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 body weighs W_(0) in air. Its apparent weights in a liquid at t_(1)""^(@)C " and " t_(2)""^(@)C " are " W_(1) " and " W_(2) respectively. If the coefficient of volume expansion of the material of the body is gamma,find the coefficient of real expansion of the liquid.

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Solution :Weight of displaced liquid at `t_(1)""^(@)C=W_(0)-W_(1)`
Let `V_(1) " and " V_(2)` be the volumes of the BODY and `d_(1) " and " d_(2)` be the densities of the liquid at `t_(1)""^(@)C "and "t_(2)""^(@)C` RESPECTIVELY.
We have, `W_(0)-W_(1)=V_(1)d_(1)g "...(1)"`
and `" " W_(0)-W_(2)=V_(2)d_(2)g "...(2)"`
DIVIDING equation (1) by equation (2) we get,
`""(W_(0)-W_(1))/(W_(0)-W_(2))=(V_(1)d_(1))/(V_(2)d_(2))`
Now, `V_(2)=V_(1)[1+gamma(t_(2)-t_(1))] " and " d_(1)=d_(2)[1+gamma_(l)(t_(2)-t_(1))]`, where `gamma_(l)` is the coefficient of real EXPANSION of the liquid.
`therefore ""(W_(0)-W_(1))/(W_(0)-W_(2))=(1+gamma_(l) times t)/(1+gamma times t) " "["where "t=(t_(2)-t_(1))]`
or, `" " 1+gamma_(l)t=(W_(0)-W_(1))/(W_(0)-W_(2)) times (1+gammat)=(W_(0)-W_(1))/(W_(0)-W_(2))+(gammat(W_(0)-W_(1)))/(W_(0)-W_(2))`
or, `""gamma_(l)t=(W_(2)-W_(1))/(W_(0)-W_(2))+(gammat(W_(0)-W_(1)))/(W_(0)-W_(2))`
or, `""(W_(0)-W_(1))(1+gammat)=(W_(0)-W_(2))(1+gamma_(l) times t)`
or, `""gamma_(l)=(W_(2)-W_(1))/((W_(0)-W_(2))t)+(gamma(W_(0)-W_(1)))/(W_(0)-W_(2)).`
2.

In SHM………….quantities are always positive.

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SOLUTION :FORCE and ACCELERATION.
3.

Three identical discs A,B and C rest on a smooth horizonal plane as shown in figure -40151. The disc A is set in motion with velocity b along the perpendicular bisector of the line BC joining the centres of the stationary discs. The distance BC joining the cnetes of the stationary disck B and C is n times the diameter of each disc. At what value of n will the disc A recoil, Stop, and move on after elastic collision. ?

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ANSWER :`"["nltsqrt2,n=sqrt2,ngtsqrt2"]"`
4.

Figure shows a cube of side length 'a' placed on a horizontal surface. A particle projected from A with speed u, at an angle theta=tan^(-1)sqrt(5) with horizontal, just touches the cube at two vertices as shown. Find projection speed u.

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`SQRT(27/8 AG)`
`sqrt(2AG)`
`sqrt(3ag)`
`sqrt(8/3 ag)`

ANSWER :C
5.

Obtain an expression for Gravitational field intensity measured with an object of unit mass.

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Solution :`vec(F_m) = m vec(E)`
Now we can equate this with Newton's second LAW `vec(F) = m vec(a)`
`m vec(a) = m vec(E)`
`vec(a) = vec(E)`
In other words, equation IMPLIES that the gravitational FIELD at a point is equivalent to the acceleration experienced by a particle at that point. HOWEVER, it is to be noted `vec(a)` that and `vec(E)` are separete physical quantities that have the same magnitude are direction.
6.

Ice in cold storage melts at the rate of 3 kg per hour. The temperature outside is 27^(@)C. Find the minimum power output of a motor used to drive the refrigerator, which just prevents ice from melting.Latent heat of ice 80 cal g^(-1), J=4.2xx10^(7) erg cal^(-1). given 1h.p. =746 W.

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ANSWER :0.0371 h.p.
7.

A load of 1kg.wt is attached to one end of a steel wire of area of cross section 3mm^2 and Young's modulus 10^11 N.m^-2. The other end is suspended vertically from a hook on a wall. Then the load is pulled horizontally and released. When the load passes through its lowest position, find the frictional charge in the length of the wire.{take g=10 m.s^-2)

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ANSWER :`0.3 TIMES 10^-4`
8.

(A) : A body can have energy without having momentum but it can have momentum without having energy.(R ) : Momentum and energy have same dimensions.

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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
(A) is false and (R ) is false

ANSWER :C
9.

If vec(A) xx vec(B) = 0 and vec(B) xx vec(C ) = 0, the angle between vec(A) and vec(C ) is :

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0
`PI`
`pi//4`
`pi//2`

ANSWER :A::B
10.

Standing is not allowed in the upper deck of a double decker bus. why ?

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Solution :When the passengers stand in the upper deck, the centre of GRAVITY of the bus is RAISED which MAKES it UNSTABLE.
11.

Explain thetotal linear momentum is conserved in anelasticcollisionand alsoexplainthe inelasticcollision and completelyelasticcollision.

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SOLUTION :Yes , because total momentumconserves as per LAW of conservation of momentum .
12.

A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest but it moves at a speed of 2 m/s as it reaches A. The work done by the person on the mass is -3J. The potential at A is

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`-3J//kg`
`-2 J//kg`
`-5J//kg`
`-1 J//kg`

ANSWER :C
13.

"It is not advisable to put on wet clothes". Why?

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Solution :Wet clothes may TAKE away too much of heat from the BODY due to HIGH value of the specific heat capacity of water and HENCE the body temperature may fall even below the NORMAL temperature.
14.

(A): The speedometer of an automobile measure the instantaneous speed of the automobile. (R) : Average velocity is equal to total distance divided by total time taken.

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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 :C
15.

How motion of any particle is described ?

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SOLUTION :We describe motion of any PARTICLE with the HELP of displacement, VELOCITY and ACCELERATION.
16.

A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 m/s.The frequency heard by the observer in Hz is (velocity of sound = 333m/s)

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409
429
517
500

Answer :D
17.

Calculate the values of the molecular mass and gamma for a gaseous mixture consisting of n_(1)=m moles of oxygen and n_(2)=3 moles of carbon dioxide. The gases are assumed to be ideal.

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ANSWER :`39.2,1.36`
18.

The slope of the position-time graph will give

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displacement
velocity
acceleration
force

ANSWER :C
19.

In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm. An object is placed at 2 cm from the objective and the final image is formed at 25 cm from the eye lens. The distance between the two lenses is … (in cm).

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6
7.75
9.25
11

Answer :D
20.

A wire of length 4 m and of cross sectional area 4mm^2 is elongated by 0.1 mm by the applications of a force. IF young's modulus for the material of the wire is 2 times 10^12 N.m^-2, then calculate the amount of energy stored in the wire.

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ANSWER :`10^-3`
21.

When the point of suspension of a pendulum is moved. Its period of oscillation :

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decreases when it moves vertically upwards with an ACCELERATION 'a' 
decreases when it moves vertically downwards with acceleration greater than 2G 
INCREASES when it moves HORIZONTALLY with acceleration 'a' 
all of the above 

Answer :A::B
22.

A physical quantity of the dimensions of length that can be formed out of c, G and (e^2)/( 4pi in_0) is [ c is velocity of light, G is universal constant of gravitation and e is charge ]:

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`(1)/( C) G (E^2)/( 4PI in_0)`
`(1)/( c^2) [ G(e^2) /( 4pi in_0) ]^(1//2)`
`e^(2) [G (e^2)/( 4pi in_0) ]^(1//2)`
`(1)/( c^2) [ (e^2)/( G4pi in_0) ]^(1//2)`

Answer :B
23.

A black body at a temperature of 1640 K has the wavelength corresponding to maximum emission equal to 1.75mu. Assuming the moon to be a perfectly black body, the temperature of the moon, if the wavelength corresponding to maximum emission is 14.35mu is

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100 K
150 K
200 K
250 K

Answer :C
24.

The variation of pressure P with volume V for an ideal monoatomic gas during an adiabatic process is shown in figure. At point A the magnitude of rate of change of pressure with volume is

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`3/5P_(0)/V_(0)`
`5/3P_(0)/V_(0)`
`3/2P_(0)/V_(0)`
`5/2P_(0)/V_(0)`

ANSWER :D
25.

Two identical balls A and B having velocities of 0.5 ms^(-1) and -0.3ms^(-1) respectively collides elastically in one dimension. The velocities of B and A after the collision will be

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`-0.5 MS^(-1) and 0.3 ms^(-1)`
`0.5 ms^(-1) and -0.3 ms^(-1)`
`-0.3 ms^(-1) and 0.5 ms^(-1)`
`0.3 ms^(-1) and 0.5 ms^(-1)`

Solution :Masses of the BALLS are same and collision is elastic, so their VELOCITY will be INTERCHANGED after collision
`:. v_A= +u_B = -0.3 ms^(-1) and v_B = u_A = 0.5 ms^(-1)`.
26.

A circular disc of radius r and mass m rotates about the axis passing through the centre and perpendicular to its plane. What will if rotational kinetic energy?

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`(1)/(4)mromega^(2)`
`(1)/(2)mromega^(2)`
`(1)/(4)mr^(2)omega^(2)`
`(1)/(2)mr^(2)omega^(2)`

Solution :Rotational kinetic ENERGY = `(1)/(2)IOMEGA^(2)`
`=(1)/(2)xx(mr^(2))/(2)xxomega^(2)`
`=(1)/(4)mr^(2)omega^(2)`
27.

The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of T/6 it passes its mean position, its

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VELOCITY will be HALF its maximum velocity 
DISPLACEMENT will be half its amplitude 
Acceleration will be NEARLY 86% of its maximum acceleration 
KE = PE

Answer :A::C
28.

Amanstartsbicyclingin themorningata temperaturearound25^(@)C hecheckedthe pressureof tirewhichis equalto be500 Kpa.Afternoonhe foundthat theabsolutepressurein thetyreis increasedto520 Kpa. Byassumingthe expansionof tyre is negligiblewhat isthetemperatureof typeat afternoon ?

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Solution :For ideal gas equation of state
`PV=nRT`
`P_(1)=500kPa, T_(1)=25^(@)C=25+273=298K, P_(2)=520kPa, T_(2)=?`
EXPANSION of tyre is negligible `(V_("CONSTANT"))`
`(P_(V)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`therefore""T_(2)=((P_(2))/(P_(1)))T_(1)=(520)/(500)xx298=(154960)/(500)-=309.92K`
`T_(2)=309.92-273, T_(2)=36.9^(@)C`
29.

The mass m of the largest stone that can be moved by flowing river depends upon the velocity v of water, density d of water and acceleration due to gravity g. Then

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`m PROP v^6`
`m prop sqrtv`
`m prop G `
`m prop (1)/(d^2)`

ANSWER :A
30.

Ventilation are provided at the top of room

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to BRING OXYGEN for breathing
so that sunlight may enter the room
to MAINTAIN CONVECTION currents to keep the air fresh in the room
to provide an OUTLET for carbondioxide

Answer :C
31.

At ordinary pressure and temperature the average distance between the molecules of a gas is around A. Here, A refers to

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`10^(-9)m`
`10^(-10)m`
`10^(-11)m`
`10^(-8)m`

ANSWER :A
32.

Two particles each of the same mass move due north and due east respectively with the same velocity 'V'. The magnitude and direction of the velocity of centre of mass is

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`(V)/(SQRT(2))NE`
`sqrt(2)` V NE
2 V SW
`(V)/(2)` SW

Answer :A
33.

The workdone in twisting a steel oflength0.25 meter and radius 1 mmthroughan angleof 45^(0) will be if eta=8xx10^(10) pascal

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15 JOULE
0.15 joule
1.5 joule
0.015

Solution :`W=(1)/(2) (eta pi R^(4))/(2l)theta^(2)`
`W=(8xx10^(10) xx 3.14 xx (10^(-3))^(4))/( 4xx0.25) xx ((3.14)/(4))^(2) =0.15` joule
HENCETHE correct answer will be (2)
34.

A point object O is placed at a distance of 30cm from a convex lens of focal length 20cm cut in to two halves each of which is displaced by 0.05cm as shown in figure. Find the positions of the image. IF more than one image is formed, find their number and distance between them.

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Solution :
considering each part as SEPARATE lens with u=-30cm and f=20cm from lens FORMULA `1/v-1/u=1/f`
we have `1/v-1/(-30)=1/20` i.e., v=60cm . So each part will form a real image of the point object O at m from the lens as shown in figure. As there are two pieces, two IMAGES are FORMED. Now in similar triangles `OI_1L_2 and OL_1L_2`
`(I_1I_2)/(L_1L_2)=(OP)/(OQ)=((u+v))/u i..,e I_1I_2=90/30 times (2 times 0.05)=0.3cm`
SO the two images formed are 0.3cm APART.
35.

The minimum number of forces of equal magnitude in a plane that can keep a particle in equilibrium is

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4
2
3
5

Answer :B
36.

Heat is a form of ………….. Which produces in use the ……….

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SOLUTION :ENERGY, SENSATION of WARMTH
37.

A lead ball dropped into a lake from a diving board 5m above the water hits the water with certain velocity and then sinks to the bottom with the same constant velocity. If it reaches the bottom in 3s after it is dropped the depth of the lake is

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30m
15m
10m
20m

Answer :D
38.

Let n number of little droplets of water of surface tension S dyn. Cm^-1, all of the same radius r cm, combine to form a single drop of radius R cm. J is joule's mechanical equivalent of heat. While using CGS system of units answer the following questions. ii. If the whole energy released is taken by the water drop formed, then rise in temperature in ^@C is

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`S/J[1/r-1/R]`
`(4S)/(J)[n/r-1/R]`
`(3S)/(J)[1/r-1/R]`
`S/J[n/r-1/R]`

ANSWER :C
39.

Let n number of little droplets of water of surface tension S dyn. Cm^-1, all of the same radius r cm, combine to form a single drop of radius R cm. J is joule's mechanical equivalent of heat. While using CGS system of units answer the following questions. iii. What is the change in excess of pressure inside the big drop formed and a small drop if the change in temperature is ignored ?

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`2S[1/r-1/R]`
`S[1/r-1/R]`
`S[n/r-1/R]`
`2S[n/r-1/R]`

ANSWER :A
40.

In the case of a freely falling spherical body , its acceleration due to gravity depends on

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massof the BODY
RADIUS of the body
density of the MATERIAL of the body
NONE of the above

Answer :D
41.

A ball of mass m moving with a speed u_(1) collides elasticity with another identical ball moving with velocity u_(2). (a) Find the velocities of the balls after collision if the impact is direct. (b) Find the angle between velocities after collision if they collide obliquely and u_(2)=0

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ANSWER :(a) `v_(1)=u_(2),v_(2)=u_(1)`; (B) `(pi)/(2)`
42.

The equation of motion of a particle started at t=0, is given by y = 5 sin(20 t + pi/3) cm. The least time after which acceleration becomes zero is:

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`pi/90` SEC
`pi/120` sec
`pi/60` sec
`pi/30` sec `

ANSWER :D
43.

A motorheal is racing towards north at 25kmh^(-1) and the water current in that region is 10kmh^(-1) in the direction of 60 cast of south. Find the resultant velocity of the boat.

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SOLUTION :`V=21.8kmh^(-1)` ANGLE with NORTH, `theta=23.4^(@)`
44.

A uniform rod of mass M and length L is hinged at its end to a wall so that it can rotate freely in a horizontal plane. When the rod is perpendicular to the wall a constant force F starts acting at the centre of the rod in a horizontal direction perpendicular to the rod. The force remains parallel to its original direction and acts at the centre of the rod as the rod rotates. (Neglect gravity). (a) With what angular speed will the rod hit the wall ? (b) At what angle theta(see figure) the hinge force will make a 45^(@)angle with the rod ?

Answer»


Answer :(a) `SQRT((3F)/(ML))`
(b) `theta = TAN^(-1) ((1)/(10))`
45.

A ball of mass m hits a floor with a speed v making an angle of incidence theta with the normal. The coefficient of restitution is 'e'. Find the speed of the reflected ball and the angle of reflection of the ball.

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Solution :Suppose the angle of REFLECTION is `theta.` and the speed after the collision is v.. The floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel COMPONENT of the velocity of the ball remains UNCHANGED. This given v. in `theta.- v SIN theta` ...(1)
For the components normal to the floor, the velocity of separation is `v. cos theta.` and the velocity of approach is `v cos theta`. HENCE `v. cos theta= e v cos theta.` ...(ii)
From (i) and (ii), `v.= v sqrt(sin^(2) theta + e^(2) cos^(2) theta) and tan theta. = (tan theta)/(e ) rArr e = (tan theta)/(tan theta^(1))`
46.

If vec(P) + vec(Q) = vec(R ) and vec(P) - vec(Q) = vec(S), then R^(2) + S^(2) is equal to

Answer»

<P>`P^(2) + Q^(2)`
`2(P^(2) - Q^(2))`
`2(P^(2) + Q^(2))`
4PQ

Answer :C
47.

If the inclination of an inclined plane is 30^(@), a block of mass m kept at rest on the surface of the plane, will just slide down. If the inclination is raised to 45^(@) what will be the acceleration of the body ?

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Solution :when `theta=30^(@)`, the block just SLIDES down
So `mu=tantheta=tan30^(@)=0.5774`
The ACCELERATION of a BODY moving down the surface of an INCLINED plane is
`a=g(sintheta-mucostheta)`
`=9.8(sin45-0.5774cos45)`
`a=2.93m//s^(2)`
48.

A ball is thrown vertically down from a height of 40m from the ground with an initial velocity 'v'. The ball hits the ground, loses one-third of its total mechanical energy and rebuounds back to the same height. If the acceleration due to gravity is 10 ms^(-2), the value of 'v' is

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`5MS^(-1)`
`10 ms^(-1)`
`15 ms^(-1)`
`20 ms^(-1)`

ANSWER :D
49.

A body collides head on and elastically with another identical body at rest then after collision

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First body comes to rest, then second body BEGINS to move with the initial velocity of first body
Both the bodies MOVES with same initial velocity of first body
First body REVERSED in DIRECTION and move with same speed but second body continues to be at rest
Bothe the bodied comes to rest after collision

Answer :A
50.

The x-z plane separates two media A and B of refractive indices, mu_1=1.5 and mu_2=2. A ray of light travels from A to B. Its directions in the two media are given by unit vectors vecmu_1=a hati +b hat j and vec mu=c hati+d hatj . Then

Answer»

Solution :so `SIN i=a/sqrt(a^2+b^2)`
and `TANR = c/d , sin r= c/sqrt(c^2+d^2)`
`mu_1 sin i =mu_2 sin r (3/2) (a/sqrt(a^2+b^2))=2 (c/sqrt(a^2+b^2))`
But as `a hati+b hat j and c hati +d hat j` are unit vectors so
`sqrt(a^2+b^2)=sqrt(c^2+d^2)=1`
HENCE `3/2a=2c, so a/c=4/3`