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(A) The acceleration due to gravity $g$ is given by the formula $g = \frac{GM}{R^2}$.Substituting the mass of the earth $M = \frac{4}{3} \pi R^3 \rho$

Vedclass · Accepted Answer

$19.6$

Vedclass · Answer

(A) The acceleration due to gravity $g$ is given by the formula $g = \frac{GM}{R^2}$.Substituting the mass of the earth $M = \frac{4}{3} \pi R^3 ho$,where $ho$ is the density,we get:$g = \frac{G}{R^2} 	imes \frac{4}{3} \pi R^3 ho = \frac{4}{3} \pi G R ho$.Since the radius $R$ is kept constant,we have $g \propto ho$.Therefore,the ratio of the new acceleration $g_2$ to the initial acceleration $g_1$ is $\frac{g_2}{g_1} = \frac{ho_2}{ho_1}$.Given that the density is doubled,$ho_2 = 2ho_1$.Thus,$g_2 = 2 	imes g_1 = 2 	imes 9.8\,m/s^2 = 19.6\,m/s^2$.

If the density of the earth is doubled keeping its radius constant,then acceleration due to gravity will be ........ $m/s^2$. $(g = 9.8\,m/s^2)$

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