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.

Two similarly and equally charged idential metal spheres A and B repel each other with a force of2 xx 10 ^(-5)) N.A third identical uncharged sphere C is touched with A and then , placed at the mid - point between A and B. Calculate the net electric force on C.

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Solution :Let charges on A and B be Q each R be the distance between them. Then as shown in we have.
` F= ( 1)/(4pi in _0) .(q^(2))/( r^(2)) =2 xx10 ^(-5)N. `
On touching C with A , charge on A as well as C will be `(q)/(2) `. Now
as shown in net force on C is
` "" F_C =|oversetto (F_(CA)) +oversetto (F_(CB))| =oversetto (F_(CB)) -oversetto (F_(CA)) =(1)/(4pi in _0) .(((q)/(2)) ((q)/(2)))/( ((r)/(2)) ) `
` =(1)/(4piin _0) [(2q^(2))/(r^(2)) -(q^(2))/(r^(2)) ]=(1)/(4 pi in _0) =2 xx 10 ^(-5) N `along CA.
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2.

Releative permiottivity and permeability of a material are which of he following value of these quantites are allwoed for a diamagneticmaterial

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`epsilon_(R)=1.5 mu_(r)=1.5`
`epsilon_(r )=0.5 mu_(r )=1.5`
`epsilon_(r )=1.5 mu_(r )=0.5`
`epsilon_(r )=0.5 mu_(r )=0.5`

ANSWER :C
3.

Explain the term 'Channel' in a communication system.

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Solution :Channel is a medium used for transmission and reception. This medium can be a wire, COAXIAL CABLE, a wave guide, an optical fibre or a RADIO link. Communication channel is also CALLED transmission medium or link.
4.

The internal energy of a system remains constant when it undergoes

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a cyclic PROCESS.
an isothermal process
an adiabatic process.
any process in which the HEAT GIVEN out by the system is equal to a WORK done on the system.

Answer :A::B::D
5.

A glass tube of 1.0 m length is filled with water, the water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500Hz is brought at the upper end of the tube and the velocity of sound is 330 mlSi then the total number of resonances obtained will bet

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4
3
2
1

Answer :B
6.

Unpolarised light is incident from air on a plane surface of a material of refractive index u! At a particular angle of incidence mu it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation ?

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`I =tan^(-1)((1)/(MU))`
Reflected light is polarised with its ELECTRIC vector parallel to the plane of incidence.
`i=sin^(-1)((1)/(mu))`
Reflected light is polarised with its electric vector perpendicular to the plane of incidence.

Solution :When reflected and REFRACTED rays are perpendicular then only `SIGMA` component are there in reflected light (which are perpendicular to incidence plane) in polarization in reflection
7.

Which of the following system of unit is also called a decimal system.

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CGS
FPS
MKS
SI

Answer :D
8.

The logic symbol shown in figure represents

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OR GATE
XOR gate
NAND gate
NOR gate

Answer :A
9.

The speed of a body that bas Mach number more than 1 is

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supersomc
SUBSONIC
300m/s
about 10 MIS

ANSWER :a
10.

Two projectiles thrown with different velocities and at different angles so as to cover the same maximum heights. The sum of the time taken by each to reach highest point is equal to:

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TOTAL TIME of FLIGHT of each
`1/2`of the total time of flight of each
`1/4` TH of the total time of flight of each
None of these.

Answer :A
11.

Show by drawing a ray diagram, how a totally reflecting prism may be used to invert the image without changing its size and direction.

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Solution :The adjoining Fig. 9.38, shows the USE of a TOTALLY reflecting prism to invert the IMAGE without changing its size and direction.
12.

Which of the graph between acceleration and time is correct ?

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A
B
C
D

Answer :C
13.

Drops of water fall at regular intervals from the rood of a building of height H=16m,the first drop striking the ground at the same moment as the fifth drop detaches from the rood.Find the distance between the successive drops.

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Solution :Step-i:Time taken by the first
Drop to touch the ground =`t=sqrt(2h)/(g)`
For h=16m,t=`sqrt(2xx16)/(g)=4sqrt(2)/(g)`
Time interval between two DROPS is
`t_(int erval)=(1)/(n-1)t=(1)/(4)t=sqrt((2)/(g))`
Where n=number of drops
Step-ii
DISTANCE between first and SECOND drops
`=S_(1)-S_(2)=(1)/(2)"gt"_("int erval")^(2)[4^(2)-3^(2)]=7M` .
Distance between second and third drops
`=S_(2)-S_(3)=(1)/(2)"gt"_("interval")^(2)[3^(2)-2^(2)]=5m`.
Distance between third and fourth drops
`=S_(3)-S_(4)=(1)/(2)"gt"_("interval")^(2)[2^(2)-1^(2)]=3m`
Distance between fourth and fifth drops.
`=S_(4)-S_(5)=(1)/(2)"gt"_("interval")^(2)[1^(2)-0]=1M`.
14.

A block of mass m kg is pushed up against a wall by a force P that makes an angle 'theta' with the horizontal as shown in figure. The coefficient of static friction between the block and the wall is mu. The minimum value of P that allows the block to remain stationary is

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<P>

Solution :`P sin THETA + F=mg`
`f=(mg-P sin theta)`
`MU N =(mg-P sin theta)`
`mu P COS theta = mg-P sin theta`
`P[sin theta+mu cos theta]=mg`
`P=(mg)/((sin theta + mu cos theta))`
15.

Two nicol prisms are oriented with their principal planes making an angle of 60^(@). Then ..... % of incident unpolarised light which will pass through the system.

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50
100
`37.5`
`12.5`

Solution :The intensity of emerging LIGHT from first NICOL prism,
`I_(1)=(I_(0))/(2)`
The intensity of emerging light from second nicol prism
`I_(2)=I_(1) cos^(2) theta`
`=(I_(0))/(2)xx cos^(2)60^(@)`
`=(I_(0))/(8) "" [ :. cos60^(@)=(1)/(2]`
`=(I_(0))/(8)xx100%xx (10)/(I_(0))=12.5%`
16.

An idial choke takes a current of 8A when connected to an a.c source of 100 volt and 50 Hz. A pure resistance under the same conditions takes a current of 10A. If two are connected in series to on a.c. Supply to 100V and 40 Hz, then the current in the series combination of above resistor and inductor is

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10A
8A
`5sqrt(2)A`
`10sqrt(2)A`

ANSWER :C
17.

The average translational energy and the r.m.s. speed of molecules of a sample of oxygen gas at 300 K" are "6 cdot 21 xx 10^(-21)J and 484 ms^(-1) respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) :

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`8 cdot 78xx10^(-21) J, 684 ms^(-1)`
`6 cdot 21 xx10^(-21)J, 968 ms^(-1)`
`12 cdot 42 xx10^(-21) J, 968 ms^(-1)`
`12 cdot 42 xx10^(-21) J, 968 ms^(-1)`

Solution :`K.E. prop` temperature
`THEREFORE (E_(2))/(E_(1))=(T_(2))/(T_(1)) rArr E_(2)=2E_(1)=12 cdot 42 xx10^(-21) J`
And `v_(rms) prop SQRT(T)`
`(v_(2))/(v_(1))=sqrt((T_(2))/(T_(1)))`
`rArr v_(2)=v_(1) sqrt((600)/(300)) =1 cdot 414 xx 484 =684 ms^(-1)`
Thus, CORRECT choice is (c).
18.

NOR gate is the combination of ….. Gate and …… gate.

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OR, AND
NOT, OR
NAND, OR
NONE of these

ANSWER :B
19.

Two sound sources of frequencies 357.9Hz and 425Hz produces waves of wavelength difference 15cm. The velocity of sound will be

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340m/s
300m/s
400m/s
450m/s

Answer :A
20.

For L-C-R, A.C. circuit resonance frequency is 600 Hz and frequencies at half power point are 550 Hz and 650 Hz. What will be the Q-factor ?

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`1/6`
`1/3`
6
3

Solution :Q-factor , `Q=(omegao)/(Deltaomega)`
`=600/(650-550)=600/100`=6
21.

Range of a function is

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Values of X for which y is REAL or defined
Values of y for which x is real or defined
Both A and B
None of these

Answer :B
22.

Define the term current density of a metallic conductor. Deduce the relation connecting current density (J) and the conductivity (sigma) of the conductor when an electric field E is applied to it.

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Solution :According to Ohm.s law, we know that `V=RI = (rho l)/(A).I rArr V/l = rho . l/A`
but `l/A = J` and `V/l = E` , hence we have
`E= rho J ` or `J/E = 1/ rho= sigma ` ,where ` sigma `is the ELECTRICAL conductivity of given material. The relation`J/E = sigma " or" J = sigma E`is YET ANOTHER form of Ohm.s law. In VECTOR notations,
` VECJ = sigma vec E`
23.

The phenomenon of polarisation of electromagnetic wave proves that the electromagnetic waves are

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LONGITUDINAL
transverse
neither longitudinal nor transverse
both longitudinal and transverse

Answer :B
24.

A circuit draws 330W from a 110V, 60Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity equal to :

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`31muF`
`54muF`
`151 MUF`
`201 muF`

ANSWER :C
25.

A block rests on a rough inclined plane making an angle of 30 ^@with the horizontal . The coefficient of static friction between the block and the plane is 0.8 If the frictional force on the block is 10 N the mass of the block is ""(g=10 ms ^(-2))

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1kg
2KG
3kg
4kg

Solution :`f=mgsintheta`
`m=f/(gsintheta)=10/(10xx1/2)=2kg`
26.

A man standing at some distance from a cliff hears the echo of sound after 2s. He walks 495 m away from the cliff. He produces a sound there and receivesthe echo after 5s. What is the speed of sound?

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330 m/s
340 m/s
390 m/s
380 m/s

ANSWER :A
27.

The intensity of electromagnetic waves i.e. energy crossing per second per area of a surface is given by

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SOLUTION :I=`1/2(E_0B_0)/mu_0`=`1/2(B^2)/(mu_0)C` =
28.

In a hydrogen atom, the vece moves in an orbit of radius 0.5xx10^(-10) m making 10^(16) r.p.s Calculate the magnetic moment associated with the orbital motion of vece.

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ANSWER :`1.256xx10^(-23)Am^2`
29.

Define Electric field .

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Solution :The electric field at the point P at a distance r from the point CHARGE q is the experienced by a unit charge and is given by
`VECE=(VECF)/(q_(0))`
The electric field is a vector QUANTITY and its SI unit is Newton per Coulomb `(NC^(-1))`
30.

In the circuit shown the resistance R is kept in a chamber whose temperature. is 20^(@)C which remains constant. The initial temperature and resistance of R is 50^(@)C and 15 Omegarespectively. The rate of change of resistance R with temperature is (1)/(2)Omega//^(@)C and the rate of decrease of temperature of R is In(3//100) times the temperature difference from the surrounding (Assume the resistance R loses heat only in accordance with Newton's law of cooling) If K is closed at t = 0, then find the (a) value of. R for which power dissipation in it is maximum. (b) temperature of R when power dissipation is maximum (c) time after which the power dissipation will be maximum

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Solution :(a) `50OMEGA`
(b) `30^(@)C`
(c) `100 SEC`
31.

A closely wound solenoid of 800 turns and area of cross section 2.5 xx 10^(-4) m^2 carries a current of 3.0A. What is its associated magnetic moment?

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`0.5 J//T`
`0.3 J//T`
`0.6 J//T`
`0.8J//T`

ANSWER :C
32.

A body is projected horizontally from a height of 78.4 m with a velocity 10 ms^(-1). Its velocity after 3 seconds is - (g = 10 ms^(-2)] (Take direction of projection on veci and vertically upward direction on vecj)

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` 10 hat i - 30 hat J`
`10 hat i + 30 hat j`
`20 hat - 30 hat j`
`10 hat i + 10 SQRT(3) hat j`

ANSWER :A
33.

At what distance from a convex mirror of focal length 2.5 m should as boy stand so that his image has a height equal to half the original height?

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2.5 m from the mirror
5 m from the mirror
7.5 m from the mirror
10 m from the mirror

Answer :A
34.

In process 1, the energy stored in the capacitor E_C and heat dissipated across resistance E_D are related by -

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` E_C=E_D`
` E_C =E_D"in" 2`
` E_C=2E_D`
` E_C=(1)/(2) E_D`

ANSWER :A
35.

Radiowaves of constant amplitude can be generated with

Answer»

Filter
Rectifier
FET
Oscillator

Answer :D
36.

An AND gate can be prepared by repetitive use of a) NOT gate b) OR gate c) NAND gate d) NOR gate

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Only a and B are CORRECT
Only C and d are correct
Only a, b and c are correct
All are correct

Answer :B
37.

A car battery has emf 12V and internal resistance 5 x 10^-2 Omega. it draws 60A current. The terminal voltage of the battery in

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3 V
5 V
9 V
15 V

38.

A light ray passes from air into a liquid as shown in figure. Findrefractive index of liquid.

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SOLUTION :`mu_(LIQUID)=SQRT(3//2)`
39.

Discuss power in AC circuit containing only capacitor.

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Solution :The instantaneous power SUPPLIED to the capacitor is,
`P_(C ) = IV = I_(m) cos ( omega t ) V_(m) sin ( omega t ) `
`= I_(m) V_(m) cos ( omegat ) sin( omega t ) `
`= ( I_(m) V_(m))/( 2) . ( 2 cos omega t sin omega t ) `
`= ( I_(m) V_(m))/( 2) sin ( 2 omega t )`
Average power,
`P_(C ) = langle(I_(m) V_(m) sin2 omega t )/(2) rangle = ( I_(m) V_(m))/( 2) lt sin e omega t gt `
Since ` sin ( 2 omega t)=0` over a complete cycle. This is explained by figure in detail.
`:.` Average power `P_(C ) = 0 `

0-1 the current i flows as shown and from the MAXIMUM at 0, reaches a zero value at 1. The plate A is charged to positive polarity while NEGATIVE CHARGE q builds up in B reaching a maximum at 1 until the current becomes zero. The voltage `V_(c ) = q//C ` is in phase with q and reaches maximum value at 1. Current and voltage are both positive. So, `P = V_(c ) I` is positive. Energy is absorbed from the source during this quarter cycle as the capacitor is charged.

1-2 the current I reverse its direction. The accumulated charge is depleted means, the capacitor is discharged during this quarter cycle. The voltage gets reduced but is still positive. The current is negative . Their product the power is negative. The energy absorbed during the `1//4` cycle 0-1 is RETURNED during this quarter.

One complete cycle of voltage / current. Note that the current leads the voltage.

2-3 As I continues to flow from A to B, the capacitor is charged to reversed polarity means the plate B acquires positive and A acquires negative charge. Both the current and the voltage are negative . Their productp is positive. The capacitor absorbs energy during this `1//4` cycle.

3-4 the current I reverse its direction at 3 and flows from B to A.The accumalated charge is deplected and the magnitude of the voltage `V_(c )` reduced. `V_(c )` becomes zero at 4 when the capacitor is fully discharged. The power is negative. Energy absorbed during 2-3 is returned to the source. Net energy absorbed is zero.
40.

In above question, if eyes if man are at 1 - 70 m above the ground, find the position of the mirror :

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0.85 m
0.65 m
0.45 m
0.25 m

Solution :(A) `H = BN_(2) = EB - EN_(2)`
`=EB - (1)/(2) EB = (1)/(2) EB`
`=(1)/(2) XX 1.7 = 0.85m.`
41.

The magnetic induction at the centre of a current carrying circular coil of radius R is

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DIRECTLY PROPORTIONAL to R
INVERSELY proportional to R
directly proportional to `R^2`
inversely proportional to `R^2`

ANSWER :B
42.

Radius of concave mirror is 40 cm and the size of image (real) is twice as that of object, then the object distance is

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60cm
20cm
40cm
30cm

Answer :D
43.

In a Young's double-slit experiment fringes are obtained on a screen placed at certain distance away from the slits. If the screen is moved by 5 cm towards the slit, the fringe width changes by 30mum. Given that the slits are 1 mm apart. Calculate the wavelength of the light red.

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Solution :Here `d=1mm=1xx10^(-3)m,` CHANGE in distance of screen from the slits `DELTAD=5cm=5XX10^(-2)m`
and change in fringe width `Deltabeta=mum=30xx10^(-6)m`.
From the relation `beta=(lamdaD)/(d)`, we have `Deltabeta=(lamdaDeltaD)/(d)`
`implies LAMDA=(dDeltabeta)/(DeltaD)=((1xx10^(-3))xx(30xx10^(-6)))/((5xx10^(-2)))=6xx10^(-7)m=600nm`
44.

A rod of length L, cross sectional area A and density rho is hanging from a rigid support by spring of stiffries k. A very small sphere of mass m is rigidly attached at the bottom of the rod. The rod is partially immersed in a liquid of density rho. Find the period of small oscillations.

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Solution :`2pi SQRT((m + L rho. A)/(K+ A rho G))`
45.

the angular resolution of a 10 cm objective diameter telescope at a wavelength of 5000overset@A is about

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`10^6` rad
`10^(-2)` rad
`6xx10^(-4)` rad
`6xx10^(-6)` rad

Answer :D
46.

Find the sum of the kinetic and the potential energies of an electron in the first Bohr orbit. Explain the meaning of the sign of the total energy

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Solution :`E= K+U= (mv^(2))/(2)- (e^(2))/(4piepsilon_(0)(a_(0)))= -13.6eV`
The minus sign MEANS that the formation of a hydrogen ATOM from a free proton and an electron is accompanied by the LIBERATION of energy equal to 13.6 eV.
47.

In above question if resistance of meter bridge wire is 1 Omega /cm then the value of current / is

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0.11 A
0.33 A
0.66 A
3.3 A

Answer :C
48.

A lunar landing module is descending toward the moon's surface at a steady velocity of 10 ms. At a height of 120 m, a small objects falls from its landing gear. Taking the moon's gravitational acceleration as 1.6 ms , at what speed (in ms) does the object strike the moon ?

Answer»

202
22
19.6
16.8

Solution :Here `V^(2)-10^(2)=2xx1.6xx120`
or `v^(2)=484` andv=22 m
49.

A parallel plate capacitor is being charged by a time varying current. Explain briefly how Ampere's circuital law is generalized to incorporate the effect due to the displacement current.

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Solution :
According to Ampere.s circuital law,
`oint vecB. VEC(dl) = mu_0 I_e`
For `C_1 = oint vecB * vec(dl) = mu_0 I_e`
For `C_2oint vecB * vec(dl)=0`
There is an inconsistency in Ampere.s circuital law. To explain this displacement current was INCORPORATED .
Using Ampere.s law, we FIND DIFFERENT value of B for the two loops.
Hence displacement current was introduced.
50.

A radioactive nucleus decay as follows.Aoverset(beta^+)toA_1 oversetalphatoA_2 If the mass number and the atomic number of A_2 are 176 and 71 respectively, what are the corresponding numbers for A_1 and A ? Which of these three elements are isobars ?

Answer»

SOLUTION :`_74A^180overset(BETA^+)to_73A_1^180 oversetalphato_71A_2^176`
The ELEMENTS A and `A_1` are ISOBARS.