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.

A container of volume 1m^(3)is divided into twoequal compartments, one of which contains an ideal gas at 300 K. The other compartment is vaccum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be

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300 K
250 K
200 K
100 K

ANSWER :A
2.

Find out the sign of work done in the following case. Work done by a man in lifting a bucket out of a well.

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SOLUTION :POSITIVE
3.

A block moves down a smooth inclined plane of inclination 45^@. Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is vin, where n is a number greater than zero. The coefficient of friction mu is given by

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`MU = (1-(1)/(N))`
`mu =(1-(1)/(n^2))`
`mu = SQRT(1-(1)/(n^2))`
`mu =sqrt(1-(1)/(n))`

SOLUTION :`mu =(1-(1)/(n^2))`
4.

h d G has the dimensions of (h= height, d= density, G= gravitational constant)

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Pressure
Power
Torque
Acceleration

Answer :D
5.

An object is flowing through the liquid. The viscous damping force acting on it is proportional to the velocity.Select the correct statement of the following.

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DIMENSION of CONSTANT of proportionality are `ML^(-2)T^(2)`
Dimension of constant of proportionality are `MLT^(-1)`
Dimension of constant of proportionality are `ML^(0)T^(-1)`
Dimension of constant of proportionality are `M^(0)LT^(-1)`

Solution :Viscous FORCE `prop` Velocity
`FpropvorF=Kv`
`[K]=([F])/([v])`
`=([MLT^(-2)])/([LT^(-1)])`
`=[ML^(0)T^(-1)]`
6.

A car is driven at constant speed on a rough road. The shock absorbers vibrate vertically. As we know, if the displacement varies sinusoidally, the oscillation will be a simple harmonic. When there is matching between the disturbance and the oscillating shockers, we call it as resonance. Consider the mass of the car and its contents M=1500 kgwith a man of mass m=75 kg, rising vertically S=0.1 m due to the peaks separated by lambda=10 m equally and the speed of the car being v and time period being T.Then: Critical speed at maximum displacement is (in ms^(-1)):

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`2.56`
`3.56`
`5.26`
`3.76`

ANSWER :B
7.

A car is driven at constant speed on a rough road. The shock absorbers vibrate vertically. As we know, if the displacement varies sinusoidally, the oscillation will be a simple harmonic. When there is matching between the disturbance and the oscillating shockers, we call it as resonance. Consider the mass of the car and its contents M=1500 kgwith a man of mass m=75 kg, rising vertically S=0.1 m due to the peaks separated by lambda=10 m equally and the speed of the car being v and time period being T.Then: The nature of oscillation of shocker will be :

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DAMPED oscillations
SIMPLE HARMONIC oscillations of constant AMPLITUDE
peridoic with increasing amplitude
information GIVEN in the passage are insufficient

Answer :A
8.

Give thedimensional formula of coefficient of viscosity .

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Solution :DIMENSIONAL formula of coefficient of VISCOSITY is `M^(1)L^(-1)T^(-1)` .
9.

A car is driven at constant speed on a rough road. The shock absorbers vibrate vertically. As we know, if the displacement varies sinusoidally, the oscillation will be a simple harmonic. When there is matching between the disturbance and the oscillating shockers, we call it as resonance. Consider the mass of the car and its contents M=1500 kgwith a man of mass m=75 kg, rising vertically S=0.1 m due to the peaks separated by lambda=10 m equally and the speed of the car being v and time period being T.Then: The frequency of oscillation of the car is proportional to:

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`sqrt((MG)/(MS))`
`sqrt((mg)/(d))`
`sqrt((3mg)/(2MS))`
`sqrt((mg)/(MD))`

ANSWER :A
10.

Two heavy spheres of mass 100 kg radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational field and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? Is so, is the equilibrium stable or unstable.

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Solution :Gravitational FIELD at the mid-point of the LINE joining the centres of the two SPHERES
`= (GM)/((r//2)^(2)) (-hat(r)) + (GM)/((r//2)^(2)) hat(r) = 0`
Gravitational potential at the mid point of the line joining the centres of the two spheres is
`V = (-GM)/(r//2) + ((-GM)/(r//2)) = (-4GM)/(r) = (-4 xx 6.67 xx 10^(-11) xx 100)/(1.0) = 2.7 xx 10^(-8) J//kg`
As the EFFECTIVE force on the body placed at md-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibirum. Hence, the body is in unstable equilibirum.
11.

Two separate air bubbles (radii 0.002 cm and 0.004 cm) formed of the same liquid (surface tension 0.070 N/m) come together to form a double bubble. Radius of curvature of the internal film surface common to both the bubble is concave towards the smaller bubble. Find the radius (in mm) of the internal film surface common to both the bubbles.

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ANSWER :4
12.

One mole of a monoatomic gas is mixed with one the mixture?

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1.5
1.54
1.4
1.45

Answer :A
13.

Choose the wrong statement :

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In the process of explosion some changes may occur in MOMENTUM of individual fragments due to internal FORCES but the motion of the centre of MASS is unaltered.
Motion of centre of mass depends upon the external force.
The location of centre of mass depends on the REFERENCE frame used locate it
The position of centre of mass depends upon the shape of the BODY and the distribution of mass.

Answer :C
14.

The dimensional formula for moment of inertia…………..

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

ANSWER :D
15.

If density (D), acceleration (a) and force (F) are taken as basic quantities, then Time period has dimensions

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`(1)/(6)` in F
`-(1)/(6)` in D
`-(2)/(3)` in a
All the above are true

Answer :D
16.

At the uppermost point of a projectile, the velocity and acceleration are at an angle of

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`0^@`
`45^@`
`90^@`
`180^@`

Answer :C
17.

(A): Elastic restoring forces may be conservative.(R): The value of strain for same stress are different while increasing the load and while decreasing the load

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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 but (R) is false
Both (A) and (R) are false

Answer :A
18.

Vernier constant is the

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VALUE of IMSD divided by total number of divisions on the main SCALE.
Value of IVSD divided by total number of divisions on the main scale.
Total number of divisions on the main scale divided by the total number of divisions on the VERNIER scale
Difference between the value of IMSD and IVSD

Answer :D
19.

The angular displacement of a body is given by theta = 2t^(2) + 5t -3.Find the value of the angular velocity and angular acceleration when t = 2s.

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SOLUTION :`OMEGA = dtheta//dt = 4T +5 = 13 "rad/s", alpha = domega//dt = 4 "rad/s"^(2)`
20.

A vertical tube has diameter 0.016 m at its bottom end from which water flows out at the rate of 1.2 kg per minute. The pressure at the end is atmospheric pressure 0.7 m of mercury. If the diameter of the tube is 0.004 m at a height of 0.3 m from the bottom end, fidn the pressure there.

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ANSWER :[0.7303 m of MERCURY]
21.

A body thrown at an angle of 15^(@) wit the horizontal with certain speed has a horizontal range of 10m. Ifthe body is thrown with the same speed but at an angle of 45^(@) with horizontal, find its horizontal range?

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ANSWER :20 m and 15M
22.

Calculate the maximum horizontal range attained by the particle in the case of oblique projection.

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Solution :We know that HORIZONTAL range ,
`R = (U^(2) sin 2 THETA)/( g)`
for R tobe maximum u and g are constant.
Therefore variable is ` ' theta'`.
i.e., `sin 2 theta = 90^(@)`
` :. theta = 45^(@)`
23.

A wire of length L and radius r fixed at one end and a force Fapplied to the other end produces an extension l. The extension produced in another wire or the same material of length 2L and radius 2r by a force 2F is .......

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`l/2`
l
2l
4l

SOLUTION :`Y= (FL )/(PI r ^(2) l ) and Y. = ((2F) (2L ))/( pi (2r) ^(2) l.) `
`THEREFORE Y= Y.`
`therefore ( FL )/( pi r ^(2) l ) = ( 4 FL )/( 4pi r ^(2) l .)`
`therefore l . =l`
24.

A man turns on rotating table with a omega angular speed. He is holding two equal masses at arms length. Without moving his arms, he hust drops the two masses. How will his angular speed change ?

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LESS than `omega`
more than `omega`
it will be equal to `omega`
it will be more than `omega` if dropped mass is more than `9.8` KG and it will be less than `omega` if the mass dropped is less than `9.8` kg

ANSWER :C
25.

A Carnot engine whose efficiency is 40% receives heat at 500K. If the efficiency isto be 50%. The source temperature for the same exhaust temerature is

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600K
700K
800K
900K

Answer :A
26.

In a planetary motion the areal velocity of a position vector of a planet depends on angular velocity .w. and distance of the planet from the sun (r). If so the correct relation for areal velocity

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`(DA)/(dt) PROP OMEGAR`
`(dA)/(dt) prop omega^(2)r`
`(dA)/(dt) prop omegar^(2)`
`(dA)/(dt) propsqrt(omegar)`

ANSWER :C
27.

Two cylindrical hollow drums of radii R and 2R and of a common height h are rotating with angular velocities omega (anti-clockwise) and omega (clockwise) respectively. Their axes, fixed are parallel and in a horizontal plane separated by 3R+delta. They are now brougth in contact (deltararr0). (a) Show the frictional forces just after contact. (b) Identify forces and torques external to the system just after contact. (c) What would be the ratio of final angular velocities when friction ceases?

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Solution :(a) Figure shows the situation given in the QUESTION.

(B) `F_(1)=F=F_(2)` where `F_(1) and F_(2)` are EXTERNAL forces through support.
`therefore F_("net")=0` (One each cylinder)
External torque = `Fxx3R`, (anti-clockwise)

(c ) LET `Omega_(1)andomega_(2)` be final angular velocities of smaller and bigger drum respectively.
Finally, there will be no friction.
Hence, `Romega_(1)=2Romega_(2)implies (omega_(1))/(omega_(2))=2`
28.

What is meant by resonance ? State an example.

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Solution :Resonance is a special case of forced vibrations where the frequency of EXTERNAL periodic force (or driving force) matches with the natural frequency of the VIBRATING body (driven). Hence the oscillatng body begins to begins to vibrate such that its amplitude increases at each step and ultiimately it has a large amplitude. Such a phenomenon is known as reasonance. The corresponding vibrations are known as resonance vibrations.
Eg:
The breaking of GLASS due to SOUND.
29.

A car accelerates from rest with 2m//s^(2) on a straight line path and then comes to res after applying brakes. Total distance travelled by the car is 100 m in 20 seconds. Then the maximum velocity attained by the car is

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`10m//s`
`20m//s`
`15m//s`
`5m//s`

ANSWER :A
30.

The molar specific heat for solid is

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3RT
2R
4R
3/2R

Answer :A
31.

In the planet Mars, the average temperature is around -53^(@)C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of the molecules in unit volume in the planet Mars? Is this greater than that in earth ?

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Solution :`T=-53^(@)C=220K`
`P=0.9xx10^(3)PA`
`V=1m^(3)`
`"NUMBER of molecules "mu=(VP)/(RT)=(0.9xx10^(3))/(8.314xx220)=0.00049xx10^(3), mu="0.49 MOL"`
32.

A uniform rod of length l rotating with an angular velocity omega, while its centre moves with linear velocity v=(omegal)/(6). If the end A of the rod is suddenly pivoted, the angular velocity of the rod will be

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`(3)/(4)omega`
`(omega)/(3)`
`(omega)/(2)`
`(2omega)/(3)`

Answer :C
33.

A spring of force constant k is stretched by a small length x. Find the work done in stretching it further by a small length y.

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Solution :Let `W_(1)` is the work done in stretching a spring of force CONSTANT .K. through a length ..X... Then `W_(1)=(1)/(2)Kx^(2)`
Let `..W_(2)..` is the workdone in stretching the spring through a length `(x+y)`. Then `W_(2)=(1)/(2)K(x+y)^(2)`
`THEREFORE` Additional work done, to increase the elongation by ..y.. is `W=W_(2)-W_(1)`
`W=(1)/(2)K(x+y)^(2)-(1)/(2)Kx^(2)`
`therefore W=(1)/(2)Ky (y+2x)`
34.

If the time period of oscillation of a pendulum is measured as 2.5 second using a stop watch with the least count (1)/(2) second, then the permissible error in the measurement is

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`10%`
`30%`
`15%`
`20%`

Solution :Here, T= 2.5 SECOND, `DeltaT= (1)/(2)` second
Therefore, the permissible error in the measurement of time period is `(DeltaT)/(T)xx 100= ((1)/(2))/(2.5)xx100= 20%`
35.

A block A of 10 kg rests on a rouch, horizontal table and is connected to another block of mass 5 kg by a string passing overfrictionless pulley. This mass hangs vertically (a) Determinem the minimum weight of a third block C which can be placed on A to prevent it from sliding if mu_(s)=0.2 (b) The block C is suddenly lifted, then what is the acceleration of the block A ? (mu_(k) = 0.18)

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ANSWER :(a) `15KG` (B) `2 MS^(-2)`
36.

A body P moving with a velocity of 20kms^(-1) collides with another body of Q of the same mass at rest. After collision the velocity of Q if P comes to rest after collision is

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`10 KMS^(-1)`
`20 Kms^(-1)`
`30Kms^(-1)`
`5 Kms^(-1)`

ANSWER :B
37.

The displacement of a particle executing shm is x = 0.5 cos (314t - 0.3) m. Find (i) amplitude (i) period.

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SOLUTION :`a=0.5 m,T=2pi//omega=2pi//314=0.02` SECOND
38.

State kirchhoff's law of blakc radiations.

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ANSWER :For any GIVEN TEMPERATURE and wavelength the ratio of the emissive POWER to the absorptive power the same for all subatances is equal to the emissive power of a perfectly blakc at the same temperature.
39.

State the law of floatation?

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SOLUTION :The law of floatation STATES that a body will FLOAT in a liquid if the weight of the liquid DISPLACED by the immersed part of the body equals the weight of the body.
40.

Average life of a human being is of the order of

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`10^7s`
`10^(-7)s`
`10^9s`
`10^(-9)s`

ANSWER :(C )
41.

A wire of length 60 cm is bent into a circle with a gap of 1 cm at its ends. On heating it by 100^(@)C, the length of the gap increases to 1.02 cm. alpha of material of wire is

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`2 xx 10^(4)//^(@)C `
`4 xx 10^(-4) //^(@) C `
`6 xx 10^(-4) //^(@)C `
`1 xx 10^(-4) //^(@)C `

ANSWER :A
42.

The acceleration due to gravity on moon is 1//6th of that on the earth. If the length of the seconds pendulum is 96m on the earth, what is its length on the moon ?

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SOLUTION :For SECONDS pendulum time period is 2s `T= 2pisqrt((l)/(g))` for SIMPLE pendulum time remains constant in both the planets `(l)/(g)` = constant `:.(l_(m))/(g_(m))= (l_(E))/(g_(e))` or`(l_(m))/((g_(e))/(6))= (96)/(g_(e))`
`l_(m)= 16 cm`
43.

An electric motor drill, rated 350 W has an efficiency of 35%. The torque produced, if it is working at 3000 rpm.

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0.25 N m
0.35 N m
0.39 N m
0.30 N m

Solution :Angular velocity of the motor drill ,
`omega = (2PI epsilon)/(60) = (2 pi xx 3000)/(60)`
Let `tau` be the torque produced by the motor.
Power produced = `tau omega = tau xx (2pi xx 3000)/(60)`
Now `tau xx (2 pi xx 3000)/(60) = 35%` of 350 W
or `tau xx 2pi xx 50 = (35)/(100) xx 350`
or `tau = (35 xx 350)/(100 xx 2pi xx 50) = 0.39 Nm`
44.

A rectangular block of glass is placed on a printed page lying on a horizontal surface. Find the minimum value of the refractive index of glass for which the letters on the page are not visible front any of the vertical faces of the block

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SOLUTION :The situation is as shown in the Fig. Light will not emerge out from the vertical face BC if at it `I GT theta_c` or
`sin I gt sin theta_c, sin I gt 1/ mu`……….(1)
`1 TIMES sin theta=mu sin r` and in `Delta OPR`
`r+i=90^@, r=90^@-i`
SO `sin theta= mu sin (90^@-i)`
`cos i= (sin theta//mu)`
But `sin i=sqrt(1-cos^2 i)=sqrt(1-(sin theta//mu)^2)`.........(2)
Substituting the value of SINI From eq (2) in (1)
`sqrt(1-(sin^2 theta)/mu^2) gt 1/mu, mu^2 gt 1+ sin 2 theta`
Now as `(sin^2 theta)_(max)=1`,
`mu^2 gt 2, mu gt sqrt2`
Hence `mu_(mi n)=sqrt2`
45.

The moment of inertia of a flywheel making 300 revolutions per minute is 0.3 kgm^2 . Find the torque required to bring it to rest in 20s.

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SOLUTION :Torque `TAU = I alpha = I((omega_(2)-omega_(1)))/(t)`
`therefore tau = (0.3xx(0-10pi))/(20) = -0.15 pi = -0.471 Nm`.
46.

Abomb travelling in a parabolic path under the effect of gravity explodes in mid air. The centre of mass of fragments will

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Move vertically UPWARDS and then vertically DOWNWARDS
Move vertically upwards
Move in an irregular path
Move in the parabolic path the UNEXPLODED bomb WOULD have travelled.

Answer :D
47.

What provides the centripetal force to a car taking a turn on a level road?

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SOLUTION :FORCE of friction between the tyre and road PROVIDES CENTRIPETAL force
48.

A metal ball of radius 'r' and density 'd' travels with a terminal velocity 'v' in a liquid of density d//4. The terminal velocity of another ball of radius '2r' and density '3d' in the same liquid is

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`(44v)/(3)`
`(22V)/(3)`
`(11v)/(3)`
`(3V)/(44)`

ANSWER :A
49.

Which of the following example represent periodic motion (a) A swimmer completing one (return) trip from one bank of a river to other bank. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow.

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Solution :(a) It is not a PERIODIC motion. Though the motion of a swimmer is to and fro but will not have a definite period. (B) It is a periodic motion because a freely suspended magnet if once displaced from N-S direction and let it go, it OSCILLATES about this position. Hence it is simple harmonic motion also. (C) It is also a periodic motion , (d) It is not a periodic motion.
50.

Liquids of different densities in tanks, having an orifice at exactly same height below liquid surface, have same efflux velocity of liquid coming out. Why?

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SOLUTION :Efflux VELOCITY is INDEPENDENT of DENSITY of the liquid.