Q72 / 90 · KVPY 2010 (SB/SX) · Physics
A certain planet completes one rotation about its axis in time $T$. The weight of an object placed at the equator on the planet's surface is a fraction $f\left(f\right.$ is close to unity) of its weight recorded at a latitude of $60^{\circ}$. The density of the planet (assumed to be a uniform perfect sphere is given by-
- Correct answerA$\frac{4-f}{1-f} \frac{3 \pi}{4 \mathrm{GT}^{2}}$
- B$\frac{4-f}{1+f} \frac{3 \pi}{4 G T^{2}}$
- C$\frac{4-3 f}{1-f} \frac{3 \pi}{4 G T^{2}}$
- D$\frac{4-2 f}{1-f} \frac{3 \pi}{4 \mathrm{GT}^{2}}$
Solution
$\quad \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ also $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$ or $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{r}^{3}}{\mathrm{GM}}}$ when $\mathrm{v}$ is replaced by $\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ now $g_{e f f}=g-\omega^{2} R_{e} \cos ^{2} \phi$
now $f=\frac{g-\omega^{2} R_{e} \cos 0^{\circ}}{g-\omega^{2} R_{e} \cos 60^{\circ}}$
ving we get $\mathrm{R}_{\mathrm{e}}=\frac{4 g(\mathrm{f}-1)}{\omega^{2}(\mathrm{f}-4)}$ or $\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^{2}}(\mathrm{f}-1)=\frac{\omega^{2} \mathrm{R}_{\mathrm{e}}}{4}(\mathrm{f}-4)$
$$
\mathrm{R}_{\mathrm{e}}{ }^{3}=\frac{4 \mathrm{GM}}{\omega^{2}} \frac{(\mathrm{f}-1)}{(\mathrm{f}-4)}
$$
now $\rho=\frac{M}{\frac{4}{3} \pi R_{e}^{3}}=\frac{3}{16} \frac{\omega^{2}}{\pi G} \frac{(f-4)}{(f-1)}$
also $\mathrm{T}=\frac{2 \pi}{\omega} \quad \therefore \mathrm{p}=\frac{3 \pi(\mathrm{f}-4)}{4 \mathrm{~T}^{2} \mathrm{G}(\mathrm{f}-1)}$
Hence, the correct answer is option: 'A'.