Q115 / 125 · BITSAT 2015 · Physics
Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$, respectively. The ratio of masses of $\mathrm{X}$ and $\mathrm{Y}$ is
- A$\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{1 / 2}$
- B$\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)$
- Correct answerC$\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}$
- D$\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)$
Solution
When a charge particle is allowed to move in a uniform magnetic field, then it describes spiral or circular path
$$
\begin{array}{l}
\text { Centripetal force, } \frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB} \quad \therefore \\
\mathrm{v}=\left(\frac{\mathrm{qB}}{\mathrm{m}} \mathrm{R}\right) \\
\text { Hence, } \sqrt{\frac{2 \mathrm{q} \mathrm{V}}{\mathrm{m}}}=\left(\frac{\mathrm{qB}}{\mathrm{m}}\right) \mathrm{R}\left[\because \mathrm{V}=\sqrt{\frac{2 \mathrm{qV}}{\mathrm{m}}}\right] \\
\Rightarrow \mathrm{R}=\left(\frac{2 \mathrm{mV}}{\mathrm{q}}\right)^{1 / 2} \times \frac{1}{\mathrm{~B}} \\
\text { or, } \mathrm{m} \propto \mathrm{R}^{2}
\end{array}
$$
$[\because \mathrm{V}, \mathrm{q}$ and $\mathrm{B}$ are constant $]$
$$
\text { or, } \frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}
$$
Hence, the correct answer is option: 'C'.