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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2020
  • Hard
  • MCQ

The mass density of a planet of radius \(R\) varies with the distance \(r\) from its centre as \(\rho( r )=\rho_{0}\left(1-\frac{ r ^{2}}{ R ^{2}}\right) .\) Then the gravitational field is maximum at

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Answer & Solution

Correct Answer

(B) \(r=\sqrt{\frac{5}{9}} R\)

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Detailed explanation

\(E 4 \pi r ^{2}=\int \rho_{0} 4 \pi r ^{2} dr\)

\(\Rightarrow Er ^{2}=4 \pi G \int \limits_{0}^{ r } \rho_{0}\left(1-\frac{ r ^{2}}{ R ^{2}}\right) r ^{2} dr\)

\(\Rightarrow E =4 \pi G \rho_{0}\left(\frac{ r ^{3}}{3}-\frac{ r ^{5}}{5 R ^{2}}\right)\)

\(\frac{ d E }{ dr }=0 \therefore r =\sqrt{\frac{5}{9}} R\)

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