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.

According to Huygen's construction, each point of secondary wavefront will emit the light waves of

Answer»

Each point of a WAVEFRONT is a SOURCE of secondary disturbance
The wavelets emanating from POINTS on wavefront spread out in all DIRECTIONS with the speed of wave
The direction of ray is always in the direction GIVEN by tangent to the wavefront
Both (1) & (2)

Answer :D
2.

What is the function of a solar cell? Briefly explain its working and draw its I-V characteristic curve.

Answer»

Solution :The function of a solar cell is to transform solar energy directly into electrical energy. A solar cellis BASICALLY a p-n junction which generates emf when solar radiations fall on the junction. In a solar cell no external bias is applied and the junction area is kept LARGE for more solar radiation to be incident.
When light (with photon energy `hv gt E_(g)`) falls at the junction, electron-hole pairs are generated which move in mutually opposite directions due to the junction field. If no external load is connected of if the circuit of solar cell is open, electrons and holes are collected at the two sides of the junction giving a photo-voltage `V_(oc )`. When external load R is connected, a photo-current `I_(L)` flows. For R = 0, the current has a maximum VALUE `I_(sc)`, which is known as short circuit current. V-I characteristics is shown in figure. The graph is in the fourth QUADRANT because a solar cell not draw current but supplies current to the load.
3.

Two concentric spherical shells of radii R and 2R carry charges Q and 2Q respectively. Change in electric potential on the outer shell when both are connected by a conducing wire is (K= (1)/( 4 pi epis_0))

Answer»

zero
`(3 KQ) /( 2R)`
`(KQ)/(R )`
`(2KQ)/(R )`

ANSWER :A
4.

A steel flywheel is made in the shape of a solid ring of 40 cm external and 30 cm internal diameter. What is its maximum design speed? At what speed will it fly apart?

Answer»


SOLUTION :We shall solve the problem in a rotating REFERENCE frame. Consider a segment of the flywheel which subtends small angle `alpha` at its centre (Fig.). The forces acting on this metallic segment are the CENTRIFUGAL force of inertia and two elastic forces. The relation between them follows from the condition for equilibrium: `I_(ct) = Talpha `. Here `T = sigma S`, where S is the cross-sectional area of the flywheel RIM, and `alpha` is the stress. The volume of the separated segment is `V = lS = alpha S R_(av)`, where `R_(av) = (R + r)//2` is the average radius of the flywheel rim. The centrifugal force of inertia is
`I_(ct) = m omega^2 R_(av) = omega^2R_(av) rho v = alpha rho S omega^2 R_(av)^2`
where `rho` is the density of the metal. Substituting into the condition for equilibrium, we obtain the dependence of the stress in the metal on the speed of rotation:
`sigma = rho omega^2 R_(av)^2`
The speed is safe when the stress does not exceed the elastic limit. In this CASE the elastic forces will return the flywheel to its original state when the speed of rotation is reduced. Therefore the maximum safe speed of rotation of the flywheel is
`omega = sqrt((sigma E)/(rho R_(av)^2)) = 2/(R + r) sqrt((sigma E)/(rho))`
The flywheel will fly apart when the stress in it reaches the breaking stress, i.e. when
`omega_(br) = 2/(R + r) sqrt((sigma tw)/(rho))`
5.

As angle of incidence increases, angle of deviation first _____minimum value and then increases.

Answer»

SOLUTION :DECREASES
6.

A : When a light wave travels from a rarer to a denser medium, it looses speed. The reduction in speed imply a reduction in energy carried by the light wave. R : The energy of a wave of proportional to frequency of wave.

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 :D
7.

Aglassprism of refractive index 1.5 is immersed in water of refractive index 1.33. In this condition what should be the limiting angle of the prism for no emergent ray?

Answer»


ANSWER :`125^(@)30'`
8.

If veca and vecb are two vector with |veca| = |vecb| and |veca + vecb|+ 2|veca|,then angle between veca and vecb.

Answer»

`0^(@)`
`90^(@)`
`60^(@)`
`180^(@)`

ANSWER :A::D
9.

A block of mass 20kg is pushed with a horizontal force of 90N. If the co-efficient of static & kinetic friction are 0.4 & 0.3, the frictional force acting on the block is (g=10 ms^(-2))

Answer»

90 N
80 N
60 N
30 N

Answer :C
10.

The thermo-electric power P of a thermocouple is given by

Answer»

`P = a theta + B theta^2`
`P = theta^(2) + b theta^(3)`
`P = a + 2 b theta`
NONE of these

Answer :C
11.

The resistance of a wire is 5 Omega at 50^@C and 6 Omega at 100^@C . The resistance of the wire at 0^@Cwill be

Answer»

`3OMEGA`
`2OMEGA`
`1Omega`
`4Omega`

SOLUTION : From the relation `R=R_0[1+ ALPHA(T-T_0)]`, we have `5 = R_0[1 + 50alpha]` and `6=R_0[1 + 100 alpha] `
12.

A small sphere A of radius r and carrying a charge q is enclosed by a concentric spherical shell B of radius 2r carrying a charge 29. When A and B are connected by a conducting wire, the charge will flow from __________ to __________ .

Answer»

Solution :A, B. Electric potential of inner SPHERE A is greater than that of outer SPHERICAL shell B and hence CHARGE flows from A to B.
13.

In a coil of self-inductance 0.5henry, the current varies at a constant rate from zero to 10A in 2 sec. The induced e.m.f. in the coil is:

Answer»

1.25V
2.5V
5.0V
10V

Answer :B
14.

Suppose the Sun contracts (collapses) to a pulsar. Estimate the minimum radius of the pulsar and its period of rotation. The period of revolution of the Sun about its axis is 25.38 days (1 day = 24 hours).

Answer»


Solution :Applying the law of conservation of momentum we can find the relation between the pulsar radius and its period of revolution. We have
`I_(o.)omega_(o.)=IOMEGA, "or " 2/5MR_(o.)^2(2pi)/(T_(o.))=2/5MR_(o.)^2(2pi)/T`
Hence
`R_(o.)^2/T_(o.)=R^2/T`
To prevent the escape of matter as the speed of rotation is increased the force of GRAVITY should exceed the centrifugal force, i.e.`momega^2RltR_(GR) ` Hence `(4pi^2mR)/(T^2)lt(gammamM_(o.))/(R^2)`
THEREFORE the second relation sought between the pulsar radius and its period of revolution is of the form
p`R^3/T^2lt(gammaM_(o.))/(4pi^2)`
Eliminating the pulsar period from both equations, we obtain
`Rge(4pi^2R_(o.)^1)/(gammaM_(o.)T_(o.)^2)=(4pi^2xx7^4xx10^(32))/(6.67xx10^(-11) xx2xx10^(30) xx2.2^(2) xx10^(12)) ~~15km`
`T ~~ R^2T _(o.)//R_(o.)^2 ~~10^(-3)8`
15.

A particle of mass m = 5 is moving with uniform speed v = 3sqrt(2) in the XOY plane along the line Y = X + 4. The magnitude of angular momentum of the particle about the origin is :

Answer»

60units
`40sqrt(2)`units
zero
7.5 units

Solution :Here momentum of the particle `= 5x3sqrt(2) = 15sqrt(2)` units.
The line of action of momentum is in XY PLANE given by the st. line y = x + 4. Comparing it with y = mx + c. The SLOPE of line tan `THETA` = 1 `:. theta` = 45° and the intercept of the line of y-axis is 4 units. Leng TH of the perpendicular distance 3 from the origin to the line of action of momentum is
`z=4sin45^@=4/sqrt(2)`
`=2sqrt(2)` units
Moment of momentum or angular momentum is given by
`L=mvz=15sqrt(2)xx2sqrt(2)=60` units
16.

Draw equipotential surfaces for a uniform electric field along z-axis.

Answer»

SOLUTION :
17.

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n-1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Answer»

Solution :for`n^(th)`orbit ` E_n= ( - 2pi^@ me ^4 ) /( ( 4 piepsi_0)^2 n^2 h^2)`
for `p^(th)` orbit`, E_p =(-2 PI ^2 m e^4 )/( ( 4 piepsi_0)^2 p^2h^2) `
frequency`, v = (E_a-E_p) /( h )= (2pi^2m e^4)/((4 pi epsi_0 )^2 h^2) [ (1)/(p^2) - (1)/( n^2)]`
forn=nand p=n -1 we GET
` v= (E_n -E_(n-1))/(h)= ( 2 pi^2 m e^4 )/( (4 piepsi_0 )^2 h^3) [ (1)/( (n-1)^2)-(1)/(n^2)]`
i.e,` v= (2 pi ^2 m e^4)/( (4 pi EPSI _0 ) ^2 h^3) . ((2n-1))/(n^2 (n-1)^2)`
forlargevaluesof `n2m-1~~2 n ` andn-1`~~` n
` thereforev= (2 pi^2m e ^4 )/( (4 pi epsi_0)h^3).( 2n )/( h^4) = ( 4 pi ^2 me ^4 )/((4 pi epsi_0)^2 n^3 h^3 )`
classicalfrequnency,` V_ C =(v)/( 2 pir)`
we havemvr`= (n h )/(2 pi )thereforev=(nh )/( 2 pi m r)`
` thereforev= (nh )/( 4 pi ^2mr^2 )`
But `r=((4 pi epsi_0 )n^2 h^2)/( 4 pi ^2 m e ^2)`
` thereforeV_c = ( 4pi ^2 m e ^4 )/(4 pi epsi_0 )^2 n^3 h^3`
thenforlargevalues of `n, upsilon = upsilon _c` hencetheproof.
thisis calledbohr .scorrespondenceprinciple
18.

Which magnetic substance has negative susceptibility.

Answer»

SOLUTION :DIAMAGNETIC SUBSTANCE.
19.

There are two charges +1muC and +5 much. The ratio of the force acting on them is :

Answer»

`1:1`
`1:2`
`1:3`
`1:4`

ANSWER :A
20.

The osicllating electric and magnetic vectors of an electromagnetic wave are oriented along

Answer»

the same DIRECTION and are in phase.
the same direction but differe in phase by `(pi)/(2)`.
mutually PERPENDICULAR DIRECTIONS and are in phase.
mutually perpendicular directions and DIFFER in phase by `(pi)/(2)`.

Answer :C
21.

Assertion : When a charge particle moves in a circular path. It produces electromagnetic wave. Reason : Charged particle has acceleration.

Answer»

If both assertion and REASON are true and reason is the CORRECT explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is FALSE.
If both assertion and reason are false.

ANSWER :A
22.

Calculate the current amplification factor beta when change in collector current is 1mA and change in base current is 20 mu A.

Answer»

50
25
75
100

Answer :A
23.

(a) A group of quantum states of the hydrogen atom has n= 5. How many values of l are possible for states in the group? (b) A subgroup of hydrogen atom states in the n = 5 group has l = 3. How many values of m_(l) are possible for states in this subgroup?

Answer»

SOLUTION :(a) 5, (B) 7
24.

If the length of a cylinder on heating increases by 2%, the area of it's base will increase by

Answer»

a)`0.5%`
b)`2%`
C)`1%`
d)`4%`

ANSWER :D
25.

Statement (A) : Moving charges produce not only an electric field but also magnetic field in space Statement (B) : The force is exerted by a magnetic field on moving charges or on a current carrying conductor only but not on stationary charges

Answer»

A is a TRUE B is false
A is false B is tme
A and B are true
A and B is false

Answer :C
26.

If a light ray incidents normally on one of the faces of the prism of refractive index '2' and emergent ray just grazes the second face of the prism, then the angle of deviation is

Answer»

`0^@`
`30^@`
`60^@`
`90^@`

ANSWER :C
27.

A cannon shell is fired to hit a target at a horizontal distance R. However, it breaks into two equal parts at its highest point. One part (A) returns to the cannon. The other part:

Answer»

Will fall at a distance of R beyond the target
Will fall at a distance of 3R beyond the target
Will hit the target
Have nine times the kinetic ENERGY of A just after explosion

Solution :`u_x`: velocity at HIGHEST POINT, and v: velocity of other part after the cannon breaks.
`impliesm u_(x) = -m/2 u_x + m/2 v IMPLIES v = 3u_x`.
`implies` Horizontal distance traveled by `2^(nd) part is 3 times the horizontal distance (R/2) traveled by the 1st part after the cannon breaks and KE of 2nd part is 9 times the KE of the FIRST part at highest point.
28.

A current I from A toB is increasinginmagnitudes as shown in figure. What is the direction of induced current in the loop.

Answer»

Solution :
The current I will produce a magnitudefield B in upward direction in theregion in which COIL is PLACED.
Since, the current is increasing in magnitude , B will also increase with time, subsequently increaseing flux.
To counter the increasing flux , incduce field `B_i` is produced opposite to direction b. POINTING right hand thumb in the direction of `B_i` and curling fingers, we can find out the direction ofinduced current to be clockwise.
29.

A point object is placed at the focus of a convex mirror, the image will be formed at

Answer»

infinity
cente of curvature
at FOCUS itself
none of these

Solution :(d) IMAGE will not FORM, because object is placed on the side from where REFLECTION is not possible.
30.

In aseries L-C-R circuit, resonance occurs, when:

Answer»

`R=X-L-X_C`
`X_L=X_C`
`X_L-X_C GT R`
`X_L-X_C LT lt R`

ANSWER :B
31.

Two short bar magnets of magnetic moments 'M' each are arranged at the opposite corners of a square of side 'd' such that their centre coincide with the corners and their axes are parallel to one side of the square. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is

Answer»

`(mu_(0))/(4PI) (M)/(d^(3))`
`(mu_(0))/(4pi) (2M)/(d ^(3))`
`(mu_(0))/(2pi) (M)/(d ^(3))`
`(mu_(0))/(2pi) (2M)/(d ^(3))`

ANSWER :A
32.

Answer the following question: (a) Long distance radio broadcasts use short wave bands. Why? (b) It is necessary to use satallites for long distance T.V. Transmission. Why? (c) Optical and radio telescope are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (d) The small ozone layer on top of the atmosphere is crucial for human survival. Why? (e) If the earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? (f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter' with a devastating effect on life on earth. What might be the basis of this prediction.

Answer»

Solution :(a) It is so because ionsophere reflects the wave in these bands.
(b) Yes, television signals being of high frequency are not reflected by the IONOSPHERE. Therefore, to reflect
them satellites are needed. That is why, satallites are used for long distance T.V. transmision.
(c) Optical and radio waves can penetrate the whereas X-rays being of much smaller wavelength
are absorbed by the atmosphere. That is why we can work with optical and radio TELESCOPES on earth's
surface but X-rays astronomical telescopes must be used on the satellite orbiting above the earth's
atmosphere.
(d) The small ozone layer present on the top of the stratosphere absorbs most of the ultraviolet radiations
from the sun which are dangerous and cause genetic damage to the living cell, prevents them from
reaching the earth's surface and thus helps in the survival of the life.
(E) The temperature of the earth would be lower because the Green house effect of the atmosphere would
be absent.
(f) The clouds by a global nuclear WAR would perhaps cover most parts of the sky preventing solar light
from reaching many parts of the globe. This would cause a 'nuclear winter'.
33.

A thin prism is made of a material having refractive indices 1.61 and 1.65 for red and violet light. The dispersive power of the material is 0.07. It is found that a beam of yellow light passing through the prism suffers a minimum deviation of 4.0^@ in favourable conditions. Calculate the angle of the prism.

Answer»

Solution :Given that : `mu_r=1.61`
`mu_v=1.65`
`omega=0.07 and delta_y=4^@`
Now `w=(mu_v-mu_r)/(mu_y-1)`
`=0.07=(1.65-1.61)/(mu_y-1)`
`mu_y-1=0.04/0.07=4/7`
Again `delta=(mu-1)A`
`RARR A=delta_y/(mu_y-1)=4/(4/7) =7`
34.

Name the field which is conservative in nature .

Answer»


ANSWER :ELECTRIC FIELD
35.

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in whichR=3Omega, L=25.48mH, C=7.96xx10^(-4)F. The impedance Z of the series LCR circuit is :

Answer»

`2OMEGA`
`5OMEGA`
`10Omega`
`12OMEGA`

ANSWER :B
36.

It is desired to make a long cylindrical conductor whose temperature coefficient of resistivity at 20^(@)C will be close to zero. If such a conductor is made by assembling alternate disks of iron and carbon, find the ratio of the thickness of a carbon disk to that an iron disk. ("For carbon", p = 3500 xx 10^(-8)Omega m and alpha= -0.50 xx 10^(-3) .^(@)C^(-1) "for iron, p"=9.68 xx 10^(-8)Omega m and alpha=6.5 xx 10^(-3).^(@)C^(-1))

Answer»

0.36
0.036
1
2

Solution :The direction of CURRENTLY DENSITY is the direction of flow of positive charge in the circuit which is POSSIBLE DUE to electric field produced by charges accumulated on the surface of wire.
37.

Give two differences between interference and diffraction.

Answer»

SOLUTION :.
38.

Direction of motion of unit the positive test charge gives direction electric field. Direction of motion of………………………... gives direction of magnetic field.

Answer»

SOLUTION :UNIT NORTH POLE
39.

Deompositionof uraeintoNH_(3) and CO_(2) isfollowedby theactionof enzyme :

Answer»

urease
pepsin
TRYPSIN
NONE of these

Solution :ACCORDING to FACT.
40.

When you have learned to integrate, derive the formula for the moment of inertia of a disk.

Answer»


Solution :The differential of MASS is equal to the mass of a RING of thickness DR. We have
`(dm)/M=(2pirdr)/(piR^2)` , form which `dm= (2Mrdr)/(R^2)`
The differential of the moment of inertia is equal to the moment of inertia of this ring:
`dI=dm.r^2=(2Mr^3dr)/(R^2)`
Hence
`I = int_0^Rdm.r^2=(2M)/(R^2)int_0^Rr^3dr=(2M)/R^2=(2M)/(R^2)[r^4/4]_0^R=(2MR^4)/(R^2xx4)=(MR^2)/2`
41.

Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45^(@) to the horizontal ?

Answer»

The speed at the top of the trajectory is zero.
The OBJECT's total speed remains constant during the entire flight.
The horizontal speed decreases on the way up an increases on the way down.
The vertical speed decreases on the way up and increases on the way down.

Solution :SINCE the ACCELERATION of the projectile is always downward (because it's gravitational acceleration),the vertical speed decreases as the projectile rises and increases as the projectile FALLS. Choice (A) is false because gravity always points down. Choice (B) is false because the projectile still has its horizontal velocity at the top of the trajectory. Choice (C ) is false because the vertical component of the velocity changes, even if the horizontal component does not, so the total speed changes.
42.

A convex lens of 40 cm focal length is combined with a concave lens of focal length 25 cm. The power of combination is

Answer»

<P>`-1.5 D`
`-6.5 D`
`+ 6.6 D`
`+ 6.5 D`

Solution :`P=P_(1)+P_(2)=(1)/(f_(1))=(100)/(40)+(100)/(-25)=-1.5D`
43.

There are three source of sound of equal intensities with frequencies 400, 401 and 402 Hz. The no. of beats per second is :

Answer»

0
1
2
3

Solution :We known that y = r sin `2 pi v t`.
` thereforey_(1) = r sin 2pi 400 t , y_(2)= r sin 2 pi401 t `
`y_(3) = r sin 2pi 402 t.`
Now y = `y_(1) + y_(3) + y_(2)`
y = r sin `2pi` 400t + r sin `2pi` 402 t + r sin `2pi` 401 t.
y = r sin `2 pi ((402 - 400)/(2)) t cos 2pi ((402 - 400)/(2) ) + r sin 2 pi 401 t`.
`y = 2r sin 2pi 401 t cos 2pi t + r sin 2 pi 401 t`.
y = `[2 cos 2 PIT + 1] r sin 2 pi 401 t`.
Now Amplitude of resultant wave is
` A = (2 cos 2pi t + 1) r. `
For A to be maximum , cos `2 pi t = + 1 `
`rArr "" 2 pi t = 2 n pi, n = 0,1,2, t = n `
Amplitude will be maximum at time t = 0 , 1,2,3 ... sec.
Time interval between successive maximum = 1 sec.
`therefore` BEAT FREQUENCY = 1 Hz.
hence the CORRECT choice is (b) .
44.

Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is :

Answer»

7
8
3
5

Solution :Use V = `sqrt(T// m ) and v_(1) = (V)/(lambda_(1)) `
and `v_(2) = V//lambda_(2)THEREFORE B = | v_(1) - v_(2) | . `
CORRECT choice is a .
45.

The binding energy of the electron in the energy of the electron in the second orbit is 3.4 eV. Its P.E. in the same orbit is_____eV.

Answer»

-1.51
-3.4
-6.8
6.8

Answer :C
46.

A: a-particles with a large angle scattered leads to the discovery of the nucleus of the atom. R: The central part of the atom is almost concentrated by positive electric charges

Answer»

Both ASSERTION and REASON are TRUE and the reason is CORRECT EXPLANATION of the assertion.
Both assertion and reason are true but reason is not correct explanation of the assertion.
Assertion is true but the reason is false.
Both assertion and reason are false

Solution :Both assertion and reason are true and the reason is correct explanation of the assertion.
47.

An object of mass 1 kg is whirled in a horizontal circle of radius 25 cm. If it performs 60 rev/min, the centripetal force acting on it is,

Answer»

10 N
9.1 N
11 N
9.8 N

Answer :D
48.

Two lenses of power + 12D and - 2D are combined together. What is their equivalent focal length?

Answer»

10 cm
12.5 cm
16.6 cm
8.33 cm

Solution :`P = P_(1) + P_(2) = + 12 - 2 = 10 D`
`F = (1)/(P) = (1)/(10)m = 10 cm`
49.

Define the term 'mutual inductance between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r_(1) and r_(2) (r_(2) > r_(1)). Total number of turns in two solenoids are N_(1) and N_(2) respectively.

Answer»

Solution :Mutual inductance of two neighbouring coils is numerically equal to the amount of magnetic FLUX linked with one coil when unit current FLOWS through the other coil.
Alternately mutual inductance of two coils is equal to the emf induced in one coil when rate of CHANGE of current through the other coil is unity.
SI unit of mutual inductance is henry (H).
Mutual inductance of two long coaxial solenoids wound one over the other : See Short ANSWER Question Number 27.
50.

A slit of width d is placedin front of a lens of focal length 0.5 m and is illuminated normally with light of wavelength 5.89 xx 10^(-7)m. The first diffraction minima on either side of the central max. Are separated by 2 xx 10^(-3)m. The width of the slit is :

Answer»

`1.47 xx 10^(-4)` m
`2.29 xx 10^(-4)` m
`1.47 xx 10^(-7)` m
`2.29 xx 10^(-7)m`

Solution :WIDTH of central max. = separation between first min. On EITHER SIDE of central max.
`2theta = (2lambda)/(d)`
LINEAR width `2y = 2((lambda)/(d))f`" `[because (lambda)/(d) = theta]`
`therefore d = (2lambda.f)/(2y) = 2.92 xx 10^(-4)m`.