If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ' \(\mathrm{g}\) ' on the earth is \(9.8 \mathrm{~ms}^{-2}\), then the value of \(\mathrm{g}^{\prime}\) on that planet is $=\ldots \ldots \ldots \mathrm{ms}^{-2}$ (A) \(4.9\) (B) \(0.98\) (C) \(0.49\) (D) 49

We are given: 1. Mass of Earth is 80 times the mass of the planet: \(M_{Earth} = 80 M_{Planet}\) 2. Diameter of Earth is 2 times the diameter of the planet: \(D_{Earth} = 2D_{Planet}\) 3. g on Earth = \(9.8 \mathrm{~ms}^{-2}\)

The formula for gravitational force is: \(F = \cfrac{GMm}{r^{2}}\) Where: F is the gravitational force between two masses, G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm^{2}kg^{-2}}\)), M and m are the two masses, r is the distance between their centers. The force acting on an object due to gravity is the gravitational force, so using F = ma, \(ma = \cfrac{GMm}{r^{2}}\) Here, a is the acceleration due to gravity (g), so: \(g = \cfrac{GM}{r^{2}}\) The value of g depends on the mass of the planet and the distance from its center (which is half of the diameter, i.e., its radius).

Let g' be the acceleration due to gravity on the planet. Using the given data, we can rewrite the relationship found earlier: \(g' = \cfrac{G(80 M_{Planet})}{(2r_{planet})^{2}}\) Simplify the equation: \(g' = \cfrac{80 GM_{Planet}}{4r_{planet}^{2}}\) \(g' = \cfrac{20 GM_{Planet}}{r_{planet}^{2}}\) Now, we have the relationship \(g_{Earth} = \cfrac{GM_{Earth}}{r_{Earth}^{2}}\) and since \(g_{Earth} = 9.8 \mathrm{~ms}^{-2}\), we can plug the values. \(9.8 = \cfrac{G(80 M_{Planet})}{r_{Earth}^{2}}\) Dividing the two equations: \(\cfrac{g'}{9.8} = \cfrac{\cfrac{20 GM_{Planet}}{r_{planet}^{2}}}{\cfrac{80 GM_{Planet}}{r_{Earth}^{2}}}\) Cancel terms: \(\cfrac{g'}{9.8} = \cfrac{20}{80}\cfrac{r_{Earth}^{2}}{r_{planet}^{2}}\) Given relationship: \(D_{Earth} = 2D_{Planet}\), so \(r_{Earth} = 2 r_{Planet}\) \(\cfrac{g'}{9.8} = \cfrac{20}{80}\left(\cfrac{2r_{planet}}{r_{planet}}\right)^{2}\) \(\cfrac{g'}{9.8} = \cfrac{20}{80} (2^{2})\) \(\cfrac{g'}{9.8} = \cfrac{1}{2}\) Finally, solve for g': \(g' = 9.8 \times \cfrac{1}{2}\) \(g' = 4.9 \mathrm{~ms^{-2}}\)

The calculated value of g' on the planet is \(4.9 \mathrm{~ms}^{-2}\), which matches option (A). Therefore, the correct choice is (A) \(4.9 \mathrm{~ms}^{-2}\).