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 Multiple Choice QuestionsMultiple Choice Questions

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1. 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
  • 3 x 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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2.

A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3 . Capacitance of the capacitor is now 

  • 1.8 pF 

  • 45 pF

  • 40.5 pF

  • 40.5 pF


C.

40.5 pF

fraction numerator Aε subscript 0 over denominator begin display style straight d subscript 1 over 3 plus straight d subscript 2 over 6 end style end fraction space equals space fraction numerator Aε subscript 0 over denominator begin display style straight d over 9 plus fraction numerator 2 straight d over denominator 18 end fraction end style end fraction space equals space fraction numerator 18 space Aε subscript 0 over denominator 4 straight d end fraction


C' = 40.5 PF
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3. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3 . Capacitance of the capacitor is now 
  • 1.8 pF
  • 40.5 pF

  • 45 pF
  • 45 pF

B.

40.5 pF

fraction numerator Aε subscript 0 over denominator begin display style straight d subscript 1 over 3 plus straight d subscript 2 over 6 end style end fraction space equals space fraction numerator Aε subscript 0 over denominator begin display style straight d over 9 plus fraction numerator 2 straight d over denominator 18 end fraction end style end fraction space equals space fraction numerator 18 space Aε subscript 0 over denominator 4 straight d end fraction

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4.

An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience

  • a torque only

  • a translational force only in the direction of the field

  • a translational force only in a direction normal to the direction of the field

  • a translational force only in a direction normal to the direction of the field


D.

a translational force only in a direction normal to the direction of the field

164 Views

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5.

A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

  • ½ (K – 1) CV2

  • CV2 (K – 1) /K

  • (K –1) CV2

  • (K –1) CV2


D.

(K –1) CV2

183 Views

6.

A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is

  • (n − 1)C

  • (n + 1)C

  • C

  • C


A.

(n − 1)C

Ceq=(n−1) C (Q all capacitors are in parallel)

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7.

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

  • 1

  • 2

  • 1/4

  • 1/4


D.

1/4

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8.

A fully charged capacitor has a capacitance ‘C’ it is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘∆T’. The potential difference V across the capacitance

  • square root of fraction numerator 2 mC increment straight T over denominator straight s end fraction end root
  • fraction numerator mC increment straight T over denominator straight s end fraction
  • fraction numerator ms increment straight T over denominator straight C end fraction
  • fraction numerator ms increment straight T over denominator straight C end fraction

D.

fraction numerator ms increment straight T over denominator straight C end fraction

Dimensionally only 4th option is correct.

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9.

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 −V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2?


(e = 1.6 × 10−19 C, me = 9.11 × 10−31 kg)

  • 32 × 10−19 m/s

  • 2.65 × 106 m/s

  • 7.02 × 1012 m/s

  • 7.02 × 1012 m/s


B.

2.65 × 106 m/s

1 half mv squared space equals space eV
straight v space equals space square root of fraction numerator 2 eV over denominator straight m end fraction end root space equals space 2.65 space straight x space 10 to the power of 6 space straight m divided by straight s
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10.

An electric charge 10-3 µC is placed at the origin (0,0) of X - Y coordinate system. Two points A and B are situated at left parenthesis space square root of 2 space comma square root of 2 right parenthesis and (2, 0) respectively. The potential difference between the points A and B will be

  • 9 volt

  • 0 volt

  • 2 volt

  • 2 volt


B.

0 volt

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