The mass density of a planet of radius R varies with the distance r from its centre as ρ (r) = ρ₀ (1 - (r²)/(R²)). Then the gravitational field is maximum at: (a) r = 1/√(3) R (b) r = R (c) r = √(3/4)R (d) r = √(5/9)R

Answer: (d). Mass within r: M(r) = ∫₀r 4π r²ρ₀(1-(r²)/(R²))dr = 4πρ₀(r³/3 - (r⁵)/(5R²)) E = (GM(r))/(r²) = 4π Gρ₀(r/3 - (r³)/(5R²)) For maximum, dE/dr = 0: 1/3 - (3r²)/(5R²) = 0 ⇒ r² = 5R²/9 So the answer is (d) r = √(5/9)R.

The mass density of a planet of radius R varies with the distance r… | JEE Main 2020 (3 Sep, Shift 2)

Q10JEE Main 2020 · 3 SEP · SHIFT 2Difficult

Gravitational Field

The mass density of a planet of radius R varies with the distance r from its centre as $\rho$ (r) = $\rho_0 (1$ $- \frac{r^2}{R^2})$ .

Then the gravitational field is maximum at:

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Answer: (d)

Mass within $r$ : $M(r) = \int_0^r 4\pi r^2\rho_0\left(1-\frac{r^2}{R^2}\right)dr = 4\pi\rho_0\left(\frac{r^3}{3} - \frac{r^5}{5R^2}\right)$
$E = \frac{GM(r)}{r^2} = 4\pi G\rho_0\left(\frac{r}{3} - \frac{r^3}{5R^2}\right)$
For maximum, $\frac{dE}{dr} = 0$ : $\frac{1}{3} - \frac{3r^2}{5R^2} = 0 \Rightarrow r^2 = \frac{5R^2}{9}$
So the answer is (d) $r = \sqrt{\frac{5}{9}}R$ .

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