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.

Consider the charges q.q and -qplaced at the vertices of an equilateral triangles as .Showin figure. What is the force on each charges?(##JYT_AJP_AIO_PHY_XII_C01_SLV_011_Q01.png" width="80%">

Answer»

Solution :The FORCES acting on charges Q at A due to charges q at B and -q at C are `F_(12) `along BA and `F_(12) `along AC respectively as shown in figure. Bythe parallelogram law. The total force ` F_1 ` on the charges q at A is given by `F_1 = F hatr_(1) "where " HATR _1 ` is a unit vector along BC.
The force of atraction or repulsion for each pair of charges has the same magnitude ` F-(q^(2))/( 4pi in _0l^(2))`
The total force ` F_2`on charges q at B is thus `F_2 =F HAT r _2.` where hat `r_2` is a unit vector along AC
Similarlythe total force on charge -at C is ` F_3= sqrt 3 hatn " where "hat n`is the unit vector along the direction bisecting the `angle BCA ` The sum of the forces on the three chargesis zero, i.e.
` F_1+F_2+F_3=0 ` (for the system)
The result follows straight from the fact that Coulomb.s law is consistant with Newton.s third law.
Discussion
2.

Consider telecommunication through optical fibres. Which of the following statements is not true?

Answer»

Optical FIBRES may have homogeneous core with a suitable cladding.
Optical fibres can be of GRADED refractive index.
Optical fibres are subject to ELECTROMAGNETIC interference from outside.
Optical fibres have extremely low transmission loss.

Answer :C
Discussion
3.

The frequency of A.C. mains in India is ….. Hz.

Answer»

30
50
60
70

Answer :C
Discussion
4.

A message signal of frequency 10 KHz and peak voltage of 10V is u sed to modulate a carrier of frequency 1 MHz and peak voltage of 20V. The frequency of the side bands produced is

Answer»

1000 KHZ, 990KHz
1010KHz,990KHz
990KHz,1000KHz
1010KHz,1000KHz

Solution :The side BONDS PRODUCED are at (1000+10)KHz =1010 kHz and (1000-10)kHz= 990kHz.
Discussion
5.

Find the current as a function of time in case of (a) charging and (b) discharging of a capacitor in a simple RC circuit

Answer»

Solution :During charging,`q = q_(0 )(1 - E^ (-t//CR)`
`i=(dq)/(dt)=-q_(0)((-1)/(CR))e^(-I//CR)=q_(0)/(CR)e^((t)/(CR)`
where `q-(o) = CV_(O). V_(0)` being the EMF of the cell
(b) During discharging. `q = q_(o) e^(t// CR)`
`i=(dq)/(dt)=-(q_(0))/(CR)e^(-t//CR)`
. where `q_(o)`=the initial charge on the CAPACITOR:
Discussion
6.

In Textual Example 1, the electron drift speed is estimated to be only a few mm s^(-1) for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed?

Answer»

Solution :Electric FIELD is established throughout the circuit, almost instantly (with the speed of light) (CAUSING at every point a LOCAL electron drift.Establishment of a current does not have to wait for ELECTRONS from one end of the conductor travelling to the other end. However it does take a little while for the current to reach its steady value.
Discussion
7.

Assertion: In common base configuration, the current gain of the transistor is less than unity. Reason: The collector terminal is reverse biased for amplification.

Answer»

If both the assertion and reason are true and reason is a true explantion of the assertion.
If both the assertion and reason are true but thereason is not true the correct explantion of the assertion.
If the assertion is true but reason false
If both the assertion and reason are false.

Solution :In COMMON base configuration input is applied between emitter and base and output is taken between collector and base. In this connection the base of the transistor is made common to both input and output terminals. For amplification, emitter terminals is forward biased and collector terminal is reverse biased. In this configuration, the CURRENT gain is ratio of change in collector current to the change in emitter current at constant collector base VOLTAGE `(V_(CB))`.
`:. alpha=((DeltaI_(C))/(DeltaI_(E)))`
Current gain `(alpha)` is less than unity because collector current `I_(C)` is always less than emitter current `(I_(E))`. Hence, option (B) is true.
Discussion
8.

Which of the following is correct for light diverging from a point source ?

Answer»

The INTENSITY decreases in PROPORATION for the DISTANCE squared.
The wavefront is parabolic.
The intensity at the wavelength does not DEPEND on the distance.
NONE of these.

Answer :A
Discussion
9.

A metallic road held horizontally along east-west direction is allowed to fall under gravity. Will there be an emf induced at it's ends? Justify your answer.

Answer»

SOLUTION :YES, because the horizontal COMPONENT of earth.s magnetic FIELD is INTERCEPTED by the rod.
Discussion
10.

when a glass rod is rubbed with silk cloth, both get charged.

Answer»


ANSWER :POSITIVE and NEGATIVE CHARGE respectvely
Discussion
11.

A transistor oscillator is (i) an amplifier with positive feedback (ii) an amplifier with reduced gain (iii) the one in which dc supply energy is converted into ac output energy . Then

Answer»

A) all (i) , (ii) and (III) are CORRECT
B) only (i) and (ii) are correct
C) only (i) and (iii) are correct
D) only (ii) and (iii) are correct

ANSWER :C
Discussion
12.

A body having a mass 100 gm is allowed to fall freely under gravity. Calculate KE after 10 seconds g=9.8 m//s^(-2).

Answer»

280.2 J
380.2 J
480.2 J
580.2 J

Answer :C
Discussion
13.

An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The magnetic field due to this current at the point M is B_(1). Now, another infinitely long straight conductor QS is connected at Q so that the current is I//2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now B_(2) = 4 mT. Find the value of B_(1) ("in" mT).

Answer»


Solution :`B_(1)=(mu_(0)I)/(4 pi R)`
`B_(2)=(mu_(0)I)/(4pi R) + (mu_(0)(I/2))/(4 pi R) = 3/4 (mu_(0)I)/(4 pi R)`
`(B_1)/(B_2) = 3/4 implies B_(2)= 3/4 B_(2) = 3/2 xx 4 = 3 MT`.
Discussion
14.

A sample of HCI gas is placed in a uniform electric field of magnitude 3xx10^(4)N.C^(-1) . The dipole moment of each HCI molecule is 3.4xx10^(-39) Cm. Calculate the maximum torque experienced each HCI molecule.

Answer»

Solution :The maximum torque experienced by the dipole is when it is aligned PERPENDICULAR to the applied field
`tau_("MAX.)=pEsin 90^(@)=3.4xx10^(-30)xx3xx10^(4)`NM
`tau_("max")=10.2xx10^(-26)`Nm
Discussion
15.

Wavelength of a light emitted from second orbit to first orbit in a hydrogen atom is

Answer»

`1.215 xx 10^(-7)` m
`1.215 xx 10^(-5)`m
`1.215 xx 10^(-4)` m
`1.215 xx 10^(-3)` m

Solution :`1/(lambda) = R [(1)/(1^(2)) - (1)/(2^(2))] = (3R)/(4)`
`lambda = (4)/(3R) = (4)/(3 xx 1.097 xx 10^(7)) = 1.215 xx 10^(-7)`m
Discussion
16.

There exists an electric field of 100 N/C along Z-direction. The flux passing through a square of 10 cm sides placed on XY plane inside the electric field is.......

Answer»

`1.0 Nm^(2)//C`
2.0 Vm
10 Vm
`4.0 Nm^(2)//C`

Solution :`E = 100 hatk N//C, l = 10 cm = 0.1 m, hatk` is for Z-axis.
`A = 0.1 xx 0.1 = 0.01 m^(2)hatk (therefore A =l^(2))`

FLUX, `phi = vecE.vecA= 100 xx 0.1 xx hatk.hatk`
`therefore phi =1 Nm^(2)//C`
Discussion
17.

Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I (refer to the figure given in previous question). The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?

Answer»

SOLUTION :The ring will stay on the CARDBOARD but it will exert a downward force on the cardboard. As the current in the solenoid is DECREASING, then according to Lenz.s law, this decrease will be resisted and this can happen if the current in the ring is in opposite direction as in the solenoid, i.e. clockwise (as seen from the top of the ring). SINCE opposite poles are developed on sides of ring and solenoid FACING each other, the ring gets attracted towards the solenoid, in the downward direction. So, the ring exerts a downward force on the cardboard.
Discussion
18.

A lighted candle is placed between two plane mirrors inclined at an angle 60^@ _____ no of images are formed.

Answer»

SOLUTION :`(360)/60 -1=5`
Discussion
19.

How is forward biasing different from reverse biasing in a p-n junction diode?

Answer»




ANSWER :D
Discussion
20.

In the circuit shown in Fig. 4.63, the terminal voltage of the battery is 6.0 V. Find the current I through the 18 Omega resistor.

Answer»


ANSWER :`0.25 A`
Discussion
21.

The allowed energy for the particle for a particular value of n is proportional to

Answer»

`a^(-2)`
`a^(-3/2)`
`a^(-1)`
`a^(2)`

Solution :For the STANDING wave, `a = n (lambda)/(2) or lambda = (2A)/(n)`
`P = (h)/(lambda) = (NH)/(2a) , E = (P^(2))/(2m) = (n^(2)h^(2))/(2a^(2)m), E infty a^(-2)`
Discussion
22.

If charge q induced on outer surface of sphere of radius R, then intensity at point P at distance S from centre is

Answer»

inversely PROPORTIONAL to `(S+R)^(2)`
inversely proportional to `R^(2)`.
inversely proportional to `S^(2)`.
DIRECTLY proportional to `S^(2)`.

Answer :C
Discussion
23.

In Q.30, the magnitude and direction of magnetic field will be

Answer»

`2.5 MUT` in X-direction
`2.5 muT` in Y-direction
`2.5 muT` in Z-direction
None of these

Answer :C
Discussion
24.

Are of the hysteresis loop is an indication of

Answer»

permeabilitty of the medium
susceptibility of the substance
energy DISSIPATED PER cycle
retentivity of the material

Answer :C
Discussion
25.

Consider the following parts of spectra1. Visible 2. Infrared 3. Ultraviolet 4. Microwave Which one of the following is the correct sequence in which their wavelengths increase ?

Answer»

4- 3- 1- 2
4- 1- 2- 3
3-2- 1- 4
3- 1- 2- 4

ANSWER :d
Discussion
26.

Read the vermier 10 division of vermier scale are matching with 9 divisionsd of main scale

Answer»

SOLUTION :10 vernier scale divisions = 9MM
1vermier scale division = 0.9mm
`rArr` least count = (Main scale division -vernier Scale division)
`=1mm -0.9mm`
`= 0.1mm`
Thickness of the object = (main scale READING) + (vernier scale Reading) (least count)
So thickness of the object = 15MM + (6) (0.1mm) = 15.6mm` .
Discussion
27.

A flat uniform circular disk (radius = 2.00 m, "mass" = 1.00 xx 10^(2) kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendic ular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m//s relative to the ground. Find the resulting angular speed of the disk (in "rad"//s) and describe the direction of the rotation.

Answer»

`0.500 "RAD"//s`
`0.250 "rad"//s`
`2.00 "rad"//s`
`1.00 "rad"//s`

ANSWER :A
Discussion
28.

A man of weight W is standing on a lift which is moving up words with an acceleration 'a' . The apparent weight of the man is

Answer»

`W(1+a/g)`
`W(1-a/g)`
`W=(1-(a^2)/(g^2))`
W

Answer :A
Discussion
29.

(a) The graphs (i) and (ii) represent the variation of the opposition offered by the circuit elements to the flow of alternating current with frequency of the applied emf. Identify the circuit element corresponding to each graph. (b) Write the expression for the impedance offered by the series combination of above two elements connected across the a.c. source. Which will be ahead in phase in this circuit, voltage or current ?

Answer»

Solution :(a) The CIRCUIT element shown in Fig. 7.17(i) is a resistor, because opposition OFFERED by it for flow of current is a constant and does not change with frequency. The circuit element shown in Fig. 7.17 (ii) is an inductor because its opposition to current is INCREASING regularly with increase in frequency and it is possible for only an inductor as `X_(L) = L omega = L 2PI v`.

(b) The impedance offered by the series combination of R and L is given by :
`Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(R^(2) + (Lomega)^(2))`
In a L-R circuit voltage is ahead in phase than current by a phase angle `phi`,where `tan phi = X_(L)/R`.
Discussion
30.

Give an expression for force acting on a charge moving in magnetic field and explain the symbols, when does the force become maximum.

Answer»

SOLUTION :`F_(m)=qVB sin theta`
Where B = Magnetic FIELD
q = Magnitude of charge
V = velocity of charged PARTICLES.
`theta`-Angle between the directions of B and V
The force is MAXIMUM when `theta=90^(0)`. Then
`F_("max")=qVB`
Discussion
31.

A ray of light is incident on a transparent plate of a material of refractive index sqrt(3) at the polarising angle. Find the angle of refraction.

Answer»


ANSWER :`30^(@)`
Discussion
32.

A double convex lens of focal length 30 cm is made of glass. When it is immersed in a liquid of refractive index 1.4, the focal length is found to be 126 cm. The critical angle between glass and the liquid is

Answer»

`SIN^(-1) (3/4)`
`sin^(-1)(4/5)`
`sin^(-1) (7/13)`
`sin^(-1) (7/8)`

ANSWER :D
Discussion
33.

A deutron and an alpha particle move in a magnetic field along the same circular path , then the ratio of their velocities is _________ .

Answer»

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

SOLUTION :N/A
Discussion
34.

Name three processes involved in the formation of solar cell.

Answer»

SOLUTION :GENERATION, SEPARATION and COLLECTION.
Discussion
35.

What is the background of the story:

Answer»

This is an imaginary event which did not happen in TWENTIETH CENTURY and being DISCUSSED in twenty fifth century
Story has a crime scene background
Both A and B
All of the above

Answer :A
Discussion
36.

When two objects are rubbed with each other approximately a charge of 50 nC can be produced in each object . Calculate the number of electrons that must be transferred to produce this charge.

Answer»

SOLUTION :Charge produced in each object q= 50 nC
`q = 50 xx10^(-9)C`
Charge of electron (e )`=1.6xx10^(-19)C`
NUMBER of electron transferred n `=(q)/(e ) = (50xx10^(-9))/(1.6xx10^(-19)) = 31.25 xx10^(-9)xx10^(19)`
`n = 31.25 xx10^(10)` electrons
Discussion
37.

Direction of magnetic dipole moment of a magnet is .....

Answer»

from NORTH POLE to SOUTH pole.
from south pole to north pole.
possible in any direction.
not decided.

Answer :B
Discussion
38.

Is a wire carrying current charged ?

Answer»

Solution :No, the CURRENT in a wire is due to FLOW of electrons in a definite directions.At any time, the wire has as MUCH negative CHARGES as the positive charges.
Discussion
39.

The magnifying power of an astronomical telescope for relaxed vision is 16 and the distance between the objective and eyelens is 34 cm. Then the focal length of objective and eyelens will be respectively

Answer»

17 cm, 17 cm
20 cm, 14 cm
32 cm, 2 cm
30 cm, 4 cm

Answer :C
Discussion
40.

Calculate the potential at a point P due to charge of 4 xx 10^(-7)C located 9 cm away. Hence obtain the work done in bringing a charge of 2 xx 10^(-9)C from infinity to the point. P. Does the answer depend on the path along which the charge is brought?

Answer»

SOLUTION :`V=(1)/(4pi epsi_(0)) Q/r= 9 XX 10^(9) xx (4 xx 10^(-7))/(0.09)=4 xx 10^(4)V`
`W=qV =2 xx 10^(-9) xx 4 xx 10^(4) =8 xx 10^(-5)J`.
No, work done will be path independent (A property of electric field).
Discussion
41.

A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the hase. If alpha and beta the base angles and theta be the angle of projection, prove that theta= tan alpha + tan beta.

Answer»

Solution :The SITUATION is shown in FIGURE. From figure, we have

`TAN alpha+ tan beta=(y)/(x)+(y)/(R-x)`
`tan alpha+ tan beta=(yR)/(x(R-x)) ""...(i)`
But EQUATION of trajectory is `y=x tan THETA[1-(x)/(R)]`
`tan theta=(yR)/(x(R-x)) ""...(ii)`
From Eqs (i) and (ii) `tan theta= tan alpha+ tan beta`
Discussion
42.

A convex lens of focal length 10 cm is used as a simple microscope. Its magnifying power when the image is formed at near point.

Answer»

3.5
0.1
10
2.5

Answer :A
Discussion
43.

A : Average velocity of gas molecules is zero, R : Duetorandommotionofgasmolecules, velocities of different molecules cancer each other.

Answer»

If bothAssertion & REASON are true and the reason is the CORRECT explanation of the ASSERTION, then Mark (1)
If both Assertion & Reason are true but the reasonis not the correct explanation of the assertion, then mark (2)
If Assertion is true STATEMENT but Reason is false, then mark (3)
If both Assertion and Reason are false statements, then mark (4).

Answer :A
Discussion
44.

Three lenses L_(1) , L_(2) , L_(3) are placed co-axially as shown in figure. Focal length's of lenses are given 30 cm , 10 cm and 5 cm respectively. If a parallel beam of light falling on lens L_(1), emerging L_(3) as a convergent beam such that it converges at the focus of L_(3) . Distance between L_(1) and L_(2) will be

Answer»

40 CM
30 cm
20 cm
10 cm

Answer :C
Discussion
45.

In Joule's heating law, when R and t are constant, if the H is taken along the y axis and I^(2) along the x axis, the graph is ………………. .

Answer»

STRAIGHT line
parabola
CIRCLE
ELLIPSE

Answer :A
Discussion
46.

भारत की लगभग कितनी प्रतिशत आबादी कृषि कार्यों में संलग्न है ?

Answer»

48 प्रतिशत
54 प्रतिशत
64 प्रतिशत
70 प्रतिशत

Answer :D
Discussion
47.

Consider sunlight incident on a slit of width 10^(4)Å. The image seen through the slit shall

Answer»

be a fine sharp slit white in colour at the center.
a bright slit white at the center diffusing to zero INTENSITIES at the edges.
a bright slit white at the center diffusing to regions of different colours.
only be a diffused slit white in colour.

Solution :be a fine sharp slit white in colour at the center.
Here average wavelength in the visible light is `lamda=(4000+8000)/(2)=6000Å` and here, width of slit is `d=10^(4)Å=10000Å`
`(lamda)/(d)=(6000)/(10000)=0.6implies(lamda)/(d)lt1`
Amount of diffraction will be quite small. Hence image of slit will be quite sharp in which central region will appear white because of superposition of constituent colours in equal proportion. Moreover, due to zero PHASE difference, the superposing waves will interfere constructively GIVING rise to central maximum with maximum intensity.
Discussion
48.

Equations a stationary and a travelling waves are as follows y_1 = a sin kx cos omega tand y_2 = a sin (omega t - kx). The phase difference between two points x_1 = (pi)/(3k) and x_2 = (3pi)/(2k) are phi_1and phi_2respectively for the two waves . The ratio phi_1/phi_2 is

Answer»

1
`5/6`
`3/4`
`6/7`

ANSWER :D
Discussion
49.

A rectangularconductorLMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10ms^(-1), calculate the emf induced in the arm. Given the resistance of the arm to the 5 Omega ( assuming that other arms are of negligible resistance) find the value of the current in the arm.

Answer»

Solution :LET ON be at some point X.
The EMF INDUCED in the loop `= epsilon`
`epsilon= (-d phi)/(DT) = (-d (Blx))/(dt)= Blv`
`= 0.5xx0.2xx10= 1V`
`:.` Current in the arm,
`I =(epsilon)/(R) = 1/5 = 0.2 A`.
Discussion
50.

If a shell fired from a canon is exploded in air then-

Answer»

MOMENTUM decreases
Momentum increases
Kinetic ENERGY increases
K.E. decreases

Answer :C
Discussion