Figure shows variation of acceleration due to | vedclass

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Question

Inside planet

$\frac{g_{0}}{4}=g_{0}\left(1-\frac{d}{R}\right)$

$\mathrm{d}=\frac{3 \mathrm{R}}{4}$

outside planet

$g^{\prime}=g_{0} \frac

Vedclass · Accepted Answer

$\frac{7R}{4}$

Vedclass · Answer

Inside planet

$\frac{g_{0}}{4}=g_{0}\left(1-\frac{d}{R}\right)$

$\mathrm{d}=\frac{3 \mathrm{R}}{4}$

outside planet

$g^{\prime}=g_{0} \frac{R^{2}}{(R+h)^{2}}$

$\mathrm{n}=\mathrm{R}$

$\mathrm{r}_{2}-\mathrm{r}_{1}=\mathrm{d}+\mathrm{h}=\mathrm{R}+\frac{3 \mathrm{R}}{4}=\frac{7 \mathrm{R}}{4}$

Figure shows variation of acceleration due to gravity with distance from centre of a uniform spherical planet, Radius of planet is $R$. What is $r_2 -r_1.$

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