Question

As shown in figure, when a spherical cavity (centered at $O$) of radius $1$ is cut out of a uniform sphere of radius $R$ (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 B $(R^2+R+1)(2-R)=1$

$M_0=\frac{4}{3}\pi R^3\rho$
$M_{\text{cavity}}=\frac{4}{3}\pi \left(\right(1{)}^3\rho )$
$M_{\text{Remaining}}=\frac{4}{3}\pi R^3\rho -\frac{4}{3}\pi (1{)}^3\rho$

By concept of COM
Remaining mass $\times (2-R)=$ Cavity mass $\times (R-1)$
$(\frac{4}{3}\pi R^3\rho -\frac{4}{3}\pi 1^3\rho )$
$(2-R)=\frac{4}{3}\pi 1^3\rho \times (R-1)$
$(R^3-1)(2-R)=R-1$
$(R^2+R+1)(2-R)=1$
Final Answer: $\left(c\right)$