If ' $R$ ' is the radius of earth \& ' $g$ ' is acceleration due to gravity on earth's surface, then mean…
- $\frac{4 \pi \mathrm{G}}{3 g R}$
- $\frac{3 \pi R}{4 g G}$
- $\frac{3 \mathrm{~g}}{4 \pi \mathrm{RG}}$
- $\frac{\pi \mathrm{RG}}{12 \mathrm{~g}}$
Solution
Substituting equation (i) in equation (ii), $g=\frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho \Rightarrow \rho=\frac{3 g}{4 \pi R G}$ :
Asked in: MHT CET 2024 (04 May Shift 1)