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.

Ina Galileo's telescope , the inverted image formed by its objective serves as a virtual object for its eyepiece .If the eyepiece has to form an inverted and magnified image of the virtual object , the eyepiece has to be a concave lens and it must be so placed that the virtual object falls

Answer»

WITHIN F
between F and 2F
at 2F
BEYOND 2F

Answer :A
Discussion
2.

If the input to the NOT gate is A = 1011, its output is

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`0100`
`1000`
`1100`
`0011`

ANSWER :A
Discussion
3.

Figure depicts an idealized cycle of a petrol internal combustion engine. The segment 1-2 corresponds to the adiabatic compression of the combustible mixture, segment 2-3, to the isochoric combustion of fuel in the course of which the working fluid receives an amount of heat Q, segment 3-4 corresponds to the adiabatic expansion of the working fluid, segment 4-1, to the isochoric exhaust of spent gases. Express the engine's efficiency in terms of the gas compression ratio x = V_2//V_1.

Answer»


Solution :The efficiency of the cycle is `eta = W//Q`. Since the work in isochoric sections is ZERO, the useful work is equal to the difference between the work of adiabatic expansion and that of adiabatic compression :
`W = m/M C_(mV) (T_3 - T_4)`-
`- m/M C_(mV) (T_2 - T_1) = `
`= m/M C_(mV) (T_3 - T_2 + T_1 - T_4)`
The WORKING MEDIUM receives heat in the process of isochoric combustion of fuel :
`Q = m/M C_(mV) (T_3 - T_2)`
Hence `eta = 1 - (T_4- T_1)/(T_3 - T_2)`
Making use of the result, express the temperatures in terms of the volumes. We have `V_2^(gamma) T_2 = V_1^(gamma - 1) T_1 and V_(2)^(gamma-1) T_3 = V_(1)^(gamma -1) T_4`. Dividing the first equality by the second, we obtain `T_2//T_3 = T_1//T_4`. Transform the expression for the efficiency and reduce it to the FOLLOWING form:
`eta = 1 - (T_4)/(T_3) cdot (1 - (T_1//T_4))/(1- (T_2//T_3))`
But the second fraction is , evidently , unity and the first fraction `T_4//T_3 = (V_2//V_1)^(gamma-1) = x^(gamma -1)`. Hence `eta= 1 - x^(gamma -1)`.
Discussion
4.

Two blocks each having mass m is attached to each other with a spring having force constant k. The system is placed on a rough horizontal surface having coefficient of friction mu. The maximum initial compression, that can be given to thespring so that when released the blocks do not move is

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`2mu mg//k`
`MU mg//k`
ZERO
not possible

ANSWER :B
Discussion
5.

A dip needle free to move in a vertical plane perpendiculr to the magneetic meridian will remain

Answer»

Horizontal
Vertical
At an ANGLE of `60^(@)` to the vertical
At an angle of `45^(@)` to the horizontal

ANSWER :B
Discussion
6.

A lift is going up. The total mass of the lift and the passengers is 1500 kg. The variation in the speed of the lift is given by the graph as shown in figure. a) What will be the tension in the rope pulling the lift at time t equal to i) 1 sec ii) 6 sec iii) 11 sec? b) What is the height to which the lift lakes the passengers? c) What will be the average velocity and the average acceleration during the course of the entire motion?

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Answer :(a) i) `1.8 m//s^(2)` , II) 0 , iii)`-1.8 m//s^(2)` ; (B) 36 m ; (C) 0
Discussion
7.

In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45^(@). The value of C is

Answer»

`(1)/(PI f(2pi f L-R))`
`(1)/(2pi (2pi fL-R))`
`(1)/(pi f(2pi fL + R))`
`(1)/(2pi (2pi f L + R))`

Solution :`tan phi = (X_(C )-X_(L))/(R )rArr tan((pi)/(4))=((1)/(omega C)-omega L)/(R )`
Now,`R = (1)/(omega C)- omega L`
`rArr (R +2pi f L)=(1)/(2pi FC)`or`C=(1)/(2pi f(R +2pi f L))`
Discussion
8.

(A): The possibility of an electric bulb fusing is higher al the time of switching on and off. (R): Inductive effects produce a surge at the time of switch-off and switch-on

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Both A and R are true and R is the CORRECT explanation
Both A and R are true but R is not the correct explanation of A
A is true but R is false
Both A and R are false

Answer :A
Discussion
9.

Maxwell's modified form of Ampere's circuital law is

Answer»

`oint barB.bardl= mu_0 I`
`oint barB.bardl= mu_0 I + mu_0epsilon_0 (dphi_z)/(DT)`
`oint barB.phi s = 0`
`oint barB.bardl= mu_0 I + 1/epsilon_0 (DQ)/(dt)`

ANSWER :B
Discussion
10.

In an experiment for the determination of focal length of the convex mirror a convex lens of focal length 20cm is placed on the optical bench and an object pin is placed at a distance 30cm from the lens. When a convex mirror is introduced in between the lens and the real and inverted image of the object, the final image of object O is formed at O itself. If the distance between the lens and the mirror is 10cm, then the focal length of the mirror is

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`10 CM`
`20 cm`
`24 cm`
`50 cm`

ANSWER :D
Discussion
11.

n-type semiconductor ………

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ENERGY LEVELS of impurity atoms `E_(D)` are nearer to valence energy level `E_(V)`.
energy levels of `E_(D)` are nearer to conduction energy levels `E_(C )`.
energy levels `E_(D)` are in MIDDLE of energy levels `E_(C ) and E_(V)`.
energy level `E_(D)` are on the TOP of energy levels.

ANSWER :B
Discussion
12.

Bohr's atom model presumes that.

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the nucleus is of INFINITE mass and is at rest.
mass of ELECTRON remains constant.
electron in a quantised orbit will not RADIATE energy.
all the above conditions.

Answer :D
Discussion
13.

A flow of 10^(7) electrons per second in a conduction wire constitutes a current of

Answer»

`1.6 XX 10^(-26)` A
`1.6 xx 10^(-2)` A
`1.6 xx 10^(-2)` A
`1.6 xx 10^(-26)` A

SOLUTION :`I = Q/t =(10^(7) xx 1.6 xx 10^(-19))/1 = 1.6 xx 10^(-12)` A
Discussion
14.

…... gave the proof for the existance of electromagnetic wave in the laboratory.

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Hertz
Ampere
GAUSS
MAXWELL

ANSWER :A
Discussion
15.

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

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SOLUTION :`varepsilon=|L.(DI)/(dt)|,""varepsilon=200V,dI=(5-0)A,dt=0.1sec`
`L=(EPSILON)/(((dI)/(dt)))=(200xx0.1)/(5)=4H`
Discussion
16.

Two racing cars of masses m_(1) and m_(2) are moving in circles of radii r_(1) and r_(2) respectively. Their speeds are such that each makes a complete circle in the same time t. The ratio of the angular speed of the first to the second car is

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`r_(1):r_(2)`
`m_(1):m_(2)`
`1:1`
`m_(1)m_(2):r_(1)r_(2)`

ANSWER :C
Discussion
17.

An electrical device draws 2 kW power from AC mains (voltage 223V (rms) V_(rms) = sqrt(50000) V ). The current differs (lags) in phase by (tan phi=-3/4) As compared to voltage. Find (i) R, (ii) X_C - X_L , and (iii) I_M. Another device has twice the values for R, X_C and X_L. How are the answers affected ?

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SOLUTION :Given , power drawn P=2 kW=2000 W
`tan phi =-3/4 , V_"RMS" =V=223` V
`I_M`= maximum current `I_m`=? , R=? , `X_C-X_L` = ?
Power `P=V^2/Z RARR Z=V^2/P=(223xx223)/(2xx10^3)`
`therefore P=24.86 W approx` 25 W
`Z=sqrt(R^2+(X_L-X_C)^2)`
`25=sqrt(R^2+(X_L-X_C)^2)`
`therefore 625=R^2+(X_L-X_C)^2`
but tan `phi=(X_L-X_C)/R`
`therefore 3/4 =(X_L-X_C)/R`
`therefore X_L-X_C=(3R)/4`...(1)
`therefore 625=R^2+((3R)/4)^2`
`therefore 625=R^2+(9R^2)/16=(25R^2)/16`
`therefore R^2=(625xx16)/25`
`therefore R^2=25xx16`
(a) `therefore R=20 Omega`
(b)`X_L-X_C=(3R)/4 = (3xx20)/4 = 15 Omega` [ `because` equ. (1)]
(c) Main current ,
`I_M=I_m=V_m/Z=(sqrt2V_(rms))/Z=(sqrt2xx223)/25`
`therefore I_M`=12.612 A `approx` 12.6 A
If R, `X_L, X_C` are all doubled , tan `phi` CHANGE from `tan phi =(X_L-X_C)/R`
In `Z=sqrt(R^2+(X_L-X_C)^2), R, X_L, X_C` are doubled , then Z is doubled and `I=V/Z` as Z is doubled current I is halved and in power P = VI, V is constant so power is also halved when Z is doubled.
Discussion
18.

The electrical conductivity of a semiconductor increase when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in eV for the semiconductor is

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0.9
0.7
0.5
1.1

Solution :`E_(G)=(HC)/(lambda_max)=(1237.5eVnm)/(2480nm)=0.5eV`
Discussion
19.

A short magnet is placed horizontally in the magnetic meridian with its north pole pointing north. It is then found that there is a neutral point 20 cm from the middle of the magnet. If the magnet is turned through 180^(@), where will the neutral point be? [Hint: When magne is placed with its north place pointing north, neutral point occurs at the broadside-on position. When it is rotated through 180^(@), the neutral point will shift to the end-on position.]

Answer»


ANSWER :`25.2 CM`
Discussion
20.

(A) : The magnetic field of an infinite length of wire carrying current has cylindrical symmetry. (R) : The magnetic field at every point on a circle of radius r,with current carrying conductor as axis is same in magnitude.

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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
'A' is false and 'R' is false

Answer :A
Discussion
21.

Which of the following acts as a circuit protection device ?

Answer»

FUSE
CONDUCTOR
inductor
seitch

Solution :fuse
Knowledge BASED question.
Discussion
22.

Three vectors vec A,vec B "and"vec C are such that vecA*vec B = 0 and vec A*vec C = 0. The vector vec A is parallel to:

Answer»

`VEC B`
`vec C`
`VECB*VECC`
`vecB XX vecC`

Answer :D
Discussion
23.

Which of the following equation represents a wave?

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`(X-vt)^(2)`
`(x+vt)^(2)`
`E^(-x(x-vt)^(2))`
`1/(x+vt)`

Answer :C
Discussion
24.

An electron beam in a TV picture tube is accelerated through a potential difference of V Volts. It passes through a region of transverse magnetic induction field (B) and follows a circular orbit of radius 'r'. The magnetic field (B) is

Answer»

` ( 2mV )/( SQRT(E r))`
`sqrt(( 2mv )/(e r ^2))`
`sqrt((2mV )/(r ))`
`sqrt((2M Vr)/(e ))`

Answer :B
Discussion
25.

For one series LCR A.C. circuit, voltages across L, C and R are each 10 V. Now, if only resistance is made half, new voltages across L, C and R will be .....,.... and ... respectively.

Answer»

10 V , 10 V and 5 V
10 V , 10 V and 10 V
20 V , 20 V and 5 V
20 V , 20 V and 10 V

Solution :At time of resonance , Z=min. =R and so `I=V/R=V/R rArr V=IR` =10 VOLT (given) =`V_R`Also `V_L=IX_L` =10 Volt and `V_C=IX_C`=10 Volt
Now, after making resistance HALF , `R.=R/2` . Hence, new current in the resonance condition is ,
`I.=V/(Z.)=V/(R.)=V/(R//2)=(2V)/R=2I`
(i)`V._R=I.R.=(2V)/RxxR/2 =V` = 10 Volt
(II)`V._L=I.X_L=(2I)(X_L)=2V_L`=20 Volt
(iii) `V._C=I.X_C=(2I)(X_C)=2V_C`= 20 Volt
Discussion
26.

An electron moves a distance of 8cm when accelerated from rest by an electric field of strength 3 xx 10^(4) NC^(-1). Calculate the tme of travel.

Answer»

Solution :Data supplied,
`S=8 xx 10^(-2), u=0, E=3 xx 10^(4) NC^(-1), q=e =1.6 xx 10^(-19)C`
`m=9 xx 10^(-31)kg`
`S=ut+1/2 at^(2)`
`F=qE =ma therefore a=(q E)/(m)`
`a=(1.6 xx 10^(-19) xx 3 xx 10^(8))/(9 xx 10^(-31))=0.533 xx 10^(16) m//s^(2) therefore S=1/2 at^(2) therefore t=sqrt((2S)/(a)`
`t=sqrt((2 xx 8 xx 10^(-2))/(0.533 xx 10^(16)))=5.48 xx 10^(-9)sec`
Discussion
27.

What is toroid? Mention an expression for magnetic field at point inside a toroid.

Answer»

Solution :Toroid : It is hallow circular ring on which a large number of turns of a WIRE are closely wound. The magnetic FIELD at a POINT inside the toroid is B=`mu_(0)nI`
Discussion
28.

Is the electrostatic field pattern shown in the figure possible? (Hint: Assume a rectangular loop inthe field and find the work done by the electric field on a test charge in moving along that loop. Did you arrive at a contradiction?)

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SOLUTION :Not POSSIBLE
Discussion
29.

A small bar of ferromagnetic substance placed in a non-magnetic field then in which direction does it move ?

Answer»

SOLUTION :From WEAK to STRONG MAGNETIC FIELD
Discussion
30.

A transistor having alpha=0.99 and V_BE=0.7V, is given in the circuit. Find the value of the collector current.

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Solution :`V_(BE)=V_(C C)-I_(B)R_(B)`
`0.7=12-I_B(10K)`
`I_B=12-0.7/(10k)=1.13mA`
`beta=alpha/(1-alpha)=0.99/(1-0.99)=99`
`beta=I_c=gtI_C=betaI_B=99xx1.13mA`
`I_C=111.87mA`
Discussion
31.

When a.c. circuit with L, C, R in series are brought into resonance, the current has large value. Why? If the capacitance C is increased, will current increase or decrease ? Explain.

Answer»

Solution :The impedance of a series of a series LCR CIRCUIT is given by
`Z = sqrt(R^2 + (omegaL - 1/(OMEGAC))^2` when the circuit is brought into resonance
Discussion
32.

Electric field outside a long wire carrying charge q is proportional to

Answer»

`1/R`
`1/(r^2)`
`1/(r^(3//5))`
`1/(r^(3//2)`

ANSWER :A
Discussion
33.

In single slit diffraction experiment, the width of the central maximum inversely proportional to

Answer»

Distance between source and screen
Slit width
Wavelength of LIGHT used
Both (1) & (3)

Answer :B
Discussion
34.

A ray of light after refraction through a concave lens travels parallel to its principal axis. By drawing a ray diagram, state the condition for it to occur ?

Answer»

Solution :Light rays directed towards the FIRST focus of the GIVEN CONCAVE lens will travel PARALLEL to its principal AXIS after refraction from the lens as shown in Fig. 9.29.
Discussion
35.

The energy of infrared waves is greater than that of

Answer»

a. VISIBLE LIGHT
B. ULTRAVIOLET waves
c. X-rays
d. micro waves

Answer :D
Discussion
36.

A short circuited coil is placed in a time varying magnetic field. Electrical power is dissipated in the form of Joule heat due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be .........

Answer»

halved
the same
doubled
quadrupled

Solution :LET `EPSILON=-N . (dphi)/(dt)`, here if `-(dphi)/(dt)` then `epsilon prop N` …(1)
and `R=(rhol)/A` is constant in `rho/pi`
`therefore R prop L/r^2` …(2)
And power `P=epsilon^2/R=(N^2r^2)/l`
`therefore P_1/P_2=(N_1/N_2)^2xx(r_1/r_2)^2xx(l_2/l_1)`...(3)
Now by constant volume , increasing the length of wire from `l_1` to `l_2` its radius BECOMES from `r_1` to `r_2`
`therefore pir_1^2l_1 = pir_2^2l^2`
`therefore (r_1/r_2)^2 =l_2/l_1`....(4)
From EQN. (3) and (4) ,
`therefore P_1/P_2=(N_1/N_2)^2xx(r_1/r_2)^4`...(5)
If `N_2=4N_1` and `r_2=r_1/2` then,
`P_1/P_2=(1/4)^2 xx(2)^4 =1:1`
Discussion
37.

A proton of charge e and mass m enters a uniform magnetic field B = Bi with an initial velocity v=v_xhati+v_yhatj. Find an expression in unit vector notation for its velocity at time t.

Answer»


ANSWER :`VECV(t)=v_(0X)hati+v_(0y)(sin OMEGAT)hatk`
Discussion
38.

{:("(i) Artificial magnet","(a) Tangent Galvanometer"),("(ii) Magnetic flux", "(b) Desired shape"),("(iii) high energy accelerator","(c ) Weber"),("(iv) Measurement of small current","(d) Cyclotron"):}

Answer»


ANSWER :(i)`RARR` (d); (ii) `rarr` (a); (iii) `rarr` (b) ; (iv)` rarr` (C)
Discussion
39.

A short bar magnet passes at a steady speed right through a long solenoid . A galvanometer is connect across the solenoid. Which graph best represents the variation of the galvanometers deflection theta with time:-

Answer»




ANSWER :a
Discussion
40.

Name the electromagnetic radiations having the wavelength ranging from 10^(-11) m to 10^(-8)m Give it's two imortant applications.

Answer»

Solution :x-rays, They are used in (i) medical diagnosis and (II) In the study of CRYSTAL STRUCTURE,
Discussion
41.

A body of mass 5 kg is at rest. Three force F_(1) = 10 N due North F_2=10N along East and F_(3) = 10 sqrt(2) Nalong N-W act on it simultaneously. The acceleration produced in the body is,

Answer»

`4 MS^(-2) ` along NORTH
`2ms^(-2)` along North
`4ms^(-2) ` along East
`2ms^(2)` along East

Answer :A
Discussion
42.

The electric field lines of a positive charge is as shown in figure:Give the sign of potential difference V_p-V_q.

Answer»

SOLUTION :POSITIVE
Discussion
43.

A marble block of mass 2 kg lying on ice when given a velocity of 6 m//s is stopped by friction in 10 s. Then the coefficient of friction is:

Answer»

0.01
0.02
0.03
0.06

Solution :Here applying `nu= U + at`
`0 = 6 + AXX 10:. a =-0.6 MS^(-2)`
Also `mu mg = ma` or`mu=(a)/(g) =( 0.6)/(10) = 0.06`
HENCE choice is (d).
Discussion
44.

Explain the use of transformer in long distance transmission of electric power .

Answer»

SOLUTION :At the transmitting point, the VOLTAGE is INCREASED and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the DESTINATION without any appreciable LOSS. At the receiving point, the voltage is decreased and the current is increased to appropriate and then it is given to consumers.
Discussion
45.

A loaded truck of mass 3000 kg moves on a level road at a constant speed of 6.00 0 m/s. The frictional force on the truck from the road is 1000 N. Assume that air drag is negligible. (c) What is the total work done by the engine in the full 20 min ?

Answer»

Solution :KEY IDEA
Adding up the results from (a) and (b) , we can find the total WORK done by engine in 20 minutes.
CALCULATIONS: The total work done by the engine in 20 minutes is
`W_("tot")= W + W. = 3.600xx 50 J + 5.604 xx 10J`
` = 5.964 xx 10J`
or about 60 MJ.
Discussion
46.

Two similar bar magnets p and q each opf magnetic moment m are taken if p is cut along its axial lineand qis cut along its equatorial lineall the four pieaesobtainedhave

Answer»

EQUAL polestrength
MAGNETIC MOMENT `(m)/(4)`
magnetic moment `(m)/(2)`
magnetic moment m

Answer :C
Discussion
47.

Match the following : {:((A),"Rocket proplusiosn",(P),"Bernoulli's principle in fluid dynamics"),((B),"Aeroplane",(Q),"To"),((C),"Fungus",(R),"Abscisic acid" II),((D),"Herring maize grains",(S),"Kinetin"),(,,(T),"Photoelectri effect"):}

Answer»

`{:(A,B,C,D),(R,Q,P,S):}`
`{:(A,B,C,D),(R,P,Q,T):}`
`{:(A,B,C,D),(T,P,Q,R):}`
`{:(A,B,C,D),(R,P,Q,S):}`

SOLUTION :(A) Newton's law of MOTION is used in rocket propulsion
(B) Bernoulli's PRINCIPLE of fluid dyamics is used in aeroplane.
( C) Totalinternal reflection of light is used in optical fibres.
(D) MAGNETIC confinement of plasma is used in fusion test reactor.
Discussion
48.

The minute hand of a wall clock measures 12 cm from its tip to the axis about which it rotates. The magnitude and angle of the displacement vector of the tip are to be determined for three time intervals. What are the (a) magnitude and (b) angle from a quarter after the hour to half past, the ( c) magnitude and (d) angle for the next half hour, and the (e) magnitude and (f) angle for the hour after that?

Answer»

SOLUTION :(a) 17 CM, (b) `-135^(@)`, ( C) 24 cm, (d) `90^(@)`, (e) zero, (F) zero
Discussion
49.

How can the sensitivity of a pontentiometer be increased?

Answer»

Solution :By INCREASING the length of the pontentiometer WISE.
By REDUCING the current in the PRIMARY circuit.(USING rheostat or a series resistor)
Discussion
50.

Theequivalent resistance of two resistors connected in series is 6 Omega and their parallel equivalent resistance is 4/3 Omega. What are the values of resistances ?

Answer»

`4Omega, 6 Omega`
`8 Omega, 1 Omega`
`4 Omega, 2 Omega`
`6 Omega, 2 Omega`

Solution :`R_(1) + R_(2) = 6`
`(R_(1)R_(2))/(R_(1) = R_(2)) = (4)/(3) RARR R_(1)R_(2) = 8 Omega rArr R_(1) = 2Omega, R_(2) = 4 Omega`
Discussion