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(A) The acceleration due to gravity $g$ is given by $g = G \cdot \frac{4}{3} \pi R \rho$. Given $\rho' = \frac{2}{3} \rho$ and $R' = \frac{1}{4} R$,th

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$3\, m, 6 : 1$

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(A) The acceleration due to gravity $g$ is given by $g = G \cdot \frac{4}{3} \pi R ho$. Given $ho' = \frac{2}{3} ho$ and $R' = \frac{1}{4} R$,the acceleration due to gravity on the moon $g'$ is:$g' = G \cdot \frac{4}{3} \pi R' ho' = G \cdot \frac{4}{3} \pi \left(\frac{R}{4}ight) \left(\frac{2}{3} hoight) = \frac{1}{6} g$.$(I)$ Maximum height $h_{\max} = \frac{u^2}{2g}$. Since the initial velocity $u$ is the same,$h_{\max} \propto \frac{1}{g}$.Therefore,$h' g' = h g \implies h' = h \left(\frac{g}{g'}ight) = 0.5 	imes 6 = 3\, m$.$(II)$ The time of flight $t = \frac{2u}{g}$. Since $u$ is constant,$t \propto \frac{1}{g}$.Therefore,$\frac{t'}{t} = \frac{g}{g'} = \frac{g}{g/6} = 6$.Thus,the ratio of time on the moon to the earth is $6 : 1$.

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