Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

The half-life of a radioactive substance is 40 years. How long will it take to reduce to one fourth of its original amount and what is the value of decay constant

40 year, 0.9173/year

90 year, 9.017/year

80 year, 0.0173 year

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) To reduce one fourth it takes time $t = 2 (T_{1 / 2})$ = 2 $\times$ 40
= 80 years.
Decay constant $λ = \frac{0.693}{T_{1 / 2}} = \frac{0.693}{40} = 0.0173 years$

● ● ● ● ●

A radioactive material has a half life of 10 days. What fraction of the material would remain after 30 days

0.5

0.25

0.125

0.33

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) $N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow \frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{30}{10}} = \frac{1}{8} = 0.125$

● ● ● ● ●

Starting with a sample of pure ${}_{66}{Cu}$ , $\frac{7}{8}$ of it decays into Zn in 15 min. The corresponding half-life is

5 min

$7 \frac{1}{2}$ min

10 min

15 min

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) $\frac{7}{8}$ part decays i.e. remaining part is $\frac{1}{8}$

$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{\frac{15}{T_{1 / 2}}} \Rightarrow T_{1 / 2} = 5 \min$

● ● ● ● ●

A radioactive material has a half-life of 8 years. The activity of the material will decrease to about 1/8 of its original value in

256 year

128 year

64 year

24 year

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $\frac{A}{A_{0}} = {(\frac{1}{2})}^{t / T_{1 / 2}} \Rightarrow \frac{1}{8} = {(\frac{1}{2})}^{t / 8} \Rightarrow t = 24 years$

● ● ● ● ●

A nucleus of mass 218 amu in free state decays to emit an $α$ -particle. Kinetic energy of the $α$ -particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is

1.0

0.5

0.25

0.125

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) Using conservation of momentum $P_{daughter} = P_{α}$
$\Rightarrow \frac{E_{d}}{E_{α}} = \frac{m_{α}}{m_{d}} \Rightarrow E_{d} = \frac{E_{α} \times m_{α}}{m_{d}} = \frac{6.7 \times 4}{214} = 0.125 MeV$

● ● ● ● ●

$C^{14}$ has half life 5700 years. At the end of 11400 years, the actual amount left is

0.5 of original amount

0.25 of original amount

0.125 of original amount

0.0625 of original amount

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $N = N_{0} \times {(\frac{1}{2})}^{11400 / 5700} = N_{0} {(\frac{1}{2})}^{2} = 0.25 N_{0}$

● ● ● ● ●

Mean life of a radioactive sample is 100 seconds. Then its half life (in minutes) is

0.693

$10^{- 4}$

1.155

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) Mean life (T) = 1/ $λ$ = 100 second
Half-life = $\frac{0.693}{λ} = \frac{0.693 \times 100}{60} = 1.155 \min$

● ● ● ● ●

${}_{86}^{222}A \to {}_{84}^{210}B$ . In this reaction how many $α$ and $β$ particles are emitted

6 $α$ , 3 $β$

3 $α$ , 4 $β$

4 $α$ , 3 $β$

3 $α$ , 6 $β$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) By using $n_{α} = \frac{A - A'}{4}$ and $n_{β} = 2 n_{α} - Z + Z'$

● ● ● ● ●

The phenomenon of radioactivity is

Exothermic change which increases or decreases with temperature

Increases on applied pressure

Nuclear process does not depend on external factors

None of the above

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

● ● ● ● ●

If half life of radium is 77 days. Its decay constant in day will be

$3 \times 10^{- 13}$ /day

$9 \times 10^{- 3}$ /day

$1 \times 10^{- 3}$ /day

$6 \times 10^{- 3}$ /day

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $λ = \frac{0.693}{T_{1 / 2}} = \frac{0.693}{77} = 9 \times 10^{- 3} / day$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET 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.