑙1, 𝑙3), (𝑙2, 𝑙4) and (𝑙3, 𝑙5)

because perpendicular distance between two lines i.e. 𝑙1 and 𝑙2 is|1|

√12+12=

1

√2 , for getting diagonal length

2 then side of square is√2.

For that we need to choose alternate pairs so three pairs in first family and 3 pairs in second family.

So total number of squares 3 × 3 = 𝟗 squares.


NSEA – 2010

24. The planet from the solar system which comes closest to the earth is

a) Mercury b) Mars c) Jupiter d) none

Sol When Venus lies between Earth and the Sun, it experiences what

is known as an inferior conjunction. It is at this point that it makes

its closest approach to Earth (and that of any planet) with an

average distance of 41 million km (25,476,219 mi).

25. When the moon is at 900 East of sun, it will rise

a) in the morning b) at noon c) in the evening d) at midnight

Sol

26. Two hollow pipes slide over each other as shown below. At the free end of one pipe, a buzzer of frequency

2000 Hz is fixed and holes are made near the buzzer end as shown. When the other pipe is moving out, a

louder sound is heard at a separation of (speed of sound: 340 m/s)

a) 68 cm b) 17 cm c) 34 cm d) 51 cm

Sol DELETED


NSEA – 2010

27. Intercalary month (Adhik masa) is the feature of

a) Solar calendar b) Lunar calendar c) Luni-Solar calendar d) Gregorian calendar

Sol Intercalation or embolism in timekeeping is the insertion of a leap day, week or month into some

calendar years to make the calendar follow the seasons or moon phases.

Luni-solar calendars may require intercalations of both days and months.

The solar year does not have a whole number of lunar months, so a luni-solar calendar must have a

variable number of months in a year.

Regular years have 12 months, but embolismic years insert a 13th "intercalary" or "leap" every

second or third year.

Whether to insert an intercalary month in a given year may be determined using regular cycles

such as the 19-year Metonic cycle (Hebrew calendar) or using calculations of lunar phases (Hindu

lunisolar and Chinese calendars).

The Buddhist calendar adds both an intercalary day and month on a usually regular cycle. The

Jewish year is arranged according to a luni-solar system.

28. BC is a line segment. Two straight lines l and m, one through B and other through C are drawn making

some angles with BC. Bisectors of these two angles are perpendicular. Therefore, BC and l, m form

a) an equilateral triangle b) an isosceles triangle c) a right-angled triangle d) no triangle

Sol Let line l passes through point B and makes an angle α and m is line

passes through point C and makes an angle β. And bisectors of these two

angles are perpendicular say(P)

Then 𝛼

2+

𝛽

2= 900 (from 𝛥𝑃𝐵𝐶)

⇒ 𝛼 + 𝛽 = 1800

Hence line l and m are parallel. So, no triangle is possible.

29. If sec A + tan A = a, then sin A can be expressed as

a) 𝑎 − 1

𝑎 + 1

b) 𝑎

𝑎 + 1c) 𝑎 − 1

𝑎2 + 1

d) 𝒂𝟐 − 𝟏

𝒂𝟐 + 𝟏

Sol Given 𝑠𝑒𝑐𝐴 + 𝑡𝑎𝑛𝐴 = 𝑎 … (1)

𝑠𝑒𝑐𝐴 − 𝑡𝑎𝑛𝐴 = 1

𝑎 … (2) (∵ 𝑠𝑒𝑐2𝐴 − 𝑡𝑎𝑛2𝐴 = 1)

Solving equation (1) and (2), we will get

𝑠𝑒𝑐𝐴 = 1

2(𝑎 +

1

𝑎)

𝑡𝑎𝑛𝐴 = 1

2(𝑎 −

1

𝑎)

Now 𝑠𝑖𝑛𝐴 =𝑡𝑎𝑛𝐴

𝑠𝑒𝑐𝐴=

1

2(𝑎−

1

𝑎)

1

2(𝑎+

1

𝑎)=

𝑎2−1

𝑎𝑎2+1

𝑎

=𝒂𝟐−𝟏

𝒂𝟐+𝟏

30. If place A has local time ahead of the standard time, then with respect to the standard meridian place A lies

to the

a) east b) south c) west d) North


NSEA – 2010

Sol • Earth makes one complete revolution of 360° in one day or 24 hours: it passes through 15° in one

hour or 1° in 4 minutes.

• The earth rotates from west to east, so every 15° we go eastwards, local time is advanced by 1

hour. Conversely, if we go westwards, local time is retarded by 1 hour.

• We may thus conclude that places east of Greenwich see the sun earlier and gain time, whereas

places west of Greenwich see the sun later and lose time.

• If we know G.M.T., to find local time, we merely have to add or subtract the difference in the

number of hours from the given longitude.

31. A circle touches Y axis at P (0, 9) and cuts X axis at A and B. If A is (3, 0), the coordinates of B are

a) (27, 0) b) (6, 0) c) (15, 0) d) (12, 0)

Sol Given A circle touches Y – axis at P (0, 9) and cuts X – axis at

A and B & A (3, 0)

Let B (x, 0) because it lies on X – axis,

Apply the power of a point theorem then,𝑂𝐴 × 𝑂𝐵 = 𝑂𝑃2

3 × 𝑥 = 92 ⇒ 𝑥 =81

3= 27

Hence coordinates of point B (27, 0).

32. If x + y = 1, the largest value of xy is

a) Zero b) 0.25 c) 0.5 d) 0.75

Sol Given 𝑥 + 𝑦 = 1, we know that 𝐴𝑀 ≥ 𝐺𝑀

⇒ 𝑥 + 𝑦 ≥ 2√𝑥𝑦 ⇒ √𝑥𝑦 ≤𝑥+𝑦

2⇒ √𝑥𝑦 ≤

1

2⇒ 𝑥𝑦 ≤ (

1

2)2⇒ 𝑥𝑦 ≤

1

4= 0.25

So maximum value of xy is 0.25

33. The event in which a smaller and nearer object passes across the disc of a larger object in the background

is known as

a) transit b) occultation c) opposition d) Obscuration

Sol In astronomy, a transit is a phenomenon when a celestial body passes directly

between a larger body and the observer.

As viewed from a particular vantage point, the transiting body appears to

move across the face (disc) of the larger body, covering a small portion of it.

The word "transit" refers to cases where the nearer object appears smaller

than the more distant object. Cases where the nearer object appears larger and

completely hides the more distant object are known as occultations.

34. Which of the following is not a leap year?

a) 1860 b) 1956 c) 𝟏𝟗𝟎𝟎 d) 2004


NSEA – 2010

Sol Leap year should be evenly divided by 4: 1900

4 = 475

If the year is evenly divided by 100, then it should also be evenly divided by 400.

1900

100 = 19 [evenly divided by 100]

1900

400 = 4.75 [not divisible by 400 evenly]

35. The north celestial pole lies in the constellation of

a) Ursa Major b) Ursa Minor c) Cassiopeia d) Draco

Sol The North Celestial Pole is the point in the sky about which all the stars seen from the Northern Hemisphere

rotate.

The North Star, also called Polaris, is located almost exactly at this point in the sky.

The north celestial pole is the imaginary point in the sky where the Earth's axis of rotation, indefinitely

extended, intersects the celestial sphere. The north and south celestial poles appear permanently directly

overhead to an observer at the Earth's North Pole and South Pole, respectively.

The north celestial pole lies in Ursa Minor constellation.

Refer to Q. No. 15 for a diagrammatic representation.

36. If α and β are the roots of the equation x2 + x + 3 = 0, then α6 + β6 is

a) −10 b) 10 c) −30 d) 30

Sol If α, β are the roots of the equation 𝑥2 + 𝑥 + 3 = 0, then by Vieta’s Relations

𝛼 + 𝛽 = −1 and 𝛼𝛽 = 3

We can write 𝛼2 + 𝛽2 = (𝛼 + 𝛽)2 − 2𝛼𝛽= (−1)2 − 2.3 = 1 − 6 = −5

⇒ 𝛼2 + 𝛽2 = −5Cube on both sides

(𝛼2 + 𝛽2)3 = (−5)3

𝛼6 + 𝛽6 + 3𝛼2𝛽2(𝛼2 + 𝛽2) = −125𝛼6 + 𝛽6 + 3(9)(−5) = −125𝛼6 + 𝛽6 − 135 = −125

𝛼6 + 𝛽6 = 𝟏𝟎

37. Choose the correct statement regarding total solar eclipse and lunar eclipse

(A) each of the two can be observed from a small region on the surface of the earth

(B) total solar eclipse can be observed from a small region while total lunar eclipse can be seen from

a large region (almost half) of the earth’s surface

(C) total lunar eclipse can be seen from a small region while total solar eclipse can be observed from a

large region on the surface of the earth

(D) each of the two can be observed from a large region (more than 20%) of the earth’s surface

Sol Total solar eclipse: moon comes between sun and earth. Only the people present in the small area of

umbra can see the solar eclipse.

Total lunar eclipse: earth comes between sun and moon. Since half of the earth is on the other side of sun,

all those who are present on the part not facing sun can see the lunar eclipse.


NSEA – 2010

38. The activity of a radioactive sample of initial mass m and molecular mass M after a time t is

(NA – Avogadro number, λ – decay constant)

a) (𝒎𝑵𝑨

𝑴) λ𝒆−𝝀𝒕 b) (

𝑀𝑁𝐴

𝑚) λ𝑒−𝜆𝑡 c) (

𝑚

𝑀𝑁𝐴) λ𝑒−𝜆𝑡 d) (

𝑚𝑁𝐴

𝑀) λ𝑒𝜆𝑡

Sol Initial mass of the radioactive sample: m

Molecular mass of the radioactive sample: M

Initial no. of molecules (atoms) in the sample: N0 = 𝑚

𝑀 NA

Radioactive decay law: N = N0 e-λt = 𝑚

𝑀 NA e-λt

Activity of a radioactive sample: R = 𝑑𝑁

𝑑𝑡= λN =

𝒎

𝑴 NA λe-λt

39. In the figure, ratio of currents in the 3 Ω and 1Ω resistors and that of the corresponding potential

difference respectively, are

a) 2/3 and 2/1 b) 3/2 and 1/2 c) 2/3 and 1/2 d) 3/2 and 2/1

Sol Let the current in 1 Ω resistor be I.

Current through 3 Ω resistor: i1 = 6

3+6I =

2

3I →

𝑖1

𝐼= 𝟐

𝟑

Potential difference across 1 Ω resistor: 1 x I = I

Potential difference across 3 Ω resistor: 3 x 2

3 I = 2I

Ratio of potential differences: 2𝐼

𝐼= 𝟐

𝟏

40. If a, b, c are in AP, x is GM of a and b while y is GM of b and c, then b2 is

a) AM of x2 and y2 b) GM of x2 and y2 c) HM of x2 and y2 d) none of these

Sol Given a, b, c are in AP.

⇒ 2𝑏 = 𝑎 + 𝑐 ……. (1)

‘x’ is GM of a, b

⇒ 𝑥 = √𝑎𝑏 , ⇒ 𝑥2 = 𝑎𝑏 ……. (2)

And, similarly 𝑦2 = 𝑏𝑐 ……. (3)

Adding (2) and (3) we will get

𝑥2 + 𝑦2 = 𝑎𝑏 + 𝑏𝑐𝑥2 + 𝑦2 = 𝑏(𝑎 + 𝑐)

𝑥2 + 𝑦2 = 2𝑏2 using (1)

𝑏2 =𝑥2 + 𝑦2

2Hence 𝑏2 is AM of 𝒙𝟐 and 𝒚𝟐

41. A particle A has charge +q and a particle B has a charge +4q with each of them having the same mass m.

When allowed to fall from rest through the same potential difference, the ratio of their speeds vA:vB will be

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


NSEA – 2010

Sol Charge on the first particle: q1 = +q

Charge on the second particle: q2 = +4q

Mass of both the particles is same: m1 = m2 = m

Potential difference applied to both the particles is same: V1 = V2

Work-Energy theorem: wnet = ∆k → qV = 1

2 mv2

qV = 1

2 mvA

2 ---- (1)

4q V = 1

2 mvB

2 ---- (2)

vA : vB = 1 : 2

42. A luminous point object is placed at a distance of 30 cm from a convex lens of focal length 20 cm. On the

other side of the lens, the distance at which a convex mirror of radius of curvature 10 cm must be placed in

order to have real image of the object coincident with it is

a) 12 cm b) 30 cm c) 50 cm d) 60 cm

Sol Object distance for the lens: u = 30 cm

Focal length of the lens: f = 20 cm

Image distance: 1

𝑣−

1

𝑢=

1

𝑓 →

1

𝑣 +

1

30=

1

20→ v = 60 cm

Radius of curvature of the convex mirror: R = 10 cm

Focal length of the convex mirror: f1 = 5 cm

For the image to coincide with the object, the light rays should pass through center of curvature as shown

in figure.

So, the distance of convex mirror: 50 cm

43. In ∆ABC, A = 900. Therefore, tan-1 (𝑏

𝑎+𝑐) + tan-1 (

𝑐

𝑎+𝑏) equals

a) 𝝅

𝟒b) 𝜋

2c) tan-1 (

𝑎

𝑏+𝑐) d) tan-1 (

𝑏𝑐

𝑎𝑏+𝑏𝑐+𝑐𝑎)

Sol ∆𝐴𝐵𝐶 is right angled triangle then 𝑎2 = 𝑏2 + 𝑐2 ……. (1)

Now 𝑡𝑎𝑛−1 (𝑏

𝑎+𝑐) + 𝑡𝑎𝑛−1 (

𝑐

𝑎+𝑏) = 𝑡𝑎𝑛−1 [

(𝑏

𝑎+𝑐)+(

𝑐

𝑎+𝑏)

1−(𝑏

𝑎+𝑐)(

𝑐

𝑎+𝑏)]

= 𝑡𝑎𝑛−1 [𝑎𝑏+𝑏2+𝑎𝑐+𝑐2

𝑎2+𝑎𝑏+𝑎𝑐+𝑏𝑐−𝑏𝑐]

= 𝑡𝑎𝑛−1 [𝑎𝑏+𝑎𝑐+𝑎2

𝑎2+𝑎𝑏+𝑎𝑐] using

equation (1)

= 𝑡𝑎𝑛−1[1] = 𝝅

𝟒

44. An excited hydrogen atom returns to the ground state. The wavelength of emitted photon is λ. The

principal quantum number of the excited state is (R – Rydberg constant)

a)

√𝝀𝑹

𝝀𝑹 − 𝟏

b) 𝜆𝑅

𝜆𝑅 + 1

c)

√𝜆𝑅

𝜆𝑅 + 1

d) 𝜆𝑅

𝜆𝑅 − 1


NSEA – 2010

Sol Ground state of the hydrogen atom: n = 1

Let the excited state of the hydrogen atom: n

Wavelength of the emitted photon: λ

Rydberg formula: 1

𝜆= R (

1

𝑛𝑖2 −

1

𝑛𝑓2)

1

𝜆= R (1 −

1

𝑛2) →

1

𝜆𝑅= 1 −

1

𝑛2 →

1

𝑛2= 1−

1

𝜆𝑅= 𝜆𝑅−1

𝜆𝑅→ n = √

𝝀𝑹

𝝀𝑹−𝟏

45. The equivalent resistance between A and B in the arrangement of resistors shown is

a) 3 Ω b) 4 Ω c) 4.5 Ω d) 5 Ω

Sol The given circuit can be redrawn into a balanced Wheatstone’s bridge as shown in figure.

Effective resistance between A and B: 8 𝑥 8

8 + 8 = 4 Ω

46. The difference between the apparent frequencies of sounds heard when the source approaches and then

recedes with the same speed is 5% of the actual frequency. If the speed of sound is 320 m/s, the speed of

source is

a) 4 m/s b) 8 m/s c) 12 m/s d) 10 m/s

Sol Apparent frequency of the approaching source: f1 = f (𝑣

𝑣−𝑣𝑠)

Apparent frequency of the receding source: f2 = f (𝑣

𝑣 + 𝑣𝑠)

Given: 𝑓1−𝑓2

𝑓 x 100 = 5

f (𝑣

𝑣−𝑣𝑠)−f (

𝑣

𝑣 + 𝑣𝑠)

𝑓 x 100 = 5 → (

𝑣

𝑣−𝑣𝑠) − (

𝑣

𝑣 + 𝑣𝑠) = 0.05

2𝑣𝑣𝑠

𝑣2−𝑣𝑠2 = 0.05 →

2𝑣𝑠

𝑣 = 0.05 → vs = 8 m/s

47. A torch bulb rated at 4.5 W, 1.5 V is connected in a circuit as shown. The

emf of the cell needed to make the bulb glow with rated power is

(a) 4.5 V b) 1.5 V c) 2.67 V d) 13.5 V

Sol Rating of the torch bulb: 4.5 W, 1.5 V

Resistance of the bulb: R = 𝑉2

𝑃= 2.25

4.5 = 0.5 Ω

Effective resistance of the circuit: R = 2.67 + 0.5 𝑥 1

0.5+1 = 3 Ω

Current through the cell: i = 𝐸

𝑅= 𝐸

3

For the bulb to glow with rated power, the pd across 1 Ω resistor should be 1.5 V

1.5 = (1) 0.5

1+0.5

𝐸

3→ E = 13.5 volt


NSEA – 2010

48. The total number of terms in the expansion of (a + b + c)73 is

a) 74 b) 146 c) 2775 d) 5476

Sol (𝑎 + 𝑏 + 𝑐)73 = [(𝑎 + 𝑏) + 𝑐]73,

Using binomial theorem,

[(𝑥 + 𝑦)𝑛 =∑𝑛𝐶𝑟𝑥𝑛−𝑟𝑦𝑟

𝑛

𝑟=0

]

Let 𝑥 = (𝑎 + 𝑏) and 𝑦 = 𝑐

Then, (𝑎 + 𝑏 + 𝑐)73 = 73𝐶0(𝑎 + 𝑏)73 + 73𝐶1(𝑎 + 𝑏)72𝑐 + . . . + 73𝐶73(𝑐)73

Number of terms = 74 + 73 + 72 + … + 1

= 𝑛(𝑛+1)

2 =

74(74+1)

2 = 𝟐𝟕𝟕𝟓

49. Two identical bar magnets each of dipole moment pm and length 𝑙 are perpendicular to

each other as shown in the figure. The dipole moment of the combination is

a) pm/2 b) pm / √2 c) pm √𝟐 d) 2pm

Sol Dipole moment of the magnetic dipole: pm

Direction of magnetic moment: S to N

The two magnets are identical: P1 = P2 = pm

The two moments are perpendicular to each other: P = √𝑃12 + 𝑃2

2 = √𝟐 pm

50. An unbalanced Wheatstone bridge is shown in the figure in which current flows from B to D. To balance

the bridge

a) R should be increased b) Q should be increased

c) S should be increased d) P should be decreased

Sol Given: Current is flowing from B to D: B is at higher potential as compared to D.

Bridge balance condition: 𝑃

𝑄 =

𝑅

𝑆 [B and D at the same potential / current through

galvanometer is zero]

Potential at B: VB = 𝑄

𝑃+𝑄 V

Potential at D: VD = 𝑆

𝑅+𝑆 V

In order to increase the potential at D, we can decrease R or increase S.

51. Three integers are chosen at random from the numbers 1 to 20. The probability that their product is even,

is

a) 2

19

b) 𝟏𝟕

𝟏𝟗

c) 13

19

d) 4

19

Sol Sample set 𝑆 = 1, 2, 3, … , 19, 20, total elements = 20

We want product of 3 integers as even. So at least one of them should be even. So same as saying all three

should not be odd.

Probability to getting odd product = 10𝐶320𝐶3

=10×9×8

3!20×19×18

3!

=10×9×8

20×19×18=

2

19

𝑃(𝐸) = 1 −2

19=

𝟏𝟕

𝟏𝟗


NSEA – 2010

52. The circuit shown in fig. contains two diodes each with a forward drop of

0.6 volt. If a battery of 5 volt is connected as shown in the circuit, the current

through D1 will be

a) 1

5 A b) 𝟏

𝟏𝟎 A c) 5

4 A d) 5

8 A

Sol Since both branches are identical, same current will flow in both.

Apply KVL along PQRSP:

−0.6 – 24i – 10 (2i) + 5 = 0

4.4 = 24i + 20i = 44i

i = 4.4

44=

𝟏

𝟏𝟎 A

53. If the sun emits radiation of wavelength λ at a rate L watt, then the number of photons of wavelength λ

emitted per second is

a) 𝐿

ℎ𝑐𝜆

b) 𝑳𝝀

𝒉𝒄

c) ℎ𝑐

𝐿𝜆

d) ℎ𝑐𝜆

𝐿

Sol Power of solar radiation: P = L

Energy of each photon of solar radiation: E = ℎ𝑐

𝜆

Total energy: E1 = nE = 𝑛ℎ𝑐

𝜆

Power of solar radiation: L = 𝐸1

𝑡 =

𝑛ℎ𝑐

𝜆 (t = 1 sec)

No. of photons emitted: n = 𝑳𝝀

𝒉𝒄

54. Two wires P and Q, 20 cm apart are at right angles to a uniform magnetic field of induction B0 = 10-5 T as

shown in figure. The wire Q carries a current of 1 A directed out of the plane of the paper and the

magnetic induction midway between P and Q is zero, the magnitude and direction of current in P must be

a) 4 A outwards b) 4 A inwards c) 6 A outwards d) 6 A inwards

Sol Strength of the magnetic field: B = 10-5 T

Strength of magnetic field created by Q at the midpoint:

BQ = 𝜇0

2𝜋

𝑖

𝑟 = 2 x 10-7 x

1

10 𝑥 10−2 = 0.2 x 10-5 T [anticlockwise]

Strength of magnetic field created by P at the midpoint:

BP = 𝜇0

2𝜋

𝑖

𝑟 = 2 x 10-7 x

𝑖

10 𝑥 10−2 = 0.2 x 10-5(i) T

Resultant field at the mid-point is zero: B = B0 + BQ + BP

0 = 10-5 – 0.2 x 10-5 + BP → BP = −0.8 x 10-5 T

0.2 x 10-5(i) = −0.8 x 10-5 → i = −4 A [inward]


NSEA – 2010

55. A cell is connected across a series combination of a resistor and a capacitor through a switch. When the

switch is closed at time t = 0,

(A) no current will ever flow through the resistor

(B) a constant current will flow through the resistor

(C) current will keep on decreasing and ultimately becomes zero

(D) current will keep on increasing and attains a constant value

Sol Let the equivalent resistance of the circuit: R

Capacitance of the capacitor: C

Emf of the cell: E

Current through the circuit: i = i0 𝒆−𝒕/𝝉

i0: steady state current

τ: time constant of the capacitive network: τ = RC

56. Let be a unit vector and = 2𝑖 − 𝑗 + , 𝑐 = 𝑖 − 3𝑗 + 2. The maximum value of [ 𝑐] is

a) √𝟑𝟓 b) 7 c) √84 d) 35

Sol We know that [𝑐] = ||. | × 𝑐|𝑐𝑜𝑠𝜃

× 𝑐 = |𝑖 𝑗 2 −1 11 −3 2

| = 𝑖 − 3𝑗 − 5

As || = 1 because is unit vector.

[𝑐] = ||. | × 𝑐|𝑐𝑜𝑠𝜃 = 1. √12 + (−3)2 + (−5)2 . 𝑐𝑜𝑠𝜃 = √35 𝑐𝑜𝑠𝜃

For maximum value 𝑐𝑜𝑠𝜃 = 1

Then, [𝑐] = √𝟑𝟓

57. The area of an equilateral triangle is 4√3. Therefore, the area of the circumcircle of the triangle is

a) 4√3 𝜋 b) 8𝜋

3

c) 𝟏𝟔𝝅

𝟑

d) 16 𝜋

Sol Let the side of given equilateral triangle is ‘a’. Then area of equilateral

triangle is given as √3

4𝑎2 .

Hence, √3

4𝑎2 = 4√3 ⇒ 𝑎2 = 42 ⇒ 𝑎 = 4

Now circumradius of an equilateral triangle 𝑅 =2

3(𝑚𝑒𝑑𝑖𝑎𝑛)

𝑅 = 2

3(√3

2𝑎) ⇒ 𝑅 =

𝑎

√3=

4

√3

Hence Area of circumcircle is = 𝜋𝑅2 = 𝜋 (4

√3)2=

𝟏𝟔𝝅

𝟑

58. The maximum uncertainties in the measurement of mass and density of a cube respectively, are 2% and

4%. The maximum uncertainty in the measurement of length will be

a) 4% b) 6% c) 2% d) 9%


NSEA – 2010

Sol Density of a material: 𝜌 = 𝑚

𝑣

Maximum uncertainty in the measurement of mass: ∆𝑚

𝑚 x 100 = 2

Maximum uncertainty in the measurement of density: ∆𝜌

𝜌 x 100 = 4

For maximum uncertainty in the measurement of length: V = 𝑚

𝜌𝑑𝑉

𝑉x 100 =

𝑑𝑚

𝑚x 100 +

𝑑𝜌

𝜌x 100 → 3

𝑑𝑙

𝑙x 100 =

𝑑𝑚

𝑚x 100 +

𝑑𝜌

𝜌 x 100

3 𝑑𝑙

𝑙x 100 = 2 + 4 = 6 →

𝑑𝑙

𝑙 x 100 = 2 %

59. The number of different ways in which all the digits of the number 623236556 can be arranged so that the

odd digits occupy even positions is

a) 60 b) 64 c) 2880 d) 7560

Sol Let the given number is 𝑁 = 623236556

In N ,‘6’ repeats 3 times, ‘2’ repeats 2 times, ‘2’ repeats 2 times, ‘5’ repeats 2 times

Number of odd digits 4 i.e. 3, 3, 5, 5 and number of even digits are 5 i.e. 2, 2, 6, 6, 6.

The odd digits 3, 3, 5, 5 occupy even positions is 4!

2!2!= 6

And the even digits 2, 2, 6, 6, 6 occupy 5 odd positions is 5!

2!3!=

5×4×3!

2!×3!= 10

Hence the total number of ways to arrange digits to get desired result is 6 × 10 = 𝟔𝟎

60. The number formed by the last two digits of the sum (1! + 2! + 3! + …..+10!) is

a) 00 b) 13 c) 23 d) 30

Sol Let the given number is N , i.e. N = (1! + 2! + 3! + …..+10!)

Number of zeros in the end of n! = [𝑛

5] + [

𝑛

52] + [

𝑛

53] + ⋯ , where [.] denotes g.i.f.

So, number of zeros in the end of 10! = [10

5] + [

10

52] + [

10

53] + ⋯ = 2 + 0 + 0 + … = 2

Hence 10! ends with 2 zeros. So, 10!, 11!, 12!, . . . , 101! Ends with minimum number of 2 zeros.

Now for 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880

For last 2 digits solve (1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 36288)(𝑚𝑜𝑑 100) = 13

61. If angle between and is θ, then the value of (. x ) is

a) P2Q cos θ b) zero c) P2Q sin θ d) P2Q sin θ cos θ

Sol Angle between the vectors: and : θ

x = QP sin θ [perpendicular to the plane containing and ]

. x = P (QP sin θ) cos 900 = 0

62. For a diatomic gas the P-V diagram is a straight line passing through origin. Then, the molar specific heat

of the gas will be (R – gas constant)

a) 4 R b) 2.5 R c) 3 R d) 1.33 R

Sol Given: P-V diagram of a diatomic gas is a straight line passing through origin: P ∝ V → PV-1 = constant

For polytropic process: PVx = constant

Molar specific heat of the gas: C = Cv + 𝑅

1−𝑥→ C =

5

2 R +

𝑅

1+1= 6

2 R = 3R


NSEA – 2010

63. Let f(x) = 𝑠𝑖𝑛−1(𝑥−3)

√9−𝑥2. The domain of f(x) is

a) −3 < x < 3 b) 2 ≤ x < 3 c) −1 ≤ x ≤ 1 d) 2 ≤ x ≤ 4

Sol sin−1(𝑥 − 3) exists if −1 ≤ (𝑥 − 3) ≤ 1 ⇒ 2 ≤ 𝑥 ≤ 4 ……..(1)

For 1

√9−𝑥2 ; 9 − 𝑥2 > 0 , ⇒ 𝑥2 − 9 < 0 ⇒ (𝑥 − 3)(𝑥 + 3) < 0 ⇒ − 3 ≤ 𝑥 ≤ 3 ……..(2)

From equation (1) and (2)

𝑥 ∈ (−3, 3) ∩ [2, 4] 𝑥 ∈ [2 , 3)

i.e. 𝟐 ≤ 𝒙 < 𝟑

64. In the expansion of 100C25, the exponent of 5 is

a) 0 b) 1 c) 2 d) 5

Sol We know that nCr =𝑛!

(𝑛−𝑟)! 𝑟!

Then 100C25 = 100!

(100−25)! 25! =

100!

(75)! 25!

Number of exponent of 5 in n! is given by = [𝑛

5] + [

𝑛

52] + [

𝑛

53] + ⋯

In 100! = [100

5] + [

100

52] + [

100

53] + ⋯ = 20 + 4 + 0 + 0 +⋯ = 24

In 25! = [25

5] + [

25

52] + [

25

53] + ⋯ = 5 + 1 + 0 + 0 +⋯ = 6

In 75! = [75

5] + [

75

52] + [

75

53] + ⋯ = 15 + 3 + 0 + 0 +⋯ = 18

Hence number of exponent of 5 in 100C25 is 24 − (6 + 18) = 24 − 24 = 𝟎

65. A point on the hypotenuse of a right-angled triangle is at a distance m

and n from the sides of the triangle. Therefore, the minimum length of

the hypotenuse is

a) (√𝑚 + √𝑛)2 b) 2 √𝑚2 + 𝑛2 c) √2 (m + n) d)

(𝒎𝟐𝟑⁄ + 𝒏

𝟐𝟑⁄ )

𝟑𝟐⁄

Sol In ∆𝐴𝐵𝐶, ∠𝐵 = 900 and ∠𝐶 = 𝜃, so ∠𝐴𝐸𝐷 = 𝜃 (∵ 𝐷𝐸 ∥ 𝐵𝐶)Let 𝐴𝐶 = 𝑦, 𝐴𝐸 = 𝑦1, 𝐸𝐶 = 𝑦2 ⇒ 𝑦 = 𝑦1 + 𝑦2 ( ∵ 𝐴𝐶 = 𝐴𝐸 + 𝐸𝐶) …(1)

In ∆𝐴𝐷𝐸, cos𝜃 = 𝑚

𝐴𝐸 ⇒ 𝐴𝐸 = 𝑚 𝑠𝑒𝑐𝜃 … (2)

In ∆𝐸𝐹𝐶, 𝑐𝑜𝑠𝜃 = 𝑛

𝐸𝐶 ⇒ 𝐸𝐶 = 𝑛 𝑐𝑜𝑠𝑒𝑐𝜃 … (3)

From equation (2) and (3) 𝐴𝐶 = 𝑦 = 𝑚 𝑠𝑒𝑐𝜃 + 𝑛 𝑐𝑜𝑠𝑒𝑐𝜃 … (4)

For minimum length of 𝑦, 𝑑𝑦

𝑑𝜃= 0

⇒𝑑𝑦

𝑑𝜃=

𝑑

𝑑𝜃(𝑚 𝑠𝑒𝑐𝜃 + 𝑛 𝑐𝑜𝑠𝑒𝑐𝜃) = 𝑚 𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 − 𝑛 𝑐𝑜𝑠𝑒𝑐𝜃𝑐𝑜𝑡𝜃 = 0

⇒ 𝑡𝑎𝑛3𝜃 =𝑛

𝑚

⇒ 𝑡𝑎𝑛𝜃 = (𝑛

𝑚)13⁄


NSEA – 2010

We know that, 𝑠𝑒𝑐2𝜃 = 1 + 𝑡𝑎𝑛2𝜃 ⇒ 𝑠𝑒𝑐𝜃 = √1 + 𝑡𝑎𝑛2𝜃 = √1 + (𝑛

𝑚)23⁄=

[𝑚23⁄ +𝑛

23⁄ ]12⁄

𝑚23⁄

…… (5)

Similarly, 𝑐𝑜𝑠𝑒𝑐𝜃 =[𝑚

23⁄ +𝑛

23⁄ ]12⁄

𝑛23⁄

…… (6)

From equation (4), (5) and (6)

𝑦𝑚𝑖𝑛 = 𝑚 𝑠𝑒𝑐𝜃 + 𝑛 𝑐𝑜𝑠𝑒𝑐𝜃 = (𝒎𝟐𝟑⁄ + 𝒏

𝟐𝟑⁄ )

𝟑𝟐⁄

66. If x + 1

𝑥= 2 cos θ, then x3 +

1

𝑥3 is equal to

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

Sol Given that 𝑥 +1

𝑥= 2𝑐𝑜𝑠𝜃, cube on both sides

(𝑥 +1

𝑥)3= (2𝑐𝑜𝑠𝜃)3 ⇒ 𝑥3 +

1

𝑥3+ 3(𝑥 +

1

𝑥) = 8𝑐𝑜𝑠3𝜃

⇒ 𝑥3 +1

𝑥3= 8𝑐𝑜𝑠3𝜃 − 6𝑐𝑜𝑠𝜃 = 2(4𝑐𝑜𝑠3𝜃 − 3𝑐𝑜𝑠𝜃) = 𝟐𝒄𝒐𝒔𝟑𝜽

67. A bob of simple pendulum of mass m is released from a position in which the angular displacement is 600.

The tension in the string when the bob crosses the equilibrium position is

a) mg b) mg/2 c) 2 mg d) Zero

Sol Initial angular position of the bob: θ = 600

Conservation of energy: mg𝑙 (1 – cos θ) = 1

2 mv2 → v2 = 2g𝑙 (1 – 0.5) = g𝑙

Tension in the string: T = mg cos θ + 𝑚𝑣2

𝑙→ T = mg cos 0 +

𝑚

𝑙 g𝑙 = mg + mg = 2 mg

68. If cos𝛼

𝑎= sin𝛼

𝑏, then a cos 2α + b sin 2α is equal to

a) a b) b c) ab d) a + b

Sol Given that

cos𝛼

𝑎= sin𝛼

𝑏⇒ 𝑡𝑎𝑛𝛼 =

𝑏

𝑎and we know that 𝑐𝑜𝑠2𝛼 =

1−𝑡𝑎𝑛2𝛼

1+𝑡𝑎𝑛2𝛼=

1−(𝑏

𝑎)2

1+(𝑏

𝑎)2 =

𝑎2−𝑏2

𝑎2+𝑏2 ,

And 𝑠𝑖𝑛2𝛼 = 2𝑡𝑎𝑛𝛼

1+𝑡𝑎𝑛2𝛼=

2(𝑏

𝑎)

1+(𝑏

𝑎)2 =

2𝑎𝑏

𝑎2+𝑏2

Now, 𝑎𝑐𝑜𝑠2𝛼 + 𝑏𝑠𝑖𝑛2𝛼 = 𝑎 (𝑎2−𝑏2

𝑎2+𝑏2) + 𝑏 (

2𝑎𝑏

𝑎2+𝑏2) =

1

𝑎2+𝑏2[𝑎3 − 𝑎𝑏2 + 2𝑎𝑏2] =

1

𝑎2+𝑏2[𝑎3 + 𝑎𝑏2]

⇒ 𝑎𝑐𝑜𝑠2𝛼 + 𝑏𝑠𝑖𝑛2𝛼 =𝑎

𝑎2 + 𝑏2[𝑎2 + 𝑏2] = 𝒂

69. The maximum value of 5 cos θ + 3 cos (𝜃 +𝜋

3) + 3 is

a) 10 b) 11 c) 14 d) 12

Sol = 5 cos θ + 3 cos (𝜃 +𝜋

3) + 3 = 5 cos θ + 3 [𝑐𝑜𝑠𝜃. 𝑐𝑜𝑠

𝜋

3− 𝑠𝑖𝑛𝜃. 𝑠𝑖𝑛

𝜋

3]+ 3

= 5 cos θ + 3 [𝑐𝑜𝑠𝜃.1

2− 𝑠𝑖𝑛𝜃.

√3

2]+ 3 = 𝑐𝑜𝑠𝜃 (5 +

3

2) + 𝑠𝑖𝑛𝜃 (

−3√3

2) + 3

= 13

2𝑐𝑜𝑠𝜃 + (

−3√3

2) 𝑠𝑖𝑛𝜃 + 3 = 𝑎𝑐𝑜𝑠𝜃 + 𝑏𝑠𝑖𝑛𝜃 + 𝑐

Maximum value of the equation is = 𝑐 + √𝑎2 + 𝑏2 = 3 +√(13

2)2+ (

−3√3

2)2

= 3 +14

2= 𝟏𝟎


NSEA – 2010

70. lim𝑛 →∞

(12+22+32+⋯…+𝑛2)2

3𝑛6 =

a) 1

3

b) 1

9

c) 𝟏

𝟐𝟕

d) 1

81

Sol 𝑙𝑖𝑚𝑛 →∞

(12+22+32+⋯…+𝑛2)2

3𝑛6

= 𝑙𝑖𝑚𝑛 →∞

(𝑛(𝑛+1)(2𝑛+1)

6)2

3𝑛6 = 𝑙𝑖𝑚

𝑛 →∞

𝑛6

36((1+

1

𝑛)(2+

1

𝑛))

2

3𝑛6=

1

36.1

3. 22 =

𝟏

𝟐𝟕

71. A boy observes the motion of a tennis ball dropped from some height. The ball bounces from the ground

several times. During the motion, the acceleration of the ball

a) is constant equal to −g b) is constant equal to +g

c) is not constant d) varies uniformly between +g and −g

Sol Downward motion (free fall): +g

Upward motion (vertical projection): −g

During the collision: A tennis ball gets deformed during the collision (inelastic) with the ground. The

forces acting on the tennis ball during the deformation and restoration are complex (impulse forces).

So, the acceleration of the ball does not remain constant.

72. If 0 < a < 1, the expression log (1 + a) + log (1 + a2) + log (1 + a4) + log (1 + a8) + …. upto infinity has the

value

a) log (1 + a) b) log (1 – a) c) log (1

1 + 𝑎) d) −log (1 – a)

Sol Given that 𝑎 ∈ (0, 1) then,

log (1 + a) + log (1 + a2) + log (1 + a4) + log (1 + a8) +… = log[(1 + a)(1 + a2)(1 + a4)(1 + a8)... ]

Multiply and divide with (1 – a)

= log [(1 – a) (1 + a)(1 + a2)(1 + a4)(1 + a8)...

(1 – a) ] = log [

(1 – a∞)

(1 – a) ] = log [

1

(1 – a) ] = −𝐥𝐨𝐠 (1 – a)

As 𝑎∞ = 0, ∵ 𝑎 ∈ (0, 1)

73. Parallel sides of a trapezium have length 25 cm and 10 cm, while its non-

parallel sides are 14 cm and 13 cm long. The area of this trapezium is

a) 125 cm2 b) 98 cm2 c) 150 cm2 d) 196 cm2

Sol As sides of ∆𝐶𝐵𝑀 are 13, 14, 15. area of ∆𝐶𝐵𝑀 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

where 𝑠 =𝑎+𝑏+𝑐

2=

13+14+15

2= 21 𝑐𝑚

Area of ∆𝐶𝐵𝑀 = √21 × 8 × 7 × 6 = 7 × 3 × 4 = 84 𝑐𝑚2

And Area of ∆𝐶𝐵𝑀 =1

2×𝑀𝐵 × 𝐶𝑁 = 84 ⇒

1

2× 15 × ℎ = 84 ⇒ ℎ =

56

5 𝑐𝑚

Now Area of parallelogram: 𝐴𝑀𝐶𝐷 = 𝐵𝑎𝑠𝑒 × 𝐻𝑒𝑖𝑔ℎ𝑡 = 10 × ℎ = 10 ×56

5= 112 𝑐𝑚2

Area of trapezium 𝐴𝐵𝐶𝐷 = 112 + 84 = 𝟏𝟗𝟔 𝒄𝒎𝟐


NSEA – 2010

74. A string under tension of 80 N has a linear mass density 5 x 10-2 kg/m. A harmonic wave of frequency 60

Hz and amplitude 6 cm propagates along the string. The rate at which energy transfers along the string is

a) 0.0512 W b) 1024 W c) 512 W d) 5.12 x 10-4 W

Sol Tension in the string: T = 80 N

Linear mass density: λ = 5 x 10-2 kg/m

Speed of the wave: v = √𝑇

𝜆 = √

80

5 𝑥 10−2 = √1600 = 40 m/s

Frequency of the harmonic wave: f = 60 Hz → ω = 2𝜋f = 2 x 3.14 x 60 = 376.8 rad/s

Amplitude of the harmonic wave: A = 6 cm = 6 x 10-2 m

Rate of energy transfer along the string: Power: P = 1

2 A2ω2 λv

P = 1

2 x 36 x 10-4 x 1.42 x 105 x 5 x 10-2 x 40 = 18 x 1.42 x 5 x 4 = 512 W

75. Temperature of water at one atmospheric pressure is varied from 00 C to 1000 C. Let ρ be the density of

water and c be its specific heat capacity. Then, over the temperature variation

(A) both ρ and c remain constant

(B) ρ passes through a maximum while c remains constant

(C) ρ and c both pass through a maximum

(D) ρ passes through a maximum while c passes through a minimum

Sol Water has maximum density (minimum volume) at 40 C. As the temperature is increased from

00 C to 1000 C, the density passes through a maximum.

Specific heat of water decreases with increase in temperature, reaches minimum at around 400 C and then

increases

76. If ax = cq = b and cy = zz = d, then

a) xy = qz* b) 𝑥

𝑦= 𝑞

𝑧c) x + y = q + z d) xy = qz

Sol Given 𝑎𝑥 = 𝑐𝑞 = 𝑏 …… (1)

And 𝑐𝑦 = 𝑎𝑧 = 𝑑 …… (2)

From equation (1), 𝑐𝑞 = 𝑏 ⇒ 𝑐 = 𝑏1𝑞⁄ and 𝑎𝑥 = 𝑐𝑞 ⇒ 𝑐 = (𝑎)

𝑥𝑞⁄ ……(3)

From equation (2), 𝑎𝑧 = 𝑑 ⇒ 𝑎 = 𝑑1𝑧⁄ and 𝑐𝑦 = 𝑎𝑧 ⇒ 𝑐 = (𝑎)

𝑧𝑦⁄ ……(4)

From equation (3) and (4) we will get, 𝑐 = (𝑎)𝑥𝑞⁄ = (𝑎)

𝑧𝑦⁄

⇒𝑥

𝑞=𝑧

𝑦⇒ 𝒙𝒚 = 𝒒𝒛


NSEA – 2010

77. The roots of 𝑥2−𝑏𝑥

𝑎𝑥−𝑐= 𝑚−1

𝑚+1 are equal and opposite. Therefore, the value of m is

a) 𝒂 − 𝒃

𝒂 + 𝒃

b) 𝑎 + 𝑏

𝑎 − 𝑏

c) 1

𝑐

d) 1

Sol Given, 𝑥2−𝑏𝑥

𝑎𝑥−𝑐= 𝑚−1

𝑚+1⇒ 𝑥2 − 𝑏𝑥 =

𝑚−1

𝑚+1(𝑎𝑥 − 𝑐) ⇒ 𝑥2 − [𝑏 + 𝑎 (

𝑚−1

𝑚+1)] 𝑥 + 𝑐 (

𝑚−1

𝑚+1) = 0

Sum of roots is zero because they are equal and opposite,

⇒ [𝑏 + 𝑎 (𝑚−1

𝑚+1)] = 0

⇒𝑚−1

𝑚+1=

−𝑏

𝑎

⇒ 𝑚 =𝒂−𝒃

𝒂+𝒃

78. In the binomial expansion of (a + b)9 the sum of binomial coefficients is

a) 81 b) 288 c) 423 d) 512

Sol We know that,

(𝑎 + 𝑏)𝑛 =∑𝑛𝐶𝑟𝑎𝑛−𝑟𝑏𝑟

𝑛

𝑟=0

Then sum of coefficients in (𝑎 + 𝑏)𝑛 is 2𝑛.So, in (𝑎 + 𝑏)9, 𝑛 = 9, sum of coefficients = 29 = 𝟓𝟏𝟐.(we will get sum of coefficients in (𝑎 + 𝑏)𝑛 when 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 1)

79. In the familiar decimal number system, the base is 10. In another number system using base 6, the

counting will proceed as 1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 20, 21…. The twenty fifth number in this

system will be

a) 33 b) 34 c) 40 d) 41

Sol Decimal number system means the number with base 10. So 25th term of

decimal number system is 24. i.e. 0, 1, 2, 3, 4, …, 24

(24)10 = 2 × 101 + 4 × 100 = 20 + 4 = 24 with base 6,

(24)10 = (40)6 = 4 × 61 + 0 × 60 = (24)10

6 24 R

6 4 0

0 4

80. The function f(x) =𝑥

log𝑥 increases in the interval

a) (0, ∞) b) (0, e) c) (e, ∞) d) (1

𝑒, 𝑒)

Sol 𝑓(𝑥) is said to be increasing/decreasing in (𝑎, 𝑏) if 𝑓′(𝑥) is positive/negative in (𝑎, 𝑏) respectively.

Given 𝑓(𝑥) =𝑥

log𝑥 ,

Then, 𝑓′(𝑥) =𝑙𝑜𝑔 𝑥.1−𝑥.(

1

𝑥)

(𝑙𝑜𝑔 𝑥)2=

𝑙𝑜𝑔 𝑥−1

(𝑙𝑜𝑔 𝑥)2

For increasing 𝑓′(𝑥) > 0

⇒𝑙𝑜𝑔 𝑥 − 1

(𝑙𝑜𝑔 𝑥)2> 0 ⇒ 𝑙𝑜𝑔 𝑥 − 1 > 0 ⇒ 𝑙𝑜𝑔 𝑥 > 1 ⇒ 𝑥 > 𝑒1

⇒ (𝒆, ∞)

MATHEMATICS CONTRIBUTOR: SANDEEP MOHAN (MSc. Maths – IITM)

kalyan’s physics challenge

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