Question

Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about yy’ axis will be-

• A. $7/2P{Q}^2$
• B. $2/7P{Q}^2$
• C. $2/5P{Q}^2$
• D. $5/2P{Q}^2$

Answer 1

The correct option is A $7/2P{Q}^2$

Moment of inertia about yy' axis
$I=I_1+I_2+I_3...\left(1\right)$
$\because I_1=I_2$ i.e., the moment of inertia of ring about the tangent

Moment of inertia of ring about its diameter is $\frac{M{R}^2}{2}$
By the parallel axis theorem
$I_1=I_2=(\frac{M{R}^2}{2}+M{R}^2)=\frac{3}{2}M{R}^2$
and $I_3=$ moment of inertia of ring about diameter
$i.e.,I_3=\frac{M{R}^2}{2}$
$\therefore$ From equation (1)
$=2\left(\frac{3}{2}P{Q}^2\right)+\left(\frac{P{Q}^2}{2}\right)\left\{\begin{array}{c}\because M\to P\\ R\to Q\end{array}\right\}$$I=\frac{7}{2}P{Q}^2$.