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.

In Fig. 27-40, the resistors have the values R_1= 700 Omega, R_2 =12.0 Omega, and R_3= 4.00Omega and the ideal battery's emf is 22.0 V. For what value of R_4 Will the rate at which the battery transfers energy to the resistors equal (a) 60.0 W, (b) the maximum possible rate P_(max) and and (c) the minimum possible rate P_("min")? What are (d) P_(max) and (e) P_("min")?

Answer»


ANSWER :(a) `1.66 OMEGA; (B) 0; (C) oo; (d) 69.1 W; (e) 48.4W`
2.

Name the optoelectronic device used for detecting optical signals and mention the biasing in which it is operated. Draw its I-V characteristics.

Answer»


ANSWER :PHOTODIODE (1/2) REVERSE BIASING (1/2)
1-V CHARACTERISTICS
3.

Match List-I with List-II

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`a-h, b-g, c-e, d-f`
`a-h, b-g, c-f, d-e`
`a-g, b-h, c-e, d-f`
`a-e, b-f, c-g, d-h`

ANSWER :A
4.

In a meter bridge the null oint is found at a distance of 33.7 cm from A. If now a resistance of 12 Omega is connected in parallel with S, the null point occurs at 51.9cm. Determine the values of R and S.

Answer»

Solution :From the first balance POINT, we have
` R/S = (33.7)/(66.3)`
After S is connected in parallel with a RESISTANCE of `12 Omega,` the ressistance across the gap charges from S to `S _(EQ),` where
`S _(eq) = (12 S)/(S +12)`
and HENCE the new BALENCE conditin now gives
`(51.9)/(48.1 ) = (R)/(S _(eq)) = (R (S + 12))/( 12 S)`
Substtuting the vblaue of R/s from Eq. `(3.87),` we get
` (51.9)/(48.1) = (S + 12)/(12) . (33.7)/(66.3)`
which gives `S = 13.5 Omega .` Using the value of R/S above, we get `R = 6.86 Omega.`
5.

A gas at NTP is slowly compressed to one-fourth of its original volume, then the final pressure is

Answer»

4 atmospheres
8 atmospheres
3/2 atmospheres
1/4 atmospheres

Answer :C
6.

The ratio of one nanometer to 1 attometer is

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`10^6`
`10^8`
`10^7`
`10^9`

ANSWER :A
7.

An LC circuit contains a 20 mH inductor and a 50 muF capacitor with an initial charge of 10 mC. The resistance of circuit is negligible. Let the instant the circuit is closed be t = 0. What is the total energy stored initially ? Is it conserved during LC oscillations ?

Answer»

Solution :(a) 1.0 J. Yes, SUM of the ENERGIES stored in L and C is conserved if R = 0.
(b) `omega=10^(3)rad s^(-1), v=159Hz`
(c ) `q=q_(0)cosomega t`
(i) Energy stored is completely electrical at `t=0, (T)/(2), T, (3T)/(2), ……`
(ii) Energy stored is completely magnetic (i.e., electrical energy is zero) at `t=(T)/(4), (3T)/(4), (5T)/(4).......`, where `T=(1)/(v)=6.3ms`.
(d) At `(T)/(8), (3T)/(8), (5T)/(8) ..........`, because `q=q_(0)"cos"(OMEGAT)/(8) =q_(0)"cos"(pi)/(4)=(q_(0))/(sqrt(2))`.
Therefore, electrical energy `=(q^(2))/(2C)=(1)/(2)((q_(0)^(2))/(2C))` which is half the total energy.
R damps out the LC oscillations eventually. The whole of the initial energy (= 1.0 J) is eventually dissipated as heat.
8.

Two isolated, charged coducting spheres of radii R_(1) and R_(2) produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is

Answer»

`(R_(1))/(R_(2))`
`(R_(2))/(R_(1))`
`(R_(1)^(2))/(R_(2)^(2))`
`(R_(2)^(2))/(R_(2)^(1))`

ANSWER :A::B
9.

In an experiment on photoelectric effect.the slope of the cutoff voltage versus frequency incident light is found to be 4.12xx10^(-15) V sec calculate the value of Planck's constant.

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`6.625xx10^(-34)JS`
`6.63xx10^(-34)Js`
`6.592xx10^(-34)Js`
`6.692xx10^(-34)Js`

ANSWER :C
10.

If at a pressure of 10^(6) dyne//cm^(2), one gram mole of nitrogen occupies 2xx10^(4) cc volume, the calculate the averageenergy of a nitrogen molecules in erg.(Given avogadro's number =6xx10^(23))

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`14xx10^(-13)`
`10xx10^(-12)`
`10^(6)`
`2xx10^(6)`

Solution : Average `KE =(1)/(2)mV^(2)_(rms)=(1)/(2)m(3p)/(p)=(1)/(2)(M)/(N)XX(3P)/(p)`
`=(1)/(2)xx(28)/(6XX10^(23))(3xx10^(6))/((1)/(2xx10^(4)))`
`=14xx10^(-13)erg`
11.

The area A of a parallel plate capacitor is divided into two equal, halves and filled with two media of dielectric constants K_1 and K_2 respectively. The capacitance of the capacitor will be

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`(epsi_0 A (K_1 + K_2))/d`
`(epsi_0A)/d ((K_1 + K_2)/2)`
`(epsi_0A)/d. (K_1K_2)/((K_1 + K_2))`
`(epsi_0A)/(2D). (K_1 K_2)/((K_1 +K_2))`

Solution :The given arrangement is a COMBINATION of two capacitors in parallel whose CAPACITANCES are
`C_1 = (K_1 epsi_0 (A/2))/d andC_2 = (K_2 epsi_0(A/2))/drArrC = C_1 + C_2 = (epsi_0A)/d ((K_1 +K_2)/2)`
12.

A ray of light incident normally on a refracting surface of the prism is totally reflected from the other refracting surface. If the prism is immersed in water how will the ray act? Refractive index of glass = 1.5, refractive index of water = 1.33.

Answer»

Solution :As the ray of light is incident on a FACE of the prism normally, so it goes straight through the surface. The ray is incident on the second refracting face and is TOTALLY reflected. So the angle of incidence of the ray at the second face is greater than the critical angle `(theta_(c))`.
`sintheta_(c) = (1)/(a^(mu)g) = (1)/(1.5) = 0.667`
`=sin41.8^(@)`
`THEREFORE "" theta_(c) = 41.8^(@)`
Now, if the prism is immersed in water, refractive index of glass with respect to water is,
`w^(mu)g = (a^(mu)g)/(a^(mu)w) = (1.5)/(1.33) = 1.128`
In this case if the critical angle is `theta_(c)`, then
`sintheta._(c) = (1)/(w^(mu)g) = (1)/(1.128) = 0.8865`
`= sin 62.44^(@)`
`or, "" theta._(c) = 62.44^(@)`
So the angle of incidence of the ray at the second face `(41.8^(@))` is lessthan the critical angle `(62.44^(@))`. Hence, instead of being totally reflected from the second face, the ray is REFRACTED throughit.
13.

We get ….. Current through rectifier without simple filter.

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CHANGING direct
CONSTANT direct
ALTERNATING CURRENT
A.C.

ANSWER :A
14.

In the caseof interference , the maximum and minimum intensities are in the ratio 16:9 .Then

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The maximum and MINIMUM amplitudes will be in the ratio `9:5`
The intensities of the individual WAVES will be in the ratio `4:3`
The amplitudes of the individual waves will be in the ratio `7:1`
The amplitudes of the individual waves will be in the ratio `4:1`

Answer :C
15.

To emit a free electron from a metal surface a minimum amount of energy must be supplied. State three methods to supply this energy to the free electrons.

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SOLUTION :FIELD EMISSION, PHOTOELECTRIC emission and THERMIONIC emission.
16.

The dimensions of sqrt(mu epsilon) are the same as that of :

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Velocity
Acceleration
1/Velocity
1/Acceleration

Answer :C
17.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Answer»

SOLUTION :Here `I_(1), = 5.0 A, I_(2) = 0.0A, 1 = 0.1 s and VAREPSILON = 200 V`
as `varepsilon=L(I_(1)-I_(2))/timplies L=(varepsilont)/(I_(1)-I_(2))=(200Vxx0.1s)/(5.0A-0.0A)=4.0H.`
18.

The coefficieents of absorption and transmission of the are 0.50 and 0.25 respectively. If 200 alories of radiant heat is incident on the surface of the body, the quantity of heat reflected will be

Answer»

140 cal
150 cal
50 cal
200 cal

Solution :`a+R+t=1`
`r=1-(a+t)=1-0.75=0.25`
`Q_(r)=rQ=0.25xx200=50cal`
19.

Did she get a formal education?

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Yes
No
Maybe
I don't know

Answer :B
20.

In the pulley system shown in figure. P and Q are fixed pulleys while A, B and C are movable pulleys each of mass 1kg. The strings are vertical and inextensible. Find the tension in the string and acceleration of frictionless pulleys A, B and C

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Solution :A single string whose ends are tied to CENTRES of A and B, passes over all the pulleys, so tension at each POINT of string is same equal to T. Weight of each pulley A, B and C is
mg = 1 g Newton.
Let `y_A ,y_B and y_C` be the DISTANCES of centres of pulleys A, B and C from FIXED pulleys at any time t. Following the string starting from end A and reaching upto end B, we have
`(y_B - y_A) + y_B + 2y_A + y_C + y_C- y_B = L`= constant
i.e.,` y_A + y_B + 2y_C = L = ` contant
DIfferentitating twice with respect to t , we get
` (d^(2)y_A)/(dt^(2)) + (d^(2)y_B)/(dt^(2)) + 2(d^(2)y_C)/(dt^(2)) =0`
i.e., `a_A + a_B + 2a_C =0` ...... (1)
where ` a_A , a_Band a_C` are acceleration of pulley A,B,C respectively .
Now equations of motion of pulleys A,B and C are
`mg + T-2T = ma_A implies mg-T =ma_A`............ (2)
`mg + T-2T = ma_B implies mg -T = ma_B`......... (3) and
`mg-2T = ma_C`
From (2) and (3) it is obvious that
and ` a_A = a_B = (g-(T)/(M)) `............ (5) and from (4) , `a_C = g-(2T)/(m)`............. (6)
susbtituting `a_A , a_B and a_C` in (1) , we get
`(g-(T)/(m)) + (g-(T)/(m)) + 2(g-(2T)/(m)) =0`
` 4g- (6T)/(m) =0 implies T = (2)/(3) mg = (2)/(3) xx 1 xx 9.8 = 6.5 N`
`:. a_A =a_B = (g-(T)/(m)) = 9.8 -(6.5)/(1) = 3.3 m//s^(2)`
From (1) , `ac=-a_A = -3.3 m//s^(2)`
21.

A sphere is moving on a smooth surface with angular velocity omega and linear velocity v. If collides elastically with another identical sphere B kept at rest. Neglecting friction everywhere, let V_(A) and omega_(A) be linear and angular velocities of sphere A after the impact and let V_(B) and omega_(B) be the corresponding parameters for spehre B. Then

Answer»

`V_(A) =0, omega_(A) = 0`
`V_(B) = V, omega_(B) = 0`
`V_(B) =0, omega_(B) = OMEGA`
`omega_(A) = 0 , omega_(B) = omega`

ANSWER :B
22.

A particle executing S.H.M. has maximum velocity alpha and maximum acceleration beta, the period of oscillation shall be

Answer»

`2PI ( alpha//beta)`
`2pi ( beta // alpha)`
`2 PI ( alpha // beta)^(1//2)`
`2 pi ( beta // alpha)^(1//2)`

ANSWER :A
23.

Estimate the porportion of boron impurity whichwill increase the conductivity of a pure silicon sample by a factor of 100.Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same tempareture is7xx10^15holes per cubic metre. Density of silicon is 5xx10^28atoms per cubic metre.

Answer»

Solution :Total number of charge carriers initially
`= 2xx7xx10^15`
`= 14xx10^15/(Cubic meter)`
finally the total number of charge carriers
`= 14xx10^(17)/m^(3)`
We know the product of the concentrations of holes and conduction electrons remains almost the same.
Let 'x' be the number of holes.
So `(7XX10^(15))xx(7xx 10^(15))`
`= x xx(14^(17)-x)`
`rArr 14X xx10^17-x^2 = 49xx10^30`
`rArr x^2-14x xx10^17-49xx10^30 = 0`
`rArr x = (14xx10^17+-(14)^2xxsqrt(10^34+4xx49xx10^30))/(2)`
` = 10^17+-sqrt(10^34+4xx49xx10^30))/(2)`
` = (28.0007)/(2)xx10^17 = 14.00035xx10^17`
= increase in number of boles or the
number of atoms of BORON added
Now, `1386.035xx10^15 ATOM of `Si` in `1 m^3`.
`1` atom of Boron is added per
`(5xx10^28)/(1386.035xx10^15)`
`3.607xx10^15xx10^13`
`3.607xx10^10`
24.

A fixed mass of a gas is taken through a process A to B to C to A . Here AtoB isisobaric. BtoC is aiabatic and CtoA is isothermal. Find eficiency of the process (take gamma=1.5)

Answer»


ANSWER :`(3-2 In 2)/3`
25.

यदि A={1, 2, 3} तथा B={3, 4, 1} हो तो (A- B)uu(B-A) होगा

Answer»

(1, 2}
(2, 3}
{1, 2, 3, 4}
{2,4}

ANSWER :D
26.

Beats are produced by two waves y_(1)=asin2000pit, and y^(2)=asin2008pit. The number of beats heard per second is :

Answer»

Zero
One
Four
Eight.

Answer :C
27.

Define the terms 'depletion layer and barrier potential' for a p-n junction. How does (i) an increase in the doping concentration, and(ii) biasing across the junction, affect the width of the depletion layer ?

Answer»

Solution :For 'DEPLETION LAYER' and barrier potential', see Short ANSWER QUESTION Number 5.
(i) The width of the depletion layer decreases on increasing the doping concentration.
(ii) In forward biasing arrangement the width of depletion layer decreases but in reverse biasing the width of depletion layer increases.
28.

A spherical cavity is made in a lead sphere of radius R such that its surface touches the outsides surface of lead sphere and passes through the centre. The shift in the centre of mass of the lead sphere as a result of this following, is

Answer»

`(R)/(7)`
`(R)/(14)`
`(R)/(2)`
R

Answer :B
29.

Find the value of current( rms and peak) through a capacitor of 10 mu Fwhen connected to a source of 110 V at 50 cycles supply. What is its capacitive reactance?

Answer»

Solution :Data supplied `C= 10 mu F = 10 xx 10^(-6) F, E_(rms) = 110 V , v = 50 Hz`
` X_C = (1)/(omega C)= (1)/(2pi v C) = (1)/(2 xx 3.14 xx 50 xx 10^(-5)) = 318.5 Omega`
` I_(rms) = (E_(rms) )/(X_C) = (110)/(318.5) = 0.345 A`
30.

In an A.C. circuit of inductance 50 mH with negligible resistance and a capacitor of capacitance 500 pF are connected in series. The resonance frequency for the given circuit is ……….

Answer»

`10^5/pi` HZ
`1/pi` Hz
`100/pi` Hz
`1000/pi` Hz

Solution :`L=50xx10^(-3) H, C= 500xx10^(-12)` F
`f=1/(2pisqrt(LC))`
`=1/(2pisqrt(50xx10^(-3)xx500xx10^(-12)))`
`=10^6/(10pi)`
`THEREFORE f=10^5/pi` Hz
31.

...... Oscillates in LC circuit.

Answer»

Inductor
Capacitor
Waves
Charge

Solution :In LC CIRCUIT, charge and current are oscillating as per following EQUATIONS.
`q=q_0 COS (omegat)`
`i=-q_0 wsin(omegat)`
32.

Consider the situation shown in the figure. If the current I in the long straight conducting wire XY is increased at a steady rate then the induced e.m.f..s in loops A and B will be

Answer»

clock WISE in A, ANTI clockwise in B
anti clock wise in A, clockwise in B
clockwise in both A and B
anti clockwise in both A and B

Answer :A
33.

Find the value of R in fig. so that there is no current in the 15 Omega resistor

Answer»

Solution :This is the WHEATSTONE bridge with the galvanometer replaced by `15 OMEGA` RESISTOR. The bridge is balanced because there is no CURRENT is 15 `Omega` resistor, hence,
`20//10 = 40//R`
`THEREFORE R= (40 xx 10 Omega)/(20) = 20 Omega`
34.

A uniform wire which connected in parallel with the 2 m long wire, will give a resistance of 2.0 Omega.

Answer»


ANSWER :20 m
35.

If V_0 be the potential at origin in an electric field E =E_x i+E_y J the potential at (x,y) is

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`V_0 = xE_x - yE_y`
`V_0+xE_x +y_y`
`xE_x + yE_y -V_0`
`SQRT(X^2+ y^2 )sqrt(E_(x)^(2) +E_(y)^(2 ) -V_0)`

ANSWER :A
36.

A : In Young.s double slit experiment the two slits are at distance d apart. Interference pattern is observed on a screen at distance D from the opposite to one of the slits, a dark fringe is observed, then the wavelength of wave is proportional to square of distance of two slits. R : For a dark fringe intensity is zero in double slit experiment.

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 and R is false
Both A and R are false

Answer :B
37.

What is the meaning of the term cajole ?

Answer»

To PERSUADE someone by being very NICE to them
To persuade someone by threatening them
Whispering into somebody's ear
None of the above

Answer :A
38.

What is nucleus ?

Answer»



SOLUTION :The central region of an atom CONTAINING whole mass and whole charge of an atom is called ATOMIC NUCLEUS.
39.

Find the radiation power developed by a non-relativistic particle with charge e and mass m, moving along a circular orbit of radius R in the field of a stationary point charge q.

Answer»

Solution :Here
`dotoversetrarr(p)=(E)/(m) xx force = (e^(2)q)/(MR^(2)) (1)/(4piepsilon_(0))`.
Thus `P = (1)/((4piepsilon_(0))^(3)) ((e^(2)q)/(mR^(2)))^(2)(2)/(3C^(2))`.
40.

What is nuclear charge ?

Answer»

Solution :The charge of the NUCLEUS=`Zxx`charge of proton =Ze,
where Z is ATOMIC NUMBER.
41.

A uniform constant magnetic field B is directed at an angle of 45^@ to the x-axis in xy-plane. PQRS is a rigid square wire frame carrying a steady current i with its centre at the origin O. At time t=0, the frame is at rest in the position shown in the figure, with its side parallel to x and y axis. Each side of the frame is of mass M and length L. (a) What is the torque vec tau about O acting on the frame due to magnetic field? (b) Find the angle by which the frame rotates under the action of this torque in a short interval of time Delta t, and the axis about which this rotation occurs (Delta t is so short that any variation in the torque during this interval may be neglected).

Answer»

Solution :` tau= (i_0L^2B)/(SQRT(2))(- hat i + hatj)`
`(B) 3/4 (i_0B)/(M)(DELTAT)^2`
42.

When a weight of 10kg is suspended from a copper wire of length 3m and diamerer 0.4 mm,it's length increases by 2.4cm if the diameter of the wire is doubledthen extension in the wire will be how much?

Answer»

Solution :`E PROP 1/r^2 THEREFOR e_1//e_2 = r_1^2//r_2^2 = ((0.2)/(0.4))^2 = 1/4`
`therefor `e_2 = e_1/4 = 2.4/4 0.6 cm.
43.

A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle theta, the spot of the light is found to move through a distance y on the scale. The angle theta is given by

Answer»

`y/x`
`(x)/(2y)`
`x/y`
`(y)/(2x)`

SOLUTION :
When plane mirror is given rotation of `theta`, then REFLECTED ray will have rotation `2THETA`. In figure,
In right angle triangle tan`2theta=y/x`
For smaller angle, `tan2theta ~~2 theta`
`therefore 2 theta=y/x therefore theta=(y)/(2x)`
44.

From sunrise to sunset, the sun subtends an angle of 180^(@) to our eyes. What will be the value of this angle to an observe under water?

Answer»


ANSWER :`98^(@)`
45.

In Bohr's model of the hydrogen atom, the pairs of quantities that are quantized among the following is

Answer»

ENERGY and linear MOMENTUM
ANGULAR and linear momentum
Energy and ANGULARMOMENTUM
None of these

Answer :C
46.

In a potentiometer, a standard cell of emf 5V of negligible internal resistance maintains a steady current through Potentiometer wire of length 5m. Two primary cells of emf E_1 and E_2arejoinedin serieswith(i)same polarity (ii) opposite polarity. The balancing point are found at length 350 cm and 50 cm in two cases respectively.(i) Draw necessarycircuitdiagram. (ii)Findthevalue ofemfE_1and E_2ofthetwo cells(ifE_1gt E_2)

Answer»


Solution :Pot. GRADIENT `k=(5)/(3)= 1 VM ^(-1) `
`L _1 = 350cm=3.5 m `
`E_1 + E_2=kl_1=3.5 "" `…(1)
` E_1- E_2 =0.5"" `…(2)
` E_1=2V, E_2 = 1.5 ` VOLT
47.

When an n-p-n transistor is used as an amplifier

Answer»

HOLES move from EMITTER to base
holes move from base to emitter
ELECTRONS move from base to COLLECTOR
electrons move from collector to base

Answer :D
48.

The potentiometer is an ……as if measures the emf of a cell very accurately

Answer»


ANSWER :IDEAL VOLTMETER
49.

A moving coil type of galvanometer is based upon the principle that a current carrying loop in a magnetic field experiences a net

Answer»

torque
force
impulse
all the above

Answer :A
50.

If copper has moduls of rigidity 12xx10^(10)N//m^(3)and Bulk muodulus 12xx10^(11)N/m and density 9 g//cm^(3)then find the velocity of longitudinal wave, when set-up in solid copper.

Answer»

4389 m/s
5000 m/s
4000 m/s
4300 m/s

Answer :A