Question

Three identical solid sphere each of mass M and radius R are placed at the corner of equilateral triangle of side $2R$ such that their centre of mass coincide with the vertex of the triangle. Moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular of the triangle is

• A. $\frac{26M{R}^2}{5}$
• B. $\frac{13M{R}^2}{5}$
• C. $\frac{27M{R}^2}{5}$
• D. $\frac{11M{R}^2}{5}$

Answer 1

The correct option is A $\frac{26M{R}^2}{5}$

$\begin{array}{c}\text{On calculating the value of d we get}\\ d=\frac{2R}{\sqrt{3}}\\ \text{we Rnow that mo ment of Inertia of a sphere about}\\ \text{its own axis is (Io)}\\ I_0=\frac{2}{5}M{R}^{2}M\to \text{massot sphere}\\ R\to \text{Radius of sphere.}\end{array}$

$\begin{array}{c}\text{Therefore By parallel axis theorem moment of}\\ \text{Inertia about the given axis will be-}\\ I^{\prime }=I_0+M{d}^2=\frac{2}{5}M{R}^2+\frac{4M{R}^2}{3}\\ I^{\prime }=\frac{26}{15}M{R}^2\\ \text{Will be}3I^{\prime }=\frac{26}{5}M{R}^2\end{array}$