Question:
A square lamina of side 'b' has same mass as a disc of radius 'R'. The moment of inertia of the two objects about an axis perpendicular to the plane and passing through the centre is equal. The ratio $\frac{b}{R}$ is
A square lamina of side 'b' has same mass as a disc of radius 'R'. The moment of inertia of the two objects about an axis perpendicular to the plane and passing through the centre is equal. The ratio $\frac{b}{R}$ is
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Logic Tip: Because the denominator for a square (6) is larger than that of a disc (2), the side $b$ must be larger than radius $R$ to maintain the same rotational inertia for the same mass.
Updated On: Apr 28, 2026
- 1:1
- $\sqrt{3}:1$
- $\sqrt{6}:1$
- $1:\sqrt{3}$
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The Correct Option is B
Solution and Explanation
Concept:
The moment of inertia for rigid bodies depends on the mass distribution relative to the axis of rotation.
Step 1: Identify the formulas for Moment of Inertia.
For a square lamina of side $b$: $I_{lamina} = \frac{Mb^{2{6}$
For a disc of radius $R$: $I_{disc} = \frac{MR^{2{2}$
Step 2: Equate the two expressions.
Given $I_{lamina} = I_{disc}$: $$\frac{Mb^{2{6}=\frac{MR^{2{2}$$
Step 3: Solve for the ratio $b/R$.
Divide both sides by $M$ and simplify: $$\frac{b^{2{R^{2=3$$ Taking the square root: $$\frac{b}{R}=\frac{\sqrt{3{1}$$
The moment of inertia for rigid bodies depends on the mass distribution relative to the axis of rotation.
Step 1: Identify the formulas for Moment of Inertia.
For a square lamina of side $b$: $I_{lamina} = \frac{Mb^{2{6}$
For a disc of radius $R$: $I_{disc} = \frac{MR^{2{2}$
Step 2: Equate the two expressions.
Given $I_{lamina} = I_{disc}$: $$\frac{Mb^{2{6}=\frac{MR^{2{2}$$
Step 3: Solve for the ratio $b/R$.
Divide both sides by $M$ and simplify: $$\frac{b^{2{R^{2=3$$ Taking the square root: $$\frac{b}{R}=\frac{\sqrt{3{1}$$
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