Question

As shown in figure, when a spherical cavity (centered at $O$) of radius $1m$ is cut out of a uniform sphere of radius (centered at $C$ ), the center of mass of remaining (shaded) part of sphere is shown by $COM$, 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

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

Step 1: Given data

Center of the cavity $=O$

Center of the solid sphere $=C$

Radius of cavity $R_c=1cm$

Radius of solid Sphere $R_s=R$

Mass of the sphere is taken to be $m_s$,which can be given as ,

$m_s=\frac{4}{3}\pi R^3\rho$

Mass of the cavity is taken to be $m_c$, which can be given as

$$\begin{gathered} m_c=\frac{4}{3}\pi \times R_c^3\rho \\ \Rightarrow m_c=\frac{4}{3}\pi \times {\left(1\right)}^3\rho \\ \Rightarrow m_c=\frac{4}{3}\pi \rho \end{gathered}$$

Where, $\rho =$ Density of the material

Step 2 : To find the equation to determine the value of $R$

By the concept of COM, $m_1{R}_1=m_2{R}_2$

Remaining mass $\times$$\left(2-R\right)$ $=$Cavity mass $\times \left(R_s-R_c\right)$

$$\begin{gathered} (m_s-m_c)(2-R)=m_c(R_s-R_c) \\ \Rightarrow \left(\frac{4}{3}\pi R^3\rho -\frac{4}{3}\pi \rho \right)\left(2-R\right)=\frac{4}{3}\pi \rho \times \left(R-1\right) \\ \Rightarrow \frac{4}{3}\pi \rho \left(R^3-1\right)\left(2-R\right)=\frac{4}{3}\pi \rho \left(R-1\right) \\ \Rightarrow \left(R^3-1\right)\left(2-R\right)=\left(R-1\right) \\ \Rightarrow \left(R-1\right)\left(R^2+R+1\right)\left(2-R\right)=\left(R-1\right) \\ \Rightarrow \left(R^2+R+1\right)\left(2-R\right)=1 \end{gathered}$$

Hence, $R$ can be determine by the equation $\left(R^2+R+1\right)\left(2-R\right)=1$.

Therefore, Option A is the correct answer.