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 circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released . The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm . Determine the torsional spring constant of the wire. (Torsional spring constant alpha is defined by the relation J=-alpha theta, whereJ is the restoring couple and theta the angle of twist).

Answer»

Solution :Here , m=10 kg , R=15 cm , T=1.5 s
MOMENT of inertia of disc, `I= 1/2 mR^2= 1/2 xx 10 xx (0.15)^2 kgm^2`
Now, `T= 2pi SQRT(I/alpha) ` so, `alpha = (4pi^2 I)/(T^2) = 4 xx ((22)/(7))^2 xx 1/2 xx (10 xx (0.15)^2)/((1.5)^2) == 1.97 N` m/rad
Discussion
2.

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released . The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm . Determine the torsional spring constant of the wire. (Torsional spring constant alphais defined by the relation J=-alpha theta , where J is the restoring couple and theta the angle of twist).

Answer»

Solution :Here, m = 10 kg , R=15 cm = 0.15 cm , T=1.5 s
Moment of inertia of DISC , `I = 1/2 mR^2 = 1/2 xx 10 xx (0.15)^2 kgm^2`
Now , `T= 2pi SQRT(T/(alpha))`
So, `alpha = (4pi^2 I)/(T^2) = 4 xx ((22)/(7))^2 xx 1/2 xx (10xx (0.15)^2)/((1.5)^2) = 1.97 N` m/rad
Discussion
3.

Assuming all contact surfaces are smooth , Find the angular frequency of wedge and block system

Answer»

SOLUTION :`SQRT((K(m+m))/(m(m+ msin^2 ALPHA)))`
Discussion
4.

Identify the increasing order of angular velocities of following(a) Earth rotating about its own axis(b) Hour's hand of clock(c ) Seconds hand of clock(d) Fly wheel of radius 2m making 300 r.p.m.

Answer»

a, B, C, d
b, c, d, a
c, d, a, b
d, a, b, c

ANSWER :A
Discussion
5.

A block of mass 5 kg is placed on a rough horizontal surface, Coefficient of static and sliding friction between the body and thesurface are 0.7 and 0.5 respectively. A horizontal force is applied on the block so that the block just starts moving. If the applied force continues to act even after the block is set in motion acceleration of the block will be [ g = 10 "m.s"^(-2)]

Answer»

1 `"m.s"^(-2)`
2 `"m.s"^(-2)`
3 `"m.s"^(-2)`
4`"m.s"^(-2)`

Answer :B
Discussion
6.

In the figure all pulleys (P1, P2, P3 …….) are massless and all the blocks (1,2,3 …..) are identical, each having mass m. The system consist of infinite number of pulleys and blocks. Strings are light and inextensible and horizontal surfaces are smooth. Pulley P1 is moved to left with a constant acceleration of a0. Find the acceleration of block1. Assume the strings to remain horizontal.

Answer»


ANSWER :`(3a_(0))/(2)`
Discussion
7.

If vec(A) = (1)/(sqrt2) cos theta hat(i) + (1)/(sqrt2) sin theta hat(j), what will be the unit vector perpendicular to vec(A)

Answer»

`COS theta i + sin theta J`
`-cos theta i + sin theta j`
`(cos theta i+ sin theta j)/(SQRT2)`
`sin theta i- cos theta j`

ANSWER :D
Discussion
8.

A carnot engine working between 300k and 600k has a work output of 800J per cycle. Find the amount of energy consumed per cycle.

Answer»

`800J`
`400J`
`1600J`
`1200J`

ANSWER :B
Discussion
9.

A temperature difference of 10^@ C is equivalent of a temperature difference .......... in Fahrenheit temperature scale.

Answer»

SOLUTION :`50^@F`
`T_F=9/5 T_C+32`
`=9/5xx10+32`
`=18+32`
`=50^@F`
Discussion
10.

Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments of inertia about the shaft are 5.0 kg m^(2) and 20.0 kg m^(2) respectively. Wheel A is made to rotate at 10 revolution per second. Wheel B, initially stationary, is now coupled to A with the help of a clutch. The rotation speed of the wheels will become

Answer»

`2 SQRT5` rps
0.5 rps
2 rps
1 rps

SOLUTION :According to the law of CONSERVATION of angular MOMENTUM ,
`(5 xx 10) + (20 xx 0) = (5 + 20) omega or omega= (50)/(25) rps = 2rps`
Discussion
11.

A wire is stretched by a force such that its length becomes double how will Y of the wire be affected?

Answer»

SOLUTION : Y is UNAFFECTED because it is INDEPENDENT of its LENGTH and ELONGATION.
Discussion
12.

(A) For a myopic person light from distant object arriving through eye lens may get converged at a point in front of retina. (R ) For myopic person eye lens produces much more divergence in the incident beam.

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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 :C
Discussion
13.

The diode D shown in the circuit is governed by the V-I relation V=(0.7 + 100 I) Volt. The current supplied by the battery is

Answer»

0.3 A
0.2 A
0.15 A
0.1 A

Answer :D
Discussion
14.

Define spring constant of a spring.

Answer»


Answer :It is the restoring force sep up in a string PER UINT EXTENSION.
Discussion
15.

A thin uniform rod of length 1 and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is omega. Its centre of mass rises to a maximum height of

Answer»

`1/6(lomega)/(G)`
`1/2(l^(2)OMEGA^(2))/(g)`
`1/6(l^(2)omega^(2))/(g)`
`1/3(l^(2)omega^(2))/(g)`

ANSWER :C
Discussion
16.

A glass rod when measured with a zinc scale, both being at 30°C appears lo be of length 100cm. If the scale shows correct reading at 0^(@) C.

Answer»

`100.078` CM
`100.054` cm
`100.024` cm
`100.027` cm

Answer :B
Discussion
17.

A container is filled up to the brim with 500 g of water and 1000 g of mercury. When 21200 cal of heat is supplied to the system, 3.52 g of water flows out of the container. Neglecting the expansion of the container, find the coefficient of real expansion of mercury. Given, volume expansion coefficient of water =1.5 times 10^(-4@)C^(-1), density of mercury =13.6g*cm^(-3), density of water 1g*cm^(-3) and specific heat capacity of mercury =0.03cal*g^(-1)*""^(@)C^(-1).

Answer»

Solution :Due to the application of heat, 3.52 g i.e., 3.52 `cm^(3)` of WATER flows out of the container.
`therefore` The TOTAL expansion of MERCURY and water `=3.52cm^(3).`
Let the rise in temperature `=t^(@)C.`
Heat absorbed by mercury + heat absorbed by water = 21200
`therefore "" 500 times 1 times t+1000 times 0.03 times t=21200 " or, " t=40^(@)C`
`therefore` Expansion of water `=500 times 1.5 times 10^(-4) times 40=3 cm^(3)`
`therefore` Expansion of mercury `=(3.52-3)=0.52 cm^(3)`
`therefore ""1000/13.6 times gamma times 40=0.52`
`""[gamma=` volume expansion COEFFICIENT of mercury`]`
`therefore "" gamma=1.768 times 10^(-4@)C^(-1).`
Discussion
18.

Each to two boats has a speed of 8 km/h . The speed of river current is 6 km/h. Starting at the same instant, thefirst boat cross the river in least time ,and the second in least path. How must later the second boat would arrive if the river is 2 km wide?

Answer»


ANSWER :7.7 MIN
Discussion
19.

Consider the equations P=underset(Deltasrarr0)Lim(F)/(DeltaS)andP_(1)-P_(2)=rhogz. In an elevator accelerating upward

Answer»

Both the EQUATIONS are VALID
The FIRST is valid but not the SECOND
The second is valid but not the first
Both are invalid

Answer :B
Discussion
20.

Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer»

SOLUTION :(a) `x GT a , 0`
(b) `-oo lt x lt oo, V_(1)`
(c ) `x lt a, x gt b, -V_(1)`
(d) `-b"/"2 lt x lt -a"/"2, a"/"2 lt x lt b"/"2, -V_(1)`.
Discussion
21.

Theheatdissipated in a resistancecanbedeterminedform therelation: H = (I^(2)Rt)/(4.2) cal. If themaximum errors in themeasurementof current , resistance, and time are 2%, 1%, and 1%, whatwould be themaximum errorin thedissipated heat ?

Answer»

0.05
0.04
0.06
`0.5%`

ANSWER :C
Discussion
22.

A source of sonic oscillations of frequency n_(0) = 1500 Hz moves at right angles to the wall with velocity u = 2.0 m//s. Two stationary receivers R_(1) and R_(2) are located on a straight line coinciding with the line of motion of the source in the sequence: R_(1) -source-R_(2)-wall. Which reciver registers the beat and what is the beat frequency? The velocity of sound, v = 340 m//s.

Answer»


ANSWER :0.44 HZ, 0
Discussion
23.

A uniform rod of mass m and length 1 can rotate in vertical plane about a smooth horizontal axis hinged at a point H. The angular acceleration alpha of the rod just after it is from initial position making an angle of 37 with the horizontal from rest 5g/bl where b is

Answer»


ANSWER :6
Discussion
24.

A uniform rod AB of mass m and length 1 is at rest on a smooth horizontal surface. An impulse J is applied to the end B perpendicular to the rod horizontal direction. Speed of particle P at a distance 1/6 from the centre towards A of the after time t=(piml)/(12J) is sqrt(x)(J)/(m) where 'x' is

Answer»


ANSWER :2
Discussion
25.

A) Polished suface is a poor absorber but good reflector B) Rough surface is a good absorber and good reflector

Answer»

A is WRONG, B is CORRECT
A is correct, B is wrong
Both A & B are correct
Both A & B are wrong

ANSWER :B
Discussion
26.

Derive an expression for the center of mass of two point masses.

Answer»

Solution :(i) Consider the point masses `m_(1) and m_(2)` which are positioned as `x_(1) and x_(2)` along X-axis. The center of mass can be found in this system in THREE ways as follows
(a) When the masses are on positive X-axis :
(i) The ORIGIN is taken arbitrarily as shown in figure.

(II) The center of mass along X-axis is,
`x_(CM) = (m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))`
(b) When the origin coincides with any one of the masses :
(i) If the orging coincide with mass `m_(1)` as shown in figure, its position `x_(1) = 0`

(ii) The center of mass along X-axis is,
`x_(CM) = (m_(1)(0) + m_(2) x_(2))/(m_(1)+m_(2)) = (m_(2)x_(2))/(m_(1)+m_(2))`
(c) When the origin coincides with the center of mass itself :
(i) If the origin coincide with center of mass as shown in figure, `X_(CM) = 0`. Hence, the position `x_(1)` is negative.

(ii) The center of mass along X-axis is,
`0 = (m_(1)(-x_(i))+m_(2)x_(2))/(m_(1)+m_(2))`
`m_(1)x_(1) = m_(2)x_(2)`
(III) The above equation is known as principle of moments.
Discussion
27.

A mercury thermometer contains 2c.c. of Hg. At 0^(@)C. Distance between 0^(@)Cand100^(@)C marks on the stem is 35cm and diameter of the bore is 0.02cm. gamma_(A) of liquid is

Answer»

0.000055/°C
0.000066/°C
0.00055/°C
0.000058/°C

Answer :A
Discussion
28.

A tunnel is dug along the diameter of the earth. The gravitational force on particle of mass m placed in the tunnel at a distance .x. from the centre.

Answer»

SOLUTION :Mass of earth `M=4/3 piR^3 rho`
Mass of imaginarysphere,
having radius x,
`M_1=4/3 PI x^3 rho` (or)`M^1/M=x^3/R^3`
`THEREFORE` Gravitational force on .m. due to `M_1`
`F=(GM^1m)/x^2` (or) `F=(Gm)/x^2 (x^3/R^3 M)="GMmx"/R^3`
Discussion
29.

The height and width of each step of a staircase are 20cm and 30m respectively. A ball rolls off the top of a stair with horizontal velocity v and hits the fifth step. The magnitude ofv is (g=10ms^(-2))

Answer»

`1.5 sqrt(5)ms^(-1)`
`3 sqrt(5)ms^(-1)`
`7.5 ms^(-1)`
`1.5 ms^(-1)`

ANSWER :A
Discussion
30.

When a wound spring dissolved in an acid, the temperature of the acid

Answer»

INCREASES
DECREASES
REMAINS same
NONE

ANSWER :A
Discussion
31.

By blowing air into a funnel through the narrow end, the filter paper inside the funnel cannot be removed. Why?

Answer»

SOLUTION :When air is blown downward into the funnel, the VELOCITY of air at the top of the paper BECOMES higher, this produces a lower pressure. Due to this pressure difference the paper GETS pushed upwards sad sticks to the wall of the funnel.
Discussion
32.

A passenger in a car moving round a curved road is thrown towards the outside of the curve. What force throws him in this direction ?

Answer»


ANSWER :CENTRIFUGAL FORCE
Discussion
33.

When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process .........

Answer»

`Q GT 0, W gt 0`
`Q lt 0 , W GT0`
`Q gt 0, W lt 0`
`Q lt 0, W lt0`

Solution :`Q gt 0, W gt 0`
Discussion
34.

(A) : The direction of velocity vecotr remains unchanged though the coordinate system is changed ( R) : The direction of real vector is independent of coordinate system

Answer»

Both (A) and ( R) are ture 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
Discussion
35.

A small hole is made at the bottom of a hollow sphere. The water enters into it when it is taken to adepth of 40 cm under water. If the surface tension of water is 0.07Nm^(-1), diameter of the hole is

Answer»

`10 //7 MM`
`10 //14MM`
`2//15 mm`
`1//14mm`

ANSWER :D
Discussion
36.

The outer surfaccs of a rectangular slab of equal thickness of iron and brass are maintained at 100^(0)C and 0^(0)C. The thermal conductivities of iron and the brass are 14 and 126" Wm"^(-1)k^(-1). The temperature of the interface in equilibrium is found to be 5x, the value of x is.

Answer»


ANSWER :2
Discussion
37.

Which one of the followingstatements is an icorrect statement ?

Answer»

The phenomenon of beats is not observed in the case of visible light waves
The decreases in the SPEED of sound at HIGH altitudes due to a FALL in pressure.
Standing waves are characterised by NODES and antinodes.
Transverse waves are characterised by crests and troughs.

SOLUTION :A change in pressure has no effect on the speed of sound.
Discussion
38.

To stimulate car accident , auto manufacturers study the collision of moving car with mounted springs of different spring constants . Consider a typical stimulation with a car of mass 800 kg moving with a speed 36.0km/h on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 xx10^(3)Nm^(-1) .What is the maximum compressionof the spring ?

Answer»

SOLUTION :`x_(m)=12.8 m `
Discussion
39.

Derive equations of uniformly acceleration motion by calculus method.

Answer»

Solution :Consider an object moving in a straight line with uniform or constant acceleration 'a'. Let u be the velocity of the object at time t=0, and v be velocity of the body at a later time t.
Velocity - time relation
(i) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
`a= (dv)/(dt)" or "dv=a.dt`
Integrating both sides with the CONDITION that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,
`int_(u)^(v) dv= int_(0)^(t) a dt = a int_(u)^(v) dt implies [v]_(u)^(v) = a [t]_(0)^(t)"""............."(1)`
`v-u = at" or " v= u+at`
DISPLACEMENT - time relation
(ii) The velocity of the body is given by the first derivative of the displacement with respect to time.
`v= (ds)/(dt)" or "ds = vdt` and since `v=u+at`
We get `ds = (u+at)dt`
Assume that initially at time t=0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time - independent, we have
`int_(0)^(s) ds= int_(0)^(t) u dt +int_(0)^(t) at dt " or "s =ut +(1)/(2) at^2"""............."(2)`
Velocity -displacement relation
(iii) The ACCELARATION is given by the first derivative of velocity with respect to time.
`a= (dv)/(dt)=(dv)/(ds)(ds)/(dt)= (dv)/(ds)v`
[ since `(ds)/(dt)=v`] where s is distance traversed]
This is rewritten as `a=(1)/(2) (dv^2)/(ds)" or " ds = (1)/(2a) d(v^2)`
Integrating the above equation, using the FACT when the velocity changes from `u^2" to " v^2`, displacement changes from 0 to s, we get
`int_(0)^(s) ds = int_(u)^(v) (1)/(2a)d (v^2)`
`:. S= (1)/(2a) (v^2-u^2)`
`:. v^2=u^2+2as"""............."(3)`
We can also derive the displacement 's' in TERMS of initial velocity 'u' and final velocity v. From equation 1, we can write
`at = v-u`
Substitute this in equation 2, we get
`S= ut +(1)/(2)(v-u)t`
`S= ((u+v)t)/(2)""".............."(4)`
The equations 1,2,3 and 4 are called kinematic equations of motion, and have a wide variety of practical applications.
Kinematic equations
`v= u+at`
`S= ut +(1)/(2) at^2`
`v^2= u^2+2as`
`s=((u+v)t)/(2)`
It is to be noted that all these kinematic equations are valid only if the motion is in a straight line with cosntant acceleration. For circular motion and oscillatory motion these equations are not applicable.
Discussion
40.

The velocities of a particle performing linear SHM are 0.13m/s and 0.12 m/s when it is at 0.12m and 0.13m from mean position, respectively. Find the period and amplitude.

Answer»

6.284 SEC, 0.1769 m
5.184 sec, 0.1863 m
2.152 sec, 0.1261 m
6.51 sec, 0.2753 m

ANSWER :A
Discussion
41.

A standard unit should be (a) consistent (b) reproducible (c) invariable (d) easily available for usage

Answer»

only a& B are true
only a& C are true
only a,c &d are true
a, b, c, d are true

Answer :D
Discussion
42.

The term contripetal acceleration was proposed by

Answer»

HUYGENS
Kepler
Newton
Galileo

Answer :C
Discussion
43.

A piece of metal weighs 46g times g in air. When immersed in a liquid of relative density 1.24, kept at 27^(@)C, its weight is 30g times g. When the temperature of the liquid is raised to 42^(@)C, the metal piece in it weighs 30.5g times g. At 42^(@)C, the relative density of the liquid is 1.20. Find the coefficient of linear expansion of the metal.

Answer»

Solution :The apparent loss in weight of the metal at `27^(@)C=` weight of an equal VOLUME of the liquid `=(46-39) g times g`,
Thus the volume of the displaced liquid at `27^(@)C=(46-30)/1.24=16/1.24 cm^(3)=` volume of the metal piece at `27^(@)C(=V_(1))`.
Similarly, the volume of the metal piece at `42^(@)C(=V_(2))=(46-30.5)/(1.20)=15.5/1.20cm^(3)`
`THEREFORE` Coefficient of volume expansion of the metal,
`gamma=(V_(2)-V_(1))/(V_(1)(t_(2)-t_(1)))=1/((t_(2)-t_(1)))(v_(2)/(v_(1))-1)`
`""=1/(42-27)(15.5/1.2 times 1.24/16-1)=1/15(961/960-1)`
`therefore` The coefficient of linear expansion of the metal piece,
`alpha=gamma/3=(6.94 times 10^(-5))/3""^(@)C^(-1)=23.15 times 10^(-6@)C^(-1).`
Discussion
44.

Explain with a simple illustration that viscous force exerted inliquid.

Answer»

Solution :A MILK ina cup is strirred with spoon , it rotatesin circular path .
If milkis stop to stir , its motion decrease and FINALLY is becomes REST .
This FACT shows that there expert a viscous force between layers of liquid.
Discussion
45.

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight. Then

Answer»

Only a & d are true
Only b
Both a &c
All are correct

Answer :1
Discussion
46.

In which type of media longitudinal waves can be transmitted ?

Answer»

SOLUTION :SOLIDS, LIQUIDS ANG GASES.
Discussion
47.

The weight of an object in the coal mine, sea level and at the top of the mountain are respectively W_(1), W_2 and W_3, then

Answer»

`W_1 LT W_(2) lt W_3`
`W_1=W_2=W_3`
`W_1 lt W_2 lt W_3`
`W_(1) gt W_2 gt W_3`

ANSWER :A
Discussion
48.

Two particles P and Q move towards with each other from rest with the velocities of 1 and 20 ms^(-1) under the mutual force of attraction. The velocity of centre of mass is

Answer»

`20 MS^(-1)`
`20MS^(-1)`
`30 ms^(-1)`
zero

Answer :D
Discussion
49.

Second lawof newton gives the definition of force

Answer»

FUNDAMENTAL
quantiative
DIMENSIONAL
both B and C

ANSWER :d
Discussion
50.

The specific heat of Argon at constant volume is 0.3122 kj/kg K. Find the specific heat of Argon at constant pressure if R= 8.314 J/Kmole K. (Molecular weight of argon = 39.95) in J/kg K is.

Answer»

520.3
530.2
230.5
302.5

Answer :A
Discussion