MHT CET · Physics · Gravitation
If ' \(R\) ' is the radius of earth \& ' \(g\) ' is acceleration due to gravity on earth's surface, then mean density of earth is
- A \(\frac{4 \pi \mathrm{G}}{3 g R}\)
- B \(\frac{3 \pi R}{4 g G}\)
- C \(\frac{3 \mathrm{~g}}{4 \pi \mathrm{RG}}\)
- D \(\frac{\pi \mathrm{RG}}{12 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \mathrm{~g}}{4 \pi \mathrm{RG}}\)
Step-by-step Solution
Detailed explanation
Mass of the Earth is given by,
\(\mathrm{M}=\mathrm{V} \rho=\frac{4}{3} \pi \mathrm{R}^3 \rho...(i)\)
Acceleration due to gravity is given by,
\(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}...(ii)\)
Substituting equation (i) in equation (ii),
\(g=\frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho \Rightarrow \rho=\frac{3 g}{4 \pi R G}\)
\(\mathrm{M}=\mathrm{V} \rho=\frac{4}{3} \pi \mathrm{R}^3 \rho...(i)\)
Acceleration due to gravity is given by,
\(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}...(ii)\)
Substituting equation (i) in equation (ii),
\(g=\frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho \Rightarrow \rho=\frac{3 g}{4 \pi R G}\)
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