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### Answers (1)

Given

\begin{aligned} & \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2} \\ & \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\frac{1}{3} \\ & \frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}=? \end{aligned}

As we know,

$\mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$

Hence,

$\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\sqrt{\frac{2 \mathrm{GM}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{A}}}}}{\sqrt{\frac{2 \mathrm{GM}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{A}} \mathrm{R}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \mathrm{R}_{\mathrm{A}}}}=\frac{1}{2}$ _____________(i)

Given

$\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}:$__________(ii)

Therefore,

\begin{aligned} \frac{g_{\mathrm{A}}}{g_{\mathrm{B}}} & =\frac{\mathrm{M}_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^2}{\mathrm{M}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}^2} \\ & =\frac{1}{4} \times \frac{1}{3} \times 9 \\ & =\frac{3}{4} \end{aligned}

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