Subject

Physics

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

Electric field at the centre of a uniformly charged semicircle of radius a is

  

  • λ2πε0a2

  • λ4 π2ε0 a2

  • λ22πε0a

  • λ2πε0a


22.

If frequency of R-L circuit is f then impedance will be

  • R2 +  2πfL 2

  • R2 + ( 2πf2 )2

  •  R2 + Lπf2 

  • R2 + (2πf )2


23.

Two  changes  +q  and  -q  are  attached  to  the  two ends of  a  light rod of length L, as shown in figure. The system is given a velocity  ν  perpendicular to magnetic  field  B. The magnetic force on the system of  charges and magnitude of force on one charge by the rod, are respectively

     

  • zero, zero

  • zero, qvB

  • 2qvB, 0

  • 2qvB,  qvB


24.

The intensity of magnetic field due to an isolated pole of strength mp at a point distant r from it will be

  • mp/r2

  • mp r2

  • r2/ mp

  • mp/r


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25.

A bulb  and  a condenser are connected in with an A>C source. On increasibg the frequency will

  • increase

  • decrease

  • sometimes increase and sometimes decrease

  • neither increase nor decrease.


26.

The  ratio of magnetic fields on the axis of a circular current carrying coil of radius a to the magnetic field at its centre will be

  • 11 + x2a232

  • 11 + a2x212

  • 11 + a2x22

  • 11 + a2x23


27.

Lumen is the unit of

  • Luminous flux

  • luminosity

  • illumination

  • quantity of light


28.

A charge q is uniformly distributed on a ring of radius r. A sphere of an equal radius is constructed with its centre lying on the periphery of the ring. The flux of electric field through the surface of the sphere will be

  • qε0

  • q2ε0

  • q3ε0

  • q4ε0


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29.

When 100 V d.c  is applied across a coil, a current of 1 A nows through it. When  100 V a.c. of 50 Hz is applied to the same coil only 0.5 A flows. The inductance of the coil is

  • 0.55 H

  • 5.5 H

  • 0.55 mH

  • 55 mH


A.

0.55 H

For dc

  V = IR  ( Resistance of inductor)

  100 = I × R

⇒  R = 100Ω

For ac

⇒    100 = 0.5 R2 + XL 2

⇒    100 = 5101002 + XL2

⇒    XL2 = ( 200 )2 - ( 100 )2

              = 300 × 100

⇒  XL3 × 1002

⇒       = 1003

⇒   ωL = 1003

⇒   2πvL = 1003

⇒   2 × π ×  50 ×  L = 1003

 L = 3π

        = 33.14

   L = 0.55 H


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30.

A bulb of 25 W. 200 V and another bulb of 100 W, 200 V are connected in series with a supply line of 220 V. Then

  • both bulbs will glow with same brightness

  • both bulbs will get fused

  • 25 W bulb will glow more brightly

  • 100 W bulb will glow more brightly


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