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.

An audio signal by e_(S)=15 sin 2pi (2000t) modulates a carrier wave given by e_(C)=60sin 2pi(100,000)t. Calculate (a)the percentage modulation (b) frequency spectrum of the modulated wave.

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Solution :a)Signla Amplitude ,B=15,Carrier amplitude,
A=60
`m=(B)/(A)=(15)/(10)=0.25`
`therefore ` PERCENTAGE modulation =`0.25xx100=25%`
b)By comparing the given EQUATIONS of signal and carrier with their standard FORM
`e_(S)=E_(s)sin omega_(s)t=E_(S)sin2pi f_(S)t` and
`e_(C)=E_(C)sin omega_(C)t=E_(C)sin 2pif_(c)t`
We have signal frequency `f_(s)`=2000 Hz and carrier frequency `f_(C)`=100,00 Hz
The frequencies present in modulated wave
i) `f_(c)`=100,000 Hz=100kHz
ii) `f_(c)-f_(s)`=100,000-2000 =98 kHz
iii)`f_(c)+f_(s)`=100 kHz+2kHz=102 kHz
Therefore,frequency spectrum of modulated wave extends from 98 kHz to 102 kHz (This is CALLED band width.)
2.

A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:

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`(2v^2)/(5g)`
`(v^2)/(g)`
`(2v^2)/(7g)`
`(v^2)/(2g)`

Solution :By conservation of L momentum
`mv_0 = 5 mv' : v' = (mv_0)/(5M) = (v_0)/5`
Conservation of M.E.
`1/2 m v_0^2 = 1/2 5m (v')^2 + mgH, ""H = (2v_0^2)/(5g)`
3.

Consider the following statements, regarding a girl who wants to jump vertically upward from a hard horizontal floor. Symbolic representation of forces is given below. vec(F)_(M) = force developed in leg muscles, vec(F)_(W) = Weight, vec(F)_(N) = Normal reaction

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Total force on FLOOR is `vec(F)_(M)+vec(f)_(w)`.
Resultant force on girl is `(-vec(F)_(M))` while girl is in contact with floor.
Floor exerts a force `vec(F)_(N)` on girl.
The harder the girl pushes down, the greater the resulting acceleration of girl.

Solution :When the girl is bending her LEGS, the total force exerted by her legs in the action, normal force by grand is the reaction. Considering FBD of girl
`vec(F)_(N)+vec(F)_(w)=mvec(a)`
`vec(F)_(N)=-vec(F)_(M)`
So force on floor is `vec(F)_(M)`
Resultant force on giel is `vec(F)_(w)-vec(F)_(M)`
NOTE: If we say the change in force of girl is `vec(F)_(M)`, then (A) and (B) WOULD also be correct.
4.

Which of the compounds HCHO (I),CH_(3)CH_(2)CHO (II), CH_(3)COCH_(3) (III) and HCOOC_(2)H_(5) (IV) will give a secondary alcohol on reaction with excess Grignard reagent, followed by hydrolysis ? Select the correct answer using the codes given below:

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II only
III only
I and IV
II and IV

Solution :
5.

How high a man be able to jump on the surface of a planet of radius 320 km, but having density same as that of the earth if he jumps 5 m on the surface of the earth? (Radius of earth = 6400 km)

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60 m
80 m
100 m
120 m

Answer :C
6.

(A):A balloon ascends from the surface of earth with constant speed.When it was at a height 50m above the ground,a packet is dropped from it .To an abserver on the balloon,the displacement of the packet,from the moment it is dropped to the moment it reaches the surface of earth is 50m. (R ):Displacement vector does not depend upon the reference frame used to measure it.

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ANSWER :D
7.

Radio waves are electromagnetic waves used for communication purpose. Name two communication system that use space wave mode of propagation.

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SOLUTION :TV, RADIO
8.

In a coil of resistance 100 Omega , a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is ……..

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250 Wb
275 Wb
200 Wb
225 Wb

Solution :`EPSILON=(DELTAPHI)/(Deltat)`
`therefore IR=(Deltaphi)/(Deltat)`
`therefore Deltaphi=IR Deltat`
`=R (I Deltat)`
`=R xx ("AREA of GRAPH" I to t)`
`=100xx1/2 xx0.5xx10`
=250 Wb
9.

4 charges are placed each at a distance .a. from origin. The distance .a. from origin. The dipole moment of configuration is :

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`2qahat(J)`
`3qahat(j)`
`2qa(HAT(i) + hat(j))`
`qahat(j)`

ANSWER :A
10.

In which of the following compound is hydroxylic proton the most acidic?

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SOLUTION :CB more STABILISED by F(1-effect)
11.

A true balance is one whose pans are of equal masses and arms are of equal lengths. When this happens, the net moment of forces about point of suspension is zero and beam remains horizontal without any weight i.e., for the true balance P_(1)=P_(2) and l_(1)=l_(2) also P_(1)l_(1)=P_(2)l_(2) (mass of beam is neglibigle). A shopkeeper uses a false balance to weigh articles. Both arms and pans of this false balance are different, but beam become horizontal without any weight. (P_(1)neP_(2) and l_(1)nel_(2) but P_(1)l_(1)=P_(2)l_(2)). Q. Shopkeeper use a weight W to weigh and article by false balance. in two weighing by using alternating pans let W_(1) and W_(2) are the weights of the article given to the customer i.e., customer gets (W_(1)+W_(2)) weight of article. Loss of gain to the shopkeeper is:

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`(W(l_(2)-l_(1))^(2))/(l_(1)l_(2))` gain
`(W(l_(2)-l_(1))^(2))/(l_(1)l_(2))` loss
No gain, no loss
Data insufficient

Solution :Since `P_(1)l_(1)=P_(2)l_(2)` and `P_(2)ltP_(1)impliesl_(2)gtl_(1)`
`Wl_(1)=W_(2)l_(2)impliesW_(2)=(Wl_(1))/(l_(2))` since `(l_(1))/(l_(2))lt1impliesW_(2)ltW`
`IMPLIES` there will be gain for shopkeeper
`DeltaW_(2)=W-W_(2)=W((l_(2)-l_(1))/(l_(2)))`
`W_(1)` be the weight of article when article put in `P_(1)`
`W_(1)l_(1)=Wl_(2)impliesW_(1)=(Wl_(2))/(l_(1))` since `(l_(1))/(l_(2))LT`
`impliesW_(1)GTW`
`implies` there will be loss for shopkeeper
`DeltaW_(1)=W_(1)-W=W((l_(2)-l_(1))/(l_(1)))`
Since weight loss is more so net loss
`=W((l_(2)l_(1))/(l_(1)))-W((l_(2)-l_(1))/(l_(2)))=W((l_(2)-l_(1))^(2))/(l_(1)l_(2))`
12.

If x=(a^m.b^(n))/(c^(p).d^(q)) then the percentageserror in the measurement of x, (Deltax)/(x)=m((Deltaa)/(A))+n((Deltab)/(b))+p((Deltac)/(c))+q((Deltad)/(c)) (R) : The above is true for all values or Deltaa, Deltab, Deltac and Deltad.

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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 but (R) is FALSE
Both (A) and (R) are false

Answer :C
13.

Uranium ""_(92)U^(238) is not suitable for chain reaction . Why ?

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Solution :Natural uranium consists of three isotopes namely `U^(233), U^(235)` and `U^(238)`. The energy required for breaking up `U^(238)` is more than 1.2 MeV, and the energy required for breaking up `U^(235)` is about 0.03 eV. The energy of emitted neutrons in a nuclear fission is of the order of 0.03 eV, thus it can produce a chain reaction with `U^(235)` and not with `U^(238)`. And EVEN if a neutron of energy more than 1.2 MeV is used to produce fission of `U^(238)`, the secondary neutrons get slowed down after colliding with `U^(238)` nuclei and unable to cause further fission. Thus `U^(238)` is not suitable for chainreaction. [The percentage of `U^(233)` in natural uranium is negligibly SMALL `~=0.006%` and `U^(235)` is `~=0.7%` and `U^(238)~=99.3%`]
14.

लार में कौन-सा एंजाइम होता है ?

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टायलिन
पेप्सिन
श्लेष्मा
ये सभी

Answer :A
15.

A conducting coil having 500 turns has cross sectional area 0.15 m^2. A magnetic field of strength 0.2 T linked perpendicular to this area changes to 1.0 T in 0.4 sec. The induced emf produced in the coil will be ….. volt.

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`10.0`
`15.0`
`75.0`
`150.0`

Solution :INDUCED EMF `=-N (Deltaphi)/(Deltat)`
`=-N ((phi_2-phi_1))/(Deltat)`
`=-500([AB_2-AB_1])/(Deltat)`
`=-500xxA((B_2-B_1))/(Deltat)`
`=(-500xx0.15xx(1.0-0.2))/0.4`
`=(-500xx0.15xx(0.8))/0.4`
`=-500xx0.12/0.4`
`=-500xx0.3`
=-150 V
`therefore |epsilon|`=150 V
16.

In Young's double sit experiment two light sources when placed at a distance d apart, then Interference pattem having fringe width w. If the distance between the sources is reduced to d/3, then fringe width would be

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`w/3`
3w
w
`w/2`

ANSWER :B
17.

At t lt 0, the capacitor is charged and the switch is opened.At t = 0 the switch is closed.The shortest time T at which the charge on the capacitor will be zero is given by:

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`PI sqrtLC`
`(3)/(2) pi SQRT(LC)`
`(pi)/(2) sqrtLC`
`2pi sqrt LC`

Solution :The time period of LC oscillations `T = 2 pi sqrt(LC)`
The time at which charge on the capactior will be ZERO is `(T)/(4)` . So, `t = (pi)/(2) sqrt(LC)`
18.

An electric bulb has a solid cylindrical filament of length l and radius r and it consumes power P when connected to a power source. Another bulb having cylindrical filament of same material, operating at same voltage and emitting the same spectrum of light consumes 8 P power. find length and radius of the new filament. Assume that the filaments do not radiate from the flat ends and radiation is the only source of heat loss from the filaments.

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ANSWER :`2l,4r`
19.

कौन- सा राज्य गेहूँ का प्रमुख उत्पादक राज्य है?

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केरल
पश्चिम बंगाल
असम
उत्तर प्रदेश

Answer :D
20.

चावल के लिए कितने से०मी० वर्षा जरूरी है?

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100 से०मी० से कम
100 से०मी० से अधिक
50 से०मी० से 80 से०मी०
90 से०मी० से कम

Answer :B
21.

The electric field of a plane electromagnetic wave varies with time of amplitude 2 Vm^(-1) propagating along z - axis. The average energy density of the magnetic field is (in Jm^(-3))

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a. `13.29xx10^(-12)`
B. `8.86xx10^(-12)`
C. `17.72xx10^(-12)`
d. `4.43xx10^(-12)`

ANSWER :B
22.

Two wires made of same material have their length are in the ratio 1:2 and their masses in the ratio 3:16. The ratio of resistance of two wires is

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

ANSWER :D
23.

Define the terms 'magnetic dip' and 'magnetic declination' with the help of relevant diagrams.

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Solution :MAGNETIC Dip : Magnetic dip `(delta)` at a place is defined as the angle which the direction of TOTAL STRENGTH of earth.s magnetic field `vecB_E` SUBTENDS with a horizontal line along the magnetic MERIDIAN as shown in Fig.

Magnetic Declination : Magnetic declination (D) at a place is defined as the angle whcih magnetic meridian at that place subtends from the geographic meridian. In fig. angle D is the magnetic declination.
24.

Two niclos are crossed to each other.Now one of them is rotated through 60^(@).What percentage of incident light will pass through the system?

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`12%`
`24%`
`37.5%`
`52%`

SOLUTION :TRANSMISSION of light for incident unpolarised light is 50%.
Intensity of light emerging from second nicol prism is
`I_(theta) = 1/2 I_(0) cos^(2)theta (Malus law)`
`theta = 90^(@) - 60^(@) = 30^(@)`
`I_(theta)/(I_(0)) = (cos^(2)30^(@))/(2) = 3/8`
% transmission = `(I_(theta))/(I _(0)) xx 100 = 3/8 xx 100`
` = 37.5 %`
25.

Two point charges are seperated by a distance r such that a medium of dielectricconstant K is occupied by a length x Now find the coulomb force between those stationary charges.

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Solution :In this CASE we removethe dielectricand its effective equivalent distance in free space can be taken as `sqrt(K)X`
Now the force between the point charges is
`F=(1)/(4pi in_(0)) (q_(1)q_(2))/({(R-x)+sqrt(K)x}^(2))`
If K = 4 and`x=r//2`
`F=(1)/(4pi in_(0)) (4q_(1)q_(2))/( 9R^(2)) =(4)/(9) (1)/(4pi in_(0)) (q_(1)q_(2))/(r^(2))`
`F=(4)/(9)F_(0)` where
`F_(0)=(1)/(4pi in_(0))(q_(1)q_(2))/(r^(2))` which is the force between the charges if they are separatedby r in vacuum.
26.

If a radio-receiver amplifies all the signal frequencies equally well, it is said to have high

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FIDELITY
DISTORTION
SENSITIVITY
SENSIBILITY

ANSWER :A
27.

A convex lens of focal length 10 cm and a concave lens of same focal length in value are placed co-axially at a separation of 20 cm as shown in figure. A point object is at infinity on the principal axis. Take the optic center of convex lens as origin of co-ordinate system and the principal axis as x-axis as shown in figure.If the concave lens is moved vertically downwards through a distance of 5 mm, then the co-ordinatesof the final image after refraction through the two lenses is

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(15 CM, -0.25 cm)
(10 cm, +0.25 cm)
(25 cm, -0.5 cm)
(15 cm, +0.5 cm)

ANSWER :A
28.

A convex lens of focal length 10 cm and a concave lens of same focal length in value are placed co-axially at a separation of 20 cm as shown in figure. A point object is at infinity on the principal axis. Take the optic center of convex lens as origin of co-ordinate system and the principal axis as x-axis as shown in figure.The co-ordinates of the image formed after refraction through both the lenses is

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(10 CM, 0)
(25 cm, 0)
(-5 cm, 0)
(15 cm, 0)

Answer :D
29.

A convex lens of focal length 10 cm and a concave lens of same focal length in value are placed co-axially at a separation of 20 cm as shown in figure. A point object is at infinity on the principal axis. Take the optic center of convex lens as origin of co-ordinate system and the principal axis as x-axis as shown in figure.In the initial position (as given in the passage) if the concave lens has a velocity of 9 cm/s towards the convex lens, the velocity of the final image will be

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10 cm/s along POSITIVE x-axis
6.75 cm/s along NEGATIVE x-axis
8 cm/s along negative x-axis
8 cm/s along positive x-axis

ANSWER :B
30.

The kinetic energy of an electron revolving around a nucleus will be

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FOUR TIMES of P.E.
Double of P.E.
Equal to P.E.
HALF of its P.E.

ANSWER :D
31.

Consider a situtation (i) two sound waves y_(1) = 0.2 sin 504 pi (t-x/300) and y_(2) = 0.6 sin 490 pi (t-x/300) are superimposed Consider another situation: (ii) two sound waves y_(1) = 0.2 sin 504 pi(t-x/300) and y_(2) = 0.4 sin 500 pi (t + x/300) are superimposed. Match the Column-I with Column -II

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ANSWER :A-qrs; B-pr; C-pr; D-qs
32.

The diameter of an optical fibre is

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`10^(-3)` cm
`10^(-4)` cm
`10^(-2)` cm
`10^(-5)` cm

ANSWER :B
33.

The question contains statement I and statement II. Of the four choices given, choose one that best describes two statements. Statement I: For free space (epsilon= 0, I = 0) oint vecE. vec(dl)=-(d)/(dt) oint vecB.vec(ds) Statement II: For steady state oint vecE. vec(dl)= oint vecB.vec(ds)=0

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STATEMENT I is true, statement II is false.
Statement I is false, statement II is true.
Statement I and II both are true but statement I is correct EXPLANATION of that of II.
Statement I and II are both true but statement I is not correct explanation of I.

Answer :C
34.

A thin film of variable thickness is illuminated by a parallel beam of white light. The colour of the film at a certain point depends upon :

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the THICKNESS of the FILM at that point
the REFRACTIVE INDEX of the film
the ANGLE of incidence
all the above factors.

Answer :D
35.

Ultraviolet radiation of 6.2 eV falls on an aluminium surface with W_0=4.5eV . The kinetic energy in Joules of the fastest electron emitted is approximately :

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`3XX10^(-21)`
`3xx10^(-19)`
`3xx10^(-17)`
`3xx10^(-15)`

ANSWER :B
36.

(A) : If an inductor coil is connected in DC source, the current supplied by it is I_(1). If the same coil is connected with an AC source of same voltage. Then current is I_(2). Then I_(2)ltI_(1) (R): In AC circuit, inductor coil offers more resistance.

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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
Both 'A' and 'R' are false

Answer :A
37.

A diffraction pattern is formed by blue light, if blue light is replaced by yellow light, then ......

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the maxima and MINIMA are BROADENED and become distinct.
the maxima and minima are narrow and more crowded.
the PATTERN does not change.
diffraction pattern disappears

Solution :Amount of diffraction depend on `(lambda)/(d)` where d is constant.
`:.` Diffraction depend on `prop lambda and lambda V lt lambda y`
By USING yellow light amount of diffraction will be more hence central maximum will be wider and more apart.
38.

निम्न मे से कौन बहुपद नहीं है ?

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`X^2+2X+5`
`X^2+4X`
`sqrtX+4X`
`X+4`

ANSWER :C
39.

निम्न मे से कौन त्रिपद बहुपद है ?

Answer»

`X^3+5X^2`
`X^2+2X+5`
`X^3+5X`
इनमे से कोई नहीं

ANSWER :B
40.

The phenomenon of interference is exhibited by

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longitudinal WAVES only
transverse waves only
electromagnetic waves only
all SORT of waves

Solution :Interference phenomenon is exhibited by all sort of waves.
41.

A 1500 kg load is hung from the free end of a horizontal aluminum rod of length 7.0 cm, diameter 9.6 cm, and negligible mass. The other end of the rod is fixed in place. The shear modulus of aluminium is 3.0xx10^(10)N//m^(2). Find (a) the shear stress on the rod and (b) the vertical deflection of the rod's free end.

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SOLUTION :(a) `2.0xx10^(6)N//m`,
(B)`4.7 10^(-6)m`.
42.

How many tigers did Maharaja kill??

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50 tigers
80 tigers
99 tigers
60 tigers

Answer :C
43.

Eight equal drops of water are falling through air with a stready velocity of 10 cm s^(-1). If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is :

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40 cm `s^(-1)`
10 cm `s^(-1)`
`30cm s^(-1)`
`80cms^(-1)`

SOLUTION :Terminal velocity v`=(2r^(2))/(9eta)(p-sigma)G`
`rArrvpropr^(2)`
When 8 DROPS COMBINE to form a single drop then
`rArrR=2r`
`therefore(v_(2))/(v_(1))=(R^(2))/(r^(2))`
`(v_(2))/(v_(1))=(4R^(2))/(r^(2))=4`
`v_(2)=4v_(1)rArrv_(1)=4xx10=40` cm/s
So correct choice is (a).
44.

An ideal battery passes a current of 5A through a resistor . When it is connected to another resistance of 10 Omega in parallel, the current is 6A . Find the resistance of the first resistor.

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Solution :For IDEAL battery internal resistance r=0
Resistance of first wire= `R_1`
Resistance of the second wire
`R_2 = 10 OMEGA`

Current through `R_1 ` in the first case `i_1 = 5A`
Current in the second case `i_2 = 6A`
Effective resistance in the second case `R= (R_1 R_2)/(R_1 + R_2)`
`V= I_1 R_1 and V= I_2 (R_1 R_2)/(R_1 + R_2)`
`I_1 R_1 = I_2(R_1 R_2)/(R_1 + R_2)`
`impliesI_1 = I_2 (R_2)/(R_1 + R_2)`
`5 = 6 xx (10)/(R_1 + 10) implies5(R_1 + 10) = 60`
`5R_1 + 50 = 60 , 5R_1 = 10`
`R_1 = (10)/(5) = 2 Omega impliesR_1 = 2 Omega`
45.

A loop ABCD containing two resistors as shown in figure is placed in a uniform magnetic field B directed outward to the plane of page. A sliding conductor EF of length l and of negligible resistance moves to the right with a uniform velocity v as shown in Fig. Determine the current in each branch.

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Solution : The magnetic field induction B, length and the velocity V of the conductor EF are mutually perpendicular, hence the emf induced in it is e = Bly (with end F of the rod at higher potential) .
` therefore ` The effective electric circuit can be redrawn as SHOWN in Fig

The resistances `R_1` and `R_2`are in parallel, so the equivalent resistance R is given by ` 1/R = (1)/(R_1) + (1)/(R_2)`
From Ohms law, the total current is `i = e/R`
`i = Bvl ((1)/(R_1) + (1)/(R_2))`
Current in AD is `i_1 = (BLV)/(R_1) , ` current in BC is `i_2 = (Blv)/(R_2)`
46.

A projectile has the maximum range of 500 m. If the projectile is now thrown up on an inclined plane of 30^(@) with the same speed, what is the distance covered by it along the inclined plane?

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Solution :`R_("max")=(u^(2))/(g) "" :. 500 =(u^(2))/(g) or u= SQRT(500g)`
`v^(2)-u^(2)= 2gs rArr 0-500g=2xx(-g 30^(@)) xx X`
`x=500 m`
47.

The working principle of the mass spectrograph is that for a given combination of accelerating potential and magnetic field. The radius of curvature of the ion beam of charge q and mass m collected at different positions on the collector will depend on the value of

Answer»

`sqrt((m)/(2q))`
`(m)/(q)`
`sqrt((m)/(q))`
`(m)/(2q)`

Solution :Let velocity with which charge enters magnetic FIELD is v
`:. (1)/(2) mv^(2) = qv_(0) rArr v= sqrt((2qv_(0))/(m))`
Charge is COLLECTED at 2r.
`:. 2r= 2(mv)/(qB)= (2m)/(qB) sqrt((2qv_(0))/(m))= sqrt((m)/(q)) ((2 sqrt2v_(0))/(B))`
48.

A uniform magnetic field Bexists along the Z-direction. A metal wire carrying current I, is placed along the curvey = b sin . (2pi)/lambdax. Calculate the magnetic force acting on a portion fromx = 0 " to " x = 5 lambda.

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Solution : The magnetic field is uniform, therefore, the magnetic force does not depend on the shape of curve. It only depends on the end points (initial or final points)
As the metal wire is PLACED perpendicular to the magnetic field, ` theta = 90^(@)` .
HENCE the force ` = BIL sin 90^(@) = BIl`.
The curve CORRESPONDING to the given equation is as given below T.

The shape of curve resembles of the shape of a wave of wavelength.
So, distance between points `Å (x = 0) " and " B (x = 5 lambda)= 5 lambda`
` :. ` Length , `l = 5 lambda`
Hence , the magnetic force ` = F = BIl = BI (5 lambda) = 5BI lambda`
49.

A parallel plate capacitor with an oil of dielectric constant 2 between the plates has a capacitance 'C'. If the oil is removed, then its capacitor becomes

Answer»

`sqrt2C`
2C
`C/sqrt2`
C/2

Answer :D
50.

A negatively charged particle is situated on a straight line joining two other stationary particles each having charge +q. The direction of the motion of the negatively charged particle will depend on

Answer»

the MAGNITUDE of CHARGE
the POSITION at which it is situated
both magnitude of charge and its position
the magnitude of +q

Answer :B