Verified Solution • 2025–26 Session moderate

QJEEM20Ph9dfe: JEE Main 2020 Physics - Class 11 JEEM20Ph9dfe Solution: The mass density of a spherical galaxy varies as over a large…

The mass density of a spherical galaxy varies as Kr{K \over r} over a large distance ‘r’ from its centre. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as :

Multiple Choice Options

  • A T2 ∝\propto R
  • B T2 ∝\propto R3
  • C T ∝\propto R
  • D T2 ∝\propto 1R3{1 \over {{R^3}}}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 2nd September Morning Slot Physics - Gravitation Question 140 English Explanation

dm = ρ\rho dv

⇒\Rightarrow dm = (kr)\left( \frac{k}{r} \right) (4 π\pi r2dr)

⇒\Rightarrow dm = 4 π\pi krdr

M = ∫0Rdm\int\limits^{R}_{0} dm = ∫0R4πkrdr\int\limits^{R}_{0} 4\pi krdr

⇒\Rightarrow M = 4πk[r22]0R4\pi k\left[ \frac{r^{2}}{2} \right]^{R}_{0}

⇒\Rightarrow M = 2 π\pi kR2

For circular motion gravitational force will provide required centripetal force.

GMmR2\frac{GMm}{R^{2}} = mv2R\frac{mv^{2}}{R}

⇒\Rightarrow G(2πkR2)mR2\frac{G\left( 2\pi kR^{2}\right) m}{R^{2}} = mv2R\frac{mv^{2}}{R}

⇒\Rightarrow v = 2πGkR\sqrt{2\pi GkR}

Time period, T = 2πRv\frac{2\pi R}{v}

= 2πR2πGkR\frac{2\pi R}{\sqrt{2\pi GkR} } ∝\propto R\sqrt{R}

⇒\Rightarrow T2 ∝\propto R

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring