The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If I_1 \over I_2 is x\pi ^2 \over 3 , then the value of x will be \underline\hspace2cm .

Rotational Motion PYQ — JEE Main 2022 · 29 Jun · Shift 2 · Q68 with Solution

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Correct answer: 8

For a uniform thin rod of length $L$ and mass $M$ , the moment of inertia about an axis perpendicular to the rod and passing through one end is

$I_1 = \frac{1}{3}ML^2.$

When the same rod is bent into a ring, its length becomes the circumference of the ring:

$L = 2\pi R \quad \Rightarrow \quad R = \frac{L}{2\pi}.$

The mass remains $M$ .

For a thin ring, the moment of inertia about an axis through its center and perpendicular to its plane is

$I_z = MR^2.$

By the perpendicular axis theorem, the moment of inertia about any diameter is

$I_2 = \frac{1}{2}MR^2.$

Substitute $R = \dfrac{L}{2\pi}$ :

= \frac{1}{2}M\cdot \frac{L^2}{4\pi^2} = \frac{ML^2}{8\pi^2}.$$ 4. **Compute the ratio** $$\frac{I_1}{I_2} = \frac{\frac{1}{3}ML^2}{\frac{ML^2}{8\pi^2}} = \frac{1}{3}\cdot 8\pi^2 = \frac{8\pi^2}{3}.$$ Given $$\frac{I_1}{I_2} = \frac{x\pi^2}{3},$$ so comparing, $$x = 8.$$ 5. **Comparison with stored answer** Stored correct answer = $8$. Our derived answer also gives $8$, so it agrees.