Question 43 of 280

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 $O{O}^{'}$ is $\dfrac{x}{4}M{a}^{2}$. Then, the value of $x$ will be _____ .

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Answer: 3

Considering four identical discs as $1,2,3$ and $4$.

Moment of inertia of disc about diagonal is ${I}_{1}={I}_{3}=\dfrac{M{R}^{2}}{4}$
Now using parallel axis theorem,
Moment of inertia of disc $2$ and $4$ is
${I}_{2}={I}_{4}=\dfrac{M{R}^{2}}{4}+M{R}^{2}=\dfrac{5}{4}M{R}^{2}$
So, net moment of inertia about $O{O}^{'}$ is
$I={I}_{1}+{I}_{2}+{I}_{3}+{I}_{4}$
$=2\left(\dfrac{M{R}^{2}}{4}\right)+2\left(\dfrac{5}{4}M{R}^{2}\right)$
$=\dfrac{M{R}^{2}}{2}+\dfrac{5}{2}M{R}^{2}$$=3M{R}^{2}$
Putting $R=\dfrac{a}{2}$
$I=\dfrac{3M{a}^{2}}{4}$, so,$x=3$