The mass density of a spherical… - JEE Main 2017 Physics PYQ

JEE Main 2017 — Physics Mechanics

2017mcqhard

The mass density of a spherical body is given by $\rho (r)=\frac{k}{r}$ for $r\leq R$ and $\rho (r)=0$ for $r>R,$ where $r$ is the distance from the center. The correct graph that describes qualitatively the acceleration, $a$ of a test particle as a function of $r$ is:

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Held on 9 Apr 2017 · Verified 6 Jul 2026.

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Solution

JEE Main 2017 Physics, Gravitation — solution figure

For Inside point $(r\leq R)$

$g=E=\frac{G}{{r}^{2}}\int _{0}^{r}(4\pi {r}^{2}dr)(\frac{K}{r})=\frac{GK}{{r}^{2}}\times \frac{4\pi {r}^{2}}{2}=2\pi GK=\mathrm{constant}$

For outside point $(r>R)$

$E=g E=\frac{GM}{{r}^{2}} E=\frac{G}{{r}^{2}}\int _{0}^{R}(4\pi {r}^{2}d{r}^{'})(\frac{K}{{r}^{'}}) E=\frac{4\pi KG}{{r}^{2}}{[\frac{{{r}^{'}}^{2}}{2}]}_{0}^{R}$

$E=\frac{2\pi KG}{{r}^{2}}{R}^{2}$

$E\propto \frac{1}{{r}^{2}}$ .

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Solution

JEE Main 2017 Physics, Gravitation — solution figure

For Inside point $(r\leq R)$

$g=E=\frac{G}{{r}^{2}}\int _{0}^{r}(4\pi {r}^{2}dr)(\frac{K}{r})=\frac{GK}{{r}^{2}}\times \frac{4\pi {r}^{2}}{2}=2\pi GK=\mathrm{constant}$

For outside point $(r>R)$

$E=g E=\frac{GM}{{r}^{2}} E=\frac{G}{{r}^{2}}\int _{0}^{R}(4\pi {r}^{2}d{r}^{'})(\frac{K}{{r}^{'}}) E=\frac{4\pi KG}{{r}^{2}}{[\frac{{{r}^{'}}^{2}}{2}]}_{0}^{R}$

$E=\frac{2\pi KG}{{r}^{2}}{R}^{2}$

$E\propto \frac{1}{{r}^{2}}$ .

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Solution

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JEE Main Mechanics in other years

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Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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