The mass density of a spherical body is given by ρ r = k r for r ≤ R and ρ r = 0 for r >…

Solution

 For Inside point $\left(r\le R\right)$

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

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

$$\begin{gathered} E=g \\ E=\frac{GM}{r^2} \\ E=\frac{G}{r^2}\underset{0}{\overset{R}{\int }}\left(4\pi r^{2}d{r}^{\text{'}}\right)\left(\frac{K}{r^{\text{'}}}\right) \\ E=\frac{4\pi KG}{r^2}{\left[\frac{{r^{\text{'}}}^2}{2}\right]}_0^R \end{gathered}$$

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

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

Asked in: JEE Main 2017 (09 Apr Online)

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