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3501.

According to moselay.s law, the frequency (vartheta) of the Kalpha line and the atomic number (z) of the element have the relation (A and B) are constants

Answer»

`(VARTHETA)/((z-A))=B`
`(sqrtvartheta)/((z-A))=B`
`vartheta(Z-A)=B`
`vartheta(Z-A)^(2)=B`

ANSWER :B
Discussion
3502.

Consider a force acting on a particle given as vec( F) = yhat( i ) + xhat( j) , where ( x, y ) is the instantaneous location of the particle. This type of force is :

Answer»

Surely a conservation force
Surely a non-conservation force
Information INSUFFICIENT to reach to a conclusion
Will be conservation only if the PARTICLE is CONSTRAINED to move in the x-y PLANE.

Answer :A
Discussion
3503.

A wire of resistance R is stretched to three times its original length. The new resistance is

Answer»

3R
9R
`R//3`
`R//9`

ANSWER :B
Discussion
3504.

Two simple harmonic motions are represented by the equations y_(1) = 0.1 sin (100 pi t + (pi)/(3)) and y_(2) = 0.1 cos pi t. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is :

Answer»

`(pi)/(6)`
`(-pi)/(3)`
`(pi)/(3)`
`(-pi)/(6)`

Solution :`y_(1) = 0.1 sin (100 pi t + pi//3)`
`therefore v_(1) = (dy_(1))/(dt) = 0.1 cos (100 pi t + pi//3 ) . 100 pi `
`y_(2) = 0.1 cos pi t = 0.1 sin ((pi)/(2) + pi t ) `
`v_(2) = (dy_(2))/(dt ) = 0.1 cos ((pi)/(2) + pi t).pi`
`therefore` initial PHASE deff. between velocity of Ist PARTICLE and 2nd particle is = `phi_(1) - phi_(2) = ((pi)/(3) - pi//2 )= - (pi)/(6) `
so CORRECT CHOICE is (d).
Discussion
3505.

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelenght is :

Answer»

`1mu m`
`30 mu m`
`100 mu m`
`300 mu m`.

SOLUTION :(b) Resolving angle of human eye
`THETA = 1.22 (lambda)/(D)`
`therefore ((y)/(25 xx 10^(-2)))`
` = (1.22 xx 500 xx 10^(-9))/(0.25 xx 2 xx 10^(-2))`
`rArr y = 30 mu m.`
Discussion
3506.

Give two examples of diamagnetic substances.

Answer»

SOLUTION :ALUMINIUM and PLATINUM.
Discussion
3507.

At some place in universe an atom consists of a positron revolving round an antiproton. The ratio of the wavelength of the corresponding spectral lines from this atom and ordinary hydrogen atoms is

Answer»


ANSWER :a
Discussion
3508.

What is the advantage of a geostationary satellite?

Answer»

Solution :The rotaional period of communication satellite (i.e., geostationary satellite) is approximately to earth. Since, the satellite is in a fixed position, earth station antennas can be MOUNTED in PERMANENT POSITIONS and POINTED at it.
Discussion
3509.

Kirchoff's first law i.e. Sigmai=0 at a junction is based on the law of conservation of:

Answer»

Charge
Energy
Momentum
Angular momentum

Answer :A
Discussion
3510.

Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

Answer»

because of energy conservation
without simultaneously releasing energy in the FORM of radiation
because of momentum conservation.
because of angular momentum conservation.

Solution :When an electron beam is made incident on free protons, each electron coming successively, experiences an attractive type of resultant Colombian force which KEEPS on changing because of change in distance as well as change in arrangement of the charges. Because of this, the stream of electrons get scattered, so as to conserve energy. In these CIRCUMSTANCES, proton and electron can not come extremely CLOSE to each other and unless they emit energy equal to magnitude of binding energy of H-atom in the form of radiation, they can not form H-atom.
`rArr` Thus, OPTIONS (A) and (B) are correct.
Discussion
3511.

A current carrying circular coil produces a magnetic field B at its center.It is then rewound into n number of turns. The magnetic field at the centre of the coil will be

Answer»

nB
2nB
`n^2B`
`B/n`

ANSWER :C
Discussion
3512.

The Bohr radius of the innermost orbit of a hydrogen atom is 5.3xx10^-11m.What are the radii of n=2 and n=3 orbits?

Answer»

SOLUTION :`r_n=n^2r_1 `
`THEREFORE r_2=2^2r_1=4xx5.3xx10^(-11)=21.2xx10^(-11)m`
` r_3=3^2r_1=9xx5.3xx10^(-11)=47.7xx10^(-11)m`
Discussion
3513.

An electric dipole is a pair of equal and opposite point charge +q and -q separated by a distance r.Write an expresion for its dipole moment.

Answer»

SOLUTION :`p=qxxr`
Discussion
3514.

Assuming the spectral distribution of thermal radiation energy of obey wien's formula u(omega,T) = A omega^(3)exp (-1 omega//T), where a = 7.64 ps.K, find for a temperature T = 2000K the most probable (a) radiation frequency, (b) radiation wavelength.

Answer»

Solution :We are GIVEN
`u(omega,T) = AOMEGA^(3) exp(-aomega//T)`
(a)Then `(du)/(d omega) = ((3)/(omega)-(a)/(T))u = 0`
so `omega_(Pr) = (3T)/(a) = (6000)/(7.64) xx 10^(12)s^(-1)`
(b) We determine the spectral distribution in wavelength.
`-overset~(u)(lambda,T)d lambda = u(omega,T)d omega`
But `omega = (2pic)/(lambda)` or `lambda = (2pic)/(omega) = (C')/(omega)`
so `d lambda =- (C')/(omega^(2))d omega, d omega=- (C')/(lambda_(2))d lambda`
(we have put a minus sign before `d lambda` to subsume just this fact `d lambda` is -ve where `d omega` is `+ve`).
`overset~(u) (lambda, T) = (C')/(lambda^(2))u ((C')/(lambda),T) = (C^(4)A)/(lambda^(5))exp(-(aC')/(lambdaT))`
This is maximum when
`(deloverset~(u))/(del lambda) =6 overset~(u) [(-5)/(lambda)+(aC')/(lambda^(2)T)]`
or `lambda_(Pr) = (aC')/(5T) = (2pica)/(5T) = 1.44mu m`
Discussion
3515.

For an adiabatic change the value of (dP)/(P) is equal to (dV=" change in vol.," gamma =(C_(p))/(C_(v))):

Answer»

`GAMMA^(1//2) cdot (dV)/(V)`
`-(dV)/(V)`
`-gamma cdot (dV)/(V)`
`-gamma^(2)(dV)/(V)`

SOLUTION :Here `(d)/(DP) (P. V^(gamma)) =(d)/(dP)` (constant)
or `V^(gamma) +gamma V^(gamma-1)P(dV)/(dP)=0`
or `V^(gamma) =-gamma V^(gamma-1)P. (dV)/(dP)`
or `(dP)/(P) =(-gamma V^(gamma-1))/(V^(gamma)). dV =-gamma. (dV)/(V).`
`therefore` Correct choice is (C ).
Discussion
3516.

A dielectric slab of thickness 't' is kept between the plates of a parallel plate capacitor separated by a distance 'd' (t lt d) . Derive the expression for the capacity of the capacitor .

Answer»

Solution :In the absence of dielectric slab , the capacitance of a PARALLEL PLATE capacitor is given by `C = (in_(0) A)/(d)`
When a dielectric slab of THICKNESS `t(t lt d)` is introduced between the PLATES without touching the plates , the electric field in air `E_(0) = (sigma)/(in_(0))` but on account of polarisation of dielectric the electric field inside the dielectric changes to `E = (E_0)/(K)` .
If potential difference between the plates of capacitor be V. now , then clearly
`V = E_0 (d- t) + E t = E_(0) (d- t) + Et = E_(0) (d - t) + (E_(0))/(K) . t = E_(0) (d - t + (t)/(K)) = (sigma)/(in_(0)) . (d - t + (t)/(K)) = (Q)/(in_(0) A) (d - t + (t)/(K))`

`therefore` New capacitance `C. = (Q)/(V.) = (Q)/((Q)/(in_(0) A (d - t + (t)/(K))) = (in_(0) A)/(d - t ((K-1)/(K))) = (in_(0) A)/(d[1 - (t)/(d) ((K-1)/(K))]) = (C)/(1 - (t)/(d) ((K-1)/(K)))`
Discussion
3517.

A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum P and energy E, then

Answer»

`P=0, E=0`
`Pne0, Ene0`
`Pne0, E=0`
`P=0, Ene0`

ANSWER :B
Discussion
3518.

A particle moves in a circle of radius 25 cm at two revolutions per second. The acceleration of the part in m/s^2 is,

Answer»

`pi^2`
`8pi^2`
`4pi^2`
`2pi^2`

ANSWER :C
Discussion
3519.

Exposure time of a camera lens at the (f)/(2.8) setting is (1)/(200) second. The correct time of exposure at (f)/(5.6) is

Answer»

0.20 second
0.40 second
0.02 second
0.04 second

Answer :B::C::D
Discussion
3520.

In the question number 66, the charge on capacitors C_(1)and C_(4) are

Answer»

`4xx10^(-3)C, 12xx10^(-3)C`
`6xx10^(-3)C,12xx10^(-3)C`
`2xx10^(-3)C,4xx10^(-3)C`
`3xx10^(-3)C, 2xx10^(-3)C`

Solution : Chargeon eachcapacitor `C_(1),C_(2),C_(3)` is same = q Letcharge on `C_(4)` be q.
`therefore(q)/(C_(1)) + (q)/(C_(2)) + (q)/(C_(3)) = 600`
`or q [(1)/(C_(1)) + (1)/(C_(2)) +(1)/(C_(3)) ] = 600`
`or(q)/(C_(s)) = 600`
`therefore "" q = 600 C_(s) = 60 xx (20)/(3) mu C = 4000 xx 10^(-6) C = 4 xx 10^(-3) C`
Also `(q)/(C_(4)) = 600`
or `q = C_(4) xx 600 MUC= 20 xx 600 mu C = 12 xx 10^(-3)C`
Discussion
3521.

A charged disc of radius 3 m has charge density sigma. Electric field at a distance of 4 m from its centre on axis is _______

Answer»

`(3sigma)/(5epsilon_(0))`
`(sigma)/(10 epsilon_(0))`
`(sigma)/(8epsilon_(0))`
NONE

Answer :B
Discussion
3522.

What does determination and perseverance mean?

Answer»

To be committed to a thing in spite of MANY difficulties
To be unrelenting
Both (A) and (B)
None of the above

Answer :C
Discussion
3523.

दक्कन का पठार किस आकार का भूभाग है?

Answer»

आयताकार
त्रिभुजाकार
गोलिय
इनमें से कोई नहीं

Answer :B
Discussion
3524.

….is the scientist postulated electromagnetic waves.

Answer»

Hertz
Faraday
MARCONI
MAXWELL

ANSWER :D
Discussion
3525.

Indicate the reasons why the following processes are forbidden: (1) Sigma^(-)rarr^^ ^(0)+pi^(-), (2) pi^(-)+prarrK^(+)+K^(-), K^(-)+n rarr Omega^(-)+K^(+)+K^(0), (6) murarr e^(-)+v_(e )+v_(mu).

Answer»

Solution :(1)` Sigma^(-)Sigma^(-)rarr^^ ^(0)+pi^(-)`,
(2) `pi^(-)+prarr K^(+)+K^(-)`,
`K^(-)+n rarr OMEGA^(-)+K^(+)+K^(0)`,
(6) `murarr e^(-)+v_(e )+v_(mu)`.
is forbidden by energy conservation. The mass difference
`M_(Sigma)-M_(^^ ^(0))= 82(MEV)/(c^(2)) LT m_(pi)^(-)`
( The PROCESS `1rarr 2+3` will be allowed only if `m_(1)gt m_(2)+m_(3)`.)
(2) `pi^(-)+prarrK^(+)K^(-)`
is disallowed by conservation of baryon number.
(3) `K^(-)+n rarr Omega^(-)+K^(+)+K^(0)`
is forbidden by conservation of charge
(4) `n+prarr Sigma^(+)^^ ^(0)`
is forbidden by strangerness conservation.
(5) `pi^(-)rarrmu^(-)+e^(-)+e^(+)`
is forbidden by conservation of muon number (or lepton number).
(6) `mu^(-)rarre^(-)+v_(e )+ overset(~)v_(mu)`
is forbidden by the seperate conservation of muon number as well as lepton number.
Discussion
3526.

-5D क्षमता के लेंस की फोकस दूरी होगी -

Answer»

0.2 मी
0.5 मी
-0.2 मी
-0.5 मी

Answer :C
Discussion
3527.

The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is

Answer»

`1/sqrt2`A
`sqrt2`A
`2A`
`2sqrt2A`

Solution :Secondary VOLTAGE `V_S`=24 V
POWER associated with secondary `P_S`=12 W
Now `P_S=I_S V_S`
`therefore I_S=P_S/V_S=12/25`=0.5 A
`therefore` Peak value of the current in the secondary
`I_m=I_Sxxsqrt2` [ `because` Here, `(I_(rms))_S=I_S`]
`therefore I_m=0.5xx1.414`=0.707
`therefore I_m=1/sqrt2A [ because 0.707 = 1/sqrt2]`
Discussion
3528.

In the given figure, a wire loop has been bent so that it has three segments. Portion OABO is quarter circle in xy plane, portion OBC is a right angle triangle in xz plane and portion OCDAO is a square in yz plane. Here are three choices for a magnetic field through the loop. (1) vecB_(1)=3hati+7hatj-5thatk (2) vecV_(2)=5thati-4hatj-15hatk (3) vecB_(3)=2hati-5thatj-12hatk If the induced current in the loop due to vecB_(1), vecB_(2) and vecB_(3) are i_(1), i_(2) and i_(3) respectively then :

Answer»

`i_(1) lt i_(2)`
`i_(1) gt i_(2)`
`i_(3) gt i_(2) gt i_(1)`
`i_(1) = i_(2) = i_(3)`

Solution :At any TIME `t`,
`OMEGA=(U sin60^(@))/(x)`
`alpha=(u sin 60^(@))/(x^(2))(-(dx)/(dt))`
`alpha=(usin60^(@))/(x^(2))(u+ucos60^(@))`
`alpha` at `t=0`
`alpha=(usin60^(@))/(a^(2))(u+ucos60^(@))=9rad//s^(2)`
Discussion
3529.

What is the energy of a photon whose wavelength is 6840 Å ?

Answer»

1.81 eV
3.6 eV
`-13.6 eV`
12.1 eV

SOLUTION :`E = h UPSILON = (HC)/(lambda) = (12400 eV Å)/(8840 Å) = 1.81 eV`
Discussion
3530.

The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 Omega is connected in parallel to the cell , the balancing length changes by 60 cm. If the internal resistance of the cell is (n)/(10) Omega the value of n is ....

Answer»

10
11
12
14

Solution :12
Let, `epsilon` is emf of battery and , is internal resistance and x is potential gradient.when onJy battery is connected, then
`epsilon =560x ""` ....(1)
After connecting resistance ,
`(epsilon XX 10)/(10 + r)` = (560x - 60) x = 500x ...(2)
From equation (1) and (2)
`(560 xx 10X)/(10 + r)= 500`x
`therefore 56 = 50 + 5r`
`therefore 6 = 5r`
`therefore r = 1.2Omega`
`therefore (n)/(10) = 1.2"" therefore n = 12 `
Discussion
3531.

In a large room a person receives direct sound waves from a source 120 m away from him. He also receives waves from the same source which reach him, being reflected from the 25m high ceiling at a point halfway between them. For which wavelengths will these two sound waves interfere constructively ?

Answer»

Solution :As shown in figure, for reflection from the ceiling :

Path SCP = SC + CP =2SC [as `ANGLE I =ANGLER, SC = CP`] or path `SCP = 2sqrt(60^2 +25^2) = 130m`
So path difference between interfering WAVES along paths SCP and SP,
`Deltax = 130 - 120 = 10 m`
Now for constructive intergerence at P, `Deltax = nlambda`,
`i.e., 10 = n lambda " or " lambda=10/n` with n =1,2,3,.......
`i.e.,lambda =10m , 5m , (10//3)m` and so on.
Discussion
3532.

A wire of variable mass per unit length mu = mu_0x, is hanging from the ceiling as shown in figure. The length of wire is l_0. A small transverse disturbance is produced at its lower end. Find the time after which the disturbance will reach to the other end.

Answer»

SOLUTION :`SQRT((8l_0)/G)`
Discussion
3533.

Choose the correct statements from the following a) in potentiometer experiment balancing length represent the potential difference across the secondary cell b) the shunt resistance between the terminals of cell in the secondary circuit of potentiometer is increased, the balancing length increases c) in potentiometer experiment the e.m.f of the cell in the secondary circuit may be greater than the e.m.f of cell in the primary circuit d) potentiometer is used to measure thermo e.m.f. also

Answer»

B and C are CORRECT 
c and d are correct 
a and d are correct 
a,b and d are correct

Answer :D
Discussion
3534.

What is the magnitude of Electron volt (ev).1ev= What ?

Answer»


ANSWER :`1.6xx10^-19C`
Discussion
3535.

In Young.s double slit experiment, the slits are 2mm apart and are illuminated by photons of two wavelengths lambda_(1)= 12000 A^(0)" and "lambda_(2)= 10,000 A^(0). At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Answer»

`3MM`
`8MM`
`6mm`
`4mm`

Answer :C
Discussion
3536.

The division marked on the scale of an a.c. ammeter are not equally spaced. Why?

Answer»

Solution :A.C. AMMETER WORKS on the principle of HEATING effect `H prop I^2`.
Discussion
3537.

The most stable particle in Baryon group is

Answer»

PROTON
lambda - particle
neutron
omega - particle

Solution :The proton is the most STABLE in the BARYON GROUP.
Discussion
3538.

A block of mass m is arranged on the wedge as shown in figure . The wedge angle istheta . If the masses of pulley and thread and negligible and friction is absent , find the acceleration of the wedge

Answer»

Solution :It is obvious that when block m moves downward along the incline of wedge , the wedge moves to the right . As the length of thread is constant , the distance traversed by wedge along the incline is equal to distance traversed by wedge to the right. This implies that acceleration of wedge to the right is equal to downward acceleration of wedge .
Let a be the acceleration of wedge to the right.
Then the force ACTING on the block m are

i) weight mg acting VERTICALLY downward
ii) normal reaction `R_1`
iii) tenstion T ( up the incline )
iv) Fictitious force ma to the left .
The free body diagram of mass m is SHOWN in figure .
For motion of block .m. on inclined plane
`mg sin theta + ma cos theta -T = ma `.......... (1)
As mass .m. does not BREAKS off the inclined plane, therefore for forces on ..m. normal to inclined plane
`R_1 + ma sin theta = mg cos theta `.......... (2)
The forces acting on the wedge are
i) weight Mg downward
ii) normal reaction of ground on wedge =`R_2`
iii) normal reaction of block on wedge `=R_1`
iv) tension (T,T) in string .
The free body diagram of wedge is shown in figure .

For maotion of wedge in horizontal direction
`R_1 sin theta = T-T cos theta = Ma`......... (3)
For vertical equilibrium of wedge on surface
`R_2 = R_1 cos theta + T sin theta + Mg `.......... (4)
From (1) ,
`T= mg sin theta + ma cos theta - ma `........ (5)
From (2) , `R_1 = mg cos theta - ma sin theta `............. (6)
substituting these values in (3) , we get
`(mg cos theta - ma sin theta ) sin theta + (mg sin theta + ma cos theta - ma) (1- cos theta ) = Ma`
`or { M + msin^2 theta + m(1- cos theta )^(2)}a`
`= ma cos thetasin theta + mg sin theta (1-cos theta )`
` a= ( mg sin theta )/( M +m sin^(2) theta + m(1- 2 cos theta + cos^(2) theta)) = ( mg sin theta )/( M + m(sin^2 theta + 1 -2 cos theta + cos^(2) theta ))`
` :. a=(mg sin theta )/( M + 2M(1-cos theta))`
Discussion
3539.

(A): In the process of nuclear fission, the fragments emit two or three neutrons as soon as they are formed and subsequently emit particles. (R): As the fragments contain an excess of neutrons over protons, emission of neutrons and particles bring their neutron/ proton ratio to stable values

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
3540.

Main scale reading is -1mm when there is no object btween the jaws In the vernier caliperse 9 main scale divisions matches with 10 vernier scale division The thickness of the object using the defected vernier calliperse will be

Answer»

Solution :Zero error - MAIN SCALE READING + (vernier scale reading) (least COUNT)
`= - 1 mm + 6 (0.1mm) = - 0.4mm`
observed reading `= 11.8mm`
So actual THICKNESS `= 11.8 - (-0.4) =12.2 mm` .
Discussion
3541.

Gate can be obtained by shorting both the input terminals of a NOR gate is …….

Answer»

OR
AND
NOT
NAND

Solution :NOT

SHORTING both terminals of NOR GATE,

Obtained `y=barA or y=barB` which becomes NOT gate.
Discussion
3542.

An emf induced in a secondary coil is 20000V when the current breaks in the primary. The mutual inductance is 5H and the current reaches to zero in 10^(-4)sin primary. Calculate the maximum current in the primary before the break.

Answer»

1A
0.2A
0.3A
0.4A

Answer :D
Discussion
3543.

The temperature Coefficient of resistivity tungsten is _____ (""^(@)C)^(-1).

Answer»

`3.9xx10^(-3)`
`4.5xx10^(-3)`
`5.0xx10^(-3)`
`4XX10^(-3)`

SOLUTION :`4.5xx10^(-3)`
Discussion
3544.

Two rods of different materials having coefficient of linear expansion alpha_(1) and alpha_(2) and Young's modulus Y_(1) and Y_(2) respectively are fixed between two rigid walls. The rods are heated to same high temperature. If alpha_(1): alpha_(2) :: 2 : 3 the thermal stress in two rods is the same. Then the rate Y_(1)//Y_(2) is :

Answer»

`2/3`
`1/1`
`3/2`
`4/9`

SOLUTION :Here STRESS is same for the TWO
Now stress = Y `xx` strain
`thereforeY_(1)xxalpha_(1)xxt_(1)=Y_(2)xxalpha_(2)xxt_(2)`
But `t_(1)=t_(2)`
`thereforeY_(1)alpha_(1)=Y_(2)alpha_(2)` or `(Y_(1))/(Y_(2))=(alpha_(2))/(alpha_(2))=3/2`
Correct choice is (C).
Discussion
3545.

Who does Lencho think sent him the money?

Answer»

God
Postmaster
Postman
NO One

Answer :A
Discussion
3546.

In a pitcher 10 kg water is contained.Total surlace area of pitcher walls is 2 xx 10^(-2) m^(2) and its wall thickness is 10^(-3) m. If surrounding temperature is 42^(@)C, find the temperature of water in the pitcher when it attains a steady value. Given that in steady state 0.1 gm water gets evaporated per second from the outer surface of pitcher through its porous walls. The thermal conductivity of the walls of pitcher is 0.8 W//m^(@)C and latent heat of vaporization of water is 2.27 x 10^(6) J//kg.

Answer»


ANSWER :`[28^(@)C]`
Discussion
3547.

32 g of O_(2) is contained in a cubical container of side 1 m and maintained at a temperature of 127^(@)C. The isothermal bulk modulus of elasticity of the gas is (universal gas constant = R)

Answer»

127 R
400 R
200 R
560 R

Answer :B
Discussion
3548.

A cell of emf epsilon and internal resistance r is connected across a variable load resistance R. The graph drawn between its terminal voltage and resistance R is

Answer»




ANSWER :C
Discussion
3549.

What is the frequency of a 20 m radiowave?

Answer»

15 MHz
12 MHz
1200 KHz
25 MHz

Answer :C
Discussion
3550.

An inductor of inductance 20 mH is connected across a charged capacitor of capacitance 5 muF & resulting L-C circuit is set oscillating at its natural frequency . The maximum charge q is 200 mu C on the capacitor . Find the potential difference across the inductor when the charge is 100 muC .

Answer»


ANSWER :20 V
Discussion