Moment of Inertia of Disc with Central Hole

❓ Question

Let $M$ and $R$ be the mass and radius of a (uniform) solid disc. A small disc of radius $R/3$ is removed from the bigger disc (hole cut out) as shown in the figure. The moment of inertia of the remaining part about an axis $AB$ passing through the centre $O$ and perpendicular to the plane of the disc is given as

$$I=\frac{x}{4}M{R}^2\mathrm{.}$$

Find the value of $x$.

(We assume the small disc is concentric with the big disc — i.e. the hole is at the center — this is the standard interpretation unless the figure indicates an off-centre cut.)

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Moment of inertia of disc
👉 Removed body concept
👉 Parallel axis theorem.

Main idea:

$$\boxed{ I_{\text{remaining}} = I_{\text{big disc}} - I_{\text{removed part}} }$$

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🔷 Step 1 — MOI of Full Disc 💯

For full disc about diameter axis through centre:

$$I_{\text{full}}=\frac14MR^2$$

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🔷 Step 2 — Mass of Removed Disc

Radius of removed disc:

$$r=\frac R3$$

Mass is proportional to area:

$$m = M\left(\frac{r^2}{R^2}\right)$$
$$= M\left(\frac{1}{9}\right)$$
$$\boxed{ m=\frac M9 }$$

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🔷 Step 3 — MOI of Removed Disc About AB

From figure:

Small disc touches outer boundary and central vertical diameter.

Hence distance of small disc centre from axis $AB$:

$$d=\frac R3$$

MOI of small disc about its own diameter:

$$I_c=\frac14mr^2$$

Using parallel axis theorem:

$$I_{\text{removed}} = \frac14mr^2+md^2$$

Substitute:

$$= \frac14\left(\frac M9\right)\left(\frac{R^2}{9}\right) + \left(\frac M9\right)\left(\frac{R^2}{9}\right)$$
$$= \frac{MR^2}{324} + \frac{MR^2}{81}$$

LCM:

$$= \frac{1+4}{324}MR^2$$
$$= \frac5{324}MR^2$$

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🔷 Step 4 — Remaining MOI

$$I= \frac14MR^2-\frac5{324}MR^2$$

Take LCM $324$:

$$I= \frac{81-5}{324}MR^2$$
$$= \frac{76}{324}MR^2$$
$$= \frac{19}{81}MR^2$$

Given:

$$I=\frac4xMR^2$$

Thus:

$$\frac4x=\frac{19}{81}$$
$$x=\frac{324}{19}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Removed disc mass ko $M/3$ le lena

❌ Parallel axis theorem apply na karna

❌ Radius ratio ko directly mass ratio maan lena

Remember:

$$\boxed{ m\propto r^2 }$$

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✅ Final Answer

$$\boxed{ x=\frac{324}{19} }$$

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