Gravitation

JEE Main level, single correct, medium

40 of 223

Question 40 of 223

The mass density of a spherical body is given by ρ(r)=kr\rho \left(r\right)=\dfrac{k}{r} for r≤Rr\leq R and ρ(r)=0\rho \left(r\right)=0 for r>R,r\gt R, where rr is the distance from the center. The correct graph that describes qualitatively the acceleration, aa of a test particle as a function of rr is:

Show answer and solution

Answer: (A)



For Inside point
(r≤R)\left(r\leq R\right)

g=E=Gr2∫0r(4πr2dr)(Kr)=GKr2×4πr22=2πGK=constantg=E=\dfrac{G}{{r}^{2}}\int _{0}^{r}\left(4\pi {r}^{2}dr\right)\left(\dfrac{K}{r}\right)=\dfrac{GK}{{r}^{2}}\times \dfrac{4\pi {r}^{2}}{2}=2\pi GK=constant

For outside point
(r>R)\left(r\gt R\right)

E=gE=GMr2E=Gr2∫0R(4πr2dr′)(Kr′)E=4πKGr2[r′22]0RE=g E=\dfrac{GM}{{r}^{2}} E=\dfrac{G}{{r}^{2}}\int _{0}^{R}(4\pi {r}^{2}d{r}^{'})(\dfrac{K}{{r}^{'}}) E=\dfrac{4\pi KG}{{r}^{2}}{[\dfrac{{{r}^{'}}^{2}}{2}]}_{0}^{R}

E=2πKGr2R2E=\dfrac{2\pi KG}{{r}^{2}}{R}^{2}

E∝1r2E\propto \dfrac{1}{{r}^{2}}.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Gravitation and orbits simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Gravitation (Gravitational field), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.