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If $\mathrm{V}$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:
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The correct answer is:
$\frac{3 V}{2}$
Potential due to sphere
$\mathrm{V}=\frac{\mathrm{GM}}{2 \mathrm{R}^3}\left(3 \mathrm{R}^2-\mathrm{r}^2\right)$
At surface $r=R \Rightarrow V=\left(\frac{G M}{R}\right)$
At centre, $r=0$
$\therefore \mathrm{V}_0=\frac{3 \mathrm{GM}}{2 \mathrm{R}}=\left(\frac{3 \mathrm{~V}}{2}\right)$
$\mathrm{V}=\frac{\mathrm{GM}}{2 \mathrm{R}^3}\left(3 \mathrm{R}^2-\mathrm{r}^2\right)$
At surface $r=R \Rightarrow V=\left(\frac{G M}{R}\right)$
At centre, $r=0$
$\therefore \mathrm{V}_0=\frac{3 \mathrm{GM}}{2 \mathrm{R}}=\left(\frac{3 \mathrm{~V}}{2}\right)$
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