BITSAT 2023Paper 2
- Exam
- BITSAT
- Year
- 2023
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Gravitation
BITSAT 2023 · Paper 2 · Physics · Gravitation
If V is the gravitational potential due to sphere of uniform...
If 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:
Options
- A
- B
V - C
- D
Correct answer
A.
Step-by-step 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=0\therefore \mmathrm{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=0\therefore \mmathrm{V}_0=\frac{3 \mathrm{GM}}{2 \mathrm{R}}=\left(\frac{3 \mathrm{~V}}{2}\right)