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.

Which of the following waves can produce photoelectric effect ?

Answer»

ultrasound
infrared
radiowaves
X-rays

Solution :ELECTROMAGNETIC radiation, being of HIGH frequency such as X-rays can produce photoelectric EFFECT.
2.

Flux associated with any point in electric field is......

Answer»

zero
NEGATIVE
positive
zero, negative or positive

SOLUTION :Area of point,
A =0
`THEREFORE ln phi = Eacostheta`,A =0
`therefore phi =0`
3.

The function of an amplitude limitter in an FM-receiver is

Answer»

to reduce the amplitude ofthe signal to SUIT IF amplifier
to amplify LOW frequency signal
to eliminate any change in amplitude of receiver FM signal
None of these

Solution :The limitter removes from the carrier all amplitude variations which may caused by changes in the transmission PATH, by man-made STATIC or natural static. This suppression of amplitude variation is necessary because FM-receives, a vary large IMPROVEMENT in S/ N results from this.
4.

What should be the distance between the object in Exercise 9.24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm^2. Would you be able to see the squares distinctly with your eyes very close to the magnifier ?

Answer»

Solution :If we want that area of each square in image is `6.25 mm^2`, it means that linear magnification:
`=sqrt((6.25 mm^(2))/(1 mm^(2))) = 2.5 rArr |v|/|u|` or `|v| 2.5 |u|`
As focal length of magnifying lens f= + 9 cm
`therefore 1/(2.5 mu) -1/u =1/9 rArr u = -5.4 cm` and HENCE, `|v| = 2.5 xx 5.4 = 13.5` cm
As now image is being formed at a DISTANCE of 13.5 cm from the eye and this distance is quite less than the least distance of distinct vision (D = 25 cm), hence, EYES will not be able to see the image DISTINCTLY.
5.

The function of a half -adder may be represented as shown in Figure. A full-adder may be constructed from two half-adders together with a single logic gate as shown in figure. Which logic gate must be used?

Answer»

`AND`
`EX-NOR`
`HAND`
`OR`

SOLUTION :When THREE bits is to be added, a full-adder MAY be CONSTRUCTED form two half-adders together with a single logic gate of `OR`.
6.

Two identical concave mirrors M_1 & M_2with principal axes perpendicular to each other and pole P_1 of mirror M_1 at origin is as shown. Let 'f' be the focal length and C be the common centre of curvature for both mirrors. A point object 'O' is at (-5f, 0) and is moving with a velocityvecV = V_0 hati m//sConsider rays incident on mirror M_1, the coordinates of the final image is

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at position (-X, 0) and MOVING along x -AXIS
at position (+x,0) and moving along x- axis
at position (-x, 0) and moving along y - axis
at position (+y,0) and moving along y - axis

Answer :A
7.

Two identical concave mirrors M_1 & M_2with principal axes perpendicular to each other and pole P_1 of mirror M_1 at origin is as shown. Let 'f' be the focal length and C be the common centre of curvature for both mirrors. A point object 'O' is at (-5f, 0) and is moving with a velocityvecV = V_0 hati m//sConsider rays incident on mirror M_2, the co-ordinates of the final image formed is

Answer»

at (-X,0) and MOVING along x - axis
at (-y,0) and moving along x - axis
at (+x, 0) and moving along x - axis
at (-y,0) and moving along y - axis

ANSWER :B
8.

Two identical concave mirrors M_1 & M_2with principal axes perpendicular to each other and pole P_1 of mirror M_1 at origin is as shown. Let 'f' be the focal length and C be the common centre of curvature for both mirrors. A point object 'O' is at (-5f, 0) and is moving with a velocityvecV = V_0 hati m//sConsider rays incident on mirror M,, the final image formed is

Answer»

REAL
virtual
may be real or virtual
all of these

ANSWER :A
9.

A resistance of 10 Omega a capacitance of 0.1muF and an inductance of 2mH are connected in series across a source of alternating emf of variable frequency. At what frequency does maximum current flow ?

Answer»

11.25 KHZ
23.76 kHz
35.46 kHz
46.72 kHz

Answer :A
10.

STATEMENT-1: A geostationary satellite must be located in the equatorial plane, i.e, at some point vertically above the equator. STATEMENT-1: The only external force acting on an earth satellite is direted towards the centre of the earth.

Answer»

Statement-1 is TRUE, Statement-2 is True, Statement-2 is a CORRECT EXPLANATION, for Statement-1.
Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is FALSE.
Statement-1 is a False, Statement-2 is True.

Answer :A
11.

As per Bohr's theory of hydrogen atom the value of Rydberg constant R is ______.

Answer»

SOLUTION :`1.03 XX 10^(7) m^(-1)`
12.

In the figure shown the magnetic field on the left on PQ is zero and on the right of PQ it is uniform.Find the time spent in the magnetic field.

Answer»

SOLUTION :The PATH will be SEMICIRCULAR TIME SPENT `=T//2=pim//qB`
13.

A coil has 1,000 turns and 500 cm^2as its area. The plane of the coil is placed at right angles to a magnetic induction field of 2xx10^(-5) wb//m^2 . The coil is rotated through 180^@in 0.2 seconds. The average e.m.f induced in the coil, in milli-volts.

Answer»

5
10
15
20

Answer :B
14.

A 10 m long wire of uniform cross-section and 20Omegaresistance is used in a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 2. If an unknown emf E is balanced at 6.0 m length of the wire, calculate(i) the potential gradient of the potentiometer wire, (ii) the value of unknown emf.

Answer»

Solution :
Here resistance of potentiometer wire `r = 20 Omega , epsi 5V , R = 480 Omega " and " L = 10 m`
` therefore ` POTENTIAL GRADIENT of the potentiometer wire
`K = ((epsi.r)/((R+r)).L)`
` = (5 xx 20)/((480 + 20) xx 10)`
`= 2 xx 10^(-2) V//m`
(ii) Value of unknown emf `E = kl = 2 xx 10^(-2) xx 6 = 0.12 V`
15.

Assertion A dish antenna is highly directional. Reason : This is because a dipole antenna is omni directional

Answer»


ANSWER :B
16.

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

Answer»

`IBL`
`(IBI)/(PI)`
`(IBI)/(2 pi)`
`(IBI)/( 4pi)`

ANSWER :C
17.

To four significant figures, find the following when the kinetic energyis 12.00 MeV: (a) gamma and (b) beta for an electron (E_(0)=0.510 998 MeV), (c ) gamma and (d) beta for a proton (E_(0)=9.38.272 MeV), and (e ) gamma and (f) betafor an alpha particle (E_(0)=3727.40 MeV).

Answer»


ANSWER :(a) 24.48; (B) 0.9992; (C ) 1.01279=1.013; (d) 0.1597; (E ) 1.0032191=1.003; (F) 0.0800
18.

Solve with due regard to significant Jigures. (i) 46.7- 10.4 = (ii) (3.0xx10^(-8))+(4.5xx10^(-6))=

Answer»

Solution :i) 46.7- 10.04
Here 46.7 has one DECIMAL place, and 10.04 has TWO decimal PLACES.
`:. 46.7-10.04 = 36.66`
The RESULT should have only one decimal place.
The result is 36.7.
`(II) 3.0xx10^(-8) +4.5xx10^(-6)`
`= 0.03xx10^(-6)+4.5xx10^(-6)`
`= 4.53xx10^(-6)`
Here `4.5xx10^(-6)` has only one decimal place and 0.03x 10 can have two decimal places. This result should be rounded off to one decimalplace.
`:. (3.0xx10^(-8))+(4.5xx10^(-6))=4.6xx10^(-6)`
19.

The natural boron of atomic weight 10.81 is found to have two isotopes .^10B and .^11B .The ratio of abundance of isotopesof natural boron should be

Answer»

`11:10`
`81:19`
`10:11`
`19:81`

Solution :Let ABUNDANCE of `B^10` be x%
`therefore` Abundance of `B^11` = (100-x)%
`therefore 10.81=((10 xx x)+11(100-x))/100`
or 1081=10x + 1100 - 11x or x=19
`therefore` Ratio of abundance =`19/(100-19)=19/81`
20.

A spherical ball of surface area 2xx10^(-3)m^(2) is suspended in a room at temperature 330 K. If the temperature of the ball is 200^(@)C, find the net rate of loss of heat from the ball if it behaves like a black body.

Answer»


ANSWER :`[4.6W]`
21.

A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If g is shot with speed 2v, the closest distance of approach would be

Answer»

`r/4`
`r/2`
2r
r

Answer :A
22.

We have achieved our political emancipation. What is the meaning of emancipation?

Answer»

FREEDOM from restriction
enslavement
slavery
both 2 and 3

Answer :A
23.

(a) Light waves from two cohernet sources having intensities I and 2I cross each other at a point with a phase difference of 60^@. What is the resultant intensity at the point ? (b) With the help of a diagram obtain an expression for finding the distance between two consecutive bright or dark fringes in the interference pattern prodcued by double-slits.

Answer»

Solution :(a) LET `I_1 = I, I_2 = 2I`
Resultant INTENSITY= `I_1 + I_2 + 2sqrt(I_1) sqrt(I_2) cos 60^@`
`= I + 2I + 2sqrt(I cdot 2I) XX 1/2`
`= 4.414 I`
24.

A particle is moving eastwards with a velocity of 5m/s. In 10s the velocity changes to 5 m/s north wards. Find the average acceleration in this time.

Answer»

Solution :`veca_(av)=(vecDeltav)/(DELTAT)=(vecv_(f)-vecv_(i))/(Deltat)`

`veca_(av)=(5hatj-5hati)/(10)=(5 sqrt(2))/(10)=(1)/(sqrt(2))m//s^(2)`. North- west direction.
25.

In the figure shows R=100Omega, L=2/(pi)H and C=8/(pi) muF are connected in series with an a.c. source of 200 volt and frequency f. V_(1) and V_(2) are two hot-wire voltmeters. If the reading of V_(1) and V_(2) are same, then

Answer»

`f=125Hz`
`f=250 piHz`
Current through `R` is `2A`
`V_(1)=V_(2)=1000` volt

Solution :`V_(1)=V_(2)impliesx_(L)=x_(C)`
`impliesf=1/(2pisqrt(LC))=125Hz`
`I_(0)=(v_(0))/R=200/100 ( :' x=0, :.z=R)`
`V_(1)=V_(2)=IX_(L)=I.(OMEGAL)=IL . 1/(sqrt(LC))=Isqrt(L/C)=2sqrt(((2//pi))/((8xx10^(-6)//pi)))=1000` volt
26.

Ship A is located 4.0 km north and 2.5 km east of ship B. Ship A has a velocity of 22 km/h toward the south, and ship B has a velocity of 40 km/h in a direction 37^(@) north of east. (a) What is the velocity of A relative to B in unit-vector notation with toward the east? (b) Write an expression (in terms of hatiandhatj) for the position of A relative to B as a function of t, where t = 0 when the ships are in the positions described above. ( c) At what time is the separation between the ships least? (d) What is that least separation?

Answer»

Solution :(a) `(-32km//H)hati-(46km//h)hatj`, (B) `(2.5-3.2t)hati+(4.0-46t)hatj`, ( c) 0.084 h, (d) `2XX10^(2)m`
27.

If a signal of frequency omega_s is used to modulate carrier wave of frequency omega_c which are the frequencies contained in the modulated signal other than omega_c ?

Answer»

SOLUTION :`omega_c + omega_s` and `omega_c - omega_s`
28.

Photo cells convert

Answer»

a. HEAT energyintoelectrical ENERGY
b. light energy into mechanical energy
C. thermal energy into mechanical energy
d. light energy into ELECTRICAL energy

Answer :D
29.

Two slabs A and B, each though similar, in all other respects, have different materials. They are placed one above the other in perfect contact and a steady difference of temperature of 36^(@)C is maintained across the combination. If the thermal conductivity ofA is twice that of B, what is the temperature of interface ?

Answer»

`16^(@)C`
`12^(@)C`
`28^(@)C`
`24^(@)C`.

SOLUTION :In steady state, rate of flow of HEAT is same
`H_(1)=H_(2)`.
`(k_(1)ADeltatheta_(1))/(d)=(k_(2)ADeltatheta_(2))/(d)`
`(Deltatheta_(1))/(Deltatheta_(2))=(k_(2))/(k_(1))=1/2`
`Deltatheta_(2)=2Deltatheta_(1)`.
Also `Deltatheta_(1)+Deltatheta_(2)=36^(@)C`
`:.""3Deltatheta_(1)=36^(@)C`
`Deltatheta_(1)=12^(@)C`
`rArr36-T=12rArrT=24^(@)C`
Thus correct choice is (d).
30.

A parallel plate capacitor C (without dielectric) is filled by dielectric slabs as shown in figure. Then the new capacitance of the capacitor is

Answer»

3.9 C
4 C
2.4 C
3 C

Answer :A
31.

A body of mass m is placed on earth surface which is taken from earth's surface to a height of h = 3R, then changein gravitational potential energy is :

Answer»

`(1)/(4)` MG R
`(3)/(4)` mg R
`(2)/(3)` mg R
`(1)/(2)` mg R.

Solution :When a body is taken to a height n times the radius of earth from the surface of earth then change in potential energy `=(n)/(n+1)mgR`.
`therefore DeltaU=(3)/(3+1)mgR rArr DeltaU=(3)/(4)mgR`.
Thus correct CHOICE is (B).
32.

An unpolarised beam of light of intensity l_0 falls on a poloroid. The intensity of the emergent beam is

Answer»

`I_0/2`
`I_0`
`I_0/4`
zero

Solution :Proof: INTENSITY of incident light with amplitude `E_0` is
`I_0 prop E_0^2`………..(1)
Intensity of transmitted light with amplitude `E_0 cos theta` (where `theta`= angle made by incident `vec E` VECTOR with the optic (or pass) axis of a given POLOROID) is,
`I prop (E_0 cos theta)^2`
`therefore I prop E_2^ cos^2 theta`..........(2)
TAKING ratio of equation (2) to equation (1) `therefore I=I_0 cos^2 theta`
Now, intensity of transmitted beam is the average of above intensity, Hence,
`lt I gt =I_0lt cos^2 theta gt `( `because I_0`= constant)
`therefore lt I gt =I_0(1/2)`
(`because` Incident light in unpolarised and so `theta` varies between 0 to `2pi` and we know average value of `cos^2 theta` over this interval is `1/2`)
`therefore lt I gt =I_0/2`
33.

A number of containers each contain initially 10,000 atoms of radioactive material with half life of 1 year are given. After 1 year

Answer»

All containers will have 5000 atoms of material
All containers will CONTAIN same NUMBER of atoms and number will be approximately 5000.
Containers will have DIFFERENT numbers of atoms of material, but their AVERAGE will be CLOSE to 5000.
None of containers can have more than5000 atoms.

Answer :C
34.

Estrogen is secreted by

Answer»

GERMINAL EPITHELIUM of ovary
Pituitary gland
Corpus luteum
Layers of graafian follicle

Answer :D
35.

Two current carrying coil having radius 'r' are seperated by distance 'd' as shown in diagram. Ifrlt ltd. Then find the force between two ring. Current in both the ring is clockwise and equal to i.

Answer»

`(6mu_(0))/(4pi)(i^(2)pi^(2)R^(4))/(d^(4))`
`(mu_(0))/(4pi)(i^(2)pi^(2)r^(4))/(d^(4))`
`(3mu_(0))/(4pi)(i^(2)pi^(2)r^(4))/(d^(4))`
`(mu_(0))/(2pi)(i^(2)pi^(2)r^(4))/(d^(4))`

Solution :`f=mu(dB)/(dx)`
`B=(2mu_(0)M)/(4pid^(3)) implies (dB)/(DD)=-(6mu_(0)M)/(4pid^(4))`
`F=M_(1)M_(2)[-(6mu_(0))/(4pid^(4))]=-(6)/(4pi)(mu_(0)M_(1)M_(2))/(d^(4))`
`-(6mu_(0)(pir^(2)i)(pi^(2)i))/(4pid^(4))`
36.

What is the resistance of an ideal (i) ammeter, (ii) voltmeter ?

Answer»

Solution :Resistance of an ideal ammeter is ZERO but that of an ideal VOLTMETER is infinite.
37.

What fraction of the total K.E. of a rolling ring is translational? (assume no slipping)

Answer»

2
1/2
3/2
2/3

Answer :B
38.

The universal property of all substance is _____.

Answer»

SOLUTION :DIAMAGNETISM
39.

Explain the Maxwell's modification of Ampere's circuital law.

Answer»

Solution :(i) The electric current PASSING through the wire is the CONDUCTION current . This current generates magnetic field around the wire connected across the capacitor.
(ii) "Therefore, when a magnetic needle is kept near the wire, deflection is observed. In order to compute the strength of magnetic field at a point, we use Ampere.s circuital law is used which states that .the line integral of the magnetic field `vecB` around any closed loop is equal to po times the net current I threading through the area enclosed by the loop. Ampere.s law in equation form is
`ointvecB*vecl=mu_0 I(t)"...(1)"`
(iii) where H, is the permeability of free space

To calculate the magnetic field at a point P near the wire an amperian loop (circular loop) which encloses the surfaces `S_1`, (circular surface), using (equation (1)),
`ointvecB*vec(dl)=mu_0 I_(C)"...(2)"`
(iv) Suppose the same loop is enclosed by balloon shaped surface `S_2`.. This means that the boundaries of two surfaces `S_1`, and `S_2`, are same but shape of the enclosing surfaces are different (first surface (`S_1`) is circular in shape and second one is balloon shaped surface (`S_2`,)). As the Ampere.s law does not DEPEND on shape of the enclosing surface, the INTEGRALS will give the same answer. But by applying (equation (1)),
`underset("enclosing"S_2)(oint)vecB*vec(dl)=0"...(3)"`
By applying equation (1)
`ointvecB*vec(dl)=0"enclosing "s_2`

(v) The right hand side of equation is zero because the surface `S_2`, no where touches the wire carrying conduction current and further, there is no current in between the PLATES of the capacitor (there is a discontinuity). So the magnetic field at a point P is zero.
(vi) Due to external sauce (battery or cell), the capacitor gets charged up because of current flowing through the capacitor. This produces an increasing electric field between the capacitor plates. The time varying electric flux between the plates of the capacitor produces a current known as displacement current.

(vii) From Gauss.s law
`phi_E = ointointvecE.dvecA = EA = q/epsi_0`
Where A is the area of the plates of capacitor.
The change in electric flux is
`(dphi_E)/(dt) = 1/epsi_0 (dq)/(dt) rArr(dq)/(dt) = I_d = epsi_d (dphi_E)/(dt)`
where `I_d` is know as displacement current.
(viii) The displacement current can be defined as the current in the region in which the electric field and the electric flux are changing with time. i.e whenever the change in electric field, displacement current is produced. Maxwell modified Ampere.s law as
`oint vecB.d vecS = mu_0 ( I_c +I_d)"...(4)"`
where `I = 1_c + I_d` which means the total current enclosed by the surface is sum of conduction current and displacement current.
(ix) When a constant current is applied, displacement current `I_d`, = 0 and hence `I_c` = I. Between the plates, the conduction current `I_c` = 0 and hence `I_d` = I.
40.

A six volt battery is connectedwith a resistance . A currentof 2 amperes flowsfor 4 minutes . Which of the following statement is wrong ?

Answer»

Resistanceis ` 3 Omega `
Heatproduced is 12 JOULES
POWER consumed is 12 Watts
Chargeflowed is 480 coulomb

Answer :B
41.

In a communication system, the signal is likely to be affected by the noice:

Answer»

At the transmitter
At the receiver
At the INFORMATION source
In the TRANSMISSION line

Answer :D
42.

State the limitations of plum pudding model of the atom.

Answer»

Solution :There are some limitations of this model it is that according to law of electrostatic stationary stable charge distributed is not possible. Because the electron of atom experience Coulomb force due to positive charge in atom. So they cannot remain stable but move with acceleration. Hence, the distribution of the electrons and positive charges are very different from that proposed in this model.
Each condensed matter at all temperature emit electromagnetic radiation in which a continuous distribution of several wavelength is present although their intensity is different.
In contrast light emitted from rarefied GASES heated in a flame has only CERTAIN discrete wavelengths. The spectrum APPEARS as a series of BRIGHT lines.
In such gases, the average spacing between atoms is large. Hence, the radiation emitted can be considered due to individual atoms rather than because of interactions between atoms or molecules.
With every element spectrum of radiation is associated which is characteristics of given element. For example, hydrogen ALWAYS gives senes of lines which are at definte distance. This fact suggested an intimate relationship between the internal structure of an atom and the spectrum of radiation.
43.

Findout the position where parallel rays will meet after coming out of the sphere and draw the appropriate ray diagram

Answer»

SOLUTION :`n_(2)/v-n_(1)/u=(n_(2)-n_(1))/R`
`rArr 1.5/v-1/oo=0.5/10`
`rArr 1.5v=0.5/10`
`rArr v=15/0.5=30`
For Iind
`1/v-1.5/10=(1-1.5)/(-10)`
`rArr v=+5`
44.

In Ingen Hauze's experiment, the wax melts up to lengths 10 and 25 cm on two identical rods of different materials. The ratio f thermal conductivities of the two materials is :

Answer»

`1:6.25`
`6.25:1`
`1:SQRT(2.5)`
`1:2.5`

ANSWER :A
45.

What type of teacher did Margie have?

Answer»

Smart
Ugly
Mechanical
Sad

Answer :C
46.

A ray of light travelling in air is incident at angle of incidence 30^@ on one surface of a slab in whichrefractive index varies with y. The light travels along the curve y, = 4x^2(y and x are in meter) in the slab. Find out the refractive index mu of the slab at y = 1/2 m in the slab.

Answer»

1.5
1.7
`(SQRT3)/(2)`
`(2)/(sqrt3)`

ANSWER :A
47.

What are radio waves?

Answer»

SOLUTION :RADIO WAVES are electromagnettic waves of WAVELENGTH `10^-3`m and HIGHER.
48.

A ball is dropped from top of the building 100m high. Simultaneously another ball is thrown upwards from bottom of tower with such a velocity that the balls collide midway. What is the speed of 2nd ball ?

Answer»

31.6 m/s
27.8 `ms^(-1)`
22.4 `ms^(-1)`
19.6 `ms^(-1)`

SOLUTION :Time taken by IST ball to fall through 50 m is
`t=sqrt((2xx50)/(10))`
`t=sqrt(10s)=3.16s`
Now durings this time 2 ball should cover disatnce 50 m
`:. 50 =uxx3.16-(1)/(2)xx10xx10impliesu=(100)/(3.16)=31.6ms^(-1)`
49.

What according to Padma is the identity of huma race?

Answer»

IDENTITY of a woman
Identity of a man
Identity of Padma
None of the above

Answer :A
50.

The diameter of steel rod is 8 xx 10^(-3) m. What force will stretch it by 0.3% of its length (Y = 20 X 10^11 N/m^2)

Answer»

a)`3 XX 10^5 N`
b)`1 xx 10^5 N`
c)`2 xx 10^5 N`
d)`4 xx 10^5 N`

ANSWER :A