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(B) The escape velocity $v_e$ from the surface of a planet of mass $M$ and radius $R$ is given by the formula:$v_e = \sqrt{\frac{2GM}{R}}$Assuming the

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$v_e \propto \sqrt{\rho} R$

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(B) The escape velocity $v_e$ from the surface of a planet of mass $M$ and radius $R$ is given by the formula:$v_e = \sqrt{\frac{2GM}{R}}$Assuming the planet is a sphere of uniform density $ho$,its mass $M$ can be expressed in terms of its volume and density:$M = 	ext{Volume} 	imes 	ext{Density} = \left( \frac{4}{3} \pi R^3 ight) ho$Substituting the expression for $M$ into the escape velocity formula:$v_e = \sqrt{\frac{2G}{R} \left( \frac{4}{3} \pi R^3 ho ight)}$Simplifying the expression:$v_e = \sqrt{\frac{8}{3} G \pi R^2 ho}$$v_e = R \sqrt{\frac{8}{3} G \pi ho}$Since $G$,$\pi$,and the constant factor are constants,we have:$v_e \propto R \sqrt{ho}$Therefore,the correct option is $B$.

If the radius of a planet is $R$ and its density is $\rho$,the escape velocity from its surface will be

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