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.

Figure8-5a shows two industrial spies sliding an initially stationary 225 kg floor safe a displacement vec(d) of magnitude 8.50 m. The push vec(F)_(1) of spy001 is 12.0 N at an angle of 30.0^(@) downward from the horizontal, the pull vec(F)_(2) of spy 002 is 10.0 N at 40.0^(@) above the horizontal. The magnitudes and directions of these forces do not change as the safe moves, and the floor and safe make frictionless contact. The safe is initially stationary. What is its speed v_(f) at the end of the 8.50 m displacement?

Answer»

Solution :KEY IDEA
The speed of the safe changes because its kinetic energy is changed when energy is transferred to it by `VEC(F)_(1)` and `vec(F)_(2)`.
CALCULATIONS: We relate the speed to the work done by combining Eqs. 8-10 ( the work-kinetic energy theorem) and 8-1 (the definition of kinetic energy):
`W=K_(f)-v_(i)=1/2 mv_(f)^(2) - 1/2 mv_(i)^(2)`.
The initial speed `v_(i)` is zero, and we now know that the work done is 153.4 J. Solving for `v_(f)` and then substituting known data, we find that
`v_(f)=sqrt((2W)/(m))=sqrt((2(153.4J))/(225g))`
`=1.17m//s`.
Discussion
2.

A gas undergoes following process :- AbtoCompressed to half of the initial volume (PpropV) BctoIsothermal expansion to intial volume CDtoAdiabatic process such that P_(D)=P_(A) Select incorrect statement

Answer»

Net heat is RELEASED
`DeltaU_(AB)LT0`
`DeltaU_(CD)gt0`
`DeltaU_(AC)gt0`

Answer :D
Discussion
3.

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed ?

Answer»

The velocity vector is tangent to the circle
The acceleration vector is tangent to the circle
The acceleration vector points to the centre of the circle
The velocity and acceleration vectors are perpendicular to each other

Solution :As the PARTICLE is MOVING with constant angular velocity the tangential COMPONENT of acceleration is ZERO.
Therefore only radial component `VEC(a_(c))` is effective.
Discussion
4.

One end of a uniform wire of length L and weight W is attached rigidly to a point in the roof and a weight W_(1) is suspended from its lower end. If S is the area of cross section of the wire, the stress in the wire at a height (3L/4) from its lower end is

Answer»

`W_(1)/S`
`([W_(1)+(W//4)])/S`
`([W_(1)+(3W//4)])/S`
`(W_(1)+W)/S`

ANSWER :C
Discussion
5.

In Young's double-slit experiment using monochromatic light of wavelength lambda, the intensity of light at a point on the screen where path difference is lambda, is K units. What is the intensity of lgight at a point where path difference is lambda/3.

Answer»

`(K)/(4)`
`(K)/(3)`
`(K)/(2)`
K

Answer :A
Discussion
6.

About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiationa.at a distance of 1m from the bulb b. at a distance of 10mAssume that the radiation is emitted isotropically and neglect reflection.

Answer»

SOLUTION :a.`A=4pi r^(2)=4xx3.14xx1^(2)m^(2)`
`I=("POWER")/("Area")=(100xx 5//100)/(4xx3.14xx1^(2))=0.4 WM^(-2)`
B.`I=(100xx 5//100)/(4xx3.14xx10^(2))=0.004 Wm^(-2)`
Discussion
7.

The process of changing some characteristic of a carrier wave in accordance with the intensity of the signal is called

Answer»

amplification
RECTIFICATION
MODULATION
NONE of these

Answer :C
Discussion
8.

Arrange the following em waves in descending order of wavelengths : γ ray, microwaves UV radiations.

Answer»

SOLUTION :MICROWAVE, U V RADIATION, γ-rays
Discussion
9.

A bullet is fired from a rifle with velocity of 750 m/s. If the length of the rifle barrel is 60 cm. Calculate average velocity of the bulet in the barrel

Answer»

Solution :Here U = 0, v = 750 m/s,
S =60 cm =0.6 m
Average velocity
= `(u+v)/2 = (0+750)/2 =375 m //s.`
Discussion
10.

What is the net power absorbed by each circuit over a complete cycle. Explain your answer.1. A 44mH is connected into 220V, 50Hz ac supply. Determine the rms value of the current in the circuit.2. A 60 μF capacitor is connected to a 110V,60Hz ac supply. Determine the rms value of the current in the circuit.

Answer»

Solution :No POWER LOSS in the an IDEAL inductor ideal capacitor
Discussion
11.

During the propagation of electromagnetic waves in a medium :

Answer»

Both electric and MAGNETIC energy densities are zero.
Electric energy density is double of the magnetic energy density.
Electric energy density is half of the magnetic energy density
Electric energy density is equal to the magnetic energy density.

Solution :`U_(av)=(1)/(2) epsilon_(0)E_(0)^(2)`
`=(1)/(2mu_(0))B_(0)^(2)`
i.e. Average electric energy density = Average magnetic energy density.
Discussion
12.

A bullet is fired from a rifle with velocity of 750 m/s. If the length of the rifle barrel is 60 cm. Calculate time taken by the bullet to travel in the barrel

Answer»

SOLUTION :`S=((U+V)/2)t or 0.6 = 375 t` of t = 0.0016s
Discussion
13.

In an oscillating LC circuit, L = 3.00 mH and C = 3.90 muF. At t=0 the charge on the capacitor is zero and the current is 1.75 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?

Answer»


Answer :(a)0.189 mC ; (B) 85.0 `MUS` ; (C ) 42.5 W
Discussion
14.

A bullet is fired from a rifle with velocity of 750 m/s. If the length of the rifle barrel is 60 cm. Calculate the average acceleration.

Answer»

SOLUTION :Again
`a = (v-u)/t = 750/0.0016 = 468.75 xx10^3 m/s^2`
Discussion
15.

If the path is circular, which force provides necessary centripetal force?

Answer»

SOLUTION :The LORENTZ MAGNETIC FORCE
Discussion
16.

Ampere's theorem helps to find the magnetic field in a region around a current carrying conductor:-Write the expression of Ampere’s theorem.

Answer»

SOLUTION :`phivecB*VEC(DL)= gamma_el`
Discussion
17.

A S.H.M. is given by y=5[sin(3pit)+sqrt(3)cos(3pit)]. What is the amplitude of the motion if y is in metres ?

Answer»

2 m
5 m
4 m
10 m.

Solution :`y=5 SIN(3pi t)+5sqrt(3) cos (3pi t)`.
`r=SQRT(r_(1)^(2)+r_(2)^(2)+2r_(1)r_(2)cos phi)`
`=sqrt(5^(2)+(5sqrt(3))^(2)+2xx5xx5sqrt(3)cos 90^(@))=10` m.
Hence correct choice is (d).
Discussion
18.

In a Young's double slit experiment, using mono-chromatic light of wavelength lambda, the intensity of light at a point on the screen where the path defference is lambda is k units. Find the intensity at a point where the path difference is lambda//3.

Answer»

Solution : A PATH difference of A, corresponds to a phase difference of `2PI`
`therefore` intensity , `I=4a^2` or `a^2=I/4`
A path difference of `lambda/3`corresponds to a phase difference of `(2pi)/3`
`therefore` Intensity =`4xxI/4. cos^2 (2pi)/3`
Discussion
19.

A magnet NS is suspended from a spring and while it oscillates, the magnet moves in and out of the coil C. The coil is connected to WA a galvanometer G. Then, as the magnet oscillates,

Answer»

G shows no deflection
G shows deflection to the left and right but the amplitude steadily decreases
G shows deflection to the left and right with CONSTANT amplitude
G shows deflection on one side.

Solution :As the SOUTH pole approaches the coil the current FLOWS clockwise lo cancel changes in the magnetic FLUX. Now the top of the coil is south pole. When the south pole moves up, the current flows anticlockwise lu PRODUCE a north pole in the coil to cancel the change of flux Flux varies and therefore amplitude will not be a constant.
Discussion
20.

Starting from rest a wheel rotates with uniform angular acceleration 2pirads^(-2). After 4s, if the angular acceleration ceases to act, its angular displacement in the next 4s is

Answer»

`8pirad`
`16pirad`
`24pirad`
`32pirad`

ANSWER :D
Discussion
21.

Continuation of Problem z Now assume that Eq, 6-29 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vertical crosssectional area of 0.040 m2 and has a drag coefficient C of 0.80.Take the air density to be 1.21 kg//m^(3) and the coefficient of kinetic friction to be 0.80. (a) In kilometers per hour, what wind speed Valong the ground is needed to maintain the stone's motion once it has started moving? Because winds along the ground are retarded by the ground, the wind speeds reported for storms are often measured at a height of 10 m. Assume wind speeds are 2.00 times those along the ground. (b) For your answer to (a), what wind speed would be reported for the storm? (c) Is that value reasonable for a high speed wind in a storm?

Answer»

Solution :`(a) 3.2xx10km//h, (b) 6.5xx10^(2) km//h` , (c) The RESULT is not resonable for a terresstrial strom. A category 5 huricane has SPEEDS
Discussion
22.

Why didn't the grandmother like music?

Answer»

It was the monopoly of harlots and BEGGARS and not meant for gentlefolk
She liked only religious prayers
She liked the traditional folk music
She THOUGHT it would distract him from studies

Answer :A
Discussion
23.

Resolving power of a telescope is given by ?

Answer»

SOLUTION :`1//d theta=a//1.22 LAMBDA`
Discussion
24.

What happen if a bar magnet is cut into two equal pieces transverse to it's length :

Answer»

Solution :In both the CASES, ONE will get TWO magnets. If the magnet has a POLE strength, m, length 2l and MAGNETIC moment m,then pole strength length and magnetic moment in casewill be m, l and m/2 while in case.
Discussion
25.

The radiation pressure (in N/m^2) of the visible light is of the order of :

Answer»

`10^(-2)`
`10^(-4)`
`10^(-6)`
`10^(-8)`

Answer :C
Discussion
26.

A person standing at a distance of 6 m from a source of sound receives sound wave in two ways, one directly from the source and other after reflection frohi a rigid boundary as shown in the figure. The maximum wavelength for which, the person will receive maximum sound intensity, is:

Answer»

4 m
`16/3` m
2 m
`8/3` m

ANSWER :A
Discussion
27.

Define the term current density.

Answer»

Solution :Current density at a point in a conductor is defined as the amount of current flowing per UNIT area of the conductor AROUNG that point proiveded the area
`J = (I)/(A)`
Current density is a vector quantity. Its direction is the direction of motion of positive CHARGE. The unit of current density is amper/`"meter"^(2), or [Am^(-2)]`
Relation between J and E:
`I = N A e v_(d)`
` = n Ae ((eE)/(m )tau )`
`= n (A e ^(2) rau E)/(m)`
`or I /A = ( n e ^(2) tau E)/( m )`
or `J = (1)/(rho) E`
`[ because J = I/A and rho = (m)/(n e ^(2) tau)]`
`J = sigma E [ because sigma = (1)/(rho)]`
Discussion
28.

A long solenoid has 500 turns. When a current of 2A is passed through it., the resultant magnetcis flux linked with each turn of the solenoid is 4xx10^-3Wb. What is the self inductance of the solenoid?

Answer»

SOLUTION :The total MAGNETIC flux(`phi`) LINKED with the SOLENOID is given by
`phi= 500xx4xx10^-3 = 2Wb `
But of a coil `phi = Li `
` therefore L = phi/i = 2Wb/2A = 1H`
Discussion
29.

The hysteresis cycle forhe material of a transformer core is

Answer»

SHORT and wide
tall and narrow
tall and wide
short and narrow

Solution :Transformer core is of SOFT iron MATERIAL which has small coercivity and large retentivity. Therefore its hystereisi LOOP is tall and narrow.
Discussion
30.

When a proton is accelerated through 1V, then its kinetic energy will be

Answer»

1 EV
13.6 eV
1840 eV
0.54 eV

Solution :`K = qV = e xx 1V = 1EV`
Discussion
31.

When the frequency of the AC voltage applied to a series LCR circuit is gradually increased from a low value, the impedance of the circuit

Answer»

monotonically increases
first increases and then DECREASES
first decreases and then increases
monotonically decreases

Solution :From the GRAPH, we CONCLUDED that with the increase in frequency, the current increases, attains a maximum value `(f=f_(r ))` and then decreases. It implies that the impedance first decreases, becomes minimum for maximum value of current and the increases.
Discussion
32.

Assertion: At ordinary temperatures. the Vibrational degrees of freedom do not contribute to the specific heat of gases Reason:The average charge corresponding to a degree of vibration is not kT and hence the molecular vibrations are not excited.

Answer»

both ASSERTION and REASON are both are both wrong
Assertion is RIGHT, Reason is wrong
both Assertion and Reasonare correct and Reason explains Assertion
both Assertion and Reason are correct, but Reasondoes not explain Assertion

ANSWER :C
Discussion
33.

Find the number of natural transverse vibration of a string length l in the frequency interval from omega to omega +domega if the propagation velocity of vibrations is equal to v. All vibrations are supposed to occur in one plane.

Answer»

Solution :Suppose the string is stretched along the `x` axis from `x=0 t o x=l` with the end points fixed. Suppose `y(x,t)` is the transverse DISPLACEMENT of the element at `x` at time `t`. Then `y(x,t)` obeys
`(del^(2)y)/(delt^(2))=V^(2)(del^(2)y)/(delx^(2))`
We look for a stationary of this equation
`y(x,t)=A `sin`(omega)/(V) x sin (omegat+delta)`
where `A &delta` are constants.. In this from `y=0 at x=0`. The further condition
`y=0 at x=l`
implies `(omega l)/(v)=NPI, N gt 0`
`N` is the number of modes of frequency `le omega`
Thus `dN=(l)/(piV)d omega`
Discussion
34.

In board jumping does it matter high you jump ? What factors determine the span of the jump?

Answer»

HIGH
LOW
Same
NONE of these

Answer :A
Discussion
35.

In Young's double slit experiment, the intensity at a point is 1/4 of maximum intensity, angular position of this point is ?

Answer»

`sin^(-1) (lambda)/(d)`
`sin^(-1)(lambda)/(2D)`
`sin^(-1)(lambda)/(3d)`
`sin^(-1)(lambda)/(4d)`

Solution :`I = (I_(0))/(4) = I_(1) = I_(2)`
`phi` is phase difference,
then `I = I_(1) + I_(2) + 2 SQRT(I_(1)I_(2)) cos phi`
`(I_(0))/(4) = (I_(0))/(4) + (I_(0))/(4) + 2.(I_(0))/(4) cos phi`
`thereforecos phi = -1/2, phi = (2pi)/(3)`
Now `Deltax = (lambda)/(2pi) phi =(lambda)/(2pi) .(2 pi)/(3) = (lambda)/(3)`
If `theta` is angular SEPARATION the d `sin theta = Deltax = (lambda)/(3)`
`therefore sin theta =(lambda)/(3d)`
`theta = sin^(-1)((lambda)/(3d))`
Discussion
36.

What is the meaning of the word "plead"?

Answer»

To MAKE a request
To make food
To GET angry
To cry

Answer :A
Discussion
37.

A star shrinks suddenly and its density increases 10 times its original value. The acceleration due to gravity increases by a factor of :

Answer»

`10^(-9)`
`10^(-6)`
`10^(6)`
`10^(9)`

ANSWER :C
Discussion
38.

The diffusion current in a p-n junction is

Answer»

<P>From N SIDE to p side
From p side to n side
From n side to p side
From p side, to n side,. if REVERSE biased

Answer :B
Discussion
39.

Who is Tarakratna?

Answer»

Priest
Landlord
Weaver
None of the above

Answer :A
Discussion
40.

A comet orbits around the Sun in an elliptical orbit. Which of the following quantities remains constant during the course of its motion?

Answer»

LINEAR velocity
Angular velocity
Angular MOMENTUM
POTENTIAL energy

Answer :C
Discussion
41.

Which crystal is used to scatter electron in the Division and Germer experiment ?

Answer»

COBALT
Nickel
Calcite
Silver

Answer :B
Discussion
42.

Match the statements in Column I labeled as (a), (b), (c ), and (d) with those in column II labeled as (p), (q), (r ), and (s). Any given statement in column I can have correct matching with one or more statements in Column II.

Answer»


ANSWER :A::B::C::D
Discussion
43.

A tangent galvanometer shows no deflection when a current is passed through it. But when the current is reversed, it gives a delfection of 180°, then the plane of the coil should have been oriented:.

Answer»

in the MAGNETIC meridian
normal to magnetic meridian
at 45° to the magnetic meridian
at ANGLE 60° the magnetic meridian

Answer :B
Discussion
44.

The total energy of a particle of mass 0.5 kg performing S.H.M. is 25J. What is the speed when the particle crosses the centre of the path

Answer»

` 5 m//s`
`7 m//s`
`10 m//s`
`12 m//s`

Answer :C
Discussion
45.

Explain the working of a forward biased p-n junction diode.

Answer»

Solution :
1-V Characteristic Curve: When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the BATTERY and n-side to the negative terminal it is SAID to be forward biased.

The direction of the applied voltage (V) is opposite to the built-in POTENTIAL V, in the semiconductor diode. As a result, the width of depletion layer decreases and the barrier height is reduced.If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, the current will be small. If the applied voltage increase to large value, the barrier height will be reduced, the current increases.
Discussion
46.

A monochromatic light passes through a glass slab (mu=1.5) of thickness 9 cm in time t_(1). If I takes a time t_(2) to travel the same distance throught water (mu=4//3). The value of (t_(1)-t_(2)) is

Answer»

`5XX10^(-11)` SEC
`5xx10^(-8)` sec
`2.5xx10^(-10)` sec
`5xx10^(-10)` sec

Answer :A
Discussion
47.

Give the expression for magnetic field at a point on the axis of a short magnetic dipole.

Answer»

Solution :
I `to` Current in the LOOP
`R to ` Radius of the loop
X-axis `to` Axis of the loop
x `to` Distance OP
dl `to` Conducting element of the loop
ACCORDING to Biot-Savart.s law, the magnetic field at P is
`db=(mu_0I|dlxxr|)/(4pir^3)`
`r^2=x^2+R^2`
`|dl xx r | =RDL` (`because` they are perpendicular )
`therefore dB=mu_0/(4pi). (IDL)/(x^2+R^2)`
i.e., dB has two components `-dB_x` and `dB_1, dB_1` is CANCELLED out and only the x-component remains.
`therefore dB_x=dBcos theta`
`cos theta =R/(x^2+R^2)^(1//2)`
`therefore dBx=(mu_0Idl)/(4pi). d/(x^2+R^2)^(3//2)`
Summation of dl over the loop is given by `2piR`.
`therefore B=(mu_0 IR^2)/(2(x^2+R^2)^(3//2))hati`
Discussion
48.

The photoelectric threshold wavelength for a metal surface is 6600 A^@. The work function for this metal is :

Answer»

0.86 E V
1.87 e V
18.7 e V
0.18 e V

Answer :B
Discussion
49.

Sky appears to be blue in clear atmosphere due tolight's

Answer»

diffraction
dispersion
SCATTERING
polarisation

Solution :SKY appears BLUE due to scattering of light by ATMOSPHERIC modecules.
Discussion
50.

Match the statements in Column I labeled as (a), (b), (c ), and (d) with those in column II labeled as (p), (q), (r ), and (s). Any given statement in column I can have correct matching with one or more statements in Column II.

Answer»


ANSWER :A::B::C::D
Discussion