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JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

Option 4 : 9 : 8

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

From question, we need to find the decayed numbers of nuclei A and B.

The number of radioactive nuclei is given by the formula:

\(N = {N_0}{\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}}\)

Where,

N_{0} = Number of nuclei at t = 0 s

t = Time

T_{1}_{/2} = Half-life of the nuclei

The number of nuclei of nuclei A is given as:

\({N_A} = {N_0}{\left( {\frac{1}{2}} \right)^{{t_A}/{T_{\left( {1/2} \right)A}}}}\)

Where,

t_{A} = Time = 60 minutes (given)

T_{(1/2)A} = Half-life of A = 10 minutes (given)

The number of nuclei of nuclei B is given as:

\({N_B} = {N_0}{\left( {\frac{1}{2}} \right)^{{t_B}/{T_{\left( {1/2} \right)B}}}}\)

Where,

t_{B} = Time = 60 minutes (given)

T_{(1/2)B} = Half-life of B = 20 minutes (given)

**Calculation:**

On substituting the values,

\( \Rightarrow {N_A} = {N_0}{\left( {\frac{1}{2}} \right)^{60/10}}\)

\(\therefore {N_A} = {N_0}{\left( {\frac{1}{2}} \right)^6}\)

\( \Rightarrow {N_B} = {N_0}{\left( {\frac{1}{2}} \right)^{60/20}}\)

\(\therefore {N_B} = {N_0}{\left( {\frac{1}{2}} \right)^3}\)

Number of decayed A nuclei:

\( \Rightarrow {N_A} = {{\rm{N}}_0} - \frac{{{N_0}}}{{{2^6}}}\)

\( \Rightarrow {N_A} = {{\rm{N}}_0} - \frac{{{N_0}}}{{64}}\)

\( \Rightarrow {N_A} = \frac{{64{N_0} - {N_0}}}{{64}}\)

\(\therefore {N_A} = \frac{{63{{\rm{N}}_0}}}{{64}}\)

Number of decayed B nuclei:

\( \Rightarrow {N_B} = {{\rm{N}}_0} - \frac{{{N_0}}}{{{2^3}}}\)

\( \Rightarrow {N_B} = {{\rm{N}}_0} - \frac{{{N_0}}}{8}\)

\( \Rightarrow {N_B} = \frac{{8{N_0} - {N_0}}}{8}\)

\(\therefore {N_B} = \frac{{7{{\rm{N}}_0}}}{8}\)

Ratio of the number of decayed nuclei is:

\( \Rightarrow \frac{{{N_A}}}{{{N_B}}} = \frac{{\left( {\frac{{63{{\rm{N}}_0}}}{{64}}} \right)}}{{\left( {\frac{{7{{\rm{N}}_0}}}{8}} \right)}}\)

\( \Rightarrow \frac{{{N_A}}}{{{N_B}}} = \frac{{63{{\rm{N}}_0}}}{{64}} \times \frac{8}{{7{N_0}}}\)

\(\therefore \frac{{{N_A}}}{{{N_B}}} = \frac{9}{8}\)