A group of N cells whose emf varies directly with the internal re

Subject

Physics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

The inward and outward electric flux from a closed surface are respectively 8 x 103 and 4 x 103 units. Then the net charge inside the closed surface is

  • − 4 × 103 coulomb

  • 4 × 103 coulomb

  • - 4 × 103ε0 coulomb

  • − 4 × 103 ε0 coulomb


22.

In the circuit as shown in the figure, the effective capacitance between A and B is

                     


23.

Capacitance of a parallel plate capacitor becomes 43 times its original value, if a dielectric slab of thickness t = d2 is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is

  • 4

  • 8

  • 2

  • 6


24.

A charged particle of mass m and charge q is released from rest in an uniform electric field E neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is

  • 2E2t2mq

  • Eq2m2t2

  • Eqmt

  • E2q2t22m


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25.

Two wires of the same dimensions but resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity of the combination is

  • ρ1 + ρ22

  • ρ1 + ρ2

  • 2 (ρ1 + ρ2)

  • ρ1 ρ2


26.

I£ a 30V, 90W bulb is to be worked in 120V line, the resistance to be connected in series with the bulb is

  • 20 Ω

  • 10 Ω

  • 40 Ω

  • 30 Ω


27.

The potential difference between the terminals of a cell in open circuit is 2.2 volt with resistance of 5 ohm across the terminals of a cell, the terminal potential difference is 1.8 volt. The internal resistance of the cell is

  • 910 ohm

  • 109 ohm

  • 712 ohm

  • 127 ohm


28.

Thirteen resistances each of resistance R ohm are connected in the circuit as shown in the figure. The effective resistance between A and B is

                    

  • 4R3 Ω

  • 2R Ω

  • R Ω

  • 2R3 Ω


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29.

A group of N cells whose emf varies directly with the internal resistance as per the equation EN = 1.5 rN are connected as shown in the figure above. The current I in the circuit is

              

  • 5.1 A

  • 0.51 A

  • 1.5 A

  • 0.15 A


C.

1.5 A

Given that,

emf EN = 1.6 rN

where rN is the internal resistance of nth cell.

Total emf E = E1 + E2 + E3 + ...... + En

                 = 1.5 [r1 + r2 + r3 + ..... + rn]

Total internal resistance

r = r1 + r2 + r3 + ..... + rn

 Current, i = Er                   i = 1.5 r1 + r2 + r3 + .... + rnr1 + r2 + r3 + .... + rnHence,        i = 1.5 A


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30.

The temperature coefficient of resistance of a wire is 0.00125/°C. Its resistance is 1 ohm at 300 K. Its resistance will be 2 ohm at

  • 1127 K

  • 1400 K

  • 1154 K

  • 1100 K


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