Subject

Physics

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

31.

Taking the earth to be a spherical conductor of diameter 12.8×103 km. Its capacity will be

  • 711 μF

  • 611 μF

  • 811 μF

  • 511 μF


32.

In junction diode, the holes are because of

  • Missing electrons

  • extra electrons

  • protons

  • neutrons


33.

The half-life of a radioactive substance is 3.6 days. How much of 20 mg of this radioactive substance will remain after 36 days?

  • 0.0019 mg

  • 1.019 mg

  • 1.109 mg

  • 0.019 mg


34.

The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10-34 Js, me =9.1×10-31 kg)

  • 5.47Ao

  • 10.9Ao

  • 2.7Ao

  • None of these


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35.

The energy in MeV is released due to transformation of 1kg mass completely into energy is (c=3×108 m/s)

  • 7.625×10MeV

  • 10.5×1029MeV

  • 2.8×10-28MeV

  • 5.625×1029MeV


36.

20 kV potential is applied across X-ray tube, the minimum wavelength of X-ray emitted will be

  • 0.62 Ao

  • 0.37 Ao

  • 1.62 Ao

  • 1.31 Ao


37.

In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt, the collector current changes by 0.05 mA. The output resistance will be

  • 10 kΩ

  • 20 kΩ

  • 5 kΩ

  • 2.5 kΩ


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38.

A student can distinctly see the object upto a distance 15 cm. He wants to see the blackboard at a distance of 3m. Focal length and power of lens used respectively will be

  • -4.8cm, -3.3D

  • -5.8cm, -4.3D

  • -7.5cm, -6.3D

  • -15.8cm, -6.33D


D.

-15.8cm, -6.33D

The rays emanating from a point actually meet at another point after reflection and/or refraction. That point is called the image of the first point. The image is real if the rays actually converge to the point, it is virtual if the rays do not actually meet but appear to diverge from the point from the point when produces backwards.

The student should use a lens which forms image at a distance of 15 cm of the object placed at 3m i.e object distance u= -3 m =-3000 cm, image distance v =-15 cm

From lens formula1v-1u=1fwe get         1-15-1-300=1f1f=1300-115=-19300f=-30019=-15.8 cmNow power of the lens isP=100fcm=100-015.8=-6.33 D


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39.

A thin glass prism (μ =15) in the position of minimum deviation deviates the monochromatic light ray by 10o, the refracting angle of prism is

  • 20o

  • 10o

  • 30o

  • 45o


40.

Two thin lenses of powers 12 D and-2D respectively are placed in contact, the power, focal length and nature
respectively will be

  • 8D, 0.8m, convex

  • 14D, 0.5m, convex

  • 5D, 0.2m, convex

  • 10D, 0.1m, convex


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