If the density of the Earth increases 4 times | vedclass

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(B) The acceleration due to gravity $g$ on the surface of the Earth is given by $g = \frac{GM}{R^2}$.Since mass $M = 	ext{density} (ho) 	imes 	ex

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(B) The acceleration due to gravity $g$ on the surface of the Earth is given by $g = \frac{GM}{R^2}$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3} \pi R^3$,we can substitute this into the formula for $g$:$g = \frac{G (ho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi G ho R$.This shows that $g \propto ho R$.Let the initial density be $ho_1$ and radius be $R_1$. Then $g_1 \propto ho_1 R_1$.Let the new density be $ho_2 = 4ho_1$ and the new radius be $R_2 = \frac{R_1}{2}$.The new acceleration due to gravity $g_2$ is proportional to $ho_2 R_2 = (4ho_1) 	imes (\frac{R_1}{2}) = 2 ho_1 R_1$.Therefore,$g_2 = 2 g_1$.Since weight $W = mg$,if $g$ doubles,the weight also doubles.

If the density of the Earth increases $4$ times and its radius becomes half of its present value,our weight will

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