Subject

Physics

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

A lamp in which 10 amp current can flow at 15 V, is connected with an alternating source of potential 220 V. The frequency of source is 50 Hz. The impedance of choke coil required to light the bulb is

  • 0.07 H

  • 0.14 H

  • 0.28 H

  • 1.07 H


22.

If the current 30A flowing in the primary coil is made zero in 0.1 sec, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is

  • 0.05 H

  • 1.05 H

  • 0.1 H

  • 0.2 H


23.

Two equal resistances R are joined with voltage source V in (i)series (ii)parallel, the ratio of electrical power consumed in two cases will be

  • 1:4

  • 4:1

  • 2:1

  • 1:2


24.

A proton enters a magnetic field of intensity 1.5 Wb/m2 with a velocity 2×107 m/s in a direction at angle 30o with the field. The force on the proton will be (charge on proton is 1.6×10-19 C)

  • 2.4×10-12 N

  • 4.8×10-12 N

  • 1.2×10-12 N

  • 7.2×10-12 N


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25.

20 ampere current is flowing in a long straight wire. The intensity of magnetic field at a distance 10 cm from the wire will be

  • 2×10-5 Wb/m2

  • 9×10-5 Wb/m2

  • 4×10-5 Wb/m2

  • 6×10-5 Wb/m2


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26.

The resistance of a galvanometer is 50 Ω. When 0.01A current flows in it, full scale deflection is obtained in galvanometer, the resistance of shunt connected to convert galvanometer into an ammeter of range 5A, will be

  • 0.1 Ω

  • 0.2 Ω

  • 0.3 Ω

  • 0.4 Ω


A.

0.1 Ω

Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt.

Here :- Ig =0.01A, G=50Ω

If resistance of galvanometer is Rg and it gives full-scale deflection when current Ig is passed through it. Then,

V=IgRg

V=0.01A × 50Ω

Let a shunt of resistance (Rs) is conected in parallel to galvanometer. If total current through the circuit is I

I =5A= Is +Ig

V=IsRs =(I-Ig)Rs

(I-Ig)Rs = IgRg

(5-0.01)Rs= 0.01×50

Rs =0.01Ω


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27.

When a resistance of α Ω is connected at the ends of a battery, its potential difference decreases from 40 V to 30V. The internal resistance of the battery is

  • 1.5Ω


28.

The equivalent resistance and potential difference between A and B for the circuit              

respectively are

  • 4Ω, 8V

  • 8Ω, 4V

  • 2Ω, 2V

  • 16Ω, 2V


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29.

A wire of resistance R is divided in equal parts, then these parts are joined in parallel, the equivalent resistance of the combination will be

  • nR

  • Rn

  • nR

  • Rn2


30.

The capacity of an air condenser is 2.0 µF. If a medium is placed between its plates, the capacity becomes 12 μF. The
dielectric constant of the medium will be

  • 5

  • 4

  • 3

  • 6


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