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.

Three identical parallel conducting plates A, B and C are placed as shown. Switches S_(1) and S_(2)are opened and connect A and C to earth when closed. +Q charge is given to B.

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If `S_1`is closed with `S_2`open, a charge of amount will PASS through `S_1`
If `S_2`is closed with `S_1`open, a charge of amount Q will pass through `S_2`
If `S_1 and S_2`are closed TOGETHER, a charge of amount `Q/3` will pass through `S_(1)`and a charge of amount `(2Q)/3` will pass through `S_2`
All the above statements are incorrect

Answer :A::B::C
2.

Obtain an expression for capacitance of this arrangement

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SOLUTION :`[C=A(epslion_0)/d]`
3.

Sodium and copper have work functions 2.3eV and 4.5eV respectively, then the ratio of the threshold wavelengths is nearest to

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`1:2`
`4:1`
`2:1`
`1:4`

ANSWER :C
4.

For three resistors of different value connected in series obtain equation of equivalent resistance. From this write equation of n resistors connected in series.

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Solution :`R_(1) , R_(2) and R_(3)` are connected with battery of V volt between A and B current in the circuit is I.

Potential difference across `R_(1) , R_(2) and R_(3) ` be `V_(1) , V_(2) and V_(3)`.
`rArr "" V_(1) = I R_(1), V_(2) = IR_(2) and V_(3) = IR_(3)`
`rArr` terminal voltage of battery,
`V = V_(1) + V_(2) + V_(3)`
`therefore V = IR_(1) + IR_(2) + IR_(3)`
`therefore (V)/(I) = R_(1) + R_(2) + R_(3)`
`rArr` But `(V)/(I)` is EQUIVALENT resistance ov resistors connected in series.
`therefore(V)/(I) = R_(EQ)`
`therefore R_(eq) = R_(1) + R_(2) + R_(3)`
If n resistance are connected in series then,
`R_(eq) = R_(1) + R_(2) + .... +R_(n)`
`rArr` If n resistors of EQUAL value R are connected in series then,
`R_(eq) = nR`
`rArr` In series connection of resistors equivalent resistance is LARGER than largest value.
5.

Draw and explain the output waveform across the load resistor R, if the input waveform is as shown in the given.

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Solution :The OUTPUT waveforms have been shown in the . It is on account of the fact that a p-n junction diode conducts only when it is in forward bias arrangement but does not conduct in the reverse bias arrangement. During the POSITIVE half-cycle of input the p-n junction diode isin forward bias. Hence, a current flows through the resistor R and an output VOLTAGE is obtained ACROSS it. During the NEGATIVE half-cycle of input the diode, being in reverse bias, does not allow flow of current and hence output voltage across load resistor is zero.
6.

Plane polarised light is incident on an analyser . The intensity then becomes three fourth . The angle of the axis of the analyser with the beam is

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`30^(@)`
`45^(@)`
`60^(@)`
`0^(@)`

Answer :C
7.

A beam of light consisting of two wavelengths 6500A^(0)" and "5200A^(0) is used to obtain interference fringes in a Young.s double slit experiment. Find the distance of the third bright fringe on the screen fromthe central maximum for wavelenght 6500A^(0).

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Solution :The distance of the NTH BRIGHT fringe from the central maximum
`y_(m)= (m LAMBDA L)/(d), y_(3)= (3lambda L)/(d)= (3XX(6500xx 10^(-10))xx1.20)/(2xx 10^(-3))= 1.17mm`
8.

A metallic rod of length 'I' is rotated with angular frequency of 'omega' with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 'l', about an axis passing through its centre and perpendicular to the plane of ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

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<P>

Solution :Consider a conducting rod PQ of length L, hinged at one end P, and other end Q at the circumference of a circular metallic ring. Let the rod rotates at a uniform angular speed `omega` normal to a uniform magnetic field B.
`therefore` Induced emf between the ends of rod i.e., between the centre and the metallic ring
`varepsilon=(dphi_(B))/dt =B|(dvecA)/(dt)|`
`=Bxx(pil^(2))/T`
(where T=time to complete one revolution `=2pi//omega))`
9.

Calculate the frequency at which the inductive reactance of 0.7 H inductor is 220 Omega.

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ANSWER :50 HZ
10.

What is photon ? Write formula for its energy.

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Solution :PHOTON. It is, an energy particle (or PACKET) EMITTED by a source of radiation. Energy of photon is given by:
`E = hv =(hc)/LAMBDA.`
11.

Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a negative charge starts on the negative x-axis at a large distance from O, moves along the x-axis, passed through O and moves far away from O. Ita acceleration a is taken as positive along its direction of motion. The particle accleration a is plotter against its x-co-ordinate. Which of the following best represents the plot?

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Solution :When the particle is to the left of the ORIGIN O, the net force F on it is towards O and hence, cause acceleration a `F= 2F_(0) cos theta`
`=2k (qQ)/((x^(2) + 1^(2))).(x)/(sqrt(x^(2) + 1^(2)))`
`=2kQ q(x)/((x^(2) + 1^(2))^(3//2))`
F is zero for large values of x and also for x=0. Thus, it must increase to a maximum and fall to zero at O. The acceleration a=F/m, must have the same nature. To the right of O, the net force is to the left while MOTION is to the right. Thus, the direction of a is OPPOSITE to the particle.s direction of motion and is taken as negative. The VARIATION of a with x will follow the same pattern. Hence, 2nd graph REPRESENTS the pattern.
12.

A piston is performing S.H.M. in the vertical direction with a frequency of 0.5 Hz. A block of 10 kg is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is.

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1.5m
1M
0.1m
0.5m

SOLUTION :When the block boxes contact withi surface, R=0 & ndistance Y.= A
So` |a| = omega^2 N IMPLIES g =omega^2A `
` A = (g)/(4pi^2f^2) = (10)/(pi^2) =1m`
13.

In a L-C-R series A.C. circuit, the voltage across each of the components, L, C and R is 50V. The voltage across the L-C combination will be ......

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50V
`50sqrt2` V
100 V
0

Solution :Here , `V_R=50V, V_L=50V, V_C=-50V`
`therefore` The VOLTAGE across L and C = `V_L+V_C`
=50-50
=ZERO
14.

Dimension of 1/(mue) is same as dimension of…….

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SQUARE of velocity
velocity
accleration
momentum

Answer :C
15.

Answer the following questions : (b) It is necessary to use satellites for long distance TV transmission. Why ?

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Solution :Generally TV waves have FREQUENCY range 100 - 200 MHz or even more and penetrate the ionosphere and are not reflected back. As space WAVE they can cover a distance of 50 - 60 km only. THEREFORE, for long distance TV transmission, we make use of SATELLITES which reflect the TV signal wave back towards the earth.
16.

Define the term power factor. What is the value of power factor for a pure resistor, pure inductor and pure capacitor?

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Solution :The power FACTOR is defined as the cosine of the ANGLE between the CURRENT and the applied voltage. For a pure capacitor and INDUCTOR, `theta=90^(@)` and HENCE the power factor is zero. For a pure resistor, `theta=0^(@)` and hence the power factory is one.
17.

How many orders will be visible if the wavelength of the incident radiation is 5000 Å and the number of lines on the grating is 101319 per metre.

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ANSWER :19
18.

Statement - I : A hallow metallic closed containes maintained at a uniform temperature can act as a source of black body radiation. Statement - II All metals act as a black body

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Statement - I is TRUE, Statement - II is true and Statement - I is CORRECT EXPLANATION for Statement - II.
Statement - I is true, Statement - II is true and Statement - II is not correct explanation of Statement - I.
Statement - I is true, Statement - II is FALSE.
Statement - I is false, Statement - II is false.

Solution :Statement I is true and statement II is false.
So correct CHOICE is (c ).
19.

Calculate the time taken by the light to travel a disance of 500m in water of refractive index of 4//3. (Given, velocity of light in vacuum =3xx10^(10)cm//s)

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`3xx10^(-10)s`
`2.22xx10^(-6)s`
`4.3xx10^(-5)s`
`3xx10^(-6)s`

SOLUTION :(B)
20.

The body weighs 100 N on the surface of the earth, then how much it would weigh at a depth of R/2 below the surface of the earth at radius R.

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SOLUTION :`g'=gR^2/((R+R//2)^2 )=4/9g
thereforeW'=mg'=4/9mg=4/9W`.
21.

In radiation mode, heat energy can be transfferred

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in PRESENCE of material MEDIUM
without any material medium
in presence of any material medium or without any material medium
none of these

Answer :C
22.

For communication which electromagnetic waves are used by cellular phones ?

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Microwaves
INFRARED WAVES
Ultraviolet waves
Ultra high FREQUENCY

ANSWER :D
23.

How resistivity of material depend on temperature. Write empirical formula .

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Solution :`rArr` Resistivity or material depend on TEMPERATURE. Different material have different dependence on temperature.
`rArr` Resistivity of metal increase with increase in temperature.
`rArr` Resistivity of semiconductor decrease with increase in temperature.
`rArr` Over a limited range of temperature that is not too LARGE the resistivity of METALLIC conductor is approximately given by.
`rho_(T) = rho_(0) [1 + alpha (T - T_(0))] `
`rho_(T)` = resistivity at temperature T
`rho_(0)` = resistivity at reference temperature `T_(0)`
`alpha` = temperature coefficient of resistivity
`rArr` Resistivity of same metals is given in table shown below.
`rArr` For metals value of `alpha` is POSITIVE.
`rArr` For semiconductor `alpha` is negative.
` rArr` For some metals value of `alpha` at `0 ""^(@) C` is shown in table below.
24.

Define the following using suitable diagrams : (i) Magnetic declination, and (ii) angle of dip. In what direction will a compass needle point when kept at the (i) poles , and (ii) equator ?

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SOLUTION :At poles a compass NEEDLE, FREE to turn in a vertical PLANE, points in the vertical direction. At equator a compass needle point in north-south direction in a horizontal plane.
25.

Answer the following questions: Two balls of the same metal having masses 5 gm and l0gm collide with a target withthe same velocity. If the total energy is used in heating the balls, which ball will attain .higher temperature ?

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SOLUTION :There will be same RISE in TEMPERATURE of both the BALLS
26.

A particle is projected form point O on the ground with velocity u = sqrt(5) m//sat anglealpha = tan^(-1) (0.5). It strikes at a point C on a fixed smooth plane AB havinginclination of37^(@) withhorizontal as shown is Fig. If the particledoes not rebound , calculate . (a) coordinates of pointC in reference t coordinatesystemas shown in the figure . (b) maximum heing fromthe groundto whichthe particlereises. (g= 10m//s^(2))

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SOLUTION :(a) `(5m,1.25m)` (B) `4.45m`
27.

The distance of the centre of mass of a hemispherical shell of radius R from its centre is

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`R/2`
`R/3`
`(2R)/2`
`(2R)/3`

ANSWER :A
28.

From a height of 100 m, a ball is thrown horizontally with an initial speed of 15 m//s. How far does it travel horizontally in the first 2 seconds ?

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SOLUTION :The question " How far does it travel horizontally .. ? " IMMEDIATELY tells us that we should use the first of the hoizontal-motion equations listed on the previous page.
` Delta x = v_(0_(x)) t = (15 m//s)(2 s)=30 m`
The information that the initial vertical position is 100 m above the ground is irrelevant (EXCEPT for the fact that it's high enough that the ball won't strike the ground before the 2 seconds have elapsed).
29.

Three charge +Q, +Q and -Q are placed at the corners of an equilateral triangle. The ratio of the force on a positive charge to the force on the negative charge will be equal to

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1
2
`SQRT(3)`
`1//sqrt(3)`

ANSWER :D
30.

An exploratory rocket of mass m is in orbit about the sun at a radius of R_(ES)//10 (one tenth of the radius of the earth's orbit about the sun). To exit this orbit, it fires its engine over a short period of time. This quickly doubles the velocity of the rocket while reducing its mass by half (due to fuel consumption). Immediately after the burn, what is the kinetic energy of the rocket?

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`10GM_(S)m//R_(ES)`
`5GM_(S)m//R_(ES)`
`20GM_(S)m//R_(ES)`
`GM_(S)m//2R_(ES)`

Solution :`KE=(1)/(2)mv^(2)=(GM_(s)m)/(2(R_(ES)//10))`
Final `KE = (1)/(2)((m)/(2))(2V)^(2)=mv^(2)=(10GMsm)/(R_(ES))`
31.

A spherical shell of radius R_(1)with a uniform charge q has a point charge q has a point charge q_(0) at its centre. Find the work performed by the electric forces during the shell expansison from radius R_(1) to radius R_(2)

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`W=(q(q_(0)+q//2))/(2piepsilon_(0))((1)/(R_(1))-(1)/(R_(2)))`
`W=(q(q_(0)+q//2))/(piepsilon_(0))((1)/(R_(1))+(1)/(R_(2)))`
`W=(q(q_(0)+q//2))/(4piepsilon_(0))((1)/(R_(1))-(1)/(R_(2)))`
`W=(q(q_(0)+q//2))/(4piepsilon_(0))((1)/(R_1)+(1)/(R_(2)))`

Answer :C
32.

If mu_(E ) and u_(B) be the time average of the electric and magnetic field energy densities at a point due to electromagnetic wave, then

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`u_(E )=mu_(B)`
`mu_(E )=2u_(B)`
`2u_(E )=mu_(B)`
none of these

Answer :A
33.

Refer to the arrangement of charges in shown figure and a Gaussian surface of radius R with Q at the centre. Then:

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total flux through the surface of the sphere is `(-Q)/(4piepsilon_(0)R^(2))`
field on the surface of the sphere is
flux through the surface of sphere DUE to 5Q is zero.
field on the surface of sphere due to -2Q is same everywhere.

Solution :Charge enlosed by sphere,
`=-2Q +Q`
`=-Q`
`therefore` flux associated with flux,
`phi = -Q/epsilon_(0)` (`therefore` According to Gauss.s law)
As SQ is outside, it doesn.tplay any role in flux.
34.

Which of the graph between velocity and time is correct ?

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A
B
C
D

Answer :B
35.

What is the refractive index of a prism whose angle A=60^(@) and angle of minimum deviationd_(m)=30^(@)?

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`sqrt(2)`
`1/(sqrt(2))`
`1`
`1/(sqrt(3))`

Solution :(a) REFRACTIVE index of PRISM
`mu=("sin" (A+delta_(m))/2)/("sin"A/2)=("sin"(60^(@)+30^(@))/2)/("sin"(60^(@))/2)`
`(sin 45^(@))/(sin 30^(@))=(1//sqrt(2))/(1//2)=1/(sqrt(2))xx2/1=sqrt(2)`
36.

Capacitance of a capacitor made by a thin metal foil is 2 muF. If the foil is foilded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm, the length of foil will be

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0.34 m
1.33 m
13.4 m
33.9 m

Solution :If LENGTH of the foil is l, then `C = (Kepsilon_(0)(lxxb))/(d)`
`rArr 2XX10^(-6) = (2.5xx8.85xx10^(-12)(lxx400xx10^(-3)))/(0.01xx10^(-3))`
`l = 33.9 m`
37.

X-ray are produced in coolidge tube. It consists of a thick glass bulb evacuated about 10^(-0) cm of Hg so that no discharge can pass in it. A cathode serves as source of electron is made up of tungsten filament heated by a low voltage current. A cylindrical metal shield S surrounds the filament and is kept at negative potential with respect to the filament is focuses the electron emitted from the filament on the target T. the target is mounted on the surface of a copper block fixed to the end of a copper rod . The rod is fitted with fins with for air cooling or cooled by circulating water A potential difference of about 20.000 V is applied across the tube to acceleration the electrons. The acceleration electrons strike to the target and produces continous and characteristic x-rays. Mark the correct graph which shows the rough variation of intensity of x-rays with wavelength lambda

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ANSWER :B
38.

A ratio nuclide with half life T=69.31 second emits beta- particles of average kinetic energy E=11.25 eV. At an instant concentration of beta- particles at distance, r=2 m from nuclide is n=3xx10^(13) per m^(3). (i) Calculate number of nuclei in the nuclide at that instant. (ii) If a small circular plate is placed at distance r from nuclide such that beta-particles strike the plate normally and come to rest, calculate pressure experienced by the due to collision of beta- particles. ("Mass of "beta-"particle"=9xx10^(-31) kg)

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ANSWER :`[(i) 9.6 PI xx10^(22), (II) 1.08xx10^(-4) Nm^(-2)]`
39.

X-ray are produced in coolidge tube. It consists of a thick glass bulb evacuated about 10^(-0) cm of Hg so that no discharge can pass in it. A cathode serves as source of electron is made up of tungsten filament heated by a low voltage current. A cylindrical metal shield S surrounds the filament and is kept at negative potential with respect to the filament is focuses the electron emitted from the filament on the target T. the target is mounted on the surface of a copper block fixed to the end of a copper rod . The rod is fitted with fins with for air cooling or cooled by circulating water A potential difference of about 20.000 V is applied across the tube to acceleration the electrons. The acceleration electrons strike to the target and produces continous and characteristic x-rays. In the coolidge tube experiment platinum is used as target material for the emission of k-series x-rays. The K,L and M energy level of this element are 7.8 xx 10^4 eV, 1.2 xx 10^4 eV and 3 xx 10^3 eV respectively C= 3 xx 10^8 m//s, 1 eV =1.6 xx 10^(-19) and h= 6.23 xx 10^(-34) j-s. The wavelength of k_(alpha) lines will be

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0.188 Å
0.376 Å
0.999 Å
0.25 Å

Answer :A
40.

A body of mass 2 kg attached to the end of the string of 2 m long is revolved in a horizontal circle. The breaking is tension of the string is 400 N. the maximum velocity of the body is,

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20 cm/sec
2 m/s
4 m/s
20 m/s

Answer :D
41.

In Young's double-slit experiment, the separation between the two slits is halved. The new fringe width will be ___ times its initial value.

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SOLUTION :TWO
[HINT: FRINGE WIDTH `betaprop1/d`]
42.

A body of mass m is projected from ground with speed u at an angle theta with horizontal. The power delivered by gravity to it at half of maximum height from ground is

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`(mgu COS THETA)/(SQRT2)`
`(mgu sin theta)/(sqrt2)`
`(mgu cos (90+theta))/(sqrt2)`
Both (2) & (3)

ANSWER :D
43.

The figure shows four arrangements of three particles of equal masses. (a) Rank the arrangements accroding to the magnitude of the net gravitational force on the particlelabeled m, greatest first. (b) In arrangement 2, is the direaction of the net force closer of the line of length d or totheline of length D ?

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Solution :1, TIE of 2 and 4, then 3 , (B) LENE of LENGTH d
44.

A transistor connected in common emitter mode, the voltage drop across the collector is 2V and betais 50,the base current if R_Cis 2KOmegais

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`40MUA`
`20muA`
`30muA`
`15muA`

ANSWER :B
45.

Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I=I_(0) sin omega t, where I_(0)=10 A and omega = 100 pi rad/s. The maximum value of e.m.f. in the second coil is

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`pi`
`5pi`
`2PI`
`4pi`

Solution :As `|e|=M(dI)/(DT)=M(d)/(dt)(I_(0)sin OMEGA t)=MI_(0)omega cos omega t`
`therefore` Maximum e.m.f. induced,
`e_(max)=0.005xx10xx100 pi xx1=5pi`.
46.

Light froma galaxy, having wavelength of 6000 Å , if found to be shifted towards red by 50 Å . Calculate the velocity of recession fo the galaxy.

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Answer :`2.5 XX 10^(6) MS^(-1)`
47.

Curie temperatureis the temperature above which

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a paramagnetic MATERIAL BECOMES diamgnetic
a ferromagnetic material becomes paramagnetic
a paramagnetic material becomes ferromagnetic
a ferromagnetic material becomes diamagnetic

ANSWER :B
48.

What is the relation between henry and weber?

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SOLUTION :1 HENRY = weber/ampere
49.

Define the term 'mobility' of charge carriers. Write its S.I. unit.

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Solution :Mobility is defined as the magnitude of the DRIFT velocity per UNIT electric field.
`mu=(|V_d|)/E=(alphatau)/E=(eE)/m TAU/E=e/m tau`
where`tau` is the average collision time for electrons.
The SI unit of mobility is `m^2//Vs` or `m^2V^(-1)S^(-1)`
50.

Two possible states for the hydrogen atom are labeled A and B. The maximum magnetic quantum number for state A is +3. For state B, the maximum value is +1. What is the ratio of the magnitudes of the orbital angular momenta, L_(A)//L_(B), of an electron in these two states ?

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ANSWER :2.45