The mass density of a planet of radius R varies with the distance r from its centre as ρ ( r ) = ρ 0 1 - r 3 R 3 . Then the gravitational field is maximum at:

g = G M in r 2 , where M in = ∫ 0 r ? ρ 4 π x 2 ⋅ d x M in = 4 π ρ 0 ∫ 0 r ? ? 1 - x 3 R 3 x 2 d x = 4 π ρ 0 r 3 3 - r 6 6 R 3 ∴ g = 4 π G ρ 0 6 2 r - r 4 R 3 For g to be maximum or minimum or constant d g d r = 0 ⇒ 2 - 4 r 3 R 3 = 0 ⇒ r = R 2 1 / 3

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $ρ (r) = ρ_{0} (1 - \frac{r^{3}}{R^{3}})$ . Then the gravitational field is maximum at:

Option 1 - <math> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>r</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mrow> </mfrac> <mi>R</mi> </math>

Option 2 - <math> <mi>r</mi> <mo>=</mo> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> <mi>R</mi> </math>

Option 3 - <math> <mo>(</mo> <mi mathvariant="normal">C</mi> <mo>)</mo> <mi mathvariant="normal"> </mi> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>r</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>

Option 4 - <math> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>r</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>

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Answered by

Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 4

Detailed Solution:

$g = \frac{G M_{in}}{r^{2}}$ , where $M_{in} = \int_{0}^{r} ? ρ 4 π x^{2} \cdot d x$

$\begin{array}{r} M_{in} = 4 π ρ_{0} \int_{0}^{r} ? ? (1 - \frac{x^{3}}{R^{3}}) x^{2} d x = 4 π ρ_{0} (\frac{r^{3}}{3} - \frac{r^{6}}{6 R^{3}}) \\ ∴ g = \frac{4 π G ρ_{0}}{6} (2 r - \frac{r^{4}}{R^{3}}) \end{array}$ $_{}_{}^{}^{}$

For $g$ to be maximum or minimum or constant $\frac{d g}{d r} = 0 \Rightarrow 2 - \frac{4 r^{3}}{R^{3}} = 0 \Rightarrow r = \frac{R}{2^{1 / 3}}$ $\frac{}{} \frac{^{}}{^{}} \frac{}{^{}}$