Subject

Physics

Class

JEE Class 12

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JEE Physics 2005 Exam Questions

Multiple Choice Questions

61.

A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 1 /2 m away, the number of electrons emitted by photocathode would

  • decrease by a factor of 4

  • increase by a factor of 4

  • decrease by a factor of 2

  • increase by a factor of 2


B.

increase by a factor of 4

On doubling the distance the intensity becomes one fourth i.e. only one-fourth of photons now strike the target in comparison to the previous number. Since the photoelectric effect is a one photon-one electron phenomenon, so only one-fourth photoelectrons are emitted out of the target hence reducing the current to one fourth the previous value.
225 Views

62.

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

  • lll

  • IV

  • I

  • II


A.

lll

increment straight E space proportional to open parentheses fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close parentheses
270 Views

63.

Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 minutes. The corresponding half-life is

  • 10 minutes

  • 15 minutes

  • 5 minutes

  • 7.5 minutes


C.

5 minutes

straight N subscript straight o space rightwards arrow with space space space space space space space space space space straight t subscript 1 divided by 2 space space space space space space end subscript on top space straight N subscript straight o over 2 rightwards arrow with space space space space space space space space space space straight t subscript 1 divided by 2 space space space space space space end subscript on top straight N subscript 0 over 4 rightwards arrow with space space space space space space space space space space straight t subscript 1 divided by 2 space space space space space space end subscript on top straight N subscript 0 over 8
3 straight t subscript 1 divided by 2 end subscript space equals space 15
straight t subscript 1 divided by 2 end subscript space equals space 5
452 Views

64.

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light = 500 nm ]

  • 5 m

  • 1 m

  • 6 m

  • 3 m


A.

5 m

fraction numerator 1.22 space straight lambda over denominator 3 space mm end fraction space equals space Resolution space limit space equals space fraction numerator left parenthesis 1 mm right parenthesis over denominator straight R end fraction
therefore space straight R space equals space 5 space straight m
275 Views

65.

A thin glass (refractive index 1.5) lens has an optical power of – 5D in air. Its optical power in a liquid medium with refractive index 1.6 will be

  • 1 D

  • 5/8 D

  • -1 D

  • 25 D


B.

5/8 D

straight P subscript straight m over straight P subscript air space equals space fraction numerator open parentheses begin display style straight mu subscript calligraphic l over straight mu subscript straight a minus 1 end style close parentheses over denominator open parentheses begin display style straight mu subscript calligraphic l over straight mu subscript straight m minus 1 end style close parentheses end fraction
straight P subscript straight m space equals space 5 divided by 8 space straight D
165 Views

66.

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor is

  • 1.1 eV

  • 2.5 eV

  • 0.5 eV

  • 0.7 eV


C.

0.5 eV

Eg = hc/λ

= 0.5eV

therefore correct answer is

0.5 eV

284 Views

67.

If radius of Al presubscript 13 presuperscript 27 nucleus is estimated to be 3.6 Fermi then the radius Te presubscript 52 presuperscript 125nucleus

  • 6 fermi

  • 8 fermi

  • 4 fermi

  • 5 fermi


A.

6 fermi

fraction numerator straight R over denominator 3.6 end fraction space equals space open parentheses 125 over 27 close parentheses to the power of 1 third end exponent
rightwards double arrow space straight R space equals space 6 space fermi
310 Views

68.

The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead, it is reduced to l/8. The thickness of lead which will reduce the intensity to l/2 will be

  • 6 mm

  • 9 mm

  • 18 mm

  • 12 mm


D.

12 mm

straight I apostrophe space equals le to the power of negative μk end exponent
rightwards double arrow space straight x equals 1 over straight mu log subscript straight e space fraction numerator straight I over denominator straight I divided by 8 end fraction space equals 3 over straight mu log subscript straight e 2 space.... left parenthesis straight i right parenthesis
straight x equals straight I over straight mu log subscript straight e space fraction numerator straight I over denominator straight I divided by 2 end fraction equals straight I over straight mu log subscript straight e 2 space..... left parenthesis ii right parenthesis
From space equation space left parenthesis straight I right parenthesis space and space left parenthesis ii right parenthesis comma space straight x equals space 12 space mm
625 Views

69.

A Young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

  • hyperbola

  • circle

  • straight line

  • parabola


C.

straight line

390 Views

70.

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is

  • 36 square root of 7
  • 36 divided by square root of 7
  • 36 square root of 5
  • 4 square root of 5

B.

36 divided by square root of 7
straight r equals fraction numerator straight h over denominator square root of straight mu squared minus 1 end root end fraction space equals space fraction numerator 36 over denominator square root of 7 end fraction
287 Views

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