Q87 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
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:
- Correct answerA$\frac{3 V}{2}$
- B
V - C$\frac{4}{3} \mathrm{~V}$
- D$\frac{\mathrm{V}}{2}$
Solution
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, $r=0$
$\therefore \mathrm{V}_0=\frac{3 \mathrm{GM}}{2 \mathrm{R}}=\left(\frac{3 \mathrm{~V}}{2}\right)$
Hence, the correct answer is option: 'A'.