Question

A point mass M is placed at centre of a sphere of radius 2a. Region of a<r<2a is filled with material of density, $\rho = \frac{\alpha}{r}$ If the field inside the region a<r<2a is constant, then value of $\alpha$ is

• A. $\frac{M}{3\pi a^2}$
• B. $\frac{M}{\pi a^2}$
• C. $\frac{M}{4\pi a^2}$
• D. $\frac{M}{2\pi a^2}$

Answer 1

Answer: (D)

$\frac{M}{2\pi a^2}$

Answer 2

To find the value of $\alpha$ such that the gravitational field is constant within the region $a < r < 2a$, we proceed as follows:

1. Calculate the mass enclosed by a sphere of radius $r$ (where $a < r < 2a$):

   The mass $dM$ of a thin spherical shell of radius $r'$ and thickness $dr'$ is given by:

   $dM = \rho(r') \cdot 4\pi r'^2 dr' = \frac{\alpha}{r'} \cdot 4\pi r'^2 dr' = 4\pi \alpha r' dr'$

   Integrating from $a$ to $r$ to find the mass $M_{\text{shell}}$ of the material in the region $a < r' < r$:

   $M_{\text{shell}} = \int_{a}^{r} 4\pi \alpha r' dr' = 4\pi \alpha \left[ \frac{r'^2}{2} \right]_{a}^{r} = 2\pi \alpha (r^2 - a^2)$

2. Total mass enclosed:

   The total mass enclosed inside the radius $r$ is the sum of the point mass $M$ at the center and the mass of the material in the region $a < r' < r$:

   $M_{\text{enc}} = M + 2\pi \alpha (r^2 - a^2)$

3. Gravitational field:

   Using Newton's law of gravitation, the gravitational field $g(r)$ at radius $r$ is:

   $g(r) = \frac{G M_{\text{enc}}}{r^2} = \frac{G [M + 2\pi \alpha (r^2 - a^2)]}{r^2}$

   Rearrange the terms:

   $g(r) = \frac{G}{r^2} [M + 2\pi \alpha r^2 - 2\pi \alpha a^2] = G \left[ 2\pi \alpha + \frac{M - 2\pi \alpha a^2}{r^2} \right]$

4. Condition for constant field:

   For $g(r)$ to be constant, it must not depend on $r$. This implies that the term involving $r$ must vanish:

   $M - 2\pi \alpha a^2 = 0$

   Solving for $\alpha$:

   $\alpha = \frac{M}{2\pi a^2}$

Therefore, the value of $\alpha$ that makes the gravitational field constant in the region $a < r < 2a$ is $\frac{M}{2\pi a^2}$.