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.

Explain streamlines and streamline flow.

Answer»

Solution :The path of motion of a FLUID particle is called a line of flow.
The path TAKEN by a fluid particle under a steady flow is a streamline.
Streamline : The curve whose tangent at any point is in the direction of the fluid VELOCITY at that point.

(a)A typical trajectory of a fluid particle.
Path of particle is shown in figure (a).It indicatesstreamline .The velocity of particle at a point is in the direction of tangent drawn at that point .
the flow for which such streamlines can be defined is called streamline flow.
In unsteady flow , flow of line can be defined but not defined streamline.
Discussion
2.

One end of a rod is maintained at 50^(0)C and the other end is kept in ice at 0^(0)C. Rate of fusion of ice is doubled. When

Answer»

Temperature of hot end is MADE `200^(0)C` and area of cross-section is doubled
Temperature of hot end is made `100^(0)C` and LENGTH of rod is made four TIMES
area of cross-section is reduced to half ad length is doubled
temperature is made `100^(0)C` and area of cross-section and length both are doubled

Answer :D
Discussion
3.

A player throws a ball upwards with an initial speed of 29.4 ms^(-1). (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 ms^(-2) and neglect air resistance).

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Solution :(a) Vertically downwards, (b) zero velocity, acceleration of `9.8 MS^(-2)` downwards, (c ) `X gt 0` (upward and DOWNWARD motion), `upsilon lt 0` (upward), `upsilon gt 0` (downward), `a gt 0` throughout, (d) 44.1 m, 6 s.
Discussion
4.

Specific heat capacity at constant pressure and at constant volume for nitrogen are respectively 1040 Jkg ^(-1) K ^(-1) and 743 Kg ^(-1) K ^(-1).Calculate the universal gas constant ?[E.Q.]

Answer»

Solution :Specidic heat capacity at CONSTANT pressure `= C_(p) = 1040 Jkg ^(-1) K ^(-1)`
Specific heat capacity at constatn volume `= C_(V) = 743 Jkg ^(-1) K ^(-1)`
`C _(p) - C_(v)= r`
SINCE the mass of gas is 1 kg, r is used
`r = C _(p) - C _(v) = 1040 - 743 = 297 J kg ^(-1) K ^(-1)`
Molar mass of `N _(2) = 28 gm = 28 xx 10 ^(-3) kg`
`R = rM = 297 xx 28 xx 10 ^(-3) = 8.316 J MOL ^(-1) K ^(-)`
Discussion
5.

Establish Newton's third law of motion from the principle of conservation of linear momentum. A bullet is fired from a gun. Which one will possess greater momentum-gun or bullet?

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Solution :Before firing the bullet, both the gun and the bullet were at rest. Hence, INITIALLY total linear MOMENTUM was zero. After firing the bullet, the gun will have an EQUAL but OPPOSITE momentum to that of the bullet so as to conserve the total linear momentum of the SYSTEM.
Discussion
6.

A solid cylinder of mass 50kg and radius 0.5m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s^(-2) is

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`50N`
`78.5N`
`157N`
`25N`

ANSWER :C
Discussion
7.

In the above problem, when the man has walked once around the platform, so that he is at his original position on it, what is his angular displacement relative to ground?

Answer»

`(6)/(5)pi`
`(5)/(6)pi`
`(4)/(5)pi`
`(5)/(4)pi`

ANSWER :B
Discussion
8.

If theta is the angle between unit vectors vec(A) and vec(B), " then " ((1-vec(A).vec(B)))/((1+vec(A).vec(B))) is equal to

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`tan^(2) (theta//2)`
`SIN^(2) (theta//2)`
`COT^(2) (theta//2)`
`cos^(2) (theta//2)`

Answer :A
Discussion
9.

A block of mass 'm' moving with a speed 'u' compresses a spring through a distance 'x' before its speed reduces by 25%. The spring constant of the spring is

Answer»

`(mv^2)/(4x^2)`
`(3mv^2)/(4x^2)`
`(7mv^2)/(16x^2)`
`(9mv^2)/(16x^2)`

ANSWER :C
Discussion
10.

A body of mass 2kg is resting on a rough horizontal surface. A force of 20N is now applied to it for 10s, parallel to the surface. If the coefficient of kinetic friction between the surfaces of contact is 0.2. Calculate change in kinetic energy of the object in 10s. (Take g=10ms^(-2)).

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Solution :Mass = 2kg
Initial velocity
`u=0`
Force `F=2N`
Coefficient of kinetic friction `mu_(k)=0.2`
Time `t=10s`
`f_(k)=mu_(k)mg`
`=0.2xx2xx10=4N`
NET force `F'=f-f_(k)`
`=20-4=16N`
Acceleration,
`a=(F')/(m)=(16)/(2)=8MS^(-1)`
Distance covered
`s=ut+(1)/(2)at^(2)`
`s=0+(1)/(2)xx8xx(10)^(2)=400m`
Change in kinetic energy
= WORK DONE by the net force
`W_(2)=F's`
`=16xx400=6400J`
Discussion
11.

A rod AB of length 1m is placed at the edge for a smooth table as shown in figure. It is hit horizontally at point B. If the displacement of centre of mass m in 1s is 5sqrt(2)m. The angular velocity of the rod is (g=10m//s^(2))

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`30 rad//s`
`20 rad//s`
`10 rad//s`
`5 rad//s`

ANSWER :A
Discussion
12.

A square plate of side / has mass M. What is its moment of inertia about one of its diagonals ?

Answer»

`(Ml^2)/(6)`
`(Ml^2)/(12)`
`(Ml^2)/(3)`
`(Ml^2)/(4)`

Solution :Moment of inertia of a SQUARE about an axis through its centre and perpendicular to its PLANE
`I_(O) = (M)/(12) (l^(2) + l^(2)) = (Ml^(2))/(6)`

ACCORDING to theorem of perpendicular axes , we get
`I_(1) + I_(2) = I_(O) = (Ml^(2))/(6)`
But `I_1 = I_2` (By SYMMETRY)
`therefore 2I_(1) = (Ml^(2))/(6) or I_(1) = (MI^(2))/(12)`
Discussion
13.

A uniform straight rod mass 50 kg and length 4 m is to be supported on twosupports equidistant from the respective ends. What can be the maximum distance of the supports so that a 30 kg counterpoising weight can be kept in equilibrium at any point on the rod?

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ANSWER :1.25 m
Discussion
14.

A bird weighs 2 kg and is inside an airtight cage of 1 kg. If its starts to fly, then what is the weight of the bird and cage assembly?

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3 kg
2 kg
1 kg
NONE of these

Solution :When the bird flies, the upthrust on it is equal and opposite to the down thrust on the CAGE. THEREFORE, the weight of the ASSEMBLY remains unchanged.
Discussion
15.

The amplitude of a damped oscillator becomes 1/(27) th of its initial value after 6 minutes. What the amplitude after 2 minutes ?

Answer»

Solution :For damped OSCILLATOR `A= A_0 e^(-bt)`
Here `(A_0)/(27) = A_0 e^(-6B)` That is `e^(-6b)` = 1/27
`A_2 = A_0 e^(-2b) = A_0 [e^(-6b)]^(1//3) = A_0 ((1)/(27))^(1/3) =A_0//3`
Discussion
16.

The cylindrical tube of a spray pump hasa cross section of 8.0 cm^2 one end of which has 40 find holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m. min^-1. What is the speed of ejection of the liquid through the holes?

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SOLUTION :`0.64 m. s^-1`
Discussion
17.

If h is the height or depth (sagitta) of a spherical surfacce and I is the mean distance between the legs of sphermoeter, then radius of curvature R of the surface is

Answer»

`R =(I^2)/(H) + (h)/(2)`
`R = (I^2)/(6H) + (h)/(2)`
`R =(I^2)/(6h) - (h)/(2)`
`R = (h^2)/(6I) + (I)/(2)`

Solution :The correct relation for R is `R = (I^2)/(6h) + (h)/(2)`
Discussion
18.

A stone is thrown upward from a moving train. Name the path followed by the stone.

Answer»

SOLUTION :PARABOLA
Discussion
19.

If the system if Figure 7.13 is released from rest in the configuration shown by solid lines, find the maximum distance h through which the weight will fall. (Neglect friction and masses of pulleys) What will happen if P = 2Q ? [Hint : Apply work-energy theorem to the compound system consisting of the three masses and ceiling. To find intantaneous velocities of the masses apply the condition that the length of the string remains constant. Also remember that gravitational pulls are the lone forces external to the system, the tension of the string and pull the internal force ceiling are of the system]

Answer»


Answer :`h=4PQl//(4Q^(2)-P^(2))`. The system will ACQUIRE TERMINAL velocity `= sqrt(gl)`
Discussion
20.

The length of a second's pendulum on the surface of earth is 1m. What will be the length of a second's pendulum on the moon?

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Solution :The time PERIOD of a simple pendulum `T= 2pi sqrt((L)/(g))`
`therefore T propto sqrt((l)/(g))""2pi` is constant
`therefore (T_m)/(T_e)= sqrt((l_m)/(g_m)xx (g_e)/(l_e))`
where `T_(e ), T_(m)` are the periodic time on earth and moon
Here, `T_(e )= T_(m) = 2s`
`l_(m), l_(e )` are the length of second.s pendulum on the surface of moon and earth respectively.
Now, `g_(m), g_(e )` are the ACCELERATION due to gravity on moon and earth respectively.
`therefore (2)/(2) = sqrt((g_e)/(g_m)xx(l_m)/(l_e))`
by squaring, `1= (g_e)/(g_m)xx(l_m)/(l_e)`
but `g_(m)= (g_e)/(6)" and "l_(e )= 1M`
`therefore 1= (g_e)/((g_e)/(6))xx (l_m)/(1)`
`therefore 1= 6xx l_(m)`
`therefore l_(m)= (1)/(6)m`.
Discussion
21.

Four simple harmonic vibrations x_(1) = 8 sin omegat, x_(2)=6 sin(omegat + pi//2), x_(3) = 4sin(omegat + pi)and x_(4) =2 sin(omegat + 3pi//2)are superimposed on each other. The resulting amplitude is

Answer»

20
`8sqrt(2)`
`4sqrt(2)`
4

Answer :C
Discussion
22.

Impulse is

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a force
a scalar
equal to RATE of CHANGE of MOMENTUM of a body
equal to change in the momentum of a body

Answer :D
Discussion
23.

The volume of an air bubble is doubled as it rises from the bottom of lake to its surface, The atmospheric pressure is 75 cm of mercury. The ratio of density of mercury to that of lake water is (40)/(30), the depth of the lake in metre is

Answer»

15
10
30
20

Answer :B
Discussion
24.

Statement I : Friction opposes motion of a body. Statement II : Static friction is self adjusting while kinetic friction is constant.

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Statement I is TRUE, statement II is true, statement II is a CORRECT explanation for statement I.
Statement I is true, statement II is true, statement II is not a correct explanation for statement I.
Statement I is true, statement II is FALSE.
Statement I is false, statement II is true.

Answer :B
Discussion
25.

A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass and of coefficient of restitution (e). The ratio of velocities of the two spheres, after collision, will be

Answer»

`e/((e + 1))`
`((1 - e)/(1 + e))`
`1/e`
`((e + 1)/(e ))`

Solution :Here, `m_1 = m_ = m, u_1 = u , u_2 = 0`
LET `v_1, v_2` be their velocities after collision.
According to PRINCIPLE of conservation of linear momentum
`m u + 0 = m (v_1 + v_2) " or " v_1 + v_2 = u ""….(i)`
by defination , `e = (v_2 - v_1)/(u - 0) " or " v_2 - v_1 = EU "".....(ii)`
Adding (i) and (ii) , we get, `v_2 = (u(1 + e))/(2)`
SUBTRACTING (ii) from (i), we get, `v_1 = ((1 - e)u)/(2) :. (v_1)/(v_2) = (1 - e)/(1 + e)`.
Discussion
26.

A ray of light travelling in a rarer medium strikes a plane boundary between the rarer medium and a denser medium at an angle of incidence I such that the reflected and the refracted rays are mutually perpendicular. Another ray of light of same frequency is incident on the same boundary from the side of denser medium. Find the maximum angle of incidence at the denser-rarer boundary so that the second ray is totally reflected.

Answer»

Solution :Figure shows incidence of a ray of the rarer-denser boundary such that reflected and refracted rays are mutually PERPENDICULAR i.e. `r+90^@+r.=180^@` or
`r.=90^@-r=90^@-i[r=i`,law of reflection]
Applying snell.s law at the boundary,
`therefore mu_R= SIN i=mu_0 sin r`
`mu_R sin i=mu_0 sin (90^@-i) =mu_0 cos i or mu_D/mu_R=tan i`...(1)
`sin theta_c=1/(Rmu_0)=1/(mu_D//mu_R)=mu_R/mu_D`.........(2)
using equation (1) `sin theta_c=1/tan i =c ot i IMPLIES theta_e= sin^-1 (cot i)`
Discussion
27.

Moment of inertia of a rigid body depends on (A)Mass of body(B) Position of axis of rotation (C ) Angular velocity of the body (D) Time period of its rotation

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A and B are CORRECT
B and C are correct
A and C are correct
C and D are correct

ANSWER :A
Discussion
28.

One litre of an ideal gas at a pressure of 6 atm undergoes an adiabatic expansion until its pressure drops to one atmosphere and volume increases to 2 litre. Find the word done during the process. [ gamma = 1.4].

Answer»

Solution :`V_(1) = 1` LITRE `= 10 6(-3) = m ^(3), P _(1) = 6 atm = 6 xx 1.013 xx 10 ^(5) Pa`
`V_(2) = 2 ` litre `= 2 xx 10 ^(-3) m^(3), P _(1) = 1atm =1 xx 1. 013 xx 10 ^(5) Pa`
Work DONE `= (1)?((gamma -1)) (P_(1) V _(1) - P _(2) V _(2))`
`= (1)/(( 1.4 -1)) [6 xx 1. 013 xx10 ^(5) xx 10 ^(-3) - 1. 013 xx 10 ^(5) xx 2xx 10 ^(-3) ] = 1013 J`
Discussion
29.

In figures (a), (b) and (c) shown, the objects A, B and C are of same mass. String, spring and pulley are massless. C strikes B with velocity u in each case and sticks it. The ratio of velocity of B in case (a) to (b) to (c) is

Answer»

(a) `1:1:1`
(B) `3:3:2`
(C) `3:2:2:`
(d) `1:2:3`

SOLUTION :`v_a=v_b=v/2`
`v_c=v/3`
`:. v_a:v_b:v_c=3:3:2`
Discussion
30.

A movable plank of mass m is placed on a horizontal surface. A solid ball of mass m and radius R is released from the incline as shown in the figure. Friction on the plank is sufficient to prevent slipping between the plank and the ball while there is no friction between the plank and the ground. The ball will leave the incline horizontally with velocity sqrt((x)/(9)"gh") where 'x' is

Answer»


ANSWER :5
Discussion
31.

A 60 kg man is inside a lift which is moving up with an acceleration of 2.45 ms^(-2). The apparent percentage change in his weight is,

Answer»

`20%`
`25%`
`50%`
`75%`

Answer :B
Discussion
32.

Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy. (J=4200 J/kcal)

Answer»

`1.302xx10^(5)J`
`2.302xx10^(5)J`
`3.302xx10^(5)J`
`4.302xx10^(5)J`

ANSWER :A
Discussion
33.

Assertion An actual gas behaves as an ideal gas most closely at low pressure and high temperature. Reason At low pressure and high temperature, real gases obey the gasl laws.

Answer»

If both Assertion and REASON are correct and Reason is the correct explanation of Assertion.
If both Assertion and Reason are correct but Reason is not the correct EXPLATION os Assertion.
If Assertionis TRUE bur Reason is false.
If Assertion is false but Reason is true.

Solution :At the given CONDITIONS, the INTERMOLECULAR force between the gas moleculses is almost zero.
Discussion
34.

A body took't' sec. to come down from top of tower. Find the time taken to cover half the height of tower

Answer»

Solution :Let AB be the TOWER of HEIGHT .h. m and the body taken .t. SECONDS to come down from A to B as shown in the figure .
Let C be the midpoint and hence `AC=h/2m`

APPLYING the above formula for both the cases, we get
`h=1/2"gt"^(2)`...(1)` , h/2=1/2"gt"^(2)`...(2)
Eliminating the unconcerned terms by DIVIDING equation (1)by (2), we get
`(1)/(2)=h/(h//2)=(1/2"gt"^(2))/(1/2"gt"^(2))rArr 2=(t/t.)^(2)rArr t^(1^(2))=t^(2)/2rArrt^(1)=t/sqrt2s`
Discussion
35.

A disc is rolling the velocity of its centre of mass is V_(CM). Which one will be correct?

Answer»

The velocity of highest point is `2V_(CM)` and point of CONTACT is zero.
The velocity of highest point in `V_(CM)` and point of contact is `V_(CM)`.
The velocity of highest point is `2V_(CM)` and point of contact is `V_(CM)`.
The velocity of highest point is `2V_(CM)` and point of contact is `2V_(CM)`

ANSWER :A
Discussion
36.

From a sphere of radius 1m, a sphere of radius 0.5m is removed from the edge. The shift in the C.M. is

Answer»

`13//16m`
`16//13m`
`14//13m`
`1//14m`

ANSWER :D
Discussion
37.

A mercury thermometer reads 99.25^(@)C when immersed in a steam bath in such a way that the entire of its stem above is 0^(@)C graduation is outside the bath. The temperature of the room is 20^(@)C. When the whole of its steam is pushed into the bath, the thermometer reads 100.5^(@)C. Calculate the coefficient of cubical expansivity of mercury. (Given theat linear expansivity of glass =8xx10^(-6)K^(-1))

Answer»


ANSWER :`18xx10^(-5)K^(-1)`
Discussion
38.

Coefficient of volume expansion of solids,liquids and gases are respectively gamma_s.gamma_l and gamma_g. Usually

Answer»

`gamma_sltgamma_lltgamma_g`
`gamma_sgtgamma_lgtgamma_g`
`gamma_lltgamma_sltgamma_g`
`gamma_lgtgamma_sgtgamma_g`

ANSWER :A
Discussion
39.

Pressure head in Bernoulli's equation is

Answer»

<P>`(PRHO)/(G)`
`(P)/(rhog)`
`rhog`
`Prhog`

ANSWER :B
Discussion
40.

The velocity of projectile at the point of projection is 2hat(i) + 3hat(j) m/s. The velocity at a point where the projectile reaches is

Answer»

`-2HAT(i) + 3HAT(J)` m/s
`2hat(i) - 3hat(j)` m/s
`2hat(i) + 3hat(j)` m/s
`-(2hat(i) + 3hat(j))` m/s

Answer :B
Discussion
41.

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

Answer»

225 J
200 J
400 J. 
175 J 

Solution :
Work done W = AREA under F - S graph
= area of trapezium ABCD = area of trapezium CEFD
`=1/2 xx (10 + 15) xx 10 + 1/2 xx (10+ 20) xx 5 = 125 + 75= 200 J`
Discussion
42.

A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall ?

Answer»

Kinetic ENERGY
Potential energy
TOTAL mechanical energy
Total linear momentum

Answer :C
Discussion
43.

Assertion: Simple harmonic motion is the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs. Reason: Simple harmonic motion is a uniform motion.

Answer»

If both assertion and RESON are true and REASON is the correct EXPLANATION of assertion
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.

Solution :SIMPLE harmonic motion is expressed by `v=omegasqrt(a^(2)-y^(2))`. With CHANGE in displacement y, velocity v changes. Thus simple harmonic motion is not a uniform motion.
Discussion
44.

(A): A steel wire of natural length .I. breaks, when elongated by length .x.. If original length is halved and elongated by x//2 it breaks. (R) : Breaking force is directly proportional to the length of the wire

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
A. is false but .R. is true

Answer :3
Discussion
45.

A circular thin disc of mass 2 kg has a diameter 0.2 m. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (in kg m^(2))

Answer»

0.01
0.03
0.02
3

Solution :`I = (MR^(2))/(2)`
About axis passing through EDGE , moment of INERTIA = I.
`therefore I. = I + MR^(2) ""` (By theorem of PARALLEL axes)
or `I. = (MR^2)/(2) + MR^(2) = (3)/(2) MR^(2) = (3)/(2) xx 2 xx (0.1)^(2)`
or `I. = 0.03 kg m^(2)`
Discussion
46.

At a given temperature the ratio of r.m.s velocities of hydrogen molecule and helium atom will be

Answer»

`SQRT2:1`
`1:sqrt2`
`1:2`
`2:1`

ANSWER :A
Discussion
47.

(A) : With increase in temperature, the surface tension of liquid decreases. (R ) : The angle of contact of a liquid decrease with increase in temperature.

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 :B
Discussion
48.

In the previous problem if q is placedat the centre of one face find total flux through the box. If q is placed at one corner of the cube, find total flux through the box and flux through the faces.

Answer»

Solution :`q/(2in_0),q/(8in_0)`,no flux is linked with the three faces at the JUNCTION of which q is KEPT, flux emerges EQUALLY through the other three front faces of the cube and EQUAL to `q/(24in_0)`
Discussion
49.

At what temperature the rms velocity of oxygen will become half that of Hydrogen NTP?

Answer»

SOLUTION :`819^0` C
Discussion
50.

A mild steel wire of length 1.0 m and cross sectional area 0.50 times 10^-2 (cm)^2 is stretched , well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid -point.

Answer»


ANSWER :0.01 m
Discussion