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.

Theresistance of 1A ammeter is 0.018ohm. To convert it into 10A ammeter, the shunt resistance required will be:

Answer»

0.18
0.0018
0.002
0.12

Answer :C
Discussion
2.

The resistance of a carbon resistor of colour code Red-Red Green Silver is (in k Omega)

Answer»

`22 00pm 5%`
`2200 PM 10%`
`220 pm 10%`
`220 pm 5%`

Answer :B
Discussion
3.

A solenoid has 50 turns and radius 2 cm, diameter of wire is 2x 10^(-4) m, current through it when it is joined with battery of 10V is ... Resistivity of material.2*10^-6 Ωm.

Answer»


Solution :1.14
I = `(V)/(R)`
`THEREFORE I = (VA)/(rho l)`
`therefore I =(V xx pi r^(2))/(rho xx 2 pi RN) "" [ because A = pi r^(2)] `
`therefore I = (10 xx (10^(-4) )^(2))/(4.4 xx 10^(-8) xx 2 xx 2 xx 10^(-2) xx 50) = (1)/(0.88)`
`thereforeI = 1.136 `A
`therefore I = 1.14A `
Discussion
4.

The magnitude of electric intensity at a distance .x. from a charge .q. is E. An identical charge is placed at a distance .2x. from it. Then the magnitude of the force it experience is

Answer»

EQ
2Eq
`(Eq)/(2)`
`(Eq)/(4)`

ANSWER :D
Discussion
5.

What is the speed of light in glass of refractive index 1.5 if the velocity of light in the free space is 3 x 10^8 ms^-1?

Answer»

`2xx10^-8`s
`4.5xx10^8 ms^-1`
`3xx10^8 ms^-1`
`3xx10^10 ms^-1`

ANSWER :A
Discussion
6.

What is the unit of potential gradient ?

Answer»

VM
V/m
`Vm^(2)`
`V//m^(2)`

ANSWER :B
Discussion
7.

Find the output of the ideal operational amplifier shown in the figure for each of the following input signals V_(I n)= 0.8sin( omega t + 70 ^@)V V_(I n ) = 0.8sin( omegat+ 75 ^@ ) V

Answer»

SOLUTION :Thegainofthe AMPLIFIER`A_v=-(R_f)/(R_(I n)) = (-39 xx 10^3)/( 15 xx 10^3)`
`=-2.6`
`V_0=-2.6 ( 4- SINOMEGAT. V) =- 10.4+ 2.6sin OMEGA t V`
Discussion
8.

In Figure, sources A and B emit long-range radio waves of wavelength 400 m, with the phase of the emission from A ahead of that from source B by 90.0^(@). The distance r_(A) from A to detector D is greater than the corresponding distance r_(B) by 150 m. What is the phase difference of the waves at D?

Answer»

SOLUTION :0.785 rad
Discussion
9.

Three capacitors are connected as shown in the figure . The potential at point O is equal to

Answer»

1.4 VOLT
4.7 volt
ZERO
1.8 volt

Answer :D
Discussion
10.

The spacebetween the electrodes of a parallel-plate capacitor is filled witha uniformpoorlyconducting mediumof conductivitysigma and permittivity epsilon. The capacitorplates shaped as rounddiscs are separtedby a distance d. Neglectingthe edgeeffects, find the magnetic fieldstrength betweenthe plates at a distance r fromtheiraxis ifan ac voltage V = V_(m) cos omega tis appliedto the capacitor.

Answer»

Solution :The electricfield between the plates can be written as,
`E = Re (V_(m))/(d) e^(I omega t)`, insteadof `(V_(m))/(d) cos omega t`.
The gives RISE to a conductioncurrent,
`j_(e) = sigmaE = Re (sigma)/(d) V_(m) e^(I omega t)`
and a displacementcurrent,
`j_(d) = (DEL D)/(del t) = Re epsilon_(0) epsilon i omega (V_(m))/(d) e^(i omega t)`
The totalcurrent is,
`j_(T) = (V_(m))/(d) sqrt(sigma^(2) + (epsilon_(0) epsilon omega)^(2)) cos (omega t + alpha)`
where, `tan alpha = (sigma)/(epsilon_(0) epsilon omega)` on takingthe real PART ofthe resultant.
The correspondingmagneticfieldis obtainedby usingcircularion THEOREM,
`H.2pi r= pi r^(2) j_(r)`
or, `H = H_(m) cos(omega t + alpha)`, where, `H_(m) = (r V_(m))/(2d) sqrt(sigma^(2) + (epsilon_(0) epsilon omega)^(2))`
Discussion
11.

The wavelength lambda_(e )of an electron and lambda_(p) of a photon of same energy E are related by

Answer»

`lambda_p prop lambda_e`
`lambda_p prop sqrtlambda_e`
`lambda_p prop 1/sqrtlambda_e`
`lambda_p prop lambda_e^2`

Solution :DE broglie wavelength of ELECTRON , `lambda_e=h/sqrt(2mE)`
`therefore`i.e. `lambda_e prop 1/sqrtE RARR lambda_e^2 prop 1/E` …(1)
de-Broglie wavelength of PROTON
`lambda_p=(hc)/E`
`lambda_p prop 1/E` …(2)
From (1) and (2)
`lambda_e^2 prop lambda_p` i.e., `lambda_p prop lambda_e^2`
Discussion
12.

Where have we come from to build our nation?

Answer»

From TOWNS and cities
From other countries
From HEAVEN and hell
None of the above

Answer :A
Discussion
13.

Liquied oxygen remains suspended between twopolefaces of a magnet because its is

Answer»

diamagnetic
PARAMAGNETIC
ferromagnetic
ANTIFERROMAGNETIC

ANSWER :B
Discussion
14.

In a biprism experiment , the distance of the 15 th bright band from the centre of the interference pattern is 6 mm . Calculate the distance of the 25th bright band and 31st dark band .

Answer»


ANSWER :10 MM , 12.2 mm
Discussion
15.

The equation of mation for a body executing S.H.M. is given by y=1.5sin(10 pi t+5). The frequency is given by :

Answer»

5 HZ
10 Hz
2.5 Hz
`5PI` Hz

Solution :`y=1.5sin(10PI t+5)`
`y=r SIN(omegat+phi)`
`omega=10pi`
`2piv=10pi`
`v=5` Hz.
Socorrectchoice is (a).
Discussion
16.

The number of photons of wavelength 13.2A^(@)in 6J of energy is (h=6.6xx10^(-34)J.s)

Answer»

`2xx10^(12)`
`4xx10^(16)`
`6xx10^(20)`
`4xx10^(24)`

SOLUTION :As `n=(E lambda)/(hc)`
`:.n=(6xx13.2xx10^(-10))/(6.6xx10^(-34)xx3xx10^(8))`
or `n=4xx10^(16)`
Discussion
17.

Momentum of photon with frequencyf will be …..(where c is velocity of light)

Answer»

`(hf)/(c^(2))`
HFC
`(F)/(c)`
`(hf)/(c )`

Solution :Momentum of photon `p=(H)/(LAMBDA)=(hf)/(c )[because (1)/(lambda)=(f)/(c )]`
Discussion
18.

In a traingle the proper length of each side equals a. Find the perimeter of this triangle in the reference frame moving relative to it with a constant velocity V along one of its (a) bisectors, (b) sides. Investigate the results obtained at V lt lt c and Vrarr c, where c is the velocity of light.

Answer»

SOLUTION :(a) In the FRAME in which the traingle is at rest the space COORDINATES of the vertices are `(000)`, `(asqrt3/2, +a/2, 0)(asqrt3/2, -a/2, 0)`, all MEASURED at the same time t. In the moving frame the CORRESPONDING coordinates at time `t^'` are
`A: (vt^', 0, 0), B: (a/2sqrt3sqrt(1-beta^2)+vt^', a/2, 0)` and `C: (a/2sqrt3sqrt(1-beta^2)+vt^', -a/2,0)`
The perimeter P is then
`P=a+2a(3/4(1-beta^2)+1/4)^(1//2)=a(1+sqrt(4-3beta^2))`
(b) The coordinates in the first frame are shown at time t. The coordinates in the moving frame are,

`A: (vt^', 0, 0), B: (a/2sqrt(1-beta^2)+vt^', asqrt3/2, 0), C: (asqrt(1-beta^2)+vt^', 0, 0)`
The perimeter P is then
`P=asqrt(1-beta^2)+a/2[1-beta^2+3]^(1//2)xx2=a(sqrt(1-beta^2)+sqrt(4-beta^2))` here `beta=V/c`
Discussion
19.

The component of a vector vec(r ) along X-axis will have maximum value only if:

Answer»

`vec(R )` acts ALONG positive Y-axis.
`vec(r )` acts along positive X-axis.
`vec(r )` acts along NEGATIVE Y-axis.
`vec(r )` acts at an angle of 45° with X-axis.

Solution :Here component of `vec(r)` along X-axis will be `vec(r)` COS `theta` which is maximum when cos `theta` is maximum i.e. `cos theta = 0°` or `vec(r)` is acting along X-axis is the +ve direction.
Discussion
20.

A uniform semicircular ring of mass m and radius r is hinged at end A so that it can rotate freely about end A in the vertical plane as shown in the figure. Calculate angle made by line AB with vertical in equilibrium position and time period of small oscillation of ring.

Answer»

SOLUTION :`tan^(-1) ((2)/(pi)), 2PI SQRT((2pi r)/(g sqrt(pi^(2) + 4)))`
Discussion
21.

The electric field at a distance R due to charge q is E. If the same charge is placed on the copper sphere of radius R, the electric field at a distance 2R from the centre of spherical conductor will be :

Answer»

0.25 E
0.5 E
E
2 E

Answer :A
Discussion
22.

The output of OR gate is 1

Answer»

if both INPUTS are zero
if either or both inputs are 1
only if both inputs are 1
if either INPUT is zero

Answer :b
Discussion
23.

A moving electron approaches another electron. What would be the change in the potential energy of this system ?

Answer»

Remains constant
Increases
Decreases
May increase or decrease

SOLUTION :Potential energy for a system of two electron
`U=(k(-e)(-e))/(R) `
`= (ke^(2))/(r)`
where `ke^(2)` is constant
`:. U PROP (1)/(r)`
`:.` As r decreses U increases.
Discussion
24.

Answer the following questions regarding earth's magnetism In which direction would a compass free to move in the vertical place point to , if located right on the geomagnetic north or south pole ?

Answer»

Solution :The geomagnetic NORTH pole is ACTUALLY a SOUTH pole .So the north pole of a compass needle WOULD point vertically downwards . Similarly , the south pole of the needle would point vertically downwards at the geomagnetic south pole .
Discussion
25.

A square coil of side 10 cm has 20 turns and carries a current of 12 A. The coil is suspended vertically. The normal to the plane of the coil makes an angle of 30^@ with the direction of a uniform horizontal magnetic field. If the torque experienced by the coil equals 0.96 N - m, find the magnitude of the magnetic field.

Answer»

Solution :Here side of square `l = 10 cm = 0.1 m`, HENCE area of coil `A = (0.1)^(2) = 0.01 m^(2)`, CURRENT `I = 12 A`, torque `tau = 0.96 N-m, theta = 30^@` and number of turns in coil N = 20.
From relation `tau = NIAB sin theta,` we have
`B = (tau)/(NIA sin theta) = (0.96)/(20 xx 12 xx 0.01 xx sin 30^(@)) = 0.8 T`.
Discussion
26.

मान लीजिए कि f(x)=3x द्वारा परिभाषित फलन F:Rrarr Rहै। सही उत्तर चुनिए

Answer»

F एकैकी आच्छादक है।
f बहुएक आच्छादक है
f एकैकी है किन्तु आच्छादक नहीं है
f न तो एकैकी है और न आच्छादक है।

Answer :A
Discussion
27.

The viscosity eta of a gas depends on the long - range attractive part of the intermolecular force, which varies with molecular separation r according to F=mur^(-n) where n is a number and mu is a constant. If eta is a function of the mass m of the molecules, their mean speed v, and the constant mu then which of the following is correct -

Answer»

`ETA PROP m^(N+1) v^(n+3)MU^(n-2)`
`eta prop m^((n+1)/(n-1))v^((n+2)/(n-1))mu^((-2)/(n-1))`
`eta prop m^(n)v^(-n)mu^(-2)`
`eta prop mv mu^(-n)`

Answer :B
Discussion
28.

An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.

Answer»


ANSWER :1 D, 50 D
Discussion
29.

(A): Exothermic chemical reactions under lie conventional energy sources. Hear the energies involved are in the range of ev. (R): Atomic energy level spacings are of the order ev.

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
A. is false and .R. is false

ANSWER :A
Discussion
30.

Which of the following is the largest when the height attained by the projectile is the greatest:

Answer»

Range
TIME of flight
Angle of PROJECTION with vertical.
None of these

Solution :When height is max. then time of flight is also max. because both INVOLVE sin `theta`.
Discussion
31.

The teacher shows a thin lens and a thick lens. a. Which of these lens forms more enlarged image ? b. Which lens has greater focal length?

Answer»

SOLUTION :a. THIN LENSB. Thin LENS
Discussion
32.

A simple pendulum with charged bob is oscillating with time period T and let theta be the angular displacement. If the uniform magnetic fieldis switched ON in a direction perpendicular to the plane of oscillation then

Answer»

TIME PERIOD will DECREASE but `theta` will REMAIN constant
time period remain constant but `theta` decrease
both t and `theta` will remain the same
both T and `theta` will decrease

Answer :A::B::D
Discussion
33.

A parallel plate capacitor is to be designed using a dielectric of dielectric constant 5, so as to have a dielectric strength of 10^(9) Vm^(-1). If the voltage rating of the capacitor is 12 kV, the minimum area of each plate required to have a capacitance of 80 pF is .........

Answer»

`10.5XX10^(-6) m^(2)`
`21.7xx10^(-6) m^(2)`
`25.0xx10^(-5) m^(2)`
`12.5xx10^(-5) m^(2)`

Solution :V =Ed
`:. D=(V)/(E)`
`C= (K in_(0)A)/(d)`
`:. A = (Cd)/(K in_(0))`
`= (CV)/(KE in_(0))`
`=(80xx10^(-12)xx12xx10^(3))/(5xx10^(9)xx8.85xx10^(-12))`
`=21.69xx10^(-6)`
`= 21.7xx10^(-6) m^(2)`
Second Method :
Q = CV
`sigma A = CV`
`:. sigma = (CV)/(A)`
`E = (E_(0))/(K) ` but `E_(0)= (sigma)/(in_(0))`
`:. E= (sigma)/(K in_(0))`
From (1)
`E = (CV)/(AK in_(0))`
`:. A =(CV)/(EK in_(0))`
`=(80xx10^(-12)xx12xx10^(3))/(5xx10^(9)xx8.85xx10^(-12))`
`:. A = 21.69xx10^(-6)`
`:. A = 21.7xx10^(-6) m^(2)`
Discussion
34.

An air bubble in a jar of water shines brightly due to phenomenon of refraction.

Answer»

SOLUTION :An air bubble in a jar of water SHINES brightly DUE to phenomenon of TOTAL internal reflection.
Discussion
35.

Identify the following electromagnetic radiations as per the wavelengths given below. Write one application of each. (a) 10^(-3) nm (b) 10^(-3)m (c ) 1nm

Answer»

Solution :(a) `GAMMA`-RAYS - used in medicine to destroy cancer cells.
(b) Microwaves - used for radar system for AIRCRAFT navigation.
(c) X-rays – used for studying crystal structure. (b) 103 m (c) 1 nm
Discussion
36.

A point mass initially at rest moves along x-axis. Its acceleration varies with time as a = 6t+ 5 ms. If it starts from origin, the distance covered by it in 1 s is

Answer»

3 m
5.5 m
3.5 m
4 m

Solution :Here ACCELERATION =6t+5
`(dv)/(dt)=6t+5`
or dv=(6t+5) dt, Intergrating
`v=(6t^(2))/(2)+5t=3T^(2)+5t`
`DS(3t^(2)+5t)dt`
`s=(3t^(3))/(3)+(5t^(2))/(2)`
`(s)_(t)=(3xx1xx1)/(3)+(5)/(2)1xx1=3.5`
Discussion
37.

A positive charge Q is uniformly distribut along a circular ring of radius R. A small tc charge q is placed at the centre of the ring per figure, Then

Answer»

If q gt 0 and is displaced away from th centre in the plane of the ring, it will b pushed back towards the centre.
If q lt 0 and is displaced away from the centr in the plane of the ring, it will never return ti the centre and will continue moving till I hits the ring.
If q lt 0, it will perform SHM for SMAI displacement along the axis.
q at the centre of the ring is in an unstabh equilibrium within the plane of the ring fo q gt 0

Solution :Positive charge Q is distributed uniformly ove the outer surface of sphere, hence E inside it i, zero.
Hence, no effect on Q by q.
TWO cases are possible due to attraction ano repulsion between Q and q. Case 1 : If q is positive then repulsion will be there between Q and q.
This will push q towards centre. Case 2 : If q is negative, then attraction will be there between Q and q. This will pull q towards surface and hence it will not TURN back to centre.
Discussion
38.

A particle of mass m is connected with a string of length 2 meter. Other end of the string is fixed with a point O at a height 1 m above the ground. The particle is thrown from some point in such a way that it strikes the ground (perfectly inelastic) with velocity v_(0) at an angle 37^(@) with vertical just below O.

Answer»

<P>P-2, Q-3, R-1, S-4
P-3, Q-2, R-4, S-1
P-4, Q-1, R-2, S-3
P-2, Q-3, R-4, S-1

Solution :(A) as the COLLISION is perfectly inelastic, velocity after collision is `v_(0)sin 37^(@)=(3)/(5)v_(0)` towards right till string becomes taut.
`R=(v^(2))/(a_(_|_))=(v_(0)^(2))/(g sin 37^(@))=(5v_(0)^(2))/(3g)`

(B)from figure, `cos theta=(1)/(2)rArr theta = 60^(@)`
after string becomes taut particle motion will seize along the string DUE to impulse. Just after impulse velocity of particle is `(3)/(5)v_(0)cos theta = (3)/(10)v_(0)` perpendicular to the string from conservation of mechanical energy , `(1)/(2)m((3)/(10)v_(0))^(2)=(1)/(2)m(sqrt(2G))^(2)+2mg(1+cos 60)` simplifying, `v_(0)=(20sqrt(2g))/(3)m//s`
(C ) `W_(G)=-2mg(1+cos 60^(@))=-3 mg`
`therefore - W_(G)=3 mg`
(D) At highest point velocity will be perpendicular to gravitational force
`therefore P = vec(F)vec(V)=0`
Discussion
39.

Two balls with charges 5muC and 10muCare at a distance of 1 m from each other. In order to reduce the distance between them to 0.5 m what amount of work should be done?

Answer»


ANSWER :0.45J
Discussion
40.

What did Bhagat Singh and others do while going towards gallows?

Answer»

SANG songs
Raised slogans
Killed the DEPUTY Magistrate
Both (a) and (B)

ANSWER :D
Discussion
41.

If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is

Answer»

Repelled by the north POLE and ATTRACTED by the south pole
Attracted by the north pole and repelled by the south pole
Attracted by both the POLES
Repelled by both the poles

Answer :D
Discussion
42.

A piece of copper and the other of germanium are cooled from the room temperature to 80 K, then which of the following would be a correct statement ?

Answer»

RESISTANCE of each INCREASES.
Resistance of each dec.reases.
Resistance of copper increases while that of germanium decreases.
Resistance of copper decreases while that of germanium increases.

Solution :As copper is a GOOD conductor of electricity and germanium is a semiconductor hence on cooling resistance of copper decreases but that of germanium increases.
Discussion
43.

When alphais the angle between plane of the coil and magnetic field vecB, flux linked with the coil is given byphi =____

Answer»

AB
`|VECAXXVECB|`
`AB cos alpha`
`vecA xx VECB`

SOLUTION :Here angle `THETA` is between `vecA` and `vecB=90^@-alpha`.
`therefore phi=vecA.vecB`
`=AB cos(90^@-alpha)`
`=AB sin alpha`
`therefore phi=|vecAxxvecB|`
Discussion
44.

Which of the following graph represents the variation of electric field E due to a thin charged spherical shell of radius R as a function of the distance r from the centre of the shell?

Answer»



SOLUTION :Charge ENCLOSED ` q= in_0( "net flux")=in_0 (phi _2-phi_1) `
Discussion
45.

Intensity of secondary waves is directly proportional to ......

Answer»

`SIN^(2) THETA`
`sin^(2)((theta)/(2))`
`COS^(2) theta`
`cos^(2)((theta)/(2))`

ANSWER :D
Discussion
46.

Find magnitude and direction of magnetic field at point P in the following cases. (a) (b) P is the centre of square. (c) P is the centre of equilateral triangle. (d) P is the centre of regular hexagon. (e) P is the centre of rectangular loop. (f) (g) A long wire carrying a current i is bent to from a plane angle theta. Find magnetic field at a point on the bisector of this angle is situated at a distance d from vertex. (h) A long, straight wire carriers a current i. Let B_(1) be the magnetic field at a point P at a distance d from the wire. consider a section of length l of this wire such that the point P lies on a perpendicular bisector of the sector. Let B_(2) be the magnetic field at this point due to this section only. find the value of d//l so that B_(2) differes from B_(1) by 1%.

Answer»

Solution :(a)
`B_(P)=(mu_(0)i)/(4pid')[cos theta_(1)+cos theta_(2)]`
`=(mu_(0)i)/(4pi(3d))[cos theta+cos theta]`
`=(mu_(0)i)/(6pid) cos theta`
`cos theta=4/5`
`B_(P)=(mu_(0)i)/(6pid) .4/5=(2mu_(0)i)/(15pid) ox`.
(b)
Magnetic field at `P` DUE to one side of SQUARE, say `AB`
`B_(1)=(mu_(0)i)/(4pid')(cos theta+cos theta)`
`d'=d, theta=45^(@)`
`=(mu_(0)i)/(4pid).2.(1)/(sqrt(2))=(mu_(0)i)/(2sqrt(2)pid)`
Magnetic field due to square loop
`B_(p) = 4B_(1) = (sqrt2mu_(0)i)/(pi d)`
Since current anticlockwise, magnetic field at `P` will be outside the plane of paper.
`B_(P)=(sqrt(2)mu_(0)i)/(pid), o.`
(c)
`tan 30^(@) =(d')/(d//2) implies d'=d/(2sqrt(3))`
`theta_(1)=theta_(2)=30^(@)`
`B_(1)=(mu_(0)i)/(4pid')[cos30^(@)+cos30^(@)]`
`=(mu_(0)i)/(4pi.d/(2sqrt(3))).2 (sqrt(3))/2=(3mu_(0)i)/(2pid)`
Magnetic field at `P` due to triangular loop
`B_(P)=3B_(1)=(9mu_(0)i)/(2pid)`
Since current is clockwise, magnetic field at `P` inside the plane of paper.
`B_(p)=(9mu_(0)i)/(2pid)`
Since current is clockwise, magnetic field at `P` inside the plane of paper.
`B_(p)=(9mu_(0)i)/(2pid), ox`
(d)
`B_(1)=(mu_(0)i)/(4pid')(cos 60^(@)+cos 60^(@))`
`=(mu_(0)i)/(4pi.d/2tan 60^(@)).2 cos 60^(@)`
`=(mu_(0)i)/(2pidsqrt(3))`
`B_(P)=6B_(1)=(sqrt(3)mu_(0)i)/(pid), ox`
(e)
From `TRIANGLEADC,AC=10d`
`cos alpha =(8D)/(10d)=4/5, cos beta=(6d)/(10d)=3/5`
`d_(1)=3d, d_(2)=4d`
`B_(1)=(mu_(0)i)/(4pid_(1))(cos alpha+cos alpha)`
`=(mu_(0)i)/(4pi.3d)xx2xx4/5=(2mu_(0)i)/(15pid)`
`B_(2)=(mu_(0)i)/(4pid_(2))(cos beta+cos beta)`
`=(mu_(0)i)/(4pi.4d)xx2xx3/5=(3mu_(0)i)/(40pid)`
`B_(P)=2(B_(1)+B_(2))=2((2mu_(0)i)/(15pid)+(3mu_(0)i)/(40pid))`
`=(2mu_(0)i)/(pid)(2/15+3/40=(16+9)/120)`
`=(5mu_(0)i)/(12pid), ox`
(f)

Magnetic field at `P`
Due to `AB:`

`B_(1)=(mu_(0)i)/(4pid_(2))(cos 90^(@)+cos 45^(@))`
`=(mu_(0)i)/(4sqrt(2)pid), ox`
Due to `BC`

`B_(2)=(mu_(0)i)/(4pid_(2))(cos 45^(@)+cos 90^(@))`
`=(mu_(0)i)/(4sqrt(2)pid), ox`
Due to `CA`

`B_(3)=(mu_(0)i)/(4pid')(cos 45^(@)+cos 45^(@))`
`=(mu_(0)i)/(4pi.d/(sqrt(2)))xx2xx1/(sqrt(2))=(mu_(0)i)/(2pid), o.`
`B_(3)gt(B_(1)+B_(2))`
`B_(P)=B_(3)-(B_(1)+B_(2))`
`=(mu_(0)i)/(2pid)-(mu_(0)i)/(2sqrt(2)pid)=(mu_(0)i)/(4pid)(2-sqrt(2)), o.`
(g)
`d'=d sin.(theta)/(2)`
`B_(1)=(mu_(0)i)/(4pid')(cos. (theta)/(2)+cos 0)`
`=(mu_(0)i)/(4pid sin.(theta)/(2))(1+cos.(theta)/(2))`
`=(mu_(0)i)/(4pid(2sin.(theta)/(4)cos.(theta)/(4)))(1+2cos^(2).(theta)/(4)-1)`
`=(mu_(0)i)/(4pid)cot.(theta)/(4)=B_(2), o.`
`B_(P)=B_(1)+B_(2)=2B_(1)`
`=(mu_(0)i)/(2pid)cot.(theta)/(4), o.`
(H)
`B_(1)=(mu_(0)i)/(2pid)`
`B_(2)=(mu_(0)i)/(4pid)(cos theta+cos theta)`
`cos theta=(l//2)/(sqrt((l//2)^(2)+d^(2))) = l/(sqrt(l^(2)+4d^(2)))`
`B_(1)gtB_(2)`
`B_(1)-B_(2)=1/100B_(1)`
`99/100 B_(1)=B_(2)`
`99/100xx(mu_(0)i)/(2pid)=(mu_(0)i)/(4pid)xx2xxl/(sqrt(l^(2)+4d^(2)))`
`(99)sqrt(l^(2)+4d^(2))=100l`
`(99)^(2)(l^(2)+4d^(2))=(100^(2))l^(2)`
`(99)^(2)4d^(2)=(100^(2)-99^(2))l^(2)`
`(d^(2))/(l^(2))=((199)(1))/((99)^(2)xx4)`
`d/l=0.07`
Discussion
47.

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A= 10cm^2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is

Answer»

`2.4 xx 10^(-5) H`
`4.8 PI xx 10^(-4) H`
`4.8 pi xx 10^(-5) H`
`2.4 pi xx 10^(-4) H`

Answer :D
Discussion
48.

Derive an expression for the energy stored in a charged parallel plate capacitor.

Answer»

SOLUTION :Let at a particular instant charge on the plate of CAPACITOR be q and its POTENTIAL difference be `q/C.` If an additional charge dq is GIVEN to the capacitor plate, work done for it is given by
`dW=(q/C).dq`
Therefore, whole process of charging from 0 to Q requires a work
`W = int_0^Q (qdp)/C = 1/C [q^2/2]_0^Q = Q^2/(2C)`
This work done is stored as the electrostatic potential ENERGY of the charged capacitor. Hence, potential energy of charged capacitor
`u = Q^2/(2C)`
But Q=CV, where V be the potential difference between the plates of capacitur, hence
`u = (Q^2)/(2C) =1/2QV =1/2 CV^2`
Discussion
49.

An electron's position is given by vecr=3.00hati-4.00t^(2)hatj+2.00hatk, with t in seconds and vecr in meters. (a) In unit-vector notation, what is the electron's velocity vecr(t)? At t = 3.00 s, what is vecv (b) in unitvector notation and as ( c) a magnitude and (d) an angle relative to the positive direction of the x axis?

Answer»

Solution :(a) `(3.00m//s)hati-(8.00tm//s)hatj,` (b) `(3.00hati-24.0hatj)m//s,` ( C) 24.2 m/s, (d) `82.9^(@)` measured CLOCKWISE from the +X direction, or `277^(@)` COUNTERCLOCKWISE from +x)
Discussion
50.

In a parallel plate capacitor with air between the plates, each has an area 8xx10^(-3)m^(2) and distance between the plates is 2 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 50 V supply, what is the charge on each plate of the capacitor? (Absolute permittivity of free space =8.85xx10^(-12)Fm^(-1))

Answer»

Solution :Given `A=8xx10^(-3) m^2, d=2xx10^(-3)`m
`C_"air"`=?
V=50 V , Q=?
`e_0=8.85xx10^(-2) Fm^(-1)`

WKT electricalcapacitanceof an air CAPACITOR
`C=(epsilon_0A)/d`
i.e.,`C=(8.85xx10^(-2) xx8xx10^(-3))/(2xx10^(-3))`
i.e.,`C=35.416xx10^(-2)` (farad)
Also,
WKT Q=chargean the capacitor =CV
i.e.,`Q=35.416xx10^(-12)xx50`
`=1770.8xx10^(-12)` C
`Q=1.771xx10^(-9)` C(Coulomb )
Discussion