MHT CET · Physics · Gravitation
The magnitude of gravitational field at distance ' \(r_1\) ' and ' \(r_2\) ' from the centre of a uniform sphere of radius ' \(R\) ' and mass ' \(M\) ' are ' \(\mathrm{F}_1\) ' and ' \(\mathrm{F}_2\) ' respectively. The ratio ' \(\left(F_1 / F_2\right)\) ' will be (if \(r_1 \gt R\) and \(r_2 \lt R\))
- A \(\frac{\mathrm{R}^2}{\mathrm{r}_1 \mathrm{r}_2}\)
- B \(\frac{\mathrm{R}^3}{\mathrm{r}_1 \mathrm{r}_2^2}\)
- C \(\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}\)
- D \(\frac{R^4}{r_1^2 r_2^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}\)
Step-by-step Solution
Detailed explanation
For \(r \gt R, F_1=\frac{G M}{r_1^2}\)
\(\begin{aligned} & \text {For } r \lt R, F_2=\frac{\mathrm{GMr}_2}{\mathrm{R}^3} \\ \therefore \quad & \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{\mathrm{GM}}{\mathrm{r}_1^2} \times \frac{\mathrm{R}^3}{\mathrm{GMr}_2}=\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}\end{aligned}\)
\(\begin{aligned} & \text {For } r \lt R, F_2=\frac{\mathrm{GMr}_2}{\mathrm{R}^3} \\ \therefore \quad & \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{\mathrm{GM}}{\mathrm{r}_1^2} \times \frac{\mathrm{R}^3}{\mathrm{GMr}_2}=\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}\end{aligned}\)
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