Question

Density of solid sphere is varied by $\rho ={\rho }_0⟮1+\frac{r}{R}⟯$ where $0\le r\le R$, R is the radius of the sphere. Moment of inertia of sphere w.r.t. axis passing through its centre will be : (${\rho }_0$ is constant)

• A. $\frac{44}{45}\pi {\rho }_0{R}^5$
• B. $\frac{44}{45}\pi {\rho }_0{R}^4$
• C. $\frac{44}{35}\pi {\rho }_0{R}^5$
• D. $\frac{48}{45}\pi {\rho }_0{R}^5$

Answer 1

The correct option is A $\frac{44}{45}\pi {\rho }_0{R}^5$

$\begin{array}{c}dM=s\left(4\pi x^{2}dx\right)\\ ={\rho }_0\left(1+\frac{x}{R}\right)\left(4\pi x^{2}dx\right)\\ dI=\frac{2}{3}dM{x}^2\\ \frac{2}{3}{\rho }_o{\int }_0^R\left(4\pi x^{4}dx+\frac{4\pi }{R}{x}^{5}dx\right)\\ =\frac{8\pi {\rho }_0}{3}\left[\frac{x^5}{5}+\frac{R^5}{6}\right]\\ =\frac{8}{3}\pi {\rho }_o\times \frac{11R^5}{30}\\ =\frac{44}{45}\pi {\rho }_0{R}^5\end{array}$

Hence,

option $\left(A\right)$ is correct answer.