JEE Mains · Physics · STD 11 - 7. gravitation
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
- A \(r =\frac{1}{\sqrt{3}} R\)
- B \(r=\sqrt{\frac{5}{9}} R\)
- C \(r=\sqrt{\frac{3}{4}} R\)
- D \(r=R\)
Answer & Solution
Correct Answer
(B) \(r=\sqrt{\frac{5}{9}} R\)
Step-by-step Solution
Detailed explanation
\(E 4 \pi r ^{2}=\int \rho_{0} 4 \pi r ^{2} dr\) \(\Rightarrow Er ^{2}=4 \pi G \int \limits_{0}^{ r } \rho_{0}\left(1-\frac{ r ^{2}}{ R ^{2}}\right) r ^{2} dr\) \(\Rightarrow E =4 \pi G \rho_{0}\left(\frac{ r ^{3}}{3}-\frac{ r ^{5}}{5 R ^{2}}\right)\)…
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