Question

Four identical discs each of mass ' $M$ ' and diameter ' $a$ ' are arranged in a small plane as shown in figure. If the moment of inertia of the system about $O{O}^{\prime }$ is $\frac{x}{4}M{a}^2$. Then, the value of $x$ will be ___

Answer 1

Answer: 3

$I_1=I_3=\frac{M^2}{4}$
$I_2=\frac{M{R}^2}{4}+M{R}^2=\frac{5}{4}M{R}^2=I_4$

So $I=I_1+I_2+I_3+I_4$
$=\frac{M{R}^2}{2}+\frac{5}{2}M{R}^2$
$=3M{R}^2,\text{ Putting R }=\frac{a}{2}$
$I=\frac{3M{a}^2}{4},\text{ So }x=3$