Question

As shown in fig. when a spherical cavity $\text{(centred at}O)$ of radius $1$ is cut out of a uniform sphere of radius $R$ $\text{(centred at}C)$ , the centre of mass of remaining (shaded) part of sphere is at $G$, i.e on the surface of the cavity. $R$ can be determined by the equation

• A. $(R^2+R+1)(2-R)=1$
• B. $(R^2-R-1)(2-R)=1$
• C. $(R^2-R+1)(2-R)=1$
• D. $(R^2+R-1)(2-R)=1$

Answer 1

The correct option is A $(R^2+R+1)(2-R)=1$

Mass of sphere = volume of sphere x density of sphere $=\frac{4}{3}\pi R^3\rho$


$\text{Mass of cavity}{M}_{cavity}=\frac{4}{3}\pi (1{)}^3\rho$

$\text{Mass of remaining}{M}_{\left(Remaining\right)}=\frac{4}{3}\pi R^3\rho -\frac{4}{3}\pi (1{)}^3\rho$

Centre of mass of remaining part(considering cavinty as negative amss and applying superposition),

$X_{COM}=\frac{M_1{r}_1+M_2{r}_2}{M_1+M_2}$

$\Rightarrow -(2-R)=\frac{\left[\frac{4}{3}\pi R^3\rho \right]0+\left[\frac{4}{3}\pi \right(1{)}^3(-\rho )][R-1]}{\frac{4}{3}\pi R^3\rho +\frac{4}{3}\pi \left(1{)}^3\right(-\rho )}$

$\Rightarrow \frac{(R-1)}{(R^3-1)}=2-R$

$\Rightarrow \frac{(R-1)}{(R-1)(R^2+R+1)}=2-R$

$\Rightarrow (R^2+R+1)(2-R)=1$