MCQ
A spherical cavity of radius $r$ is curved out of a uniform solid sphere of radius $R$ as shown in the figure below. The distance of the centre of mass of the resulting body from that of the solid sphere is given by
  • A
    $\frac{R-r}{2}$
  • B
    $\frac{R+r}{2}$
  • C
    $0$
  • $\frac{r^3}{R^2+R r+r^2}$

Answer

Correct option: D.
$\frac{r^3}{R^2+R r+r^2}$
d
(d)

Centre of mass of solid sphere is at origin (centre of sphere).

Now, centre of mass of sphere with cavity is calculated as follows.

$x_{ CM }=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$

$=\frac{\left(\frac{4}{3} \pi R^3 \cdot \rho\right)(0)-\left(\frac{4}{3} \pi r^3 \cdot \rho\right)(R-r)}{\frac{4}{3} \pi R^3 \rho-\frac{4}{3} \pi r^3 \rho}$

where, $\rho=$ density of material of sphere.

$\Rightarrow \quad x_{ CM }=-\frac{r^3(R-r)}{\left(R^3-r^3\right)}$

$=\frac{-r^3(R-r)}{(R-r)\left(R^2+R r+r^2\right)}$

$=\frac{-r^3}{R^2+R r+r^2}$

So, distance between two mass centres is

$d=\frac{r^3}{R^2+R r+r^2}$

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