Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Moment of inertia of a ring of mass M and radius R about a tangent to the circle of the ring is

Answer»

`MR^(2)`
`(1)/(2) MR^(2)`
`(3)/(2) MR^(2)`
`(7)/(2) MR^(2)`

ANSWER :C
Discussion
2.

Radiant energy from the sun strikes the earth at a rate of 1.4 xx 10^(3) "watt/m"^(2). Calculate the temperature of the surface of the sun. Average distance of earth from sun = 1.5 xx 10^(11) m, Stefan's constant = 5.7 xx 10^(-8) "Wm"^(-2) K^(-4) ( Radius of sun= 7 xx10^(8) m )

Answer»

5801K
2880 K
3080 K
4801 K

Answer :A
Discussion
3.

The displacements of a particle executing SHM at three consecutive seconds are 6 cm, 10 cm and 6 cm respectively. Find out the frequency of oscillation of the particle.

Answer»

SOLUTION :DISPLACEMENT, `x=Acos(omegat+THETA)`
LET initially at t = 0, displacement = 6 cm.
So, at t = 1 s, displacement = 10 cm
and at t = 2 s, displacement is = 6 cm
`therefore" "6=Acostheta""...(1)`
`10=Acos(omega+theta)""...(2)`
`6=Acos(2omega+theta)""...(3)`
From (1) and (3) we get,
`cos(2omega+theta)=costheta=cos(-theta)`
or, `2omega+theta=-thetaor,omega=-theta`
`therefore" "` From (2), we get
`10=Acos0or,A=10cm.`
From (1), we get
`6=10cos(-omega)or,cosomega=3/5`
or, `omega=0.927"rad"*s^(-1)`
`therefore"Frequency"=omega/(2pi)=0.927/(2xx3.14)=0.148s^(-1)`.
Discussion
4.

Which of the following statements are not correct? a) Shearing stress is possible in liquids but not in gases b) Elastomers are class of solids that donot obey Hooke.s law c) Elastic limit is the property of material of body where as elasticity is the property of a body d) Bulk modulus of most of the solid material is zero.

Answer»

Both a and b
only C and d
only b and c
only a, c and d

Answer :4
Discussion
5.

The acceleration of aparticle varie with time as shown in the figure Find the displacement of the particle in the interval from 1 = 2s to t = 4s Assume that v = 0 at t=0

Answer»

6.67 m
5.67 m
4.67 m
3.67 m

Answer :A
Discussion
6.

A glass rod weighs 90g times g in air. It weighs 49.6g times g when immersed in a liquid at 12^(@)C, and 51.9g times g at 97^(@)C. Find the real expansion coefficient of the liquid. [Volume expansion coefficientof glass =2.4 times 10^(-5@)C^(-1)]

Answer»

Solution :LET the VOLUME of the glass rod be `V_(1) " at " 12 ^(@)C`, and density of the liquid be `rho_(1).`
The mass of the displaced liquid at that temperature
`""=90-49.6=40.4g`
`THEREFORE` Volume of the glass rod at that temperature,
`"" V_(1)=40.4/rho_(1) "...(1)"`
Again at `97^(@)C`, mass of the displaced liquid
`""=90-51.9=38.1g`
Let at `97^(@)C` the volume of the glass rod be `V_(2)` and density of the liquid be `rho_(2)`.
`therefore "" V_(2)=38.1/rho_(2) "...(2)"`
From (1) and (2),
`"" V_(2)/V_(1)=rho_(1)/rho_(2) times 38.1/40.4 "...(3)"`
Now for the glass rod, `V_(2)=V_(1)[1+2.4 times 10^(-5) times 85]`
`therefore"" V_(2)/V_(1)=1+2.4 times 10^(-5) times 85=1.00204`
In case of liquid, `rho_(1)=rho_(2)[1+gamma times 85]`
`therefore "" rho_(1)/rho_(2)=1+gamma times 85`
Fromequation (3),
`"" 1.00204=(1+gamma times 85) times 38.1/40.4`
or, `"" 1+85gamma=(1.00204 times 40.4)/38.1`
or,`""85gamma=(1.00204 times 40.4)/38.1-1=2.382/38.1`
or, `"" gamma=7.35 times 10^(-4@)C^(-1).`
Discussion
7.

Four spheres each of diameter 2a and mass mare placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.

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SOLUTION :`I_(1)=(2)/(5)Ma^(2)`
`I_(2)=(2)/(5)Ma^(2)+mb^(2)`
`I_(3)=(2)/(5)ma^(2)++mb^(2)`
`I_(4)=(2)/(5)ma^(2)`
`:.` Moment of Inertia of the system I
`=I_(1)+I_(2)+I_(3)+I_(4)`
`=(2)/(5)ma^(2)+(2)/(5)ma^(2)+mb^(2)+(2)/(5)ma^(2)+mb^(2)+(2)/(5)ma^(2)`
`I=(8)/(6)ma^(2)+2MB^(2)`
Discussion
8.

A small block slides with velocity 0.5 sqrt(gr) on the horizontal frictionless surface as shown in fig. The block leaves the surface at point C. The angle theta it the fig is:

Answer»

`COS^(-1)(4//9)`
`cos^(-1)(3//4)`
`cos^(-1)(1//2)`
NONE of the above

Answer :B
Discussion
9.

A particle of mass m is placed upon the smooth face of a prism, also of mass m, which particle is placed , is inclined at an angle alpha to the horizontal . Show that when the particle has moved a distance x down the face of the prism the prism itself will have moved a distance of (1)/(2)x "cos" alpha

Answer»


ANSWER :N//A
Discussion
10.

A body is whirled in a horizontal circle of radius vector vec(r ). It has an angular velocity of vec(omega). The velocity at any point on circular path is

Answer»

`V=R OMEGA`
`V=(omega)/(r )`
`V= (r )/(omega)`
`V=m""(omega)/(r )`

ANSWER :A
Discussion
11.

A boy and a man carry a uniform pole of length 8 m and of mass 60 kg by supporting it on their shoulders. They are located at the ends of the pole. Where should a load of 90 kg be suspended from boy.s end so that boy carries only one - third of the total load ?

Answer»

Solution :The MASS of pole acts from its centre .C.. Let the load be placed at a distance x. from the boy.s end

If `N_(1),N_(2)` are the normal reactions on man and boy respectively then
`N_(1)=(2)/(3)(60+90)g=100gN`
`N_(2)=(1)/(3)(60+90)g=50gN`
Taking turning effects about boy.s end .B., we have.
`N_(1)xx8-60gxx4-90gxxx=0`
or `100gxx8-60gxx4-90gxxx=0` on solving x=6.22m
Discussion
12.

Four point masses each of value m, are placed at the corners of a square ABCD of side l, The moment of inertia of the system about an axis passing throught A and parallel to BD is

Answer»

`sqrt(3) ml^(2)`
`3 ml^(2)`
`ml^(2)`
`2 ml^(2)`

SOLUTION :AS it is clear from the FIGURE,
`AC=BD=sqrt(l^(2)+l^(2))=sqrt(2)l`
Moment of inertia of four point masses about BD
`I_(BD)=m((lsqrt(2))/(2))^(2)+mxx0+m((lsqrt(2))/(2))^(2)+mxx0`
`I_(BD) = ml^(2)`
Applying the theorem of parallel axis,
`I_(xy)=I_(BD)+M(AO)^(2)=ml^(2)+4m((l)/(sqrt(2)))^(2)=3ml^(2)`.
Discussion
13.

When mg of water at 10^@Cis mixed with mg of ice at 0^@Cwhich of the following statements are false ?

Answer»

The temperature of the system will be given by equation, `MXX 80+ mxx1 XX (T-0) = mxx 1XX(10-T)`
Whole of the ice will melt and temperature will be more than `0^@C`but less than `10^@C`
Whole of ice will melt and temperature will be `0^@C`
Whole of ice will not melt and temperature will be`0^@C`

Answer :A::B::C
Discussion
14.

At what temperature, hydrogen molecules will escape from the earth.s surface ? (Take mass of hydrogen molecule = 0.34 xx 10^(-6) kg, Boltzmann constant = 1.38 xx 10^(-23) KJ^(-1), Radius of earth= 6.4 xx 10^6 m and acceleration due to gravity=9.8 ms^(-2))

Answer»

10 KJ
`10^(2)` K
10 K
`10^(4)` K

ANSWER :D
Discussion
15.

A block slides on ice with a velocity of 5m//s comes to rest after moving through a distance of 13.5m . Find the coefficient of friction ?

Answer»


Solution :`a=v^(2)-u^(2)//2s=0-25//2xx13.5=-0.926ms^(-2)`, `mu_(K)=F_(k)//N=ma//mg=0.926//9.8=0.094`
Discussion
16.

A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperature. Assume that the emissivity of both the spheres is the same. Find the ratio of (a) the rate of heat from the aluminium sphere to the rate of heat loss from the copper sphere and (b) the rate of fall of temperature of the the aluminium sphere to the rate of fall temperature of teh copper sphere. The specific heat capacity of aluminium = 900 J//kg -^(@)C and that of copper = 390 J//kg-^(@)C. The density of copper is 3.4 times the density of aluminium.

Answer»

Solution :(a) `(u_(1))/(u_(2)) = (e sigma A_(1) T^(4))/(e sigma A_(2) T^(4)) = (r_(1)^(2))/(r_(2)^(2)) = ((r )/(2R))^(2) = (1)/(4)`
(B) Rate of fall of temperature
`(dT)/(dt) = (e sigma A)/(ms) (T^(4) - T_(0)^(4))`
`= (e sigma 4 pi r^(2))/(RHO (4)/(3) pi r^(3) s) (T^(4) - T_(0)^(4))`
`= (3 e sigma)/(rho RS) (T^(4) - T_(0)^(4))`
`((dT//dt))/((dT//dt)) = (rho_(2) r_(2) s_(2))/(rho_(1) r_(1) s_(1)) = (3.4) ((2 r)/(r )) ((390)/(900))`
` = (2.95)/(1)`

Discussion
17.

(A) The sun looks reddish at the time of sunrise and sunset because of dispersion of light. ( R) The wavelength of red colour is less than the wavelength of blue colour.

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Both A and R are true and R is the CORRECT explanation of A
Both A and R are true and R is not the correct explanation of A
A is true and R is false
Both A and R is false

Answer :D
Discussion
18.

A disc of mass 4 kg and radius 6 metre is free to rotate in horizontal plane about a vertical fixed axis passing through its centre. There is asmooth groove along the diameter of the disc and two small blocks of masses 2 kg each are placed in it on either side of the centre of the disc as shown in figure. The disc is given initial angular velocity omega_(0) =12 rad/sec and released. Find the angular speed of disc (in radian/sec) when the blocks reach the ends of the disc.

Answer»


ANSWER :4
Discussion
19.

(A): The v-t graph perpendicular to time axis is not possible in practice (R) : Infinite acceleration can be realized in practice.

Answer»

Both (A) and (R) are true and (R) is the CORRECT explanation of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :C
Discussion
20.

The coefficient of friction between a car wheels and a roadway is 0.5 The least distance in which the car can accelerate from rest to a speed of 72 kmph is (g=10 ms^(-2))

Answer»

10m
20m
30m
40m

Solution :`V^(2)-U^(2)=2AS, a=mu_(k)g`
Discussion
21.

A 2kg block is connected with two springs of force constants K_(1) = 100 N/m and K_(2) = 300 N/m as show in figure. The block is released from rest with the springs unstreched. Find the acceleration of the block in its lowest position (g = 10 m//s^(2))

Answer»

Solution :LET .X. be the maximum displacement of block downwards. Then from conservation of MECHANICAL enery : decrease in potential energy of 2 kg block = increase in ELASTIC potential energy of both the springs
`thereforemg x =(1)/(2)(k_(1)+k_(2))x^(2)` `orx = (2mg)/(k_(1)+k_(2))=((2)(2)(10))/(100+300)=0.1m`
ACCELERATION of block in this position is
`a=((k_(1)+k_(2))x-mg)/(m)("upwards")`
`=((400)(0.1)-(2)(10))/(2)=10m//s^(2)"upwards"`
Discussion
22.

The strain stress curves of three wires of different material as shown in fig. A,B,C are the elastic limits of the wire. The figure show that

Answer»

Elasticity of wire A is maximum
Elasticity of wire B is maximum
Elasticity of wire C is maximum
NONE of the above is true

Answer :A
Discussion
23.

A small particle of mass m attached with a light inextensible thread of length L is moving in a vertical circle. In the given case the particle is moving in a complete vertical circle and ration of its maximum to minimum velocity is 2 : 1. Minimum velocity of the particle is

Answer»

`4 SQRT((gL)/(3))`
`2 sqrt((gL)/(3))`
` sqrt((gL)/(3))`
`3 sqrt((gL)/(3))`

Answer :B
Discussion
24.

A small particle of mass m attached with a light inextensible thread of length L is moving in a vertical circle. In the given case the particle is moving in a complete vertical circle and ration of its maximum to minimum velocity is 2 : 1. Velocity of the particle when it is moving vertically downward is

Answer»

`SQRT((10gL)/(3))`
`2sqrt((GL)/(3))`
`sqrt((8gL)/(3))`
`sqrt((13gL)/(3))`

ANSWER :A
Discussion
25.

A small particle of mass m attached with a light inextensible thread of length L is moving in a vertical circle. In the given case the particle is moving in a complete vertical circle and ration of its maximum to minimum velocity is 2 : 1. The kinetic energy of the particle at the lowest position is

Answer»

`(4mgL)/(3)`
2 MGL
`(8mgL)/(3)`
`(2mgL)/(3)`

ANSWER :C
Discussion
26.

Two satellite A and B are orbiting around the Earth in circular orbits of the same radius. The mass of A is 16 times that of B. What I sthe ratio of the period of revolution of B to that A ?

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Solution :`T = (2 PI r)/(upsilon)` and `upsilon = sqrt(Gm//r)`, where `M` is the
mass of the Earth. It means `T` is independent of mass of the satellite but depends upon the RADIUS of the ORBITAL PATH. So
`T_(B)//T_(A) = 1: 1`
Discussion
27.

Calculate the kinetic energy per molecule and also rms velocity of a gas at 127^@C . Given k_B = 1.38 xx10^(-23) J "molecule"^(-1) K^(-1)and mass per molecule of the gas =6.4xx10^(-27) kg .

Answer»

SOLUTION :Here, T = 127 + 23 = 400 K,
`m =6.4 xx10^(-27) KG`
(i) Kinetic energy per molecule
`=1/2mv_(rms)^2=3/2k_BT`
`=3/2xx1.38xx10^(-23) xx400`
`=8.28xx10^(-21)J`
(ii) Now `1/2mv_(rms)^2=8.28xx10^(-21) J`
`:.v_(rms)=SQRT(2xx8.28xx10^(-21))/m`
`=sqrt((2xx8.28xx10^(-21))/(6.4xx10^(-27)))`
`=1.608xx10^3 MS^(-1)`
Discussion
28.

A wire increase by 10 ^(-6)times its original length when a stress of 10 ^(8) Nm ^(-2) is applied to it, calculate its Young's modulus.

Answer»

SOLUTION :`10^(14) NM ^(2)`
`Y= ("STRESS")/((Delta l )/(l )) = (10 ^(8))/( 10 ^(-6)) = 10 ^(14) Nm ^(-2)`
Discussion
29.

An object of mass 10 kg moving with a speed of 15 ms- hits the wall and comes to res within (a) 0.03 second (b) 10 second. Calculate the impulse and average force acting on the object in both the cases

Answer»

<P>

Solution :Initial momentum of the OBJECT `P_i= 10 xx 15= 150 kg ms^(-1)`
Final momentum of the object `P_f = 0`
`Delta P = 150 - 0 = 150 kg MG^(-1)`
(a) Impulse J= `DeltaP` = 150 N s and Average force `F_(avg) = (Delta P)/(Delta t) = (150)/(0.03) = 5000 N`
(B) Impulse`J = Delta P = 150 N ` s and average force `F_(avg) = 150/10 = 15N`
We see that, impulse is the same in both CASES, but the average force is different.
Discussion
30.

Which of the following statements is correct for any thermodynamics system

Answer»

The internal energy CHANGE in all process
Internal energy and ENTROPY are state function
The change in entropy can never be zero
The WORK done on an adiabatic process is always zero

Answer :B
Discussion
31.

A body of mass 100 gm is suspended from a massless spring of natural 1m. If the spring streaches through a vertical distance of 20 cm, the potential energy stored in the spring is (g= 10 ms^(-2))

Answer»

`1.2 J`
`0.6 J`
`0.1 J`
`0.2 J`

ANSWER :C
Discussion
32.

A uniform rope of length L, resting on a frictionless horizontal table is pulled at one end by a force F. What is the tension in the rope at a distance x from the end where the force is applied ?

Answer»

F (L - X)
Fx/L
(L - x)/F
F(1-x/L)

ANSWER :D
Discussion
33.

The range of distance can be measured by the direct method is………….

Answer»

102 to 10-5m`
10-2 to 102 m`
102 to 105m`
10-2 to 105 m`

SOLUTION :`10^(2) "to" 10^(-5) m`
Discussion
34.

The time period of simple pendulum at the surface of earth is T. If it is taken to a height equal to the radius of earth. Find the period of new oscillations

Answer»

Solution :At a HEIGHT R
new `G^(1)+(GM)/(r^(2))=(GM)/((2R)^(2))=g/4:.Tprop1/(SQRT(g))`
`(T^(1))/(T_(0))=sqrt(g/(g_(1)))=sqrt(g/(g/4))=2` (or) `T^(1)=2T_(0)`
Discussion
35.

The second law of thermodynamics.

Answer»

GIVES the definition of temperature
determines the direction of FLOW of heat during exchange between bodies
is another form of the principle of conservation of heat and other forms of energy.
helps in computing the EFFICIENCY of the Carnot's engine.

Answer :A,B,D
Discussion
36.

A physical relation is epsilon=epsilon_(0)epsilon_(r) where epsilon= electric permittivity of a medium epsilon_(0)= electric permittivity of vacuum epsilon_(r)= relative permittivity of medium What are dimensions of relative permittivity?

Answer»

`[ML^(2)T^(-2)]`
`[M^(0)L^(2)T^(-3)]`
`[M^(0)L^(0)T^(0)]`
`[M^(0)L^(0)T^(-1)]`

Solution :(C) `epsilon=epsilon_(0)epsilon_(r)impliesepsilon_(r)=(epsilon)/(epsilon_(0))`
RELATIVE permittivity is the RATIO of `epsilon` and `epsilon_(0)` hence it is dimensionsless .
i.e. `[M^(0)L^(0)T^(0)]`
Discussion
37.

We open a file keeping one flap on a table. Each flap is square, of mass m and dimension (30 xx 30) cm^(2). Assume that the file does not slip on thetable -

Answer»

The maximum height attained by the centre of mass of the COMPLETE file is `7.5 cm`
The maximum height attained by the centre of mass the complete file is `15 cm`
The minimum FORCE required to hold the flap at REST at `60^@` to horizontal is `(mg)/(4)`.
The minimum force required to hold the flap at rest at `60^@` to the horizontal is `(mg)/(2)`.

Answer :A::C
Discussion
38.

To measure the atmospheric pressure, four different tubes of length 1m, 2m , 3m and 4m are used. If the height of the mercury column in the tubes is h_(1),h_(2),h_(3),h_(4) respectively in the four cases, then h_(1):h_(2):h_(3):h_(4)

Answer»

1:2:3:4
4:3:2:1
1:2:2:1
1:1:1:1

Answer :B
Discussion
39.

Use the result of the above problem to show thata light ray reflected from three mutually perpendicular plane mirrors in succession reverses its direction.

Answer»


ANSWER :NA
Discussion
40.

What are the essential requirements for the choice of standerd unit ? Who decides the units ?

Answer»

SOLUTION :1 (B) 7
Discussion
41.

What is the difference between an echo and a reverberation?

Answer»

SOLUTION :An ECHO is produced when SOUND reflection from a DISTANT obstacle comes back after an interval 1/10 second or more. in an echo, the origintal and reflected sounds are heard separtely
Discussion
42.

The first law of thermodynamics is a statement of

Answer»

CONSERVATION of HEAT
Conservation of work
Conservation of momentum
Conservation of energy

Answer :D
Discussion
43.

When a bouler of mass 240kg is placed on a floating iceberg, it is found that the iceberg just sinks. What is the mass of iceberg? (Given R.D. of ice=0.90 and density of sea water=1.02).

Answer»

Solution :MASS of bouler =240kg
Let the mass of iceberg =M
`rho_("ice")=90kgm^(-3)""rhg_(w)=1020kgm^(-3)`
APPLYING (240+M)g= `vrho_(p_(w))g.` we get
`240+M=(M)/(rho)S_(w) 240=M((rho_(w))/(rho)-1)`
hence `M=(240)/((rho_(w))/(rho)-1) =(240)/((1020)/(900)-1)=(240 xx 45)/(rho)`
M=1800kg
Discussion
44.

Two identical bodies A and B moving with velocities 13 m/s and - 15 m/s respectively collide head-on elastically. What will be their velocities after collision?

Answer»

SOLUTION :When masses are equal, velocities are just interchanged after an elastic head-on COLLISION.
`:.` After collision, velocity of` A = - 15 m//s`
velocity of `B = 13 m//s `
Discussion
45.

Three solid spheres each of mass 1 kg and radius 2m are arranged at the three corners of an equilateral triangle of side 10m, such that centres of the spheres coincide with the corners of the triangle. When they are release form that position, the speed of any one sphere at the time of collision would be (G is universal gravitation).

Answer»

`sqrt((3G)/10)`
`sqrt((10G)/10)`
`sqrt(30G)`
`sqrt(3G)`

ANSWER :A
Discussion
46.

A satellite can be in a geostationary orbit around earth at a distance r from the centrer. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around if its distance from the centre is

Answer»

`(R )/(2)`
`(r )/(2sqrt(2))`
`(r )/((4)^(1//3))`
`(r )/((2)^(1//3))`

SOLUTION :Let angular velocity is `omega RARR mr omega^(2)=(GMm)/(r^(2))`
`omega^(2)=(GM)/(r^(3))` so `omega_(1)^(2)r_(1)^(3)=omega_(2)^(2)r_(2)^(3)`
`r_(2)^(3)=r_(1)^(3)((omega_(1))/(omega_(2)))^(2) rArr r_(2)^(3)=(r_(1)^(3))/(4) rArr r_(2)=(r_(1))/(4^(1//3))`
Discussion
47.

Explain the freely falling apple on Earth using the concept of gravitational potential V(r)?

Answer»

Solution :The gravitational POTENTIAL V (R) at a point of HEIGHT h from the SURFACE of the Earth is given by,
`V(r = R + h) = (GM_E)/((R + h))`
The gravitational potential V(r) on the surface of Earth is given by
`V(r = R) = ( GM_E)/R`
Thus we SEE that
`V (r = R) < V(r = R + h)`
Discussion
48.

Torque due to a force is the product of………….and……………..of line of action…………. .

Answer»

Solution :FORCE , PERPENDICULAR distance , of force from the AXIS of ROTATION.
Discussion
49.

A boat going down stream crosses a float at a point A. 't_1'minutes later the boat reverses its direction and in the next 't_2'minutes it crosses the float at a distance L from the point A. The velocity of the river is

Answer»

`2L//t_1`
`L//(t_1 + 2t_2)`
`2L//(2t_1 + t_2)`
`L//2t_1`

ANSWER :D
Discussion
50.

The centre of gravity of a body on the earth coinides with its centre of mass for a 'small' object whereas for an 'extended' object it may not. What is the qualitative meaning of 'small' and 'extended' in this regard ? For which of the following the two coincide ? A building a pond, a lake, a mountain ?

Answer»

Solution :When VERTICAL height of an object is very small compared to the radius of EARTH, the object is said to be small. Otherwise, the object is said to be extended.
A BUILDING and a pond are treated as small objects. In their cases, the centre of GRAVITY coincides with the centre of mass.
A DEEP lake and a mountain are taken as extended objects. In their cases, the centre of gravity may not coinide with the centre of mass.
Discussion