1. 3.92 m s ⁻² 2. 19.6 m s ⁻² 3. 9.8 m s ⁻² 4. 4.9 m s ⁻²

Answer: 1. 3.92 m s ⁻² Key idea: g=GM/R² . For the planet M_p=M/10 and R_p=R/2 , so g_p=G(M/10)/(R/2)²=4/10 g=0.4×9.8=3.92\ m s⁻² Option 1.

The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is :

1. 3.92 m s$^{-2}$

2. 19.6 m s$^{-2}$

3. 9.8 m s$^{-2}$

4. 4.9 m s$^{-2}$

Doubtly Asked 30 Oct 2026 NEET UG 2024 Practice test neet Save WhatsApp Report

Best answer

Answer: 1. 3.92 m s$^{-2}$

Key idea: $g=\frac{GM}{R^2}$.

For the planet $M_p=\frac{M}{10}$ and $R_p=\frac R2$, so

$$g_p=\frac{G(M/10)}{(R/2)^2}=\frac{4}{10}\,g=0.4\times9.8=3.92\ \text{m s}^{-2}$$

Option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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