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 Multiple Choice QuestionsMultiple Choice Questions

1.

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:


  • 240N/C

  • 360N/C

  • 420N/C

  • 480N/C


C.

420N/C

Resultant circuit,



As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m

fraction numerator 9 space straight x space 10 cubed straight x space 42 space straight x space 10 to the power of negative 6 end exponent over denominator 30 space straight x space 30 end fraction space equals space 420 space straight N divided by straight C

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2.

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to

  • 6 x 10-7 C/m2

  • 3  x 10-7 C/m2

  • 3 x 104 C/m2

  • 6x 104 C/m2


A.

6 x 10-7 C/m2

When free space between parallel plate capacitor, straight E space equals straight sigma over straight epsilon subscript 0
When dielectric is introduced between parallel plates of capacitor, 

straight E to the power of apostrophe space equals space straight sigma over Kε subscript 0
Electric field inside dielectric 

straight sigma over Kε subscript 0 space equals space 3 space straight x space 10 to the power of 4
where, K = dielectric constant of medium = -2.2
εo = permitivity of free space = 8.85 x 10-12
σ = 2.2 x 8.85 x 10-12 x 3 x 104
 = 6.6 x 8.85 x 10-8 = 5.841 x10-7
 = 6 x 10-7 C/m2

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3.

The electrostatic potential inside a charged spherical ball is given by Φ = ar2 + b where r is the distance from the centre; a,b are constants. Then the charge density inside the ball is

  • -6aε0r

  • -24πaε0

  • -6aε0

  • -24πaε0r


C.

-6aε0

Electric field, 
straight E space equals space minus space dϕ over dt space equals space minus space 2 ar
By space Gauss apostrophe straight s space theorem comma
straight E space left parenthesis 4 πr squared right parenthesis space equals space straight q over straight epsilon subscript 0
straight q space equals space minus 8 πε subscript 0 ar squared
Now comma space the space charge space density comma space straight rho space equals space straight q over straight v space equals space fraction numerator negative 8 πε subscript 0 ar cubed space over denominator begin display style 4 over 3 end style πr squared end fraction
space equals space minus 6 straight epsilon subscript 0 straight a

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4.

A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is

  • straight pi over 4 square root of LC
  • 2 straight pi space square root of LC
  • square root of LC
  • straight pi square root of LC

A.

straight pi over 4 square root of LC

As initially, charge is maximum
∴ q = qocos ωt
Current, 

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5.

Assume that an electric field straight E equals space 30 space straight x squared space straight i with hat on top exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is:

  • 120 J

  • -120 J

  • -80 J

  • 80 J


C.

-80 J

As we know, potential difference VA -Vo is 
dV = - Edx
integral subscript straight V subscript straight o end subscript superscript straight V subscript straight A end superscript dV space equals negative integral subscript 0 superscript 2 30 space straight x squared dx
straight V subscript straight A space minus straight V subscript straight O space equals space minus space 30 space straight x space open square brackets straight x cubed over 3 close square brackets squared space equals negative 10 space straight x left square bracket 2 cubed minus 0 cubed right square bracket
space equals space minus 10 space straight x 8 space equals space minus 80 space straight J

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6.

A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its surface. For this sphere, the equipotential surfaces with potentials 3Vo/2, 5Vo/4, 3V/4 and Vo/4 have radius R1, R2,R3 and R4 respectively. Then

  • R1 = 0 and R2>(R4-R3)

  • R1≠0 and (R2-R1)>(R4-R3)

  • R1 = 0 and R2<(R4-R3)

  • 2R<R4


C.

R1 = 0 and R2<(R4-R3)

The potential at the surface of the charged sphere.

straight V subscript 0 space equals space KQ over straight R
straight V space equals space KQ over straight r comma straight r space greater or equal than space straight R
equals space fraction numerator KQ over denominator 2 straight R cubed end fraction left parenthesis 3 straight R squared minus straight r squared right parenthesis semicolon space straight r less or equal than straight R
straight V subscript centre space equals space Vc space equals space fraction numerator KQ over denominator 2 straight R cubed end fraction straight x 3 straight R squared space equals space fraction numerator 2 KQ over denominator 2 straight R end fraction space space fraction numerator 3 straight V subscript 0 over denominator 2 end fraction
As potential decreases for outside points. Thus, according to the question, we can write

straight V subscript straight R subscript 2 end subscript space equals fraction numerator 5 straight V subscript straight o over denominator 4 end fraction space equals fraction numerator KQ over denominator 2 straight R cubed end fraction left parenthesis 3 straight R squared minus straight R subscript 2 superscript 2 right parenthesis
fraction numerator 5 straight V subscript 0 over denominator 4 end fraction space space equals space fraction numerator straight V subscript 0 over denominator 2 straight R squared end fraction left parenthesis 3 straight R squared minus straight R subscript 2 superscript 2 right parenthesis
or space 5 over 2 space equals space 3 minus open parentheses straight R subscript 2 over straight R close parentheses squared
open parentheses straight R subscript 2 over straight R close parentheses squared space equals space 3 minus 5 over 2 space equals space 1 half
or space straight R subscript 2 space equals space fraction numerator straight R over denominator square root of 2 end fraction
Similarly,
straight V subscript straight R subscript 3 end subscript space equals space fraction numerator 3 straight V subscript 0 over denominator 4 end fraction
KQ over straight R subscript 3 space equals space 3 over 4 space straight x space KQ over straight R
or space
straight R subscript 3 space equals space 4 over 3 straight R
straight V subscript straight R subscript 4 end subscript space equals space KQ over straight R subscript 4 space equals space straight V subscript 0 over 4
KQ over straight R subscript 4 space equals space 1 fourth straight x KQ over straight R
straight R subscript 4 space equals space 4 straight R

1121 Views

7.

The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant τ of this circuit lies between

  • 150 sec and 200 sec

  • 0 and 50 sec

  • 50 sec and 100 sec

  • 100 sec and 150 sec


D.

100 sec and 150 sec

Time constant τ is the duration when the value of potential drops by 63% of its initial maximum value (i.e, Vo/e)
Here, 37% of 25 V = 9.25 V which lies between 100s to 150 s in the graphs.

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8.

Charge +q and -q are placed at points A and B respectively which are distance 2L apart, C is the midpoint between A and B.The work done in moving a charge + Q along the semicircle CED is:

  • fraction numerator qQ over denominator 4 πε subscript straight o straight L end fraction
  • fraction numerator qQ over denominator 2 πε subscript straight o straight L end fraction
  • fraction numerator qQ over denominator 6 πε subscript straight o straight L end fraction
  • fraction numerator negative qQ over denominator 6 πε subscript straight o straight L end fraction

D.

fraction numerator negative qQ over denominator 6 πε subscript straight o straight L end fraction

work done is equal to change in potential energy.

In Ist case, when charge + Q is situated at C. 


Electric potential energy of system

straight U subscript 1 space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction space fraction numerator left parenthesis straight q right parenthesis left parenthesis negative straight q right parenthesis over denominator 2 straight L end fraction space plus fraction numerator 1 over denominator 4 πε subscript straight o end fraction fraction numerator left parenthesis negative straight q right parenthesis straight Q over denominator straight L end fraction space plus fraction numerator 1 over denominator 4 πε subscript straight o end fraction. qQ over straight L
In IInd case, when charge + Q is moved from C to D.



Electric potential energy of system in that case

fraction numerator 1 over denominator 4 πε subscript 0 end fraction. space fraction numerator left parenthesis straight q right parenthesis left parenthesis negative straight q right parenthesis over denominator 2 straight L end fraction space plus fraction numerator 1 over denominator 4 πε subscript 0 end fraction. fraction numerator qQ over denominator 3 straight L end fraction space plus space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator left parenthesis negative straight q right parenthesis left parenthesis straight Q right parenthesis over denominator straight L end fraction
therefore comma
work space done space equals space increment straight U space equals space straight U subscript 2 minus straight U subscript 1
equals space open parentheses fraction numerator negative 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q squared over denominator 2 straight L end fraction plus fraction numerator 1 over denominator 4 πε subscript 0 end fraction. fraction numerator qQ over denominator 3 straight L end fraction minus fraction numerator 1 over denominator 4 πε subscript 0 end fraction qQ over straight L close parentheses

open parentheses fraction numerator negative 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q squared over denominator 2 straight L end fraction minus fraction numerator 1 over denominator 4 πε subscript 0 end fraction. fraction numerator qQ over denominator 3 straight L end fraction plus fraction numerator 1 over denominator 4 πε subscript 0 end fraction qQ over straight L close parentheses
fraction numerator qQ over denominator 4 πε subscript 0 end fraction. space open square brackets fraction numerator 1 over denominator 3 straight L end fraction minus 1 over straight L close square brackets space equals space fraction numerator qQ over denominator 4 πε subscript straight o end fraction fraction numerator left parenthesis 1 minus 3 right parenthesis over denominator 3 straight L end fraction
fraction numerator space minus 2 qQ over denominator 12 πε subscript 0 straight L end fraction space equals space fraction numerator space minus qQ over denominator 6 πε subscript 0 straight L end fraction

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9.

In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF.  Q2 as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)


A.

From the given circuit,
Qc= Q1+Q2
or
C (E-V) = 1 x V + 2 XV
Or V (C+3) = CE
straight V space equals space fraction numerator CE over denominator 3 plus straight C end fraction
straight Q subscript 2 space equals space straight C subscript 2 space left parenthesis straight V right parenthesis
space equals space fraction numerator 2 CE over denominator 3 plus straight C end fraction
equals space fraction numerator 2 straight E over denominator 1 plus 3 divided by straight C end fraction
As, C1 varied from 1μF to 3μF, charge increases with decreasing slope.


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10.

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then

  • 5C1 = 3C2

  • 3C1 = 5C2

  • 3C1 = 5C2 = 0

  • 9C1 = 4C2


B.

3C1 = 5C2

q1 = q2
C1V1 = C2V2
120C1 = 200C2
⇒ 3C1 = 5C2

418 Views