A sphere of radius R has its centre at the origin. It has a uniform mass density ρ0 except that there is a spherical hole of radius r=R2 whose centre is at x=12R as shown in the figure. The magnitude of gravitational field at a point distance X on the x- axis (where X > R) is

2 G π R 3 ρ 0 3 [ 1 X 2 − 1 8 ( X − R 2 ) 2 ]

A sphere of radius R has its centre at the origin. It has a uniform mass density ρ0 except that there is a spherical hole of radius r=R2 whose centre is at x=12R as shown in the figure. The magnitude of gravitational field at a point distance X on the x- axis (where X > R) is

2 G π R 3 ρ 0 3 [ 1 X 2 − 1 8 ( X − R 2 ) 2 ]

G π R 3 ρ 0 3 [ 1 X 2 − 1 8 ( X − R 2 ) 2 ]

4 G π R 3 ρ 0 3 [ 1 X 2 − 1 8 ( X − R 2 ) 2 ]

4 G π R 3 ρ 0 3 [ 2 X 2 − 1 8 ( X − R 2 ) 2 ]

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A sphere of radius R has its centre at the origin. It has a uniform mass density $ρ_{0}$ except that there is a spherical hole of radius $r = \frac{R}{2}$ whose centre is at $x = \frac{1}{2} R$ as shown in the figure. The magnitude of gravitational field at a point distance X on the x- axis (where X > R) is

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$\frac{2 G π R^{3} ρ_{0}}{3} [\frac{1}{X^{2}} - \frac{1}{8 {(X - \frac{R}{2})}^{2}}]$

$\frac{G π R^{3} ρ_{0}}{3} [\frac{1}{X^{2}} - \frac{1}{8 {(X - \frac{R}{2})}^{2}}]$

$\frac{4 G π R^{3} ρ_{0}}{3} [\frac{1}{X^{2}} - \frac{1}{8 {(X - \frac{R}{2})}^{2}}]$

$\frac{4 G π R^{3} ρ_{0}}{3} [\frac{2}{X^{2}} - \frac{1}{8 {(X - \frac{R}{2})}^{2}}]$

answer is A.

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Detailed Solution

At point P, $| E_{g 1} - E_{g 2} | = E_{g}$

$m_{1} = \frac{4}{3} π R^{3} ρ_{0}$

$m_{2} = \frac{4}{3} π {(\frac{R}{2})}^{3} ρ_{0}$

$= \frac{4}{3} π R^{3} ρ_{0} \frac{1}{8}$

$E_{g 1} = \frac{G m_{1}}{X^{2}} = \frac{G \frac{4}{3} π R^{3} ρ_{0}}{X^{2}}$

$E_{g 2} = \frac{G m_{2}}{{(X - \frac{R}{2})}^{2}} = \frac{G \frac{4}{3} π R^{3} ρ_{0} \frac{1}{8}}{{(X - \frac{R}{2})}^{2}}$

$∴ E_{g} = G \frac{4}{3} π R^{3} ρ_{0} [\frac{1}{X^{2}} - \frac{1}{8 {(X - \frac{R}{2})}^{2}}]$

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