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Physics MCQs for NEET — Practice Questions with Answers

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A chain reaction is continuous due to

Large mass defect

Large energy

Production of more neutrons in fission

None of these

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Explanation

3. Production of more neutrons in fission

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Density (D) of nucleus related to mass no. (A) as

$D \propto A$

$D \propto A^{2}$

$D \propto A^{0}$

$D \propto A^{- 1}$

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Explanation

$(3) Density = \frac{mass}{volume} = \frac{A M_{p}}{\frac{4}{3} {πR}_{0}^{3} A} [M_{p} : M a s s o f a n u c l e o n] S o d e n s i t y i s i n d e p e n d e n t o f A$

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Heavier nuclei are unstable. The correct statement is :

binding energy per nucleon is small

get stability by nuclear fission

get stability by nuclear fusion

both (1) and (2)

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Explanation

Fon heavier nuclei, binding energy per nucleon is minimum.

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$binding energy per nucleon for deuteron ({}_{1}H_{2}) and helium ({}_{2}{He}_{4}) are 1.1 MeV and 7.0 MeV respectively . Energy released when two deuteron fuse to form a He nucleus is$

47.12 MeV

34.4 MeV

11.8 MeV

23.6 MeV

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Explanation

$(4) {}_{1}H_{2} + {}_{1}H_{2} \to {}_{2}{He}_{2} B . E . of reactant = 2 x 1.1 x 2 = 4.4 MeV B . E . of product = 7.0 x 4 = 28.0 MeV Energy released = 28.0 - 4.4 = 23.6 MeV$

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The initial activity of a certain radioactive isotope was measured as 16000 counts per minute. That the only activity measured was due to this and that its activity after 12 hours was 2000 counts/minute. Its half-life, in hours, is nearest to-

9.0

6.0

4.0

3.0

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Explanation

$(3) 2000 = 16000 {(\frac{1}{2})}^{n} n = 3 T_{H} = \frac{12}{3} = 4$

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A radioactivity nuclide can decay simultaneously by two different processes that have decay constants $λ_{1}$ and $λ_{2}$ . The effective decay constant of the nuclide is $λ .$ Then-

$λ = λ_{1} + λ_{2}$

$λ = \frac{1}{2} (λ_{1} + λ_{2})$

$\frac{1}{λ} = \frac{1}{λ_{1}} + \frac{1}{λ_{2}}$

$λ = \sqrt{λ_{1} v_{2}}$

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Explanation

Conceptual

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Which of the following is deflected by electric field

X-rays

Gamma rays

Neutrons

α-particles

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Explanation

α-particles are charged particles, so they can deflect by electric field.

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Radioactive material A has decay constant 8 $λ$ and material B has decay constant $λ$ Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{e}$ ?

$\frac{1}{λ}$

$\frac{1}{7 λ}$

$\frac{1}{8 λ}$

$\frac{1}{9 λ}$

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Explanation

(b) Let initial number of nuclei in A and B is $N_{O}$

Number of nuclei of A after time t is

$N_{A} = N_{0 e^{- 8 λ t}}$ ...(i)

Similarly, number of nuclei of A after time t is

$N_{B} = N_{0} e^{- λ t}$ ...(ii)

It is given that

$\frac{N_{A}}{N_{B}} = \frac{1}{e} |∴ N_{B} > N_{A}|$

Now,from Eqs.(i) and (ii)

$\frac{e^{- B λ t}}{e^{- λ t}} = \frac{1}{e}$

Rearranging

$\Rightarrow e^{- 1} = e^{- 7 λ t} \Rightarrow 7 λ t = 1$

$\Rightarrow T i m e (t) = \frac{1}{7 λ}$

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The half-life of a radioactive substance is 30 minutes. The time (in minute) taken between 40% decay and 85% decay of the same radioactive substance is

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Explanation

(d) Key idea Half-life of a radioactive substance is $T_{\frac{1}{2}}$ $\propto$ log $(\frac{N_{o}}{N})$

Given, $N_{1} = 0.6 N_{0} (∵ 40 % d e c a y) N_{2} = 0.15 N_{0} (∵ 85 % d e c a y)$

Putting these in the formula

$\frac{N_{2}}{N_{1}} = \frac{0.15 N_{0}}{0.6 N_{0}} = \frac{1}{4} = {(\frac{1}{2})}^{2}$

So, two half-life periods has passed.

Thus, time taken=2 $\times t_{1 / 2} = 2 \times 30 = 60$ min

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$I f r a d i u s o f t h e_{13} A l^{27} n u c l e u s i s t a k e n t o b e R_{A l} t h e n t h e r a d i u s o f_{53} T e^{125} n u c l e u s i s n e a r l y -$

(53/13)^1/3 R_Al

5/3 R_Al

3/5 R_Al

(13/53)^1/3R_Al

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Explanation

(b)

$R a d i u s o f n u c l e u s i s g i v e n b y - R = R_{0} A^{1 / 3} \Rightarrow R α A^{1 / 3} \frac{R_{A l}}{R_{T e}} = {(\frac{27}{125})}^{1 / 3} = \frac{3}{5} \Rightarrow R_{T e} = \frac{5}{3} R_{A l}$

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