Question

A uniform solid sphere of radius $R$ has a cavity of radius $1\text{m}$ cut from it. If the centre of mass of the system lies at the periphery of the cavity, then:

• 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 $(R^2+R+1)(2-R)=1$

Assume, $\rho =$ Density of the sphere
So, $\text{mass= volume}\times \text{density}$

$m_1=$ mass of parent sphere $=\frac{4}{3}\pi R^3\times \rho$
$m_2=$ mass of removed sphere $=\frac{4}{3}\pi (1{)}^3\times \rho$

$x_{com}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$
(-ve because mass is removed)
$x_2=$ COM of the removed mass $=R-1$
[taking origin at the centre of parent sphere]

$\Rightarrow x_{com}=\frac{\left[\frac{4}{3}\pi R^3\rho \right]\times 0-\left[\frac{4}{3}\pi \right(1{)}^3\rho ][R-1]}{\frac{4}{3}\pi R^3\rho -\frac{4}{3}\pi (1{)}^3\rho }=-(2-R)$
$\Rightarrow \frac{(R-1)}{(R^3-1)}=(2-R)$
$\Rightarrow \frac{(R-1)}{(R-1)(R^2+R+1)}=2-R[R\ne 1]$
[using identity $(a^3-b^3)=(a-b)(a^2+ab+b^2)$]

or $(R^2+R+1)(2-R)=1$