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 quartz vessel contains dry air with pressure 750 mm of Hg at 251 K. The vessel can withstand up to a pressure of 1150 mm of Hg without breaking. To which temperature the air can be heated without breaking the vessel. (Pressure coefficient of air is 0.0036l^(0)C).

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<P>

Solution :Initial TEMPERATURE `t_(1)=251-273=22^(0)C` , Initial PRESSURE `P_(1)=750` mm of Hg
Final pressure, `P_(2)` =1150 mm of Hg ,Pressure coefficient, `beta=0.00366l^(@)C`
Final temperature `t_(2)=?`
`beta=(P_(2)-P_(1))0.00366=(1150-750)/(750 times t_(2)-1150(-22))=400/(750t_(2)+25300)`
`therefore t_(2)=(400-0.00366 times 25300)/(0.00366 times 750)=112^(0)C`
2.

{:(,"List - 1",,"List - 2"),((a),"Position of centre of mass",(e ),"is zero"),((b),"the algebraic sum of moments of all the masses about centre of mass",(f),"in non, uniform gravitational field"),((c ),"Centre of mass and centre of gravity coincide",(g),"is independent of frame of reference"),((d),"Centre of mass and centre of gravity do not coincide",(h),"in uniform gravitational field"):}

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a-e, b-g, C-f, d-h
a-g, b-e, c-f, d-h
a-g, b-e, c-h, d-f
a-h, b-e, c-f, d-g

Answer :C
3.

The work done against friction in moving a body from A to B is 200 J. If the body is again brought back to A along the same path, the total work done for the trip against friction is

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0 J
400 J
200 J
800 J

Answer :B
4.

The dimensional formula for heat capacity..............

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`[ML^(2)T^(-2)]`
`[ML^(2)K^(-1)]`
`[ML^(2)T^(2)K^(-1)]`
`[ML^(2)T^(-2)K^(-1)]`

SOLUTION :`[ML^(2)T^(2)K^(-1)]`
5.

The measured mass and volume of a body are 22.42g and 4.7cm^3respectively with errors 0.01g and 0.1cc. The maximum error in density is

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0.002
0.02
0.05
0.0217

Answer :D
6.

Three point masses m_(1) =4 kg, m_2 = 3 kg and m_3 = 2 kg,are placed along a straight line along the y-axis as shown in Fig. What is the moment of inertia pf:the system about the x - axis ?

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SOLUTION :`I = MR^(2), I= 4 xx 4^(2) + 3 xx (-2)^(2) + (-4)^(2) -92 kg m^(2)`.
7.

Two particles execute SHM of same amplitude along the same straight line. They cross each other when going in opposite directions, each time their displacement is half of their amplitude. Calculate the phase difference between them.

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ANSWER :`120^(@)`
8.

When a capillary tube is dipped in water, water rises up to a height of 0.1 m inside the tube. If that capillary tube is dipped in mercury, then the mercury level falls down through 0.342 m. If the specific gravity of mercury is 13.6, angle of contact of water is 0^@ and that of mercury is 135^@, then compare the surface tension of water and mercury.

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SOLUTION :`1.52:1`
9.

What should be the velocity of a source of sound so that the apparent frequency to a listener will be twice the actual frequency of the source? Velocity of sound in air = v .

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Solution :Apparent frequency , `N.= (v = u_(o))/(v - u_(s)) XX n`
In this case n. = 2N . The listener is stationary, so ` u_(o) = 0 `
the apparent frequency is higher, i.e., the SOURCE is approaching the listenger. So, the velocity of the source ` u_(s) ` is positive . Therefore,
` 2n = (v)/(v-u_(s)) xx n " or " , v = 2 (v - u_(s)) " or " = v/2 `
i.e., the source is approaching the listener with half the velocity of SOUND .
10.

Satellite A is following a circular path of radius a around the earth another satellite B follows an elliptical path around the earth. The two satellites have same mechanical energy and their orbits intersect. Find the speed of satellite B at the point where its path intersects with the circular orbit of A. Take mass of earth to be M.

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ANSWER :`SQRT((GM)/a)`
11.

At what speed will the velocity of a stream of water be equal to 20 cm of mercury column? (Taking, g =10 ms^(-2))

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SOLUTION :` V = sqrt( 2gh ) = sqrt(2xx 9.8 xx 0.4 )= 2.8 ms ^(-1)`
12.

A solid sphere of radius r r is rolling on a horizontal surface. The ratiobetween the rotational kinetic energy and total energy.

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`(5)/(7)`
`(2)/(7)`
`(1)/(2)`
`(1)/(7)`

Solution :The rotation kinetic energy, `K_(r)=(1)/(2)IOMEGA^(2)`
where, `I=` moment of inertia of solid sphere. The TOTAL kinetic energy and translational kinetic energy
`K_(T)=(1)/(2)mv^(2)+(1)/(2) Iomega^(2)`
`=(1)/(2)m omega^(2)r^(2)+(1)/(2) I omega^(2)""( :. "For pure rolling" v=omegar)`
Now, `(K_(r))/(K_(T))=((1)/(2)I omega^(2))/((1)/(2) omega^(2)(mr^(2)+I))=(I)/(mr^(2)+I)`
As `I=(2)/(5)mr^(2)`(for solid sphere)
`RARR(K_(r))/(K_(T))=((2)/(5)mr^(2))/(mr^(2)+(2)/(5)mr^(2))=((2)/(5)mr^(2))/((7)/(5)mr^(2))=(2)/(5)XX(5)/(7)=(2)/(7)`.
13.

When the axis of rotation passes through its centre of gravity, then the moment of inertia of a rigid body is .......

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REDUCED, to its MINIMUM VALUE
zero
ncreased to its MAXIMUM value
infinity

SOLUTION :reduced, to its minimum value
14.

A satellite is orbiting the earth lose to its surface. Aparticle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is v_(e). Its speed with respect to the respect to the satellite

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will be LESS than `v_(E)`
will be more than `v_(e)`
will be EQUAL to `v_(e)`
will DEPEND on direction of projection

Answer :A
15.

A vertical glass capillary tube, open at both ends, contains some water. Which of the following shapes may be taken by the water in the tube?

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ANSWER :D
16.

Air expands from 5 litres to 10 litres at 2 atm pressure. External workdone is

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10J
1000J
3000 J
300 J

Answer :B
17.

Two wires of different materials, each 2m long and of diameter 2mm, are joined in series to form a composite wire. What force will produce a total extension of 0.9mm. (Y_(1) - 2 xx 10^(11) "pa" & Y_(2) = 6 xx 10^(11) "Pa" )

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`282.6` N
`212` N
319.8 N
382.6 N

Answer :B
18.

Assertion : A projectile should have two component velocities in two mutually perpendicular directions . Reason : A body is said to be projectile if it has motion in two dimensions.

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ANSWER :C
19.

A closed organ pipe has length l. The air in it is vibrating in 3rd overtone with a maximum amplitude of A. Find the amplitude at a distance of l//4 from closed end of the pipe

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A
zero
`A//SQRT(2)`
`sqrt(3)A//2`

ANSWER :C
20.

A solid sphere rolls down without slipping from rest on a 30^(0) incline. Its linear acceleration is

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`5g//7`
`5g//14`
`2g//3`
`g//3`

ANSWER :B
21.

A balloon starts from rest, moves vertically upwards with an acceleration g//8ms^(-2). A stone falls from the balloon after 8s from the start. Find the time taken by the stone to reach the ground (g=9.8ms^(-2))

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Solution :Step-1 : To FIND the distance of the stone above the ground about which it begins to fall from the balloon.
`S=ut+(1)/(2)at^(2)," here, "s=h,u=0,a=g//8`
`h=(1)/(2)((g)/(8))8^(2)=4g`
Step-2 : The velocity of the balloon at this height can be obtained from `v=u+at`
`V=0+((g)/(8))8=g`
This becomes the initial velocity `(u^(1))` of the stone as the stone falls from the balloon at the height h.
`therefore u^(1)=g`
Step-3 : For the total motion of the stone `h=(1)/(2)g t^(2)-u^(1)t`
Here, `h=4g,u^(1)=g,t` = time of TRAVEL of stone.
`therefore -4g=g t-(1)/(2)g t^(2)`
`therefore t^(2)-2t-8=0`
solving for .t. we GET `t=4 and -2s`.
Ignoring negative value of time, `t=4s`.
22.

Locate ,the centre of mass of the uniform then L-shaped body as shown in Fig. Assume its thickness to be t and density rho.

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SOLUTION :With respect to ORIGIN the C.M. of the rectangle A is `x_(A)=1.0 m`
`y_(A) = 0.10` m and that of the rectangle B is `x_(B) = (1.80 + 0.10) = 1.90 m`,
`y_(B)=-0.74 m, m_(A) =2 XX 0.2 xx trho = 0.4 rho tm^(2), m_(B) = 1.48 xx 0.2 xx rhot = -0.26 rho tm^(2)`
`x_( cm) = m_(A)x_(A) + m_(B)x_(B)//m_(A) + m_(B) = 0.4 rho t xx 1 + 0.26 rhoxx 1.9//0.4 rhot+ 0.26 rhot = 1.35 m`
23.

Suppose that on early morning when the themperature is 100^(@) C, a driver of an automobile gets his gasoline tank which is made of steel, filled with 75 litre of gasoline, which is also at 10^(@) C. During the day, the temperature rises to 30^(@) C. How much gasoline will overflow? Given alpha for steel = 1.2 xx 10^(-5)//""^(@) C and gamma for gasoline = 9.5 xx 10^(-4)//""^(@) C.

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ANSWER :1.37 LITRE
24.

From a uniform disc of radius R, a circular disc of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of original disc. Locate the centre of gravity of the resultant flat body.

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SOLUTION :Center of MASS of resulting portion lies at R/6 from the CENTRE of the ORIGINAL DISC in a direction opposite to the centre of the cut out portion
25.

A body is projected with a velocity 50 "ms"^(-1). Distance travelled in 6th second is [g=10 ms^(-2)]

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5m
10m
15m
20m

Answer :A
26.

Which of the following quantities are always negative

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`BAR F. BARA`
`barV.barr`
`bara.barr`
`barF.barV`

ANSWER :C
27.

Calculate pressure inside a small air bubble of radius 10^(-4)m radius found just below the surface of water (given S.T. of water =0.072 Nm^(-1) and" 1 atm "=1.013 xx 10^(5)Pa).

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Solution :Given, `R=10^(-4)m""T=0.072Nm^(-1)""P_(o)=1.013 XX 10^(5)Pa`
Excess of pressure inside an air BUBBLE `P-P_(o)=(2T)/(R)`
i.e, `P=P_(o)+(2T)/(R)=(1.013 xx 10^(5))+(2 xx 0.072)/(10^(-4))=(1.013+0.0144)xx 10^(5)`
`P=1.0274 xx 10^(5) NM^(-2) or Pa`.
28.

The root mean square velocity of molecules of a gas whose density is 1.4 kg m^(-3) at a pressure of 76 cm of mercury (specific gravity of mercury=13.6 and g=9.81 ms^(-2) ) is

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`4.66 xx 10^(2) "MS"^(-1)`
`2.5 xx 10^(2) "ms"^(-1)`
`2.33 xx 10^(2) "ms"^(-1)`
`6.66 xx 10^(2) "ms"^(-1)`

Answer :A
29.

Two identical hollow sphere roll two inclined planes of same height but of different angles of inclination. Then they reach the bottom

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with same speed and in same time
with DIFFERENT SPEEDS and in different times
with same speed but in different times
with different speeds in same times

Answer :C
30.

A copper bar 2n long has a circular cross section of radius 1 cm. One end is kept at 100^(0)C and the other at 0^(0)C. It is assumed that negligible heat is lost through the surface. (a) The thermal resistance of the bar is (b) The temperature gradient is (c) The temperature at 25 cm from the hot end will be (K = 400 SI units)

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159 K/w, 50 K/m, `8.75^(0)C`
15.9 K/w, 50 K/m, `87.5^(0)C`
1.59 K/w, 500 K/m, `8.75^(0)C`
159 K/w, 5 K/m, `8.75^(0)C`

ANSWER :B
31.

A solid sphere of radius 2.45m is rotating with an angular speed of 10 rad//s. When this rotating sphere is placed on a rough horizontal surface then after sometime it starts pure rolling. Find the linear speed of the sphere after it starts pure rolling.

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Solution :m=20 kg, angular speed = `omega = 100 s^(-1), r = 0.25` m
M.I. of the CYLINDER about its own axis `= I = 1/2 mr^(2)`
ROTATIONAL K.E. =`1/2 xx 1/2 mr^(2) xx omega^(2)`
`=1/4 xx 20 xx (0.25)^(2) xx (100)^(2) = 3,125` J
Angular momentum `L = I omega`
`=1/2 mr^(2) xx omega = 1/2 xx 20 xx (0.25)^(2) xx 100 = 62.5` Js
32.

A particle of mass 200g is attached to an ideal string of length 1.30 m whose upper end is fixed to ceiling. The particle is made to revolve in a horizontal circle of radius 50 cm. The tension in the string is (g = 10 m/s^2)

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Solution :Moment of INERTIA of the particle `L = NR^(2)`
Given `m = 200 = 0.200 KG, r=2 cm = 0.20 m`
`l=mr^(2) =0.200 xx 0.20 = 0.04 kg m^(2)`
33.

The length of a vector is _______

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ALWAYS a NEGATIVE quantity
always a POSITIVE quantity
either positive or negative
None of these

Answer :B
34.

Let vec(A) = (hat(i) + hat(j)) and, vec(B) = (2hat(i) - hat(j)). The magnitude of a coplanar vector vec(C) such that vec(A)vec(C) = vec(B)vec(C) = vec(A)vec(B), is given by :

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`sqrt((10)/(9))`
`sqrt((5)/(9))`
`sqrt((20)/(9))`
`sqrt((9)/(12))`

Answer :B
35.

A body is thrown vertically upward from a point 'A' 125 m above the ground. It goes up to a maximum height of 250 m above the ground and passes through 'A' on its downward journey. The velocity of the body when it is at a height of 70 m above the ground is

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`50 MS^(-1)`
`60ms^(-1)`
`80 ms^(-1)`
`20MS^(-1)`

Answer :B
36.

A body 'x' with a momentum 'p' collides with another identical stationary body 'y' one dimensionally. During the collision 'y' gives an impulse 'J' to the body 'x'. Then the coefficient of restitution is

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<P>`(2J)/(p)-1`
`(J)/(p)+1`
`(J)/(p)-1`
`(J)/(2P)-1`

Answer :A
37.

By sucking through a straw a student can reduce the pressure in his lungs to 750 mm of Hg (density =13.6g//cm^(3)).Using the straw , he can drink water from a glass upto a maximum depth of ……

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10 cm
75 cm
13.6 cm
1.36 cm

Solution :Pressure difference between lungs and ATMOSPHERE ,
`=760mm-750mm=10mm=1cm` of Hg
Pressure difference `=1xx13.6xxg=hxx1xxgthereforeh=13.6`cmthat is ONE can DRAW from a depth of 13.6 cm of water.
38.

Two pesons P and Q crosses the river starting from point A on side to excatly opposite point B on the other bank of the river. The person P crosses the river in the shortest path. The persons Q crosses the river in shortest time and walks back to point B. Velocity of river is 3 kmph and speed of each boat is 5kmph w.r.t river. If the two persons reach the point B in the same time, then the speed of walk of Q is

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Solution :
`t_(p) = (d)/(sqrt(V_(B)^(2)-V_(W)^(2)))=(d)/(sqrt(5^(2)-3^(2)))=(d)/(4)`
`t_(Q)=(d)/(V_(B))=(d)/(5), t_(P)=t_(Q)+Deltat`
`(d)/(4)=(d)/(5)+(X)/(V_(man)), "" But x=V_(W)(d)/(V_(B))`
`(d)/(4)=(d)/(5)+(V_(W)d)/(V_(B)V_(man)),""(canceld)/(4)=(canceld)/(5)+(3canceld)/((5)V_(man))`
`(1)/(4)-(1)/(5)=(3)/(5V_(man)),""(1)/(20)=(3)/(5V_(man))`
`V_(man) = ((3)(20))/(5)=12kmph`
39.

A bullet fired into a fixed targe loses half of its velocity in penetrating 15 cm. How much further it will penetrate before coming to rest?

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5 cm
15 cm
7.5 cm
10 cm

Answer :A
40.

(A): Air quickly leaking out of a baloon becomes cool (R ): The leaking air undergoes adiabatic expansion.

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

Answer :A
41.

Arrange the following liquids in increasing order of surface tension .Water ,Mercury ,Soap solution.

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SOLUTION :SOAP solution ,WATER ,MERCURY .
42.

How much will the length of a simple pendulum be if its time period is 3s?

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SOLUTION :`L= 2.24 m = 224 CM`
43.

Equation of a projectile is given by y = Px - Qx^(2) where P, Q are constants, the ratio of maximum height to Range of the projectile is

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`(P^(2))/(Q)`
4P
`(P)/(4)`
`(Q^(2))/(2P)`

ANSWER :C
44.

The velocity acquired by a body in falling to the surface of earth from a point at a height equal to the radius of the earth is approximately:

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`4 KMS^(-1)`
`8 kms^(-1)`
`4sqrt(2) kms^(-1)`
`8sqrt(2) kms^(-1)`

ANSWER :B
45.

Centre of mass of a system is a unique point which is very helpful in solving the problems related to the motion of rigid bodies. In columns I and II, some statements regarding centre of mass are given. Match the Columns :{:(,"Column - I",,"Column - II"),((A),"In the absence of enternal forces, internal forces",(P),"Inside the material of the body."),((B),"Centre of mass of a body (can be)",(Q),"May affect the motion of centre of mass"),((C ),"C.M. of a ring",(R ),"Does not affect the motion of centre of mass"),((D),"Centre of mass of solid cylinder is",(S),"Outside the material of the body"):}

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ANSWER :A-R; B-P, S; C-S; D-P
46.

The gravitational potential energy of the Moon with respect to Earth is

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ALWAYS POSITIVE
always negative
can be positive or negative
always zero

Answer :B
47.

A body stetches a spring by a certain amount at the equator, Assuming earth to be a perfect sphere of radius R, g the acceleration due to gravity on the surface of the earth and omega, the angular speed of the earth, the height at which it will stretch the spring by the same amount at the poles is

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R
`R[1-SQRT(1-(OMEGA^(2)R//g))]`
`(R)/(sqrt(1-(omega^(2)R//g)))`
`[(1)/(sqrt(1-(omega^(2)R//g)))-1]`

ANSWER :D
48.

In oblique projection, maximum height attained by the projectile is

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`(t)/(ucostheta)`
`(usintheta)/(2G)`
`(2g)/(usintheta)`
`(u^(2)SIN^(2)theta)/(2g)`

ANSWER :A::B
49.

What will be acceleration due to gravity at the earth's surface of latitude 30^(@) ?

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ANSWER :`9.77 m//s^(2)`
50.

In he process PV^2= constant, if temperature of gas is increased

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change in INTERNAL energy of gas is positive
WORK DONE by gas is positive
heat is given to the gas
heat is taken out from the gas

ANSWER :A::C