The mass density of a planet of radius R varies with the distance r from its centre as ρ(r)=ρo(1−r2R2). Then the gravitational field is maximum at :

Mass of small element of planet of radius x and thickness dm=ρ×4πx2dx=ρ0(1−x2R2)×4πx2dxMass of the planet from r=0 to rM=4πρ0∫0r(x2−x4R2)dx ⇒M=4πρ0|r33−r55R2|Gravitation field E=GMr2=Gr2×4πρ0(r33−r55R2) ⇒E=4πGρ0(r3−r35R2)E is maximum when dEdr=0⇒dEdr=4πGρ0(13−3r25R2)=0⇒r=53R

The mass density of a planet of radius R varies with the distance r from its centre as ρ(r)=ρo(1−r2R2). Then the gravitational field is maximum at :

Mass of small element of planet of radius x and thickness dm=ρ×4πx2dx=ρ0(1−x2R2)×4πx2dxMass of the planet from r=0 to rM=4πρ0∫0r(x2−x4R2)dx ⇒M=4πρ0|r33−r55R2|Gravitation field E=GMr2=Gr2×4πρ0(r33−r55R2) ⇒E=4πGρ0(r3−r35R2)E is maximum when dEdr=0⇒dEdr=4πGρ0(13−3r25R2)=0⇒r=53R

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The mass density of a planet of radius R varies with the distance r from its centre as $ρ (r) = ρ_{o} (1 - \frac{r^{2}}{R^{2}})$ . Then the gravitational field is maximum at :

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$r = \sqrt{\frac{3}{4}} R$

$r = R$

$r = \frac{1}{\sqrt{3}} R$

$r = \sqrt{\frac{5}{9}} R$

answer is D.

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Mass of small element of planet of radius x and thickness
$d m = ρ \times 4 π x^{2} d x = ρ_{0} (1 - \frac{x^{2}}{R^{2}}) \times 4 π x^{2} d x$
Mass of the planet from $r = 0 t o r$
$M = 4 π ρ_{0} \int_{0}^{r} (x^{2} - \frac{x^{4}}{R^{2}}) d x$ $\Rightarrow M = 4 π ρ_{0} | \frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}} |$

Gravitation field $E = \frac{G M}{r^{2}} = \frac{G}{r^{2}} \times 4 π ρ_{0} (\frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}})$ $\Rightarrow E = 4 π G ρ_{0} (\frac{r}{3} - \frac{r^{3}}{5 R^{2}})$
E is maximum when $\frac{d E}{d r} = 0$

$\begin{array}{l} \Rightarrow \frac{d E}{d r} = 4 π G ρ_{0} (\frac{1}{3} - \frac{3 r^{2}}{5 R^{2}}) = 0 \\ \Rightarrow r = \frac{\sqrt{5}}{3} R \end{array}$