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Physics MCQs for NEET — Practice Questions with Answers

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A neutron with velocity V strikes a stationary deuterium atom. Its kinetic energy changes by a factor of

$\frac{15}{16}$

$\frac{1}{2}$

$\frac{2}{1}$

None of these

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The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is 1.4 $kilowatt / m^{2}$ .The average earth- sun distance is $1.5 \times 10^{11}$ metres. The mass lost by the sun per day is
(1 day = 86400 seconds)

$4.4 \times 10^{9} kg$

$7.6 \times 10^{14} kg$

$3.8 \times 10^{12} kg$

$3.8 \times 10^{14} kg$

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Explanation

(d) Energy radiated = 1.4 $kW / m^{2}$

$= 1.4 kJ / \sec m^{2} = \frac{1.4 kJ}{\frac{1}{86400} day m^{2}} = \frac{1.4 \times 86400}{day m^{2}}$

Total energy radiated/day

$= \frac{4 π \times {(1.5 \times 10^{11})}^{2} \times 1.4 \times 86400}{1} \frac{kJ}{day} = E$

$∴ E = {mc}^{2} \Rightarrow m = \frac{E}{c^{2}}$

$= \frac{4 π \times {(1.5 \times 10^{11})}^{2} \times 1.4 \times 86400}{{(3 \times 10^{8})}^{2}} = 3.8 \times 10^{14} kg$

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The binding energy per nucleon of $O^{16}$ is 7.97 MeV and that of $O^{17}$ is 7.75 MeV. The energy (in MeV) required to remove a neutron from $O^{17}$ is

3.52

3.64

4.23

7.86

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Explanation

(c) The equation is $O^{17} \to {}_{0}^{1}n + O^{16}$
$∴$ Energy required = B.E. of $O^{17}$ – B.E. of $O^{16}$
= 17 $\times$ 7.75 – 16 $\times$ 7.97 = 4.23 MeV

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The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5 c. The change in its energy will be

0.026 MeV

0.051 MeV

0.079 MeV

0.105 MeV

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Explanation

$= m_{0} c^{2} (\frac{1}{\sqrt{1 - (v^{2} / c^{2})}} - 1) = 0.511 (\frac{1}{\sqrt{0.75}} - 1)$

= 0.079 MeV

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For uranium nucleus how does its mass vary with volume

$m \propto V$

$m \propto 1 / V$

$m \propto \sqrt{V}$

$m \propto V^{2}$

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Explanation

(a) Since nuclear density is constant hence mass ∝ volume.

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The rest mass of an electron as well as that of positron is 0.51 MeV. When an electron and positron are annihilated, they produce gamma-rays of wavelength(s)-

0.012 Å

0.024 Å

0.012 Å to $\infty$

0.024 Å to $\infty$

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Explanation

(a) Since electron and positron combine-
$λ = \frac{hc}{E_{Total}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{(0.51 + 0.51) \times 10^{6} \times 1.6 \times 10^{- 19}} = 1.21 \times 10^{- 12} m = 0.012 \overset{0}{A}$

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In the nuclear fusion reaction ${}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}{He} + n$ given that the repulsive potential energy between the two nuclei is $- 7.7 \times 10^{- 14} J$ , the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s constant $k = 1.38 \times 10^{- 23} J / K$ )

$10^{9} K$

$10^{7} K$

$10^{5} K$

$10^{3} K$

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Explanation

(a) Kinetic energy of the molecules of a gas at a temp. T is $\frac{3}{2} kT$
$∴$ To initiate the reaction $\frac{3}{2} kT = 7.7 \times 10^{- 14} J$

$\Rightarrow \frac{3}{2} \times 1.38 \times 10^{- 23} T = 7.7 \times 10^{- 14} \Rightarrow T = 3.7 \times 10^{9} K$
.

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A nucleus with mass number 220 initially at rest emits an $α$ -particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the $α$ -particle

4.4 MeV

5.4 MeV

5.6 MeV

6.5 MeV

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Explanation

The Q value of a nuclear reaction is the difference between the sum of the masses of the reactants and the sum of the masses of the products. In this case, the Q value is positive, indicating an exothermic reaction. The kinetic energy of the emitted alpha particle is equal to the Q value.

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The half life of radioactive Radon is 3.8 days. The time at the end of which 1/20 th of the Radon sample will remain undecayed is
(Given $\log_{10} e = 0.4343$ )

3.8 days

16.5 days

33 days

76 days

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Explanation

(b) By the formula $N = N_{0} e^{- λt}$
Given $\frac{N}{N_{0}} = \frac{1}{20}$ and $λ = \frac{0.6931}{3.8} \Rightarrow 20 = e^{\frac{0.6931 \times t}{3.8}}$
Taking log of both sides
or $\log 20 = \frac{0.6931 \times t}{3.8} \log_{10} e$
or $1.3010 = \frac{0.6931 \times t \times 0.4343}{3.8} \Rightarrow t = 16.5 days$

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If 10% of a radioactive material decays in 5 days, then the amount of original material left after 20 days is approximately

60%

65%

70%

75%

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Explanation

(b) $N = N_{0} e^{- λt}$

$∴ 0.9 N_{0} = N_{0} e^{- λ \times 5} \Rightarrow 5 λ = \log_{e} \frac{1}{0.9}$ ....(i)

and ${xN}_{0} = N_{0} e^{- λ \times 20} \Rightarrow 20 λ = \log_{e} (\frac{1}{x})$ ....(ii)

Dividing (i) and (ii) , we get

$\frac{1}{4} = \frac{\log_{e} (1 / 0.9)}{\log_{e} (1 / x)} = \frac{\log_{10} (1 / 0.9)}{\log_{10} (1 / x)} = \frac{\log_{10} 0.9}{\log_{10} x} \Rightarrow \log_{10} x = 4 \log_{10} 0.9 \Rightarrow x = 0.658 = 65.8 %$

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