The figure shows the variation of acceleratio | vedclass

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(B) For a point inside the planet at distance $r_1$ from the center,the acceleration due to gravity is given by $g(r_1) = g_0 \frac{r_1}{R}$.Given $g(

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$\frac{7R}{4}$

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(B) For a point inside the planet at distance $r_1$ from the center,the acceleration due to gravity is given by $g(r_1) = g_0 \frac{r_1}{R}$.Given $g(r_1) = \frac{g_0}{4}$,we have $\frac{g_0}{4} = g_0 \frac{r_1}{R}$,which gives $r_1 = \frac{R}{4}$.For a point outside the planet at distance $r_2$ from the center,the acceleration due to gravity is given by $g(r_2) = g_0 \frac{R^2}{r_2^2}$.Given $g(r_2) = \frac{g_0}{4}$,we have $\frac{g_0}{4} = g_0 \frac{R^2}{r_2^2}$,which implies $r_2^2 = 4R^2$,so $r_2 = 2R$.Therefore,$r_2 - r_1 = 2R - \frac{R}{4} = \frac{7R}{4}$.

The figure shows the variation of acceleration due to gravity with distance from the center of a uniform spherical planet of radius $R$. What is $r_2 - r_1$?

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