Q133 / 150 · MHT CET 2025 (23 Apr Shift 1) · Physics

A uniform solid sphere of mass ' m ' and radius ' r ' is surrounded by a uniform thin spherical shell of radius ' $2 r$ ' and mass ' $m$ ' then the gravitational field

  • A
    at a distance of 15 r from the centre is $\frac{2}{9} \frac{\mathrm{Gm}}{\mathrm{r}^2}$
  • Correct answer
    B
    at a distance of $(2 \cdot 5) \mathrm{r}$ from the centre is $\frac{8}{25} \frac{\mathrm{Gm}}{\mathrm{r}^2}$
  • C
    at a distance of $(1 \cdot 5) \mathrm{r}$ from the centre is zero.
  • D
    between the sphere and spherical shell is uniform.

Solution

The gravitational field at any point results from superposition of contributions from the solid sphere and spherical shell, each with mass $m$. For a solid sphere of radius $r$: inside ($d < r$), $E = \frac{GMd}{r^3}$; outside ($d \ge r$), $E = \frac{GM}{d^2}$. For a spherical shell of radius $2r$: inside ($d < 2r$), $E = 0$; outside ($d \ge 2r$), $E = \frac{GM}{d^2}$.

At distance $1.5r$ from center, the point lies outside the sphere ($1.5r > r$) and inside the shell ($1.5r < 2r$). Thus $E_{\text{total}} = \frac{Gm}{(1.5r)^2} + 0 = \frac{4}{9} \frac{Gm}{r^2}$, rejecting options A and C.

At distance $2.5r$ from center, the point lies outside both distributions. Hence $E_{\text{total}} = \frac{Gm}{(2.5r)^2} + \frac{Gm}{(2.5r)^2} = \frac{8}{25} \frac{Gm}{r^2}$, confirming option B.

In the region between sphere and shell ($r < d < 2r$), the field varies as $E = \frac{Gm}{d^2}$ due to the sphere alone, which depends on $d$ and is therefore non-uniform, rejecting option D.

Final answer: $\boxed{B}$

Hence, the correct answer is option: 'B'.

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