Question

A small part $A$ is cut by a plane at a distance $\frac{R}{2}$ from the centre of a solid sphere as shown in the figure. The $Y$-coordinate of centre of masss of part $A$ is: [Consider point $c$ as the origin]

• A. $\frac{40R}{27}$
  
• B. $\frac{13R}{27}$
  
• C. $\frac{27R}{40}$
  
• D. $\frac{13R}{40}$
  

Answer 1

The correct option is C $\frac{27R}{40}$


Let us consider a small elementary disc of width $dx$ at a distance $x$ from the centre $c$.
Radius of elemental disc is:
$r=\sqrt{R^2-x^2}$

Let $\rho$ be the density of substance, then the mass of elemental disc is
$dm=\rho \left(\pi r^2\right)dx=\rho \pi (R^2-x^2)dx$

Mass of the part $A,M={\int }_{R/2}^{R}dm$

$\Rightarrow M={\int }_{R/2}^R\rho \pi (R^2-x^2)dx=\rho \pi {(R^{2}x-\frac{x^3}{3})}_{R/2}^R$

$\Rightarrow M=\frac{5}{24}\rho \pi R^3$

Now,
$y_{com}=\frac{1}{M}{\int }_{R/2}^{R}dmx=\frac{24}{5\rho \pi R^3}{\int }_{R/2}^R\rho \pi (R^2-x^2)xdx$

$\Rightarrow y_{com}=\frac{24}{5R^3}{(\frac{R^2{x}^2}{2}-\frac{x^4}{4})}_{R/2}^R$

$\Rightarrow y_{com}=\frac{27}{40}R$

Note: By Symmetry $x_{com}=0$