Q157 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Physics
The radius and mass number of nucleus ' 1 ' is $R_1$ and $A_1$ respectively. The radius and mass number of nucleus ' 2 ' is $R_2$ and $A_2$ respectively. If $A_2$ is larger than $A_1$ by $2 \%$, then $R_2$ is larger than $R_1$ by
- Correct answerA$\frac{2}{3} \%$
- B$1 \%$
- C$8 \%$
- D$\frac{3}{2} \%$
Solution
For nucleus 1
$\mathrm{R}_1=\mathrm{R}_0 \mathrm{~A}_1^{1 / 3}$
For nucleus 2
$\mathrm{R}_2=\mathrm{R}_0 \mathrm{~A}_2^{1 / 3}$
Divide equation (ii) by (i)
$\begin{array}{ll}<br />\frac{\mathrm{R}_2}{\mathrm{R}_1}=\left(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right)^{1 / 3} & \left(\because \frac{\mathrm{A}_2}{\mathrm{~A}_1}=2 \%\right) \\<br />\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{1}{3} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}=\frac{2}{3} &<br />\end{array}$
$\mathrm{R}_1=\mathrm{R}_0 \mathrm{~A}_1^{1 / 3}$
For nucleus 2
$\mathrm{R}_2=\mathrm{R}_0 \mathrm{~A}_2^{1 / 3}$
Divide equation (ii) by (i)
$\begin{array}{ll}<br />\frac{\mathrm{R}_2}{\mathrm{R}_1}=\left(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right)^{1 / 3} & \left(\because \frac{\mathrm{A}_2}{\mathrm{~A}_1}=2 \%\right) \\<br />\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{1}{3} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}=\frac{2}{3} &<br />\end{array}$
Hence, the correct answer is option: 'A'.