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.


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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


A.

11 + x2a232

Consider a circular radius of a, carrying a current I. Consider a point P, which is at a distance x from the  centre of coil. We can consider that the loop is made p of a large number of short element, generating small magnetic fields. So the total field at P will be the sum of the contributions of all these elements. At the centre of coil, the field will be uniform.

By Biot- Savart’s law, the field dB due to a small element dl of the circle, centered at A is given by

       dB = μ04πdlx2 +a2

This can be resolved into two components, one along the axis OP, and other PS, which is perpendicular to OP. PS is exactly cancelled by the perpendicular component PS’ of the field due to a current and centered at A’. So, the total magnetic field at a point which is at a distance x away from the axis of a circular coil of radius a is given by,

       dB = μ0 I2 a2x2 + a232

 μ0, I are constants and by divide and multiply by a2 on denominator and numerator we get

       dB = 11 + x2a232


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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


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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