v_e and v_p denote the escape velocity from t | vedclass

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(B) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} = ho 	imes \frac{4}{3}

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$v_e = v_p/2$

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(B) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} = ho 	imes \frac{4}{3}\pi R^3$,we substitute $M$ into the formula:$v = \sqrt{\frac{2G(ho \cdot \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G ho R^2} = R \sqrt{\frac{8}{3}\pi G ho}$.Given that the mean density $ho$ is constant,we find that $v \propto R$.For the Earth $(e)$ and the planet $(p)$,we have $\frac{v_e}{v_p} = \frac{R_e}{R_p}$.Given $R_p = 2R_e$,we get $\frac{v_e}{v_p} = \frac{R_e}{2R_e} = \frac{1}{2}$.Therefore,$v_e = \frac{v_p}{2}$.

Column-$I$	Column-$II$
$(1)$ Value of escape velocity on the surface of the Earth	$(a)$ $2.38 \, km \, s^{-1}$
$(2)$ Value of escape velocity on the surface of the Moon	$(b)$ $7.92 \, km \, s^{-1}$
	$(c)$ $11.2 \, km \, s^{-1}$

$v_e$ and $v_p$ denote the escape velocity from the Earth and another planet having twice the radius and the same mean density as the Earth. Then:

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Value of escape velocity on the surface of the Earth $(a)$ $2.38 \, km \, s^{-1}$

$(2)$ Value of escape velocity on the surface of the Moon $(b)$ $7.92 \, km \, s^{-1}$

$(c)$ $11.2 \, km \, s^{-1}$

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