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

Gravitation PYQ — JEE Main 2017 · 9 Apr · Shift 1 · Q56 with Solution

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Correct answer: B

Given density distribution

The spherical body has density

\rho(r)=\frac{k}{r}, \quad r\le R

and

\rho(r)=0, \quad r>R.

We need the qualitative graph of gravitational acceleration $a(r)$ versus distance $r$ from the centre.

Find mass enclosed within radius $r$ for $r\le R$

Using spherical shells,

dM=\rho(r')\,4\pi r'^2 dr'.

So,

dM=\frac{k}{r'}\cdot 4\pi r'^2 dr' = 4\pi k r'\,dr'.

Hence mass enclosed up to radius $r$ is

M(r)=\int_0^r 4\pi k r'\,dr' = 4\pi k\left[\frac{r'^2}{2}\right]_0^r = 2\pi k r^2.

Acceleration inside the sphere $(r\le R)$

By shell theorem, only enclosed mass contributes:

a(r)=\frac{G M(r)}{r^2}.

Substitute $M(r)=2\pi k r^2$ :

a(r)=\frac{G(2\pi k r^2)}{r^2}=2\pi Gk.

So for all $0<r\le R$ , the acceleration is constant.

Also, at the centre,

M(r)\propto r^2 \implies a(r)=\frac{GM(r)}{r^2}

remains finite and equals the same constant. Thus the graph inside is a horizontal line.

Acceleration outside the sphere $(r>R)$

Total mass of the sphere is

M = M(R)=2\pi k R^2.

So for $r>R$ ,

a(r)=\frac{GM}{r^2}=\frac{2\pi Gk R^2}{r^2}.

Thus outside the sphere, acceleration decreases as

a(r)\propto \frac{1}{r^2}.

At $r=R$ ,

a(R^-)=2\pi Gk,

and

a(R^+)=\frac{2\pi GkR^2}{R^2}=2\pi Gk.

So the graph is continuous at $r=R$ .

Qualitative shape of the graph

From $r=0$ to $r=R$ : constant acceleration
For $r>R$ : falls as $1/r^2$

So the correct graph is: a horizontal line inside the sphere, smoothly joining into a decreasing inverse-square curve outside.

Matching with options

This corresponds to Option B.

Final Answer

\boxed{\text{B}}

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View written solutionFree

Correct answer: B

Given density distribution

The spherical body has density

\rho(r)=\frac{k}{r}, \quad r\le R

and

\rho(r)=0, \quad r>R.

We need the qualitative graph of gravitational acceleration $a(r)$ versus distance $r$ from the centre.

Find mass enclosed within radius $r$ for $r\le R$

Using spherical shells,

dM=\rho(r')\,4\pi r'^2 dr'.

So,

dM=\frac{k}{r'}\cdot 4\pi r'^2 dr' = 4\pi k r'\,dr'.

Hence mass enclosed up to radius $r$ is

M(r)=\int_0^r 4\pi k r'\,dr' = 4\pi k\left[\frac{r'^2}{2}\right]_0^r = 2\pi k r^2.

Acceleration inside the sphere $(r\le R)$

By shell theorem, only enclosed mass contributes:

a(r)=\frac{G M(r)}{r^2}.

Substitute $M(r)=2\pi k r^2$ :

a(r)=\frac{G(2\pi k r^2)}{r^2}=2\pi Gk.

So for all $0<r\le R$ , the acceleration is constant.

Also, at the centre,

M(r)\propto r^2 \implies a(r)=\frac{GM(r)}{r^2}

remains finite and equals the same constant. Thus the graph inside is a horizontal line.

Acceleration outside the sphere $(r>R)$

Total mass of the sphere is

M = M(R)=2\pi k R^2.

So for $r>R$ ,

a(r)=\frac{GM}{r^2}=\frac{2\pi Gk R^2}{r^2}.

Thus outside the sphere, acceleration decreases as

a(r)\propto \frac{1}{r^2}.

At $r=R$ ,

a(R^-)=2\pi Gk,

and

a(R^+)=\frac{2\pi GkR^2}{R^2}=2\pi Gk.

So the graph is continuous at $r=R$ .

Qualitative shape of the graph

From $r=0$ to $r=R$ : constant acceleration
For $r>R$ : falls as $1/r^2$

So the correct graph is: a horizontal line inside the sphere, smoothly joining into a decreasing inverse-square curve outside.

Matching with options

This corresponds to Option B.

Final Answer

\boxed{\text{B}}

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