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 Multiple Choice QuestionsMultiple Choice Questions

81.

In a decay process XZA changes into YZ-1A. Which process is this?

  • β-decay

  • β+ decay

  • α- decay

  • γ- decay.


82.

Which of the following pairs represent isotones?

  • As3377,Se34 78

  • Pt78195,Os76190

  • Ag47108,Cd48112

  • Hf72178,Ba56137


83.

Which of the following arrangement is possible?

  • n l m s
    5 2 2 +1/2
  • 2 2 0 -1/2
  • 3 -2 1 +1/2
  • 0 0 1 +1/2

84.

If the mass defect of 40X is 0.090 u , then binding energy per nucleon is :

(1 u = 931.5 MeV)

  • 9.315 MeV

  • 931.5 MeV

  • 83.0 MeV

  • 8.38 MeV


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85.

27Co60 is radioactive because :

  • its atomic number is high

  • it has high pn ratio

  • it has high np ratio

  • None of the above


86.

Assertion : For Balmer series of hydrogen spectrum, the value n1 = 2 and n2 = 3, 4, 5... .

Reason : The value of n2 for a line in Balmer series of hydrogen spectrum having the highest wavelength is 6.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


87.

In nuclear explosion, the energy is released in the form of :

  • kinetic energy

  • electrical energy

  • potential energy

  • None of these


88.

The number of atoms in 52 g of He is

  • 78.299 x 1024atoms

  • 7.820 x 10-24atoms

  • 7.829 x 1024 atoms

  • 78.234 x 1025 atoms


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89.

Threshold frequency of a metal is 5 × 1013 s-1 upon which 1 x 1014s-1 frequency light is focused. Then the maximum kinetic energy of emitted electron is

  • 3.3 × 10-21

  • 3.3 × 10-20

  • 6.6 × 10-23

  • 6.6 × 10-20


B.

3.3 × 10-20

Following the conservation of energy principle,

Kinetic energy 12mev2 = h(v-v)

                    =(6.626 × 10-34) (1014s-1 - 5 × 1013 s-1)

                    =(6.626 x 10-34J s) (5 x 1013 s-1)             

                    =3.313 x 10-20J.


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90.

From elementary molecular orbital theory we can deduce the electronic configuration of the singly positive nitrogen molecular ion as :

  • σ1s2σ*1s2σ2s2σ*2s2π2p4σ2p1

  • σ1s2σ*1s2σ2s2σ*2s2σ2p2π2p3

  • σ1s2σ*1s2σ2s2σ*2s2σ2p3 , π2p2

  • σ1s2σ*1s2σ2s2σ*2s2σ2p2π2p4


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