The correct answer is (C) 0.75% Decrease

Given weight =

\( 1.5 = \frac{\frac{GM}{R^{2}} – \frac{GM}{(R + h)^{2}}}{\frac{GM}{R^{2}}} \) \( \Rightarrow \frac{R^{2} + 2Rh + h^{2} – R^{2}}{R + h^{2}} \) \( \Rightarrow \frac{2Rh + h^{2}}{R + h^{2}} \) \( \Rightarrow \frac{1.5}{100} \)

At depth\( {h}’ and {g}’ = g (1 – \frac{h}{R}) \)

when g decreases =\( \frac{g – {g}’}{g} = \frac{h}{R} \) \( \Rightarrow \frac{\frac{2h}{R} + (\frac{h}{R})^{2}}(1 +\frac{h}{R})^{2}} \) \( \Rightarrow \frac{1.5}{100} \) \( \Rightarrow \frac{2h}{R} = \frac{1.5}{100} \) \( \Rightarrow \frac{h}{R} = \frac{0.75}{100} \)

since h < < < R

At depth\( {h}’ \) there will be a decrease of g by 0.75%

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