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 Multiple Choice QuestionsMultiple Choice Questions

91.

In this diagram, the PD between A and B is 60 V. The PD across 6 µF capacitor is

            

  • 4 V

  • 10 V

  • 5 V

  • 20 V


92.

A capacitor of capacitance 10 µF is charged to 10 V. The energy stored in it is

  • 1 µJ

  • 100 µJ

  • 500 µJ

  • 1000 µJ


93.

Which of the following graphs correctly represents the variation of heat energy (U) produced in a metallic conductor in a given time as a function of potential difference (V) across the conductor ?


94.

The concentric spheres of radii R and r have positive charges q1 and q2 with equal surface charge densities. What is the electric potential at their common centre ?

  • σε0 R + r

  • σε0 R - r

  • σε0 1R + 1r

  • σε0 1R


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95.

When an additional charge of 2C is given to a capacitor, energy stored in it is increased by 21%. The original charge of the capacitor is

  • 30 C

  • 40 C

  • 10 C

  • 20 C


96.

When a potential difference of 103 V is applied between A and B, a charge of 0. 75 mC is value of C is (in μF)

               

  • 12

  • 2

  • 2.5

  • 3


97.

See the diagram. Area of each plate is 2.0 m2 and d = 2 x 10-3 m. A charge of 8.85 x 10-8 C is given to Q. Then the potential of Q becomes

             

  • 13 V

  • 10 V

  • 6.67 V

  • 8.825 V


98.

What is the electric potential at a distance of 9 cm from 3 nC ?

  • 270 V

  • 3 V

  • 300 V

  • 30 V


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99.

A voltmeter reads 4 V when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration, voltmeter reads 2 V. What is the dielectric constant of the material ?

  • 0.5

  • 2

  • 8

  • 10


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100.

Two capacitors of 10 pF and 20 pF are connected to 200 V and 100 V sources respectively. If they are connected by the wire, what is the common potential of the capacitors ?

  • 133.3 V

  • 150 V

  • 300 V

  • 400 V


A.

133.3 V

Given,

C1 = 10 pF = 10 × 10-12 F

C2 = 20 pF = 20 × 10-12 F

V1 = 200 V, V2 = 100 V

C1 = Capacitance of 1st capacitor

C2 = Capacitance of IInd capacitor

V1 = Voltage across Ist capacitor

V2 = Volatge across IInd capacitor

We know that,

     V1 = q1C1  and  V2 = q2C2 q1 = V1C1                             ........ (i)      q2 = V2C2                             ........ (ii)

So, common potential of capacitors

V = q1 + q2C1 + C2 = V1C1 + V2C2C1 + C2    = 200 × 10 × 10-12 + 100 × 20 × 10-1210 × 10-12 + 20 × 10-12    = 200 × 10 + 100 × 2010 + 20    = 2000 + 200030    = 400030    = 133.3 V


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