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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

2001.

The differential equation of charging of a capacitor isas given below: 1/K_1 (dq)/(dt)+K_2q = K_3 The time constant and steady state charge are respectively

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`1/K_1` and `K_3`
`1/(K_1K_2)` and `K_3/K_2`
`K_2/K_1` and `K_2K_3`
`1/(K_1K_2)` and `K_3/K_1`

Solution :Consider the circuit
`epsilon-Q/C-IR=0`
`I=(dq)/(DT)`
Now, `epsilonC=(dq)/(dt). RC +q`

`rArr RC(dq)/(dt)+q = epsilonC` [where , `q_0=Cepsilon`] …(i)
Comparing EQ. (i) with `1/K_1 . (dq)/(dt) + K_2q = K_3`, we get
`tau=1/(K_1K_2) , q_0 = K_3/K_2`
Discussion
2002.

Magnetic field lines always form closed loops. comment.

Answer»

Solution :Magnetic POLES ALWAYS exist in PAIRS. Magnetic monopoles does not exist. That is why magnetic field LINES form closed loops.
Discussion
2003.

The direction of projectile at certain instant is inclined at angle a to the horizontal. After t sec, if it is inclined at an angle beta then the horizontal component of velocity is

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`g/(tan ALPHA - tan beta)`
`(g t)/(tan alpha - tan beta)`
`t/(g(tan alpha - tan beta))`
`(g t)/((tan alpha + tan beta))`

ANSWER :B
Discussion
2004.

Frequency of light changes when source move away from stationary observer, this can b explained by .......

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DOPPLER effect
interference
diffraction
NONE of these

Answer :A
Discussion
2005.

Tranmission of light in optical fibre is due to

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SCATTERING
diffaction
POLARISATION
MULTIPLE TOTAL INTERNAL reflactions

Answer :D
Discussion
2006.

The ground state energy of hydrogen atom is -13.6 eV. (i) What is the kinetic energy of an electron in the second excited state ? (ii) If the electron jumps to the ground state from the second excited state, calculate the wavelength of the spectral line emitted.

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Solution :(i) Groundstate ( n =1) energyof HYDROGENATOM`E_(1) = - 13.6 eV`
`therefore ` Energy of electron in the second excited state `(n=3) E _(3) = (-13.6)/((3)^(2)) eV = - 1.51 eV`
`therefore ` Kinetic energy of electron in the second excited state `k_(3) =+ 1.51 eV`
(ii) Energyof the photon emiited ` E_(1) + E_(2) = - 1.51 - (-13.6) = 12.09 eV = 12.09 xx eJ` .
`therefore ` Wavelenghtof thespectral line emitted .
`lambda = (hc)/(E) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(120.9 xx 1.6 xx 10^(-19)) m = 1.025 xx 10^(-19) m = 102.5 NM`
Discussion
2007.

Assume that charges were initially different. If the charges on them were equally distributed would the angle of divergenceincrease, decrease, or stay the same ?

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Solution :INCREASES because `((q_(1) + q_(2))/(2))^(2) GE q_(1)q_(2)`
Discussion
2008.

Aball of mass 'm' and radius 'r' is released in viscous liquid. The value of it's terminal velocity is proportional to

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1/r only
m/r
`(m/r)^(1/2)`
m only

Answer :D
Discussion
2009.

The figure shows three straight current carrying conductors having current I_(1), I_(2) and I_(3) respectively. Calculate line integral of magnetic induction field vec(B) along the closed path ABCDEFA

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ANSWER :`mu_(0)(I_(3)-I_(2))`
Discussion
2010.

A 44 mH inductor is connected to 220V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

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Solution :`X_(L) = OMEGA L `
`= 2pi f L `
`= ( 2) ( 3.14) ( 50) ( 44 XX 10^(-3))`
`= 13.816 Omega`
`I_(RMS) = ( V_(rms))/( |Z|)`
`= ( V_(rms))/( X_(L))( :. ` Here , `|Z | = X_(L))`
`= ( 220)/( 13.816)`
`= 15.92 A`
Discussion
2011.

A force of 0.01 N is exerted on a charge of 1.2 xx 10^(-5) C at a certain point. The electric field at that point is

Answer»

`5.3 xx 10^(4) NC^(-1)`
`8.3 xx 10^(4) NC^(-1)`
`5.3 xx 10^(4) NC^(1)`
`8.3 xx 10^(2) NC^(-1)`

SOLUTION :E = `(F)/(Q) = (0.01)/(1.2xx 10^(-1)) -8.3 xx 10^(2) NC^(1)`
Discussion
2012.

Answer the following questions:How the internal energy of an ideal gas be changed?

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SOLUTION :The internal energy of an ideal GAS can be changed,(increased or decreased) by heating or cooling the gas in a closed VESSEL i.e, by COMPRESSING or by adiabatic expansion
Discussion
2013.

It has been experimentally proved that in a huge portion of the earth's atmosphere, electric field intensity is acting vertically downwards. Field intensity is "60 V.m"^(-1) at an altitude of 200 m from the earth's surface and "120 V.m"^(-1) at an altitude of 100 m. Calculate the net electric charge enclosed by a cube of side 100 m. The cube is placed within the altitude range of 100 m to 200 m.

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ANSWER :`5.31xx10^(-6)C`
Discussion
2014.

Assertion : When lightpasses from one medium to anotherof different density the only quantity which is unchanged is its wavelength . Reason : The wavelength of light is not related to the refractive index of the medium .

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If both the assertion and reason are TRUE statement andreason is correct explanation of the assertion .
If both the assertion and reason are true statement but reason is not a correct explanation of the assertion .
If the assertion is true but the reason is a false statement.
If both assertion and reason are false statements.

Solution :In the ACT of refraction it is only the frequency that remains UNCHANGED . Wavelenght is realted to the REFRACTIVE index.Thus , both the assertionand the reason are wrong.
Discussion
2015.

A bar magnet of length l, pole strength 'p' and magnetic moment overset(to)(m) is split (l)/(2) · to two equal pieces each of length. The magnetic moment and pole strength of each piece is respectively ....... and ....... .

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`OVERSET(to)(m), (p)/(2)`
`(overset(to)(m) )/( 2), p`
`(overset(to)(m) )/(2) , (p)/(2)`
`overset(to)(m) , p`

Solution :When a bar magnet of LENGTH L is broken into two parts each of length `(l)/(2)`, each part (fragment) of the magnet BEHAVES as an independent magnet. So the pole strength p does not CHANGE but its magnetic dipole moment becomes halved.
Discussion
2016.

Two conducting spheres, each given a charge q are kept far apart as shown. The amount of charge that crosses the swithc S_(1) when it is closed , is ( the connection between the spheres is conducting

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`(Q)/(3)`
`(2q)/(3 )`
`(3q)/(4)`
zero

ANSWER :A
Discussion
2017.

Chromosome number is halved during

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FORMATION of FIRST POLAR body
Formation of SECOND polar body
Meiosis II
Division of SECONDARY oocyte

Answer :A
Discussion
2018.

A stone is projected horizontally with a velocity 9.8ms^(-1) from a tower of height 100m. Its velocity one second after projection is

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`9.8 ms^(-1)`
`4.9 ms^(-1)`
`9.8 SQRT(2) ms^(-1)`
`4.9 sqrt(2) ms^(-1)`

ANSWER :C
Discussion
2019.

A : At constant volume on increasing temperature the collision frequency increases. R : Collision frequency prop temperature of gas.

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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 :C
Discussion
2020.

ABC is right angled triangular plane of uniform thickness The sides are such that AB gt BC as sshown in figure I_(1),I_(2),I_(3) are moments of inertia about AB, BC and AC, respectively.Then which of the following relations is correct?

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`l_(1)=I_(2)=l_(3)`
`l_(2)gtl_(1)gtl_(3)`
`l_(3)ltl_(2)ltl_(1)`
`l_(3)gtl_(1)gtl_(2)`

Solution :The MOMENT of inertia of a body about an axis depends not onluy on the MASS of the body but also on the distribution of mass about the axis . Fo a given body mass is same so it will depend only on the distribution of mass about the axis .The mass is farthest from axis BC,So `I_(2)` is maximum mass is nearest to axis AC, so `I_(3)` is minimum.
Hence the correct sequence will be `I_(2)gtI_(1)gtI_(3)`
Discussion
2021.

If the refractive index of a material of an equilateral prism is sqrt3, then angle of minimum deviation will be ..... .

Answer»

`30^@`
`40^@`
`50^@`
`60^@`

SOLUTION :A=60,A+`delta_m=x`,n=`SQRT3`
For rquilateral prism n=`(SIN(A+delta_m)/(2))/(sin(A/2))`
`therefore sqrt3=(sin(x//2))/(sin(60^@/2))`
`implies sin30sqrt3=sin(x//2)`
`implies sin(x//2)=(sqrt3)/(2)`
`thereforex/2=sin^(-1)((sqrt3)/(2))=60^@`
`thereforex=A+delta_m=120impliesdelta_m=120-60=60^@`
Discussion
2022.

If the kinetic energy of a free particle is much greater than its rest energy then its kinetic energy is proportional to

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The MAGNITUDE of its MOMENTUM
The SQUARE of the magnitude of its momentum
The square root of the magnitude of its momentum
The RECIPROCAL of the magnitude of its momentum

Answer :A
Discussion
2023.

In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 xx 10^5 V/m exists in it. What should be the minimum kinetic energy of a conduction electron which can diffuse from the n - side to the p - side?

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0.1 EV
0.3 eV
0.2 eV
0.4 eV

Answer :C
Discussion
2024.

An effective resistance R is obtained by connecting 'n' identical resistors each resistance 'r' in series.When connected in parallel,the effective resistance of the combination is

Answer»

`n^2/R`
`r^2/R`
`Rr^2`
`Rn^2`

ANSWER :B
Discussion
2025.

In an AC circuit containing only capacitance, the current ……………… .

Answer»

LEADS the VOLTAGE by `180^@`
remains in phase with the voltage
leads the voltage by `90^@`
LAGS the voltage by `90^@`

SOLUTION :leads the voltage by `90^@`
Discussion
2026.

Total Ke of a rolling body is given by _____ + _____.

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SOLUTION :[`(1)/2 MV^2 + (1)/2 Iw^2`]
Discussion
2027.

Powers received at focal spot F from A and B are -

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`10^(-6)` WATT and `4 XX 10^(-6)` watt
`10^(-7)` watt and `4 xx 10^(-7)` watt
`10^(-5)` watt and `4 xx 10^(-5)` watt
`10^(-4)` watt and `4 xx 10^(-4)` watt

Answer :A
Discussion
2028.

Figure 29-64 shows a cross section of a long cylindrical conductor of radius a=4.00 cm containing a long cylindrical hole of radius b=1.50 cm. The central axes of the cylinder and hole are parallel and are distance d=2.00 cm apart, current i=5.25 A is uniformly distributed over the tinted area. (a) What is the magnitude of the magnetic field at the centre of the hole ? (b) Discuss the two special cases b=0 and d=0.

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Answer :(a) `15.3muT;(b)if b=0,THENB=(mu_(0)id)/(2pia^(2))`
Now for d=0, the formula GIVES B=0
Discussion
2029.

Which one of the following involves digital quantities?

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TEMPERATURE of a room
Current FLOWING in a conductor
Sand GRAINS on a sea-beach
Altitude of an AIR craft

Answer :B
Discussion
2030.

Communication through free space using radio waves takes place over frequency range of

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Few TENS of MHZ to a few GHz
Few tens of KHZ to a few HUNDREDS of KHz
Few tens of GHz to a few hundreds of GHz
Few hundreds of GHz to few THOUSANDS of GHz

Answer :A
Discussion
2031.

Find velocity of particle with 1 mg mass and having same wavelength that of electron moving with speed of 3xx10^(6) m/s. Mass of electron 9.1xx10^(-31) kg

Answer»

`2.7xx10^(-18) ms^(-1)`
`9xx10^(-2)ms^(-1)`
`3xx10^(-31) ms^(-1)`
`2.7xx10^(-21)ms^(-1)`

Solution :WAVELENGTH of particle ,
`(lambda_(1))=(h)/(mv)=(h)/((1xx10^(-3))xxv)`
Where V=velocity of particle
wavelength of electron ,
`(lambda_(2))=(h)/((9.1xx10^(-31))xx(3xx10^(6)))`but `lambda_(1)=lambda_(2)`
`therefore (h)/((1xx10^(-3))xxv)=(h)/((9.1xx10^(-31))xx(3xx10^(6)))` `therefore v=(9.1xx10^(-31)xx3xx10^(6))/(10^(-3))`
`=2.73xx10^(-21)ms^(-1)`
Discussion
2032.

A maximum current of 0.5 mA can be passed through a galvometer of resitanced 20 Omega. Calculate the resitance to be connected in series to convert it into a voltmeter of range0 - 5V .

Answer»

Solution :`R = G(n-1) , " where " n = (V_2)/(V_1)`
`V_2 = 5 V , V_1 = i G = 0.5 xx 10^(-3) xx 20 = 10^(-2) V`
`THEREFORE n = 500`
and R = 20 (500 -1)
`= 9980 Omega`
Discussion
2033.

Two coherent sources of intensity ratio a interfere. The value of (I_("max")-I_(min))/(I_("max")+I_(min)) is ....

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`2sqrt((ALPHA)/(1+alpha))`
`(2 SQRT(alpha))/(1+alpha)`
`(1+alpha)/(2sqrt(alpha))`
`(1-alpha)/(1+alpha)`

SOLUTION :Do the CALCULATION yourself.
Discussion
2034.

A ray of light enters from a denser medium into rarer medim. The speed of light in the rarer medium is twice that in denser medium. What is the critical angle for total intenal reflection to take place ?

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`60^@`
`45^@`
`30^@`
none of the above

Answer :C
Discussion
2035.

Two physical quantities, of which one is a vector and the other is a scalar, having same dimension are :

Answer»

MOMENT and momentum
power and pressure
impulse and momentum
TORQUE and WORK.

Solution :Torque and work have the same dimensions. Torque is an axial vector while work is a scalar. `(d)` is the correct CHOICE.
Discussion
2036.

A student is asked to measure the wavelength of monochromatic light. He sets up the apparatus sketched below S_(1),S_(2),S_(3) are narrow parallel slits, L is a sodium lamp and M is a micrometer eye-piece. The student fails to observe interference fringes.You would advise him to:

Answer»

Increase the WIDTH of `S_(1)`
Decrease the DISTANCE between `S_(2)` and `S_(3)`
Replace L with a WHITE light source
Replace M with a telescope

Answer :B
Discussion
2037.

A ball is thrown vertically up (taken as + z-axis) from the ground. The correct momentum-height h diagram is:

Answer»




SOLUTION :`V^(2)=u^(2)-2gh rArr m^(2)v^(2)=m^(2)u^(2)-2mgh`
`rArr p^(2)=A-Bh` ( where `A=m^(2)u^(2), b=2mg`)
Cleary it is parabolic FUNCTIONS.
Starting from GROUND momentum first decreases and then becomes zero athighest position. Therefore it increases iin -ve DIRECTION. Hence `3^(rd)` option is correct.
Discussion
2038.

In case of a concave mirror what is the shape of the u ~v graph?

Answer»


ANSWER :RECTANGULAR HYPERBOLA
Discussion
2039.

The frequency of na electromagnetic wave in free space is 2 MHz. When it passes through a region of relative permittivity epsilon_(r )=4.0, then its wavelength …… and frequency ……

Answer»

becomes double, becomes half.
becomes double, remains constant.
becomes half, becomes double.
becomes half, remains constant.

SOLUTION :`("velocity of light in vacuum c")/("velocity of light in medium V")=sqrt((mu_(0)EPSILON)/(mu_(0)epsilon_(0)))` (accepting `mu_(0)` is not CHANGING)
`therefore (c )/(v)=sqrt((epsilon)/(epsilon_(0)))=sqrt(epsilon_(r ))=sqrt(4)=2 ""`…(1)
If medium is CHANGED then also frequency of light f remains constant.
`therefore c = f lambda, "" v = f lambda.`
`therefore (c )/(v)=(lambda)/(lambda.)=2 ""`(From EQUATION (1))
`therefore lambda.=(lambda)/(2)`
As frequency f remains constant and wavelength becomes half.
Discussion
2040.

{:("COLUMN - I","COLUMN - II"),("A) Infinite plane sheet of change","p)"0),("B) Infinite plane sheet of uniform thickness","q)"(p)/(2epsilon_(0))),("C)Non - conducting charged solid sphere of radius R at its surface","r)"(Rp)/(3epsilon_(0))),("D) Non - conducting charge solid sphere of radius R at its centre","s)"(p)/(epsilon_(0))):}

Answer»


ANSWER :A-q; B-q; C-R; D-P
Discussion
2041.

Pushing force making an angle o to the horizontal is applied on the block of weight W placed on the horizontal table. If o is the angle of friction, the magnitude of the force required to move the body is equal to :

Answer»

`(Wcos phi)/(COS(theta-phi))`
`(Wsin phi)/(SIN(theta-phi))`
`W sin phi//cos(theta-phi)`
`W tan theta//sin(theta-phi)`

Solution :`F_(r)`= force of friction. Then `tan phi =(F_(r))/(R)`

or `F_(r) = R tan phi` where phi is angle of friction. The BLOCK will be just at the point of motion when
`F cos theta =F_(r) = R tanphi`and `F sin theta+ R = W`
Now `R=W-F sin theta`
Or `F cos theta = (W-F sin theta) tanphi`
`F cos theta+F sin theta (sin theta)/(cos phi)=(W sin phi)/(cos phi)`
or` F cos theta cos phi +F sin theta sin phi = W sin phi`
`F(cos theta cos phi + sin theta sin phi) = W sin phi`
or`F=(Wsin theta phi)/(cos (theta-phi))`
Discussion
2042.

Magnitude of an electromagnetic wave is E=4.24 sin [(7.54xx10^(6))(t-(x)/(3xx10^(8)))](mV)/(m).Then energy density associated with magnetic field of this wave is ……. (a) 796xx10^(-19) J (b) 796xx10^(-19) J//m^(3) (c) 7.96xx10^(-19) J//m^(3) (d)796xx10^(-19) W

Answer»

`796xx10^(-19)J`
`796xx10^(-19)J//m^(3)`
`7.96xx10^(-19)J//m^(3)`
`796xx10^(-19)W`

Solution :Here, `E_(m)=4.24(MV)/(m)`
`therefore E_(rms)=(E_(m))/(sqrt(2))=(4.24)/(1.414)=2.998 (mV)/(m)`
`therefore E_(rms)~~ 3xx10^(-3)(V)/(m)`
Now energy density,
`RHO = epsilon_(0)E_(rms)^(2)`
`=8.85xx10^(-12)xx(3xx10^(-3))^(2)`
`= 7.9.65xx10^(-18)~~ 796xx10^(-19)J//m^(3)`
Discussion
2043.

A ball moving with a velocity ef 10 impinges on a vertical wall at an angle of 45° with the normal to the wall. After the collision, the ball moves on the other side of the normal at an angle of 45° with the normal. The coefficient of restitution is 0.5. The velocity of the ball after the collision will be

Answer»

2 `ms^(-1)`
3`ms^(-1)`
5 `ms^(-1)`
50 `ms^(-1)`

Solution :Here the COEFFICIENT of RESTITUTION
`e=("Normal component of R.V after collison")/("Normal component of R.V before collison")`
`e=0.5=(v COS 45^(@))/(10 cos 45^(@))=(v)/(10)`
or `v=10xx0.5=5 ms^(-1)`
Discussion
2044.

Differentiate between Fresnedl and Fraunhofer diffraction.

Answer»

SOLUTION :
Discussion
2045.

What is photoelectric effect?

Answer»

Solution :When clean metal surface is IRRADIATED with RADIATION of sifficiently high frequency,electrons are emitted from metal surface.This PHENOMENA is CALLED photoelectric effect.
Discussion
2046.

A person walks along a straight road from his house to a market 2.5 kms away with a speed reaches his house with a speed of 7.5 kms/hr.The average speed of the person during the time interval 0 to 50 minutes in (in m/sec)

Answer»

`4(2)/(3)`
`(5)/(3)`
`(5)/(6)`
`(1)/(3)`

Answer :B
Discussion
2047.

Assertion : Two concentric spherical shells A and B are charged bycharges q_(A) and q_(B). If q_(A) is negative, then V_(A)-V_(B) is also negative. Reason : V_(A)-V_(B) =(q_(A)/(4pi epsi_(0)))(1/R_(A)-1/r_(B)) therefore, if q_(A) is negative, then V_(A)-V_(B) is also negative.

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 correct explanation of Assertion.
If Assertion is true but Reason is false.
If Assertion is false but Reason is true.

Answer :A
Discussion
2048.

A depolarizer essentially acts an agent which can prevent a) the covering of copper plates by a layer of hydrogen b) the back emf

Answer»

Only a is TRUE 
Only B is true 
Both a and b are true 
NEITHER a nor b 

ANSWER :C
Discussion
2049.

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's "shadow region." Previously, television signals had a wavelength of about 50 cm, but digital television signals that are transmitted from towers have a wavelength of about 10 mm. (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of 6.0 m width between two adjacent buildings. What is the angular spread of the central diffraction maximum out to the first minima) for wavelengths of (b) 50 cm and (c) 10 mm?

Answer»

Solution :(a) DIFFRACTION EFFECTS in GENERAL WOULD decrease, (b) 9.6, (c) 0.010 m.
Discussion
2050.

A 220 V - 1000 W electric heater is connected in parallel with a 220 V - 60 W electric heater is connected in parallel with a 220 V - 60 W electric lamp and , the co

Answer»


ANSWER :no CHANGE
Discussion