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.

The surface charge density of sphere of radius r is sigma. It is placed at a point A. The electric field intensity at B due to this sphere is E. Another charged sphere os radius 2r is placed at B. If the intensity at the centre of A and B is E/2 then the surface charge density of B is

Answer»

`(SIGMA)/(2)`
`-(sigma)/(2)`
`(-7sigma)/(32)`
`(7sigma)/(32)`

ANSWER :D
2.

Two waves of intensities I and 4I superpose, then the maximum and minimum intensities are :

Answer»

5I, 3I
9I, I
9I, 3I
5I, I

Answer :B
3.

0.bar5 को p/q के रूप में व्यक्त कीजिए

Answer»

2/9
3/9
5/9
1/5

Answer :C
4.

Give two definitions of mutual inductance, give its units and write factors on which its value depends.

Answer»

Solution :The flux `Phi_2` linked with the coil-2 when current through coil-1 is `I_1` is `Phi_2=M_21I_1`.
Taking `I_1`=1 unit, `Phi_21`. Thus, "The magnetic flux linked with one of the coils of a system of two coils per unit current passing through the other coil is called mutual inductance of the system formed by the two coils."
Mutual emf produced in coil-2 is given by,
`epsilon_2=-M_21(dI_1)/(dt)`
When `(dI_1)/(dt)=1` unit in the equation (2), `epsilon_2=M_21`. Thus,
"The mutual emf generated in one of the two coils due to a unit rate of change of current in the other coil is called mutual inductance of the system of two coils".
The unit of mutual inductance is `WbA^(-1)=(V.s)/A` = henry(H). Dimensional formula of mutual inductace is `M^1 L^2 T^(-2) A^(-2)`
The value of mutual inductance two coils depends UPON :
(1) Shape of coils
(2) Size of coils
(3) NUMBER of turns in two coils
(4) Distance between them
(5) ANGLE of mutual inclination
(6) The material on which they are WOUND .
5.

(a) Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. Mark the regions where the nuclear force is (i) attractive, and (ii) repulsive. (b) In the nuclear reaction n+ " "_(92)^(235)U to " "_(54)^(a)Xe+ " "_(b)^(97)Sr + 2n determine the values of ‘a’ and ‘b’.

Answer»

Solution :(b) Considering conservation of ATOMIC number as well as MASS number, the nuclear reaction is written as :
`n+" "_(92)^(235)UTO " "_(54)^(137)Xe + " "_(38)^(97)Sr + 2n`
So, a=137 and b=38.
6.

An electron moving with a speed of 2.5xx10^(-7)ms^(-1) enters a space between two horizontal metal plates which a uniformelectric field of 1600 Vm^(-1) is maintained. What will be its deflection from its straight line path on emerging out if the length of each plate is 10 cm and the electron enters in a direction perpendiular to the electric field ?

Answer»


ANSWER :2.25 MM
7.

Angle which the vector vec(A)=2hati + 3hatj makes with the y-axis is given by :

Answer»

`TAN^(-2)(3/2)`
`tan^(-2)(2/3)`
`SIN^(-1)(2/3)`
`COS^(-1)[3/2]`

Solution :Here `A_(X)`
`A_(y)=3`
`tantheta=A_(x)/A_(y)=2/3`
`:.theta=tan^(-1)2/3`
8.

A lensof negligible thickness of focal length (f) has as aperture (d). If forms an image of intensity I. Now, the central partof aperture upto diameter d/2 is blocked by opaque paper. The focal length and image intensity will change to :

Answer»

`(F)/(2),(I)/(2)`
`(3f)/(4),(I)/(2)`
`f,(I)/(4)`
`f,(3I)/(4)`

SOLUTION :(d) The focal length of lens does not change if a part of it is blocked, if central part of aperture upto `(d)/(2)` is blocked.
`therefore` EXPOSED area of aperture reduces by one FOURTH the initial areal.
`therefore (pi(d//2)^(2))/(pid^(2))=(1)/(4)`
Intensity of image reduced factor of 4.
`therefore "" I. = I - (1)/(4) I = (3I)/(4).`
9.

The equation Y = acos^(2)(2pi nt - (2pix)/lambda) represents a wave with:

Answer»

amplitude a, frequency N and wavelength `lambda`
amplitude a, frequency 2n and wavelength `2lambda`
amplitude a/2, frequency 2n and wavelength `lambda`
amplitude a/2, FREQUNCY 2n and wavelength `lambda//2`

ANSWER :D
10.

When a current carrying loop is replaced by an equivalent magnetic dipole ........

Answer»

the distance l between the poles is fixed
the pole strength p of each pole is fixed
the dipole moment is reversed.
the product pl is fixed.

Solution :Since a current carrying loop acts as a MAGNET and its magnetic dipole moment is `m=IA` where, I = current and A is the area of loop. If bar magnet of the same dipole is TAKING its magnetic dipole moment is,
`m= 2pl`,
`therefore m ALPHA pl`
`m alpha pl` Here m is CONSTANT so the product of pl is also constant.
11.

Biot Savart law in vector form is represented as doversettoB.

Answer»

SOLUTION :`mu_@/(4PI)(I(oversettodlxxoversettor))/r^3`
12.

Sketch the pattern of electric field lines due to an electric dipole.

Answer»

SOLUTION :
13.

When radiation of 400 nm wavelength Is incident on photocell ,emitter produces photoelectron with maximum kinetic energy of 1.68 eV work function of emitter will be………. [hc=1240 eVnm]

Answer»

3.09 eV
1.42 eV
1.51 eV
1.68 V

Solution :From Einstein.s equation of photoelectric EFFECT,
`h_(F)=hf_(0)+(1)/(2)mv_(max)^(2)`
`therefore (h_(c))/(lambda)=PHI+(1)/(2)mv_(max)^(2)`
`(1240)/(400)=phi+1.68`
`therefore phi=3.1-1.68 therefore phi=1.42 eV`
14.

An ideal gas is initiaily at temperature T and volumoeV. lts volume is increased by Delta V due to an increase in temperature Delta T, pressure remaining constant. The quantity k = AV//(V Delta T) varies with temperature as:

Answer»

<P>


Solution :For ideal gas `PV=nRT`
At const. pressure.
`Pdv =nRdT`
`(dV)/(dT) =(NR)/(P)`
`AB.P =(nRT)/(V)`
`therefore (dV)/(dT) =(nR)/(nRT) cdot V`
`rArr (1)/(V) cdot (dV)/(dT) =(1)/(T)`
`rArr k=(1)/(T) rArr k prop (1)/(T)`
THUS, CORRECT choice is (c ).
15.

Two separate monochromatic light beams A and B of the same intensity are falling normally on a unit area of a metallic surface. Their wave lengths are lamda_(A) and lamda_(B), respectively. Assuming that all the incident light is used in ejecting the photoelectrons the ratio of the photoelectrons from the beam A to that from B is :

Answer»

`(lamda_(A)//lamda_(B))^(2)`
`(lamda_(B)//lamda_(A))^(2)`
`(lamda_(A)//lamda_(B))`
`(lamda_(B)//lamda_(A))1`

Answer :C
16.

If the accelerating voltage across an x-ray tube is doubled

Answer»

The wavelength of characteristic LINES are halved.
The minimum wavelength of x-rays is halved
The x-rays are PROBABLY BECOMES more penetrating
The target METAL will get heat up earlier.

Answer :B::C::D
17.

Figure shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the process.

Answer»

<P>

Solution :Since the process is cyclic , HENCE the change in internal energy is zero.
The heat given to the system is then equal to the work done by it.
The work done in part `AB` is `W_(1) = 0` (the volume remains constant ) . The part `BC` represents an ISOTHERMAL process so that the work done by the gas during this part is `W_(2) = nRT_(2) "ln"(V_(2))/(V_(1))`
During the part `CA: Vprop T ""` So,`V//T` is constant and hence , `P=(nRT)/(V)` is constant
The work done by the gas during the part `CA "is" W_(3) = P(V_(1)-V_(2)) = nRT_(1) -nRT_(2) = -nR(T_(2)-T_(1))`.
The net work done by the gas in the process `ABCA` is `W= W_(1) + W_(2) + W_(3) = "n"R[T_(2)"ln"(V_(2))/(V_(1)) - (T_(2)-T_(1))]`
The same amount of heat is given to the gas.
18.

Diffraction phenomenon is common in sound but not common in light. Why is it so?

Answer»

Solution :Basic condition for diffraction to take place is that of the SIZE of the obstacle/ aperture shoud be of the ORDER of wavelength of waves used. For sound wavelength is LARGE and we get OBSTACLES of large size easily. Hence difraction is common in sound. In the CASE of light, wavelength is very small and small size obstacle is not available easily.Hence diffraction is not common in light.
19.

When a cell is conneted in a circuit, a current 1_(1) flows in the circuit. When one more identicalcell is conneted in series with the first one, a crrent I_(2) is found to flow in the circuit. When same cell is connected in parallel with the firstone, the current is found to be I_(3). Show that 3I_(2)I_(3)=2I_(1)(I_(2)+I_(3)).

Answer»
20.

Phasor's method is used for ……..

Answer»

to obtain high A.C. voltage
to obtain high frequency of A.C.
to obtain ADDITION of TWO harmonic functions
to obtain multiplication of two harmonic functions

Solution :Using Phasor.s method, we can find out resultant of given harmonic functions. Here PHASORS, representing given harmonic function are placed tail to head. Then their resultant is a vector, joining tail of first phasor to the head of LAST phasor.
Here, (i) Length of resultant phasor gives its amplitude.
(ii) Angle made by resultant phasor with reference line (usually + X-axis) gives phase of resultant phasor and
(iii) Component of resultant phasor either on X-axis or on Y-axis, gives instantaneous VALUE of resultant phasor.
21.

A generator produces a voltage that is given by V=240 sin 120t, where t is in seconds. The frequency and rms voltage are (1)

Answer»

60 HZ and 240 V
19 Hz and 120 V
9 Hz and 170 V
754 Hz and 70 V

Answer :C
22.

Which of the following have zero average value in a plane electromagnetic wave ?

Answer»

ELECTRIC FIELD
magnetic potential
electric energy
magnetic energy.

Solution :Electric or magnetic field has ZERO average value in e.m.f. waves
23.

When radio waves passes through ionosphere, phase difference between space current and capacitive displacement current is

Answer»

0 RAD
`(3 pi // 2 )` rad
`(pi//2)` rad
`pi `rad

ANSWER :A
24.

Two point charges of +Q each have been placed at the positions (-a //2, 0) and (a // 2, 0,0). The locus of the points where - Q charge can be placed such the that total electrostatic potential energy of the system can become equal to zero, is represented by which of the following equations ?

Answer»

`Z^(2)+(Y-a)^(2)=2a`
`Z^(2)+(Y-a)^(2)=27A^(2)//4`
`Z^(2)+Y^(2)=15a^(2)//4`
None

Answer :C
25.

In the circuit if the energy of 8 mu F condenser is E, the energy of 4muF condenser is

Answer»

`(E )/(2)`
`2E `
`(E )/(18)`
`(5E )/(18)`

ANSWER :C
26.

Three equal resistors connected in series across a source of emf together dissipate 10 watt of power. What would be the power dissipated if the same resistances are connected in parallel acrsoss the same source of emf?

Answer»

SOLUTION :The power consumed by a RESISTANCE R when CONNECTED across a source of emf V is given by
`P=(V^(2)//R)`
Now, if r is the resistance of each resistor, the resistance of COMBINATION, in series will be
`R_(s)=r+r+r+i.e…,"R_(S)=3r`
while in paralle, `(1)/(R_(p))=(1)/(r)+(1)/(r)+(1)/(r)"i.e...,"R_(p)=(r)/(3)`
So power consumption in series will be,
`P_(S)=(V^(2))/(3r)["as"R_(S)=3r]`
while in parallel will be,
`P_(p)=(V^(2))/((r//3))=3[(V^(2))/(r)][" as "R_(p)=(r)/(3)]`
`"i.e..,"(P_(p))/(P_(s))=3[(V^(2))/(r)]xx(3r)/(V^(2)]=9`
And as here `P_(s)=10W`
`P_(P)=9xx(P_(S))=9xx10=90W.`
27.

The figure shows a surface XY separating two transparent media, medium - 1 and medium - 2.The lines ab and cd represent wavefronts of a light wave travelling in medium -1 and incident on XY. Then lines ef and gh represent wavefronts of the light wave in medium- medium after refraction. 84. Light travels as a

Answer»

parallel beam in each medium
convrgent beam in each medium
divergent beam in each medium
divergent beam in ONE medium and convergent beam in the other medium.

Solution :The WAVEFRONTS in both media are parallel. So the light be a parallel beam in each medium.
28.

Suggest an ideal to convert a full wave bridge rectifier to a half wave rectifier by changing the connecting wire/s. Draw the diagram and explain your answer.

Answer»

Solution :The CIRCUIT diagram of a FULL wave rectifier is given here. In one half cycle we get output due to `D_(1)` because it is in forward bias and `D_(2)` is in REVERSE bias. However, in next half cyclewe get output due to `D_(2)` because now it is in forwardbias and `D_(1)` is in reverse bias.

If in the circuit shown Y point is directly connected to point B, the arrangement will behave as a half wave rectifier. It is because in one half cycle when `D_(1)` is in forward bias, we SHALL get output across `R_(L)`. However in second half cycle we do not get any output because now circuit is not completed. So the arrangement acts as a half wave rectifier only.
29.

Light waves of wave length lambda propagate in a medium. If M and N are two points on the wave front and they are separated by a distance lambda//4, the phase difference between them will be (in radian)

Answer»

`pi//2`
`pi//8`
`pi//4`
Zero

Answer :A
30.

"To emit a free electron from a metal surface a minimum amount of energy must be supplied"Give three method to supply energy to a free electron

Answer»

SOLUTION :GIVE LIGHT ENERGY or HEAT energy
31.

The intensity of sun lighth (in W//m^(2)) at the solar surface will be:

Answer»

<P>`5.6xx10^(6)`
`5.6xx10^(7)`
`4.2xx10^(6)`
`4.2xx10^(7)`

Solution :Intensity is GIVEN by,
`(P)/(A)=(3.9xx10^(26))/(4PI^(2))`
`=5.62xx10^(7)W//m^(2)`
32.

The equivalent resistance of the combination across AB is .

Answer»

`((3+sqrt(17))/4)`
`3+sqrt(17)`
`(3+sqrt(17))/2`
`2(3+sqrt(17))`

Solution :a. The equivalent circuit can be REDRAWN as shown in Figure.

Let `R = (2R xx 2)/(2R + 2) + 2 = (2R)/(R+1) + 2`
or `R = (2R+2R+2)/(R+1) = (2+4R)/(R+1)`
` R^2 - 3R - 2 or R = (3+(sqrt(17)))/(2)`
`R_(AB) = R/2 = (3+sqrt(17))/4` .
33.

A body is in pure rotation. The linear speed v of the particle, the distance r of the particle from the axis and the angular velocity omega of the body are related as omega=(v)/(r). Thus

Answer»

`omegaalpha(1)/(R)`
`omegaalphar`
V
`OMEGA` is INDEPENDENT of r

Answer :D
34.

A potential is given by V(x)=(x+alpha)^(2//2) for x lt 0 and V(X)=k(X-alpha)^(2//2) for xgt0. The schematic variation of oscillation pertiod (T) for a performing periodic motion in this potential as a function of its energy E is:

Answer»




ANSWER :B
35.

Unpolarised ligh of intensity 32 W"/"m^(2) passes through a polariser and analyer which are at ain angle of 30^(@) with respect to each other. The intensity of the light coming from analyser is

Answer»

`16 sqrt(3) W"/"m^(2)`
`12 W"/"m^(2)`
`16W"/"m^(2)`
none

Answer :B
36.

The atomic mass of m_("electron")=0.00055" amu ",m_("proton")=1.007593" amu "m_("neutron")=1.008982 amu. The binding energy per nucleon is ______ MeV (nearly)

Answer»

6
7
8
5

Answer :C
37.

Two concentric circular coils X and Y of radii 16 cm and 10 cm , respectively, lie in the same vertical plane conatining the north to south direction. Coil X has 20 turns and carries a current of 16 A, coil Y has 25 turns and carries a current of 18 A . The sense of the current inX is anticlockwise, and clockwise in Y, for an observe looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Answer»

Solution :Field at centre, `B = (mu_0 nI)/(2R)`
`B_(1) = (4 pi xx 10^(-7) xx 20 xx 16)/(2 xx 16 xx 10^(-2)) = 9 pi xx 10^(-4)`
Net field `= (9pi- 4PI) xx 10^(-4) =5 pi 10^(-4)` T TOWARDS west
38.

A wire of length L is strech such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 Omega, its new resistanc would be

Answer»

`40 Omega`
`80 Omega`
`120 Omega`
`160 Omega`

Solution :`160 Omega`
SUPPOSE diameter of wire of length `l_(1) ` is `d_(1) ` and wire of length `l_(2) ` is `d_(2) ` .
`therefore PI d_(1)^(2) = pi d_(2)^(2) l_(2)`
`therefore ((d_(1))/(d_(2)))^(2) = (l_(2))/(l_(1))`
`therefore (2)^(2) = (l_(2))/(l_(1))"" therefore 4 = (l_(2))/(l_(1))`
Now R. = `((l_(2))/(l_(1)))^(2) .R = (4)^(2) XX 10 = 160 Omega `
39.

A reel of massless thread unrolls itself falling down under gravity. The acceleration of its fall is :

Answer»

g
`(g )/(2)`
`(2)/(3)g`
`(4)/(5)g`.

SOLUTION :Here
`mgh=(1)/(2)mv^(2) +(1)/(2)Iomega^(2)`
`=(1)/(2)mv^(2)+(1)/(2)xx(MR^(2))/(2)xx OMEGA^(2)`
`=(1)/(2)mv^(2)+(1)/(4)mv^(2)=(3)/(4)mv^(2)`
`v^(2)=(4)/(3)gh" Now " v^(2)-u^(2)=2as`
`(4)/(3) gh-0=2ah or a=(2)/(3)g`.
Correct choice is (d).
40.

Which is correctly matched is a normal menstrual cycle ?

Answer»

Endometrium REGENERATES -5 to 10
Release of egg 5th day
Endometrium secretes nutrients for implantation -11 to 18 days
Rise in progesterone LEVEL -1 to 15 days

Answer :A
41.

A swimmer is capable of swimming 1.65 ms^(-1) in still water. If she swims directly across a 180m wide river whose current is 0.85 m/s, how far downstream (from a point opposite her starting point) will she reach?

Answer»

`92.7 m`
`40 m`
48 m
20 m

Answer :A
42.

The specific resistance of a wire is rho, it's volume is 3m^3 and its resistance is 3 ohm, then it's length will be:

Answer»

`1/sqrtrho`
`3/sqrtrho`
`rhosqrt3`
`rhosqrt(1/3)`

ANSWER :B
43.

Derive the expression from drift velocity of free electron in terms of relaxation time and electric field applied across a conductor.

Answer»

Solution :In the absence of an external electric field E, the conduction electron electron in a conductor MOVE randomly and their net velocity `vecu=0`.
In the presence of an electric field `vecE`, electrons EXPERIENCE an acceleration `veca=-(evecE)/(m).` If `t_(1), t_(2), t_(3)`,... be the times before two successive collisions for different electrons, then the FINAL VELOCITIES acquired by different electrons are
`vec(v_(1))=vec(u_(1))+vecat_(1), vec(v_(2))=vec(u_(2))+veca(t_(2), ... vec(v_(n))=vec(u_(n))+vecat_(n))`
`THEREFORE` Mean value of electron velocity in the presence of an electric field is called drift velocity `vec(v_(d))`.
Then
`vec(v_(d))=(vec(v_(1))+vec(v_(2))+vec(v_(3))+....+vec(v_(n)))/(n)=((vec(u_(1))+vec(u_(2))+vec(u_(3))+...+vec(u_(n)))/(n))+veca((t_(1)+t_(2)+t_(3)+....t_(n))/(n))`
`=veca tau=-(evecE)/(m)tau`, where `tua=(t_(1)+t_(2)+t_(3)+...+t_(n))/(n)="relaxation time"`
44.

A wooden ball of density p is dropped from rest from a height 'h' into the lake of water density sigma (sigma > p). Neglecting viscosity, the maximum depth to which the body sinks before returning to float is :

Answer»

<P>`(hp)/(p-SIGMA)`
`(H(p-sigma))/p`
`(h(sigma-p))/(sigma)`
`(hp)/(sigma-p)`

SOLUTION :Here P.E. of the ball
mgh`=[m/pxxsigmaxxg-mg]`H, where His the depth to which it sinks. Then
`h=[(sigma)/p-1]`H or H`=(PH)/(sigma-p)`
Correct choice is (d).
45.

In the figure shown, what is the potential difference between the points A and B between a and C respectively in steady state?

Answer»

`V_(AB)=V_(BC)=100 V`
`V_(AB)=75 V, V_(BC)=25 V`
`V_(AB)=25 V, V_(BC)=75 V`
`V_(AB)=V_(BC)=50 V`

Answer :C
46.

Above the Curie temperature, thesusceptility of aferromagentic substance varies

Answer»


Solution :Above the Curie temperature, ferrromagnetic susbstance BECOME paramagnetic in NATURE.
47.

What we call the number of electrons/protons in an atom ?

Answer»

SOLUTION :ATOMIC NUMBER
48.

How do we define photoelectric work function and how it is related to threshold frequency ?

Answer»

Solution :The minimum AMOUNT of radiant ENERGY needed to pull an electron (without imparting it any kinetic energy) from a metallic SURFACE is called work function of the METAL. The relation between work function `W_0` and threshold frequency `v_0` is `W_0 = hv_0`
49.

In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes, this water is pumped into a holding tank. Assume that the efficiency of the overally system is 25% ( that is, 80% of the incident solar energy is lost from the system ). What collector area is necessary to raise the temperature of 200L of water in the tank from 20^(@)C to 40^(@) in 1.0 h when the intensity of incident sunlight is 750W// m^(2) ?

Answer»


ANSWER :`25M^(2)`
50.

How are the different emfs generated in a three-plane AC generator? Show the graphical representation of theses emfs.

Answer»

Solution :i. In some AC generators, there are three separate COILS, which would give three-phase AC generators.
ii.In the simplified construction of three-phase AC generator, the armature core has 6 slots, cut on its inner rim. Each slot is `60^(@)` away from one another. Six armature conductors are mounted in these slots. The conductor 1 and 4 are joined in series TOFORM coil 1. The conductors 3 and 6 from coil 2 while the coil are retangular in shape and are `120^(@)` apart from one another (Figure).

iii. The initial position of the field magnet is HORIZONTAL and field direction is perpendicular to the plane of the coil 1. As it is seen in single phase AC generator, when field magnet is ROTATED from that position in clockwise direction, alternating emf `omega1` in coil 1 begin a cycle from origin O. This is show in Figure.

iv. The corresponding cycle for alternating emf `omega_(2)` in coil 2 starts at point A after field magnet has rotated through `120^(@).` Therefore, the phase difference between `omega_(1)andomega_(2)" is "120^(@).` Similarly, emf `omega_(3)` in coil 3 would begin its cycle at point B after `240^(@)` rotation of field magnet from initial position. Thus these emf produced in the difference between one another.