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.

Bulltes of 0.03 kg mass each hit a plate at the rate of 200 bullets per second, with a velocity of 50 m/s and reflect back with a velocity of 30 m/s. The average force acting on one plate (in Newton) is

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120
180
300
480

Answer :D
2.

Two equal and opposite charges of magnitude 0.2 mu C are 15 cm apart, the magnitude and direction of the resultant electric intensity at a point midway between the charges is

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`6.4 XX 10^(5)N//C` TOWARDS -ve CHARGE
`6.4 xx 10^(5)N//C` towards +ve charge
zero
infinity

Answer :A
3.

Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the parts of the eletromagnetic spectrum to which each belongs. 21 cm (Wavelength emitted by atomic bydrogen in interstellar space)

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Solution :RADIO WAVES of short WAVELENGTH END.
4.

How wll you distinguish between a compound microscope and a telescope simply by seeing them?

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SOLUTION :The APERTURE of the objective of a compound MICROSCOPE is much SMALLER than that of the eye PIECE. However, the objective of a telescope is much larger than that of the eyepiece.
5.

एक इलेक्ट्रॉन एवं प्रोटॉन एक-दूसरे से 1A दूरी पर रखे हैं। इस निकाय का वैद्युत द्विध्रुव आघूर्ण होगा -

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`1.6 XX 10^(-19)`C.m
`1.6 xx 10^(-29)`C. m
`3.2 xx 10^(-19)`C. m
`3.2 xx 10^(-29)`C. m.

Answer :B
6.

Domain of y=sin^(-1) X is-

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`(-1,1)`
`{-1,1}`
`[-1,1]`
NONE of these

Answer :C
7.

The specific resistance of the material of a sonducting wire is rho and current through cross sectional area of the wire (I,e., current decsity ) is J . What is the power cousumed per unit volume of the wire ?

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Solution :Let length of the wire = l , cross sectianal area of the wire = A and current through the wire = I .
So , resistance , `R=rho.(l)/(A),"currentdensity ", J = (I)/(A),`VOLUME of the wire = IA
Power consumed , `P=I^(2)R`
So , power consumed per unit volume ,
`(P)/(lA)=(I^(2)R)/(lA)=(l^(2))/(lA).rho(l)/(A)=((I)/(A))^(2).rho=J^(2)rho`
8.

Assuming the moment of inertia of the pulley to be Iand its radius to be r.

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Solution :SINCE the pulley rotates with an acceleration, there MUST be a torque acting on it due to the difference in the TENSIONS of the left and the right parts of the rope (see Fig) It follows from the fundamental equation of dynamics that
`m_2g-T_2=m_2ga,-m_1g+T_1=m_1a,M=IDeltaomega//Deltat`
The torque is `M=(T_2-T_1)r.` The variation of the angular VELOCITY is
`Deltaomega=omega_2-omega_1=(v_2)/r-(v_1)/r= (aDeltat)/r`
We `m_2g-T_2=m_2a,-m_1g+T_1=m_1a_1,""T_2-T_1=(Ia)/(r^2)`
Hence
`a=((m_2-m_1)G)/(m_1+m_2+I//r^2),T_1=(2m_1m_2g(1+I//2m_2r^2))/(m_1+m_2+I//r^2)`
`T_2=(2m_1m_2g(1+I//2m_1r^2))/(m_1+m_2+I//r^2)`
For `I//r^2 lt lt m_1+m_2` we obtain the answer to Problem 3.2
9.

On increasing the pressure, the melting point of ice decreases while with the increase of pressure themeltingpointofwaxis increased. Explain why ?

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Solution :When ice melts into WAFER, there is a REDUCTION in VOLUME. Now increasing the PRESSURE means helping the reduction in volume. So the ice now melts at lower ‘ temperature. The CASE is opposite in case of wax.]
10.

When a n electric dipole is in st able equilibrium, in uniform electric field, its electrostatic potential energy is ......

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`-PE`
pE
0
`prop`

SOLUTION :For stable EQUILIBRIUM we should have
`U= U_("min") implies ` When `theta=0^(@)` we have
`U =- pE = U_("min")`
11.

A 50 kg man stuck in flood is being lifted vertically by an army helicopter with the help of light rope which can bear a maximum tension of 70 kg-wt. The maximum acceleration with which helicopter can rise so that rope does not breaks is (g = 9.8 ms^-2)

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`4.00 ms^-2`
`3.92 ms^-2`
`3.52 ms^-2`
`3.00 ms^-2`

ANSWER :d
12.

What would be the radius of second orbit of He^+ion?

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1.058Å
3.023Å
2.068Å
4.458Å

Answer :A
13.

Modern train are based on maglev technology in which trains are magntically levitated, which runs its EDS maglev system. There are coils on both sides of wheels. Due to motion of the train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If train lowers down then due to Lenz's law, repulsive force increase due to which train gets uplifted and if it goes much higher then there is a net downward force due to gravity. The advantage of Maglev trains in that there is no friction between the train and the track, thereby reducting power consumption and enabling the train to attain very high speeds. Disavantage of Maglev train is that as it slows down, the electromagnetic forces decreases and it becomes difficult to keep it levitaited and as it moves forward according to Lenz's law there ia an electromagnetic drag force. Which force causes the train to elevate up?

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Electrostatic force
Time VARYING ELECTING field
Magnetic force
Induced electric field

Solution :For ANSWERS, refer to the paragrapg given in the question.
14.

Modern train are based on maglev technology in which trains are magntically levitated, which runs its EDS maglev system. There are coils on both sides of wheels. Due to motion of the train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If train lowers down then due to Lenz's law, repulsive force increase due to which train gets uplifted and if it goes much higher then there is a net downward force due to gravity. The advantage of Maglev trains in that there is no friction between the train and the track, thereby reducting power consumption and enabling the train to attain very high speeds. Disavantage of Maglev train is that as it slows down, the electromagnetic forces decreases and it becomes difficult to keep it levitaited and as it moves forward according to Lenz's law there ia an electromagnetic drag force. What is the advantage of this system?

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No friction HENCE no POWER consumption
No electric power is zero
Graviatation FORCE is zero
Electrostatic force draws the train

Solution :For answers, REFER to the paragrapg given in the QUESTION.
15.

Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the parts of the eletromagnetic spectrum to which each belongs. 1057 MHz ( frequency of radiation arising from the two close enerfgy levels in hydrogen known as Lamb shift).

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SOLUTION :RADIO WAVES ( Short WAVELENGTH END).
16.

Modern train are based on maglev technology in which trains are magntically levitated, which runs its EDS maglev system. There are coils on both sides of wheels. Due to motion of the train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If train lowers down then due to Lenz's law, repulsive force increase due to which train gets uplifted and if it goes much higher then there is a net downward force due to gravity. The advantage of Maglev trains in that there is no friction between the train and the track, thereby reducting power consumption and enabling the train to attain very high speeds. Disavantage of Maglev train is that as it slows down, the electromagnetic forces decreases and it becomes difficult to keep it levitaited and as it moves forward according to Lenz's law there ia an electromagnetic drag force. What is the disadvantage of this system?

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Train EXPERIENCES UPWARD force according to Lenz's LAW
Friction force CREATES a drag on the train
Retardation
By Lenz's law, the train experiences a drag

Solution :For answers, REFER to the paragrapg given in the question.
17.

A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of sound in a tissue is 1.7 km/s. The wavelength of sound in the tissue is close to

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`4 xx 10^(-3)` m
`8 xx 10^(-3) `m
`4 xx 10^(-4)` m
`8 xx 10^(-4)` m

Solution :Frequency `(V) = 4.2 MHZ = 4.2 xx 10^(6) Hz` and SPEED of SOUND `(v) = 1.7 km//s = 1.77 xx 10^(3)` m//s
Wavelength of sound in tissue `(lambda) = v/u =(1.7 xx 10^(3))/(4.2 xx 10^(6))`
`=4 xx 10^(-4)` m
18.

An alpha particle is accelerated through a potential difference of 100 volts. Its kinetic energy will be

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1 MeV
2 MeV
4 MeV
8 MeV

Solution :KINETIC ENERGY `K = QV = (+ 2E)xx (10^6V) = 2 xx 10^6 eV= 2MeV`
19.

A man holding a lighted candle in front of thick glass mirror and viewing it obliquely sees a number of images of the candle. What is the origin of the multiple images ?

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Solution :Images are FORMED :
REFLECTION at front surface.
Reflection at BACK surface.
Multiple reflections at back and front SURFACES. So a NUMBER of images are seen.
20.

The temperature at which the speed of sound in air becomes double of its value at 0^(@)C

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273 K
546 K
1092 K
0 K

Answer :C
21.

A medium shows relation between i and r as shown. If speed of light in the medium is nc then value of n is

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1.5
2
`2^(-1)`
`3^(-1//2)`

ANSWER :d
22.

A plumb line is hanging from the ceiling of a train. If the train moves along the horizontal track with a uniform acceleration, the plumb line gets inclined to the vertical at an angle:

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`tan^(-1)((a)/(G))`
`tan^(-1)((g)/(a))`
`SIN^(-1)((a)/(g))`
`COS^(-1)((g)/(a))`

Answer :A
23.

Potential difference between the terminals of a cell/battery is equal to its emf when the cell/battery is____.

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SOLUTION :in an OPEN CIRCUIT
24.

An electric charge will experience a force in uniform electric field. Similarly a moving charge experiences a magnetic force (Lorentz) in magnetic field. The SI unit of magnetic field intensity is defined in terms of Lorentz force. a. Write the expression for magnetic Lorentz force. b. Mention any two differences between electric and magnetic field. c. Give an account of work done by Lorentz force on a moving charge and corresponding change in K.E.

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Solution : ` VEC F =q( vec V xx vec B)`
b. Electric field is DUE to a charge, either in motion or at rest. Magnetic field is due to the motion of charge. Direction of electric force is COLLINEAR to electric field, direction of magnetic force is perpendicular to magnetic field.
c. As Lorentz force is perpendicular to displacement (velocity), WORK DONE is zero andhence there is no change in K.E.
25.

A prism of refractive index mu_1 = 1.42 and prism angle 10° is arranged with another prism of refractive index mu_2 = 1.7. If this combination gives the dispersion without deviation, then what should be the prism angle of second prism?

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`6^@`
`8^@`
`10^@`
`4^@`

Solution :For dispersion WITHOUT deviation `delta=0`
`therefore delta=delta_1+delta_2`
`0=(mu-1)A+(mu.-1)A.`
`therefore A.=-((mu-1)A)/(mu.-1)`
`=-((1.42-1)xx10)/((1.7-1))=-(0.42xx10)/(0.7)=-6^@`
Negative sign represents that PRISM angles of prisms are opposite.
`therefore` Prism ANGLE of second prism = `6^@`
26.

Find the expession for the equivalent emf & internal resistanceof the series combination of cells.

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Solution :(i) Suppose n cells, each of emf `xi` volts and internal resistance r ohms are connected in series with an external resistance R.

(ii) The total emf of the battery `n xi`
The total resistance in the CIRCUIT `NR+R` by Ohm's law, the currentin the circuit is
`I=("total emf")/("total resistance")=(n xi)/(nr+R)`
Case (a) if `r lt lt R`, then,
`I=(n xi)/(R)~~nI_(1)`
(iii) Where `I_(1)` is the current due to a single cell `(I_(1)=(xi)/(R))`
Thus, if r is negligible when COMPAREDTO R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If ` r lt lt R, I=( n xi)/(nr)~~(xi)/(r)`
(iv) It is the current due to a single cell. That is , current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.
(V) Thus series connection of cells is advantageous only when the effective internal resistanceof the cell is negaligibly small compared with R.
27.

What was the ball?

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A duck
A rat
A basketball
All of the above

Answer :A
28.

Energy corresponding to threshold frequency of metal is 6.2 eV.For given radiation stopping potential is 5 V then incident radiation will be in …region.

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X-ray
ultraviolet
infrared
visible

Solution :`E=K_(m)+PHI=eV_(0)+phi=(5+6.2)EV=11.2 eV`
`(hc)/(lambda)=11.2xx1.6xx10^(-19)J`
`therefore lambda =(6.62xx10^(-34)xx3xx10^(8))/(11.2xx1.6xx10^(-19))`
`=1.108xx10^(-7)m=1108 Å`
This WAVELENGTH is in ultraviolet REGION.
29.

The magnetic field due to a current-carrying loop of radius 3 cm at a point on its axis at the distance of 4 cm from the centre is 54 mu T. What will be its value at the centre of the loop?

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`250 mu T `
`150 mu T`
`125 MUT `
`75 mu T `

ANSWER :A
30.

Three identical rods are joined together to form an equilateral triangle frame. Three axes AA', BB' and CC' lie in the plane of the frame as shown in the figure. Then the moment of inertia is least about the axis _________ whereas maximum about the axis _____________.

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ANSWER :BB', AA'
31.

Why do crystalline state most stable ?

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SOLUTION :Crystalline STATE is a state of minimum ENERGY . So crystalline state is most stable .
32.

If a function f is defined from A to B and a function g is defined from C to D then gof is defined if

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A=B
B=C
C=D
D=A

Answer :B
33.

Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

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Solution :(f) If no. of ATOMIC dipoles PER UNIT volume is same in paramagnetic and FERROMAGNETIC materials then saturation magnetisation (or maximum possible magnetisation) obtained in paramagnetic substance is A/most of the same order as that found in ferromagnetic substance.
34.

Write the unit of magnetic dipole moment.

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`Am^(-2)`
`Am^(-1)`
`JT^(-1)`
`J^(-1) T`

Solution :TORQUE `tau = m B SIN theta`
`therefore m = (tau )/( B sin theta) therefore` Unit of `m= (J)/(T) = JT^(-1)`
35.

Compare I_("max") and I_("min") with amplitudes of the interfering waves.

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SOLUTION :`(I_("max"))/(I_("min"))=((r+1)/(r-1))^2` where `=(A_1)/(A_2). A_1 & A_2` are the amplitudes of the INDIVIDUAL interfering WAVES.
36.

An EM wave radiates outwards from a dipole antenna, with E_(0) as the amplitude of its electric field vector. The electric field E_(0) which transports significant energy from the source falls off as

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`(1)/(r^(3))`
`(1)/(r^(2))`
`(1)/(r )`
remains constant

Solution :Electromagnetic waves EMITTED from dipole antenna PROPAGATE in outward direction in form of RAYS.
At very large DISTANCE amplitude of electromagnetic wave is inversely PROPORTIONAL to its distance from the source or antenna.
37.

If v_gamma, v_x and v_m are the speeds of gamma rays, x-rays and microwaves respectively in vacuum then :

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`v_gamma GT v_x gt v_m`
`v_gamma LT v_x lt v_m`
`v_gamma gt v_x gt v_m`
`v_gamma = v_x = v_m`

ANSWER :D
38.

A stick has a length of l12.132 cm and another stick has a length of 12.4 cm. (a) If the two sticks are placed end to end, what is their total length ? (b) If the two sticks are placed side by side, what is the difference in their lengths ?

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Solution :(a) Let lengths of the sticks ARCE named as `I_(1)= 12.132cm. I_(2) = 124 CM `
Here `I_(2)` has one decimal place and `I_(1)`has to be rounded off to have only two decimal places
`I_(1)+I_(2)=12.13+12.4=24.53`
This is to be rounded off to have one decimal place only.
This is to be rounded off to have one decimal place only. The total length is 24.5cm
(b) `I_(1)=12.132 , I_(2) = 12.4 . I_(1) -I_(2)=12.4 -12.132`
Here 12.4 has only decimal place and hence 12.132 should have only two decimal places.
`:. I_(1)-I_(2) =12.4 -12.13 = 0.27`
This should be rounded off to have only one decimal place .
`:. l_(1)-l_(2)=0.3`
Hence difference of their lengths is 0.3 cm
39.

(a) The peak voltage of ac supply is 600V. What is its rms voltage? (b) The rms value of current in an ac circuit is 20 A. What is its peak current?

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SOLUTION :(a) Here `E_(0)=600V`
`therefore E_("RMS")=(E_(0))/(SQRT2)=(600)/(sqrt2)=424.3V`
(B) Here `I_("rms")=20A`.
We know `I_("rms")=(I_(0))/(sqrt2)""therefore I_(0)=sqrt2I_("rms")`
`or I_(0)=sqrt2xx20=1.414xx20=28.28A`,
40.

A jar has a mixture of H_(2) and O_(2) gas in the ratio of 1: 5. The ratio of mean K.E. of H_(2) and O_(2) molecules is :

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`1:1`
`1:5`
`1:4`
`4:1`

Solution :K.E. of gas molecules is given by
`1/2mv^(2)=3/2kT`
SINCE in MIXTURE temperature is same.
`therefore` K.E. of both the gas molecules will also be same. THUS correct choice is (a).
41.

The clouds generally appear white due to .....

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reflection of LIGHT
scattering of light
diffraction of light
dispersion of light

Solution :The ability of an object to scatter inciden waves depends UPON the dimensions of th object compared with the wavelength of light Here sunlight scattering will be DIFUSION scattering. For particles much larger than thewavelength of light scattering `(alpha(1)/(lambda^4))` become uniform for all wavelengths. Since cloud: CONTAIN water DROPLETS that are much larger than `lambda`. They scatter all wavelengths of light Hence clouds appear white.
42.

Which of following waves are used for sonography ?

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Microwave
Infrared WAVES
RADIO waves
Ultrasonic waves

ANSWER :D
43.

Calculate the resonent frequency of Q-factor (Quality factor) of a series L-C-R circuit containing a pure inductor of inductance 4H, capacitor of capacitance 27 muF and resister of resistance 8.4Omega.

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Solution :`v_(0)=(1)/(2pisqrt(LC))`
Finding `v_(0)=(1)/(2xx3.142sqrt(4xx27xx10^(-6)))=15.31Hz`
`Q=(X_(L))/(R)=(omega_(0)L)/(R)=(2piv_(0)L)/(R)`
ARRIVING at `Q=(2xx3.142xx15.31xx4)/(8.4)=48.8`
Detailed Answer:
Solution, Here, `L=4H,C=27xx10^(-6)F`, R=`8.4-2`
Resonant frequency
`f_(0)=(1)/(2pisqrt(LC))`
`f_(0)=(1)/(3.142xxsqrt(4xx27xx10^(-6)))`
`f_(0)=(1xx10^(3))/(2xx3.142xxsqrt(108))=(1xx10^(3))/(6.284xx10.4)`
`f_(0)=(1xx10^(3))/(65.353)=15.30Hz`
Q-factor `Q=(1)/(R)sqrt((L)/(C))=(1)/(8.4)xxsqrt((4)/(27xx10^(-6)))=(1)/(8.4)=(2xx10^(3))/(3sqrt(3))`
`Q=(2000)/(43.65)=45.81`
44.

Bismuth is diamagnetic but copper is paramagnetic.

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SOLUTION :FALSE - BISMUTH and COPPER both are DIAMAGNETIC.
45.

A square loop is placed near a long current carrying straight wire as shown in the figure . Match the following table

Answer»

<P>

SOLUTION :`A rarr (Q,s) , B rarr (p,r) , C rarr (p,r) , D rarr (q,s) `
46.

Magnetic field strength due to a short bar magnet on its axial line at a distance x is B. What is its value at the same distance on the equatorial line ?

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`B/2`
B
2 B
4 B

Answer :B
47.

The distance between the poles of horse shoe magnet is 10 cm and its pole strength is 10^4 A.m . Calculate the field at the point P midway between the poles.

Answer»


ANSWER :`0.8T`
48.

Which of the following semi-conducting devices is used as voltage regulator?

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Photo diode
Laser diode
Zener diode
SOLAR CELL

Solution :Zener diode is used as a VOLTAGE regulator.
49.

The magnetic flux linked with a large circular coil of radius R is 0.5 xx 10^(-3) Wb, when current of 0.5A flows through a small neighbouring coil of radius r. Calculate the coefficient of mutual inductance for the given pair of coils. If the current through the small coil suddenly falls to zero, what would be the effect in the larger coil.

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SOLUTION :`M = 1 m H`.
If the current through small coil suddenly FALLS to zero, [ as , `e_2= - M (di_1)/(dt)`] so initially large current is induced in larger coil, which soon BECOMES zero.
50.

A freely falling body traveles ____ of total distance in 5^(th) second

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0.08
0.12
0.25
0.36

Answer :D