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 circuit contains a resistance of 40 Omega and an inductance of 0.68 H, and an alternating effective e.m.f. of 500 V at a frequency of 120 Hz is applied to it. Find the value of the effective current in the circuit and power factor.

Answer»


ANSWER :0.99A, 1/128
Discussion
2.

In Young's experiment, the distance between the slits is double and the distance of the screen and slit is halved. The fringes width is :

Answer»

DOUBLE
BECOMES ONE fourth
halved
remains the same

Answer :A
Discussion
3.

Can a body have energy without having momentum. Can a body have momentum without having energy ? Explain .

Answer»

Solution :A BODY may not have momentum but may have POTENTIAL ENERGY by virtue of its position. No , a body which has momentum always has KINETIC energy.
Discussion
4.

An infinitely long thin straight wire has uniform linear charge density of 1/3 "coul.m"^(-1). Then the magnitude of the electric intensity at a point 18 cm away is

Answer»

`0.33 XX 10^(11) NC^(-1)`
`3 xx 10^(11) NC^(-1)`
`0.66 xx 10^(11) NC^(-1)`
`1.32 xx 10^(11) NC^(-1)`

ANSWER :A
Discussion
5.

If the equation for graph shown in figure is y=ae^(x) then value of a is:-

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`-5`
`5`
`1`
cannot be determined

Answer :B
Discussion
6.

The power of a AM transmitter is 100 W. If the modulation index is 0.5 and the transmission is having single side band, the percentage of useful power is

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1.1 W
11 W
2.2 W
22 W

ANSWER :C
Discussion
7.

When a bar magnet is dropped from the top of a building its magnetism decreases . Why ?

Answer»

Solution :The regular ARRANGEMENT of the MOLECULAR MAGNETS in a MAGNET is disturbed.
Discussion
8.

Which of the following statements are correct with respect to the electromagnetic waves in an isotropic medium ?

Answer»

Energy DUE to electric field is equal to that due to magnetic field .
Electric vector E and magnetic vector B are in phase .
For a given amplitude of E vector the intensity increases as the FIRST POWER of FREQUENCY f .
For cylindrical wave fronts the amplitude of the wave varies in proportion to `1//r^(2)` where r is the radius of the wave front.

Answer :A::B
Discussion
9.

Identity the wrong statement in the following Coulomb's lasw correctly described the electric force that

Answer»

binds the electrons of an atom to its nucleus
binds the PROTONS and neutrons in the nucleus of an atom
binds atoms TOGETHER to form MOLECULES
binds atoms and molecules to form solids

Solution :(b) Coulomb's law is applicable for charged particles, it is not responsible to blind the protons and neutrons in the nucleus of an atom.
Discussion
10.

The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value ?

Answer»

`0.02s`
`5 xx 10^(-3) S `
`10 xx 10^(-3) S `
`2.5 xx 10^(-3) S`

SOLUTION :`i=i_0 sinomega t`
` at= (PI)/(4) ,t_1 =(1)/( 8 f)`
Att `=t_2, I =i_0 ,impliest_2(1)/(4f)`
` Deltat=t_2 -t_1 = 2.5ms`
Discussion
11.

A spherical ball contracts in volume by 0.01% when subjected to a uniform pressure of 100 atmospheres. The bulk modulus of material is: (one atmosphere =10^(5)Nm^(-2))

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`10^(10)Nm^(-2)`
`10^(13)Nm^(-2)`
`10^(11)Nm^(-2)`
`2XX10^(11)Nm^(-2)`

ANSWER :C
Discussion
12.

The work function of a substance is 12.4 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately

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540 nm
400 nm
310 nm
100 nm

Answer :D
Discussion
13.

What is optically denser and optically rarer medium ?

Answer»

Solution :If `n_(21) gt 1` i.e, REFRACTIVE index of medium-2 is greater than of medium-1, then medium-2 is said to be denser than medium-1.
Snell.s LAW `(SIN i)/(sinr)=n_(21)`
`therefore(sini)/(sin r) gt 1`
`therefore` sin I `gt` sin r
`therefore` i `gt` r
Hence, REFRACTED ray bends toward normal. Here, medium-2 is said to be denser as compared to medium-1.
If `n_(21) lt` 1, i.e. refractive index of medium-2 is less than medium-1, then medium-2 is said to be RARER than that of medium-1.
`(sin" "i)/(sin" "r)=n_(21)` (Snell.s law)
`therefore` sin i `lt` sin r
`therefore` i `lt` r

Hence, refracted rays goes away from normal. Here, medium-2 is said to be rarer as compared to medium-1.
Discussion
14.

Is the resistance of a voltmeter greater than or less that of the galvanometer of which it is formed?

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SOLUTION :It is ALWAYS GREATER than the RESISTANCE of the GALVANOMETER.
Discussion
15.

A planet is revolving round the sun is elliptical orbit. Velocity at perigee position ( nearest) is v_(1) and at apogee position ( farthest)is v_(2). Both these velocities are perpendicularto the line joining centre of sun and planet. r_(1) is the minimum distance and r_(2) the maximum distance. When the planet is at perigee position, it wants to revolve in a circular orbit by itselff. For this, value of G

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should increase
should DECREASE
will not DEPENDS on the VALUE of G
DATA is insufficient

Answer :A
Discussion
16.

Nuclear reactions are given as (i) square (n, p)_(15) p^(32) (ii) square (p, alpha)_(8) O^(16) (iii) ._(7)square^(4) (p) ._(6)C^(14) missing particle or nuclide (in box square ) in these reactions are respectively

Answer»

`S^(32), F^(19), ._(0)N^(1)`
`F^(19), S^(32), ._(0)n^(1)`
`Be, F^(19), ._(0)n^(1)`
NONE of these

Answer :A
Discussion
17.

Two different coils have self-inductances L_1 = 16 mH and L_2 = 12mH. At a certain instant, the current in the two coils is increasing at the same rate and power supplied to the two coils is the same. Find the ratio of i) induced voltage ii) current iii) energy stored in the two coils at that instant.

Answer»

SOLUTION :`i) V_1 = L_1 (dI)/(DT) , V_2 = L_2 (dI)/(dt)`
`V_1/V_2 = L_1/L_2 = 16/12 = 4/3`
`II) P = V_1I_1 = V_2I_2 rArr I_1/I_2 = V_2/V_1 = 3/4`
`iii) U_1/U_2 = (1/2 L_1I_1^2)/(1/2 L_2I_2^2) = (L_1/L_2) (I_1/I_2)^2 = 4/3 (3/4)^2 = 3/4`
Discussion
18.

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Answer»

`(2GmM)/(3R)`
`(GMM)/(2R)`
`(GmM)/(3R)`
`(5GmM)/(6R)`

SOLUTION :Work done = FINAL total energy- Initial total energy
`=(-GMm)/(6R)-((-GMm)/(R ))`
`=(5GMm)/(6R)`
Thus, correct CHOICE is (d).
Discussion
19.

Molar heat capacity of a gas at constant volume.

Answer»

Solution :Molar heat capacity of a GAS at constant volume is defined as the QUANTITY of heat required to RAISE the TEMPERATUE of one mole of the gas through one degree ` ( 1^(@) or 1K)`, when its volumeis kept CONTANT .
Discussion
20.

A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.

Answer»

Solution :For convex lens `f_(1) = 10`cm and `u_1 = - 30 cm`, hence from lens formula `1/v -1/u =1/f`,we have:
`1/v_(1) = 1/u_(1) + 1/f_(1) =1/(-30)+ 1/10 =1/15` or `v_(1) = 15 cm`

Thus, as shown in Fig. 9.43, convex lens `L_(1)`forms REAL image `I_(1)` , of given object O at a distance 15 cm from `L_(1)`or 15 - 5 = 10 cm from concave lens `L_2`. For concave lens,behaves as a virtual object i.e.,`u_2 = + 10` cm and `f_2 = - 10 cm`. Hence, we have,
`1/v_(2) -1/(+10) = 1/(-10) rArr 1/v_(2) =-1/10 +1/10 =0` or `v_(2) = infty`
Thus, final image is FORMED at infinity only.
Discussion
21.

A conducting loop in the form of a circule is placed in a uniform magnetic field with its place perpendicular to the direction of the field. An e.m.f. will be induced in the loop if

Answer»

it is TRANSLATED PARALLEL to itself.
it is rotated about one of its diameters.
it is rotated about its own axis which is parallel to the field.
the LOOP is deformed from the original shape.

Solution :An emf will be INDUCED in the loop if there is change in magnetic FLUX through it. Flux changes through the loop only in options (b) and (d).
Discussion
22.

A digital signal possesses

Answer»

continuously varying VALUES
only two DISCRETE values
only for discrete values
none of the above

Answer :B
Discussion
23.

Two circular rings A and B each of radius a = 30 cm are placed coaxially with their axis horizontal in a uniform electric field E = 10^(5) NC^(-1)directed vertically upward as shown in figure. Distance between centers of the rings A and B (C_A and C_B)is 40 cm. Ring A has positive charge q_A = 10muCand B has a negative charge q_B = -20muC. A particle of mass m and charge q = 10muC is released from rest at the center of ring A. If particle moves along C_AC_B, then Speed of particle when it reaches at center of B is

Answer»

`6sqrt2 m//s`
`12sqrt2m//s`
`2sqrt6 m//s`
`4sqrt6m//s`

Solution :`W=q(V_(2)-V_(1))=9xx10^(-2)[((10)/(0.3)-(20)/(05))-((10)/(0.5)-(20)/(0.3))]`
`=3.6J`
SINCE only conservative forces act on the system, potential energy CHANGES to kinetic energy.
`3.6=(1)/(2)mv^(2)` or `v^(2)=72` or `v=6sqrt(2)MS^(-1)`
Discussion
24.

An element hobbyist is building a radio which reequires 150 Omega in her circuit, but she has only 220Omega, 79 Omega and 92 Omega resistors available. How can she connect the available resistor to get desired value of resistance?

Answer»

Solution :REQUIRED effectiveresistance `=150Omega`
Given resistors of RESISTANCE, `R_(1)=220Omega, R_(2)=79Omega, R_(3)=92Omega`
Parallel combination of `R_(1) and R_(2)`
`(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2))=(1)/(220)+(1)/(79)=(79+220)/(220xx79)`
`R_(p)=58Omega`

Parallel combination of `R_(p) and R_(3)`
`R_(s)=R_(p)+R_(3)=58+92`
`R_(s)=150Omega`
Parallel combination of `220Omega and 79Omega` in series with `92Omega`
Discussion
25.

The electric intensity E, current density 'j' and conductivity sigma are related as

Answer»

`J=SIGMA E`
`j=(E )/(sigma)`
`jE=s`
`j=sigma^(2)s`

ANSWER :A
Discussion
26.

A planet is revolving round the sun is elliptical orbit. Velocity at perigee position ( nearest) is v_(1) and at apogee position ( farthest)is v_(2). Both these velocities are perpendicularto the line joining centre of sun and planet. r_(1) is the minimum distance and r_(2) the maximum distance. At apogee position suppose speed of planet is slightly decreased from v_(2), then what will happen to minimum distance r_(1) and maximum distance r_(2) in the subsequent motion.

Answer»

`r_(1)` and `r_(2)` both will decrease
`r_(1)` and `r_(2)` both will INCREASE
`r_(2)` will REMAIN as it is while `r_(1)` will increase
`r_(2)` will remain as it is while `r_(1)` will decrease

ANSWER :D
Discussion
27.

An electromagnetic wave going through vacuum is described by: E=E_(0) sin(kx - omega t) B=B_(0) sin(kx - omega t)

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`E_(0)B_(0)=omega K`
`E_(0) omega=B_(0)k`
`E_(0)k=B_(0)omega`
None of these

Answer :C
Discussion
28.

{:("List-I","List-2"),((a)"Pressure ",(e)ML^(2)T^(-2)I^(-1)),((b)"Latent heat", (f) M^(0)L^(0)T^(-1)),((c)"Velocity gradient" ,(g) ML^(-1)T^(-2)),((d) "Magnetic flux" ,(h) M^(0)L^(2)T^(-2)):}

Answer»

a - H`""` B- f `""`C -G `""` d-E
a - g`""` b- h `""`c -e `""` d-f
a - g`""` b- h `""`c -f `""` d-e
a - f`""` b- g `""`c -e`""` d-h

Answer :C
Discussion
29.

The mass of a ""_(7)^(3)Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ""_(7)^(3)Li nucleus is nearly

Answer»

46 MeV
5.6 MeV
3.9 MeV
23 MeV

Solution :If `m = 1 u, C = 3 XX 10^(8) ms^(-1)` then,E = 931 MeV
`1 u = 931 MeV`
Binding energy `= 0.42 xx 931`
`= 39.10 MeV`
Binding energy PER nucleon = `(B.E)/(A) = (39.10)/(7) = 5.58 = 5.6 MeV`
Discussion
30.

What is ap-typesemiconductor ? What are the majority and minority and minority charge carriers in it ?

Answer»

Solution :If a trivalent impurity is ADDED to a tetravlent SEMICONDUCTOR, it is CALLED p-typesemiconductor.
In p-typesemicondcutor majority charge carriers are HOLES and minority charge carriers are electrons.
Discussion
31.

Two capacitors of 2muF and 3muFare charged to 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5mu Ffails to the free ends of the wire. Then

Answer»

charge on the `1.5 MUF`capacitors is `180 muF`
charge on the `2muF`capacitor is `120 muC`
charge FLOWS through A from right to left.
charge flows through A from left to right.

Answer :A::B::C
Discussion
32.

In which phenomenon, white light gets divided into its element colours ?

Answer»

REFLECTION
REFRACTION
DISPERSION
Scattering

SOLUTION : Dispersion
Discussion
33.

A body projected vertically with a velocity 'u' from the ground. Its velocity (a)At half of maximum height u/2 b) At 3//4^(th) of maximum height u/sqrt2 (c) At 1//3^(rd) of maximum height sqrt(2/3)u (d) At 1//4^(th) of maximum height sqrt3/2u

Answer»

a and B correct
b is correct
c and d correct
a is correct

Answer :C
Discussion
34.

A bar magnet is kept in the earth's magnetic field with its north pole pointingearth's north. The distance between the null point is 20 cm. If earth's horizontal magnetic field is4 xx 10^(-5)T, then find the magnetic moment of the magnet.

Answer»

Solution :The null points are formed on the equatorial line.
At null point the RESULTANT field is zero .
`:. B_(e) = (mu_0)/( 4pi) (M)/(d^(3))`
` = > 4 xx 10^(-5) = (4 pi xx 10^(-7))/(4pi) xx (M)/(10^(-3))`
`IMPLIES M = (4 xx 10^(-5) xx 10^(-3))/(10^(-7)) = 0.4 Am^(2)`
Discussion
35.

In triode valve, for a grid voltage V_(g) = -1.2 V, the plate current I_(p) (in mA) ad the plate voltage are given by the relation I_(p) = -50 +0.1 V_(p), when the grid voltage is changed to -3.2 V and the plate voltage is kept at 150 V, plate current of 5 mA is observed. Calculate the valve constant and the voltage amplification for 20 k Omega load in the plate circuit.

Answer»


ANSWER :`10^(4) Omega, 2.5 XX 10^(-3) Omega^(-1),25 ; 16.67`
Discussion
36.

ABCDEF is regular hexagon with point O as centre. The value ofoverset(rarr)AB+overset(rarr)AC+overset(rarr)AD+overset(rarr)AE+overset(rarrAF) n xx overset(rarr)AOis. Find n.

Answer»


SOLUTION :`BAR(AB)+barAF=barAO RARR barAB=barAO-barAF`
`barAC=barAB+barAO,barAD=bar2AO,barAE=barAO+barAF`
`-5barAO+barAB+barAF=bar5AO+barAO=6barAO`
Discussion
37.

In the above question if V_(S)(t)=220sqrt2 sin (2pi 50 t), find (a)l(t) ,(b)v_(R) and (c )v_(C)(t)

Answer»

SOLUTION :(a)`i(t)=I_(m)sin(omegat+phi)` =`sqrt2sin(2pi50t+45^(@))`
(b)`V_(R)=i_(R).R =i(t)R =sqrt2xx100 sin(100 pit+45^(@))`
( c)`V_(C)(t)=i_(C)X_(C)` (with a PHASE LAG of `90^(@)`) =`sqrt2xx100 sin (100pit+45-90)`
Discussion
38.

Two bodies one hot an the other cold are kept in vacuum. What will happen to the temperature of the hot body after some time?

Answer»

REMAIN the same
decrease DUE to radiation
increase due to radiation
increase due to convection

Answer :B
Discussion
39.

What is the maximum ac voltage in our household circuit?

Answer»

Solution :For the household circuit, 230V is the rms value of AC and its MAXIMUM value is `230 XX sqrt2 =325V` , which is equivalent to 230V DC SOURCE.
Discussion
40.

The power factor of an a.c. circuit is 0.5. What will be the phase difference between voltage and current in this circuit ?

Answer»

Solution :`therefore COS PHI = 0.5`, HENCE, `phi = cos^(-1)(0.5) = 60^(@)` or `pi/3` RAD.
Discussion
41.

Find the proportion of work done in breaking a big spherical drop of water into n small spherical drops eac h of radius r

Answer»

SOLUTION :`4/3 pi R^3 = N xx 4/3 pir^3`
`therefore r = R/n^(1/3)`
`therefore r^2 = R^2/(n^2/3)`
WORK done = `[4pinR^2/n^(2/3) - 4piR^2] T`
W = `4piR^2T[n^(1/3) - 1]`
`therefore` W = `n^(1/3)` -1
Discussion
42.

In Young's double-slit experiment, one of the slit is wider than other so that the amplitude of the light from one slit is double of that from other slit. If I_m be the maximum intensity, the resultant intensity I when they interfere at phase difference phi is given by:

Answer»

`(I_m)/(9)(1 + 8 COS^2 (theta)/2)`
`(I_m)/(9)(4 + 5 cos^2 phi)`
`(I_m)/(3)(1 + 2 cos^2 (phi)/2)`
`(I_m)/(5)(1 + 3 cos^2 (phi)/2)`

Solution :LET `I_1 and I_2` be intensities which are coming from SLITS `S_1 and S_2`
The resultant intensity
`:. I = I_1 + I_2 + 2sqrt(I_1I_2) cos theta`
`I_2 = K(2A)^2`
`= 4KA^2`
`:. I_2 = 4I_1`

`I_m = I_1 + I_2 + 2sqrt(I_1I_2) cos 0^@`
`= I_1 + 4I_1 + 2sqrt(4I_1^2)`
`= I_1 + 4I_1 + 2 xx 2I_1`
`I_m = 9I_1 implies I_1 = (I_m)/9 implies I_2 = (4I_m)/(9)`
`I = (I_m)/9 + (4I_m)/9 + 2 SQRT((I_m)/(9) xx (4I_m)/(9)) cos theta`
`= (I_m)/(9) + (4I_m)/9 + 2sqrt((4I_m^2)/(81)) cos theta`
`= (I_m)/(9) (5 + 4 cos theta) = (I_m)/9 (1 + 8 cos^2 theta//2)`
Discussion
43.

In problem 22, if a capacitor of capacitance C is also connected in the larger loop as shows in Fig . , find the charge on the capactior as a function of time.

Answer»

Solution :`q = C varepsilon(1 - E^(-t//RC))` where `varepsilon = (mu_(0)pia^(2))/(B)*`
Discussion
44.

Three identical capacitors are first connected in series and then the first and the last condutors of the combination areconnected to earth. A change Q is communicated to the secong conductor of the first capacitor. Prove that the potential of'this conductor is 2Q//3C where C is the capacitor.

Answer»


ANSWER :N//A
Discussion
45.

In a wire of circular cross-section with radius 'r' free electrons travel with a drift velocity v, when a current I flows through the wire. The current in another wire of half the radius and of the same material when the drift velocity is

Answer»

`2I`
`I`
`( I )/(2)`
`(I )/(4)`

ANSWER :C
Discussion
46.

In Young.s double slit experiment with a monochromatic light of wavelength 4000 A^(0), the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength 5000 A^(0) the fringe width will be

Answer»

`0.32 MM`
`0.5 mm`
`0.6 mm`
`0.8 mm`

ANSWER :B
Discussion
47.

A long, straight conductor lies along the axis of a ring. Both carry current I. The force on the ring is proportional to

Answer»

`I`
`I^(3//2)`
`I^(2)`
zero

Answer :A
Discussion
48.

A point light source is attached to the base of an empty trough. A transparent liquid of refractive index 1.35 is filled in the trough to a height of 50 cm. Calculate the total area on the surface of the liquid through which light will emerge.

Answer»

SOLUTION :`0.9533m^(2)`
Discussion
49.

In a binary star system one star has thrice the mass of other. The stars rotate about their common centre of mass then :

Answer»

Both stars have same angular momentum about COMMON CENTRE of mass
Both stars have angular momentum of same MAGNITUDE about common centre of mass
Both stars have same angular speeds
Both stars have same linear speeds

Solution :In a binary star system
`omega_(1)=omega_(2)`
Discussion
50.

Optical and radiotelescopes are built on the ground but X - ray astronomy is possible only from satellites orbiting the earth. Why ?

Answer»

Solution :Because, Earth.s atmosphere is TRANSPARENT to radio WAVES and visible light. But X - rays get ABSORBED by Earth.s atmosphere. Hence we take the help of ..Geo - stationary satellite.., orbitting around the Earth, approximately at the height 36000 km, for the STUDY of ..X - ray astronomy... At this great height atmosphere is quite thin and so X - rays are not absorbed.
Discussion