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 Multiple Choice QuestionsMultiple Choice Questions

1.

When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8 , the emitted particles will be

  • neutrons

  • alpha particles

  • beta particles

  • gamma photons


D.

gamma photons

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2.

The binding energy per nucleon of deuteron (21H) and helium nucleus (42He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

  • 13.9 MeV

  • 26.9 MeV

  • 23.6 MeV

  • 19.2 MeV


C.

23.6 MeV

As given,straight H presubscript 1 superscript 2 space plus space straight H presubscript 1 superscript 2 space rightwards arrow for space of space subscript 2 He to the power of 4 space plus space energy
The binding energy per nucleon of a deuteron (1H2 ) = 1.1 MeV
∴ Total binding energy = 2× 1.1 = 2.2 MeV

The binding energy per nucleon of helium (2He4 ) = 7 MeV
∴ Total binding energy = 4× 7 = 28 MeV
Hence, energy released in the above process = 28 - 2× 2.2 = 28 - 4.4 = 23.6 MeV

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3.

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
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4.

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
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5.

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be

  • 21/3:1

  • 1:31/2

  • 31/2:1

  • 1:21/3


B.

1:31/2

Law of conservation of momentum gives
m1 v1 = m2 v2

straight m subscript 1 over straight m subscript 2 space equals straight v subscript 2 over straight v subscript 1
space But space space straight m space equals space 4 over 3 πr cubed straight rho
or space straight m proportional to space straight r cubed
therefore space straight m subscript 1 over straight m subscript 2 space equals space fraction numerator straight r subscript 1 superscript 3 over denominator straight r subscript 2 superscript 3 end fraction space equals space straight v subscript 2 over straight v subscript 1
straight r subscript 1 over straight r subscript 2 space equals space open parentheses 1 half close parentheses to the power of 1 divided by 3 end exponent
straight r subscript 1 space colon straight r subscript 2 space equals space 1 space colon 2 to the power of 1 divided by 3 end exponent

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6.

An α-particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of

  • 1 Å

  • 10−10 cm

  • 10−12 cm

  • 10−15 cm


C.

10−12 cm

At closest approach, all the kinetic energy of the α-particle will converted into the potential energy of the system, K.E. = P.E.

straight i. straight e. space 1 half space mv squared space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r end fraction
5 space MeV space equals space fraction numerator 9 space straight x space 10 to the power of 9 space straight x space left parenthesis 2 straight e right parenthesis space straight x space 92 straight e over denominator straight r end fraction space open parentheses therefore space 1 half space mv squared space equals space 5 space MeV close parentheses

rightwards double arrow space straight r space equals space fraction numerator 9 space straight x space 10 to the power of 9 space straight x space 2 space straight x 92 space straight x space left parenthesis 1.6 space straight x space 10 to the power of negative 19 end exponent right parenthesis squared over denominator 5 space straight x space 10 to the power of 6 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction
straight r space equals space 5.3 space straight x space 10 to the power of negative 14 end exponent space straight m space equals space 10 to the power of negative 12 space cm end exponent

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7.

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
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8.

A nuclear transformation is denoted by X(n, α) Li presubscript 3 presuperscript 7. Which of the following is the nucleus of element X?

  • straight C presuperscript 12 subscript 6
  • straight B presubscript 5 presuperscript 10
  • straight B presubscript 5 presuperscript 9
  • Be presubscript 4 presuperscript 11

B.

straight B presubscript 5 presuperscript 10
straight X space plus space straight n presubscript 0 presuperscript 1 space rightwards arrow with space space on top space He presubscript 2 presuperscript 4 space plus space Li presubscript 3 presuperscript 7
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9.

Two spherical conductor B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion, between B and C is

  • F/4

  • 3F/4

  • 3F/8

  • F/8


C.

3F/8

straight F apostrophe space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator left parenthesis straight q divided by 2 right parenthesis left parenthesis 3 straight q divided by 4 right parenthesis over denominator straight d squared end fraction space equals space fraction numerator 3 straight F over denominator 8 end fraction
199 Views

10.

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.
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