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Physics MCQs for NEET — Practice Questions with Answers

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Atomic mass number of an element thorium is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of alpha and beta particles emitted is

$α$ =3, $β$ = 3

$α$ = 6, $β$ = 4

$α$ =6, $β$ =0

$α$ =4, $β$ =6

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Explanation

(b) $n_{α} = \frac{A - A'}{4} = \frac{232 - 208}{4} = 6$

and $n_{β} = (2 n_{α} - Z + Z') = (2 \times 6 - 90 + 82) = 4$

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A sample contains 16 gm of a radioactive material, the half life of which is two days. After 32 days, the amount of radioactive material left in the sample is

Less than 1 mg

$\frac{1}{4} gm$

$\frac{1}{2} gm$

1 gm

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Explanation

(a) Remaining amount

$= 16 \times {(\frac{1}{2})}^{32 / 2} = 16 \times {(\frac{1}{2})}^{16} = {(\frac{1}{2})}^{12} < 1 mg$

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Half life of a radioactive substance is T. The time taken for all the nuclei to disintegrate will be

$T^{2}$

Uncertain

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Explanation

(d) Uncertain, because it is infinite. No radioactive element can be disintegrated fully.

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The half life of polonium is 140 days. After how many days, 16 gm polonium will be reduced to 1 gm (or 15g will decay)

700 days

280 days

560 days

420 days

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Explanation

$\Rightarrow \frac{t}{140} = 4 \Rightarrow t = 560 days$

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A radioactive element emits 200 particles per second. After three hours 25 particles per second are emitted. The half life period of element will be

50 minutes

60 minutes

70 minutes

80 minutes

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Explanation

(b) $R = \frac{dN}{dt} \propto N \Rightarrow \frac{R_{2}}{R_{1}} = \frac{N_{2}}{N_{1}}$

But $\frac{N_{2}}{N_{1}} = {(\frac{1}{2})}^{t_{1 / 2}} \Rightarrow \frac{25}{200} = \frac{1}{8} = {(\frac{1}{2})}^{3} \Rightarrow \frac{t}{t_{1 / 2}} = 3$

$∴ t_{1 / 2} = \frac{t}{3} = \frac{3}{3} = 1 hour = 60 minutes$

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The half life of the isotope ${}_{11}^{24}{Na}$ is 15 hrs. How much time does it take for $\frac{7}{8} th$ of a sample of this isotope to decay

75 hrs

65 hrs

55 hrs

45 hrs

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Explanation

(d) Undecayed isotope $= 1 - \frac{7}{8} = \frac{1}{8}$

$∴ \frac{N}{N_{0}} = {(\frac{1}{2})}^{t / T} \Rightarrow (\frac{1}{8}) = {(\frac{1}{2})}^{t / 15} \Rightarrow \frac{t}{15} = 3$

or t = 45 hours

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If the half life of a radioactive sample is 10 hours,
its mean life is

14.4 hours

7.2 hours

20 hours

6.93 hours

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Explanation

(a) $Mean life = \frac{Half life}{0.6931} = \frac{10}{0.6931} = 14.4 hours$

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If 20 gm of a radioactive substance due to radioactive decay reduces to 10 gm in 4 minutes, then in what time 80 gm of the same substance will reduce to 10 gm

In 8 minutes

In 12 minutes

In 16 minutes

In 20 minutes

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Explanation

(b) 20 gm substance reduces to 10 gm (i.e. becomes half in 4 min. So $T_{1 / 2}$ = 4 min . Again

$M = M_{0} {(\frac{1}{2})}^{t / T_{1 / 2}}$

$\Rightarrow 10 = 80 {(\frac{1}{2})}^{t / 4} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{t / 4} \Rightarrow$ t = 12 min.

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16 gm sample of a radioactive element is taken from Bombay to Delhi in 2 hour and it was found that 1 gm of the element remained (undisintegrated). Half life of the element is

2 hour

1 hour

$\frac{1}{2}$ hour

$\frac{1}{4} hour$

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Explanation

(c) $N = N_{0} {(\frac{1}{2})}^{t / T_{1 / 2}} \Rightarrow 1 = 16 {(\frac{1}{2})}^{\frac{2}{T_{1 / 2}}} \Rightarrow T_{1 / 2} = \frac{1}{2} hour$

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Unit of radioactivity is Rutherford. Its value is

$3.7 \times 10^{10}$ disintegrations/sec

$3.7 \times 10^{6}$ disintegrations/sec

$1.0 \times 10^{10}$ disintegrations/sec

$1.0 \times 10^{6}$ disintegrations/sec

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Explanation

(d)

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.