QNEET-2024-Ph-75: NEET 2024 Physics - Class 11 NEET-2024-Ph-75 Solution: The mass of a planet is \( {1}{10}\)th that of the earth and…

The acceleration due to gravity (g) on a planet is given by the formula: g = G M R 2 where: G $G$ G is the universal gravitational constant, M $M$ M is the mass of the planet, and R $R$ R is the radius of the planet. In this question, we know that: The mass of the planet M p $M_p$ M p _ is 1 10 $\frac{1}{10}$ 10 1 _ of the mass of the earth M e $M_e$ M e _ , so M p = M e 10 $M_p = \frac{M_e}{10}$ M p _ = 10 M e _ _ . The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius R p $R_p$ R p _ is also half the radius of the earth R e $R_e$ R e _ . Therefore, R p = R e 2 $R_p = \frac{R_e}{2}$ R p _ = 2 R e _ _ . Using the formula for acceleration due to gravity and substituting the above information: g p = G M p R p 2 = G M e 10 ( R e 2 ) 2 Simplify the expression: g p = G M e 10 _ 4 R e 2 = 4 10 G M e R e 2 Knowing that G M e R e 2 $G \frac{M_e}{R_e^2}$ G R e 2 _ M e _ _ is the acceleration due to gravity on Earth g e � 9.8 m / s 2 $g_e \approx 9.8 \, \mathrm{m/s}^2$ g e _ � 9.8 m/s 2 , we substitute this value into the equation: g p = 4 10 _ 9.8 m / s 2 = 3.92 m / s 2 Thus, the acceleration due to gravity on the planet is 3.92 m / s 2 $3.92 \, \mathrm{m/s}^2$ 3.92 m/s 2 , which corresponds to: Option D: 3.92   m / s 2