A circular disc X of radius R is made from an iron plate of thickness t , and another disc Y of radius 4R is made from an iron plate of thickness t/4 - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4 R is made from an iron plate of thickness t /4 . The relation between the moment of inertia I_{X} and I_{Y} is:

Solution & Step-by-step Explanation

Moment of inertia of a disc is I = \frac{1}{2} m R^{2} .Mass m = Volume \times density = (π R^{2} t) ρ .So, I = \frac{1}{2} (π R^{2} tρ) R^{2} = \frac{π ρt R ^{4}}{2} .For disc X : I_{X} = \frac{π ρt R ^{4}}{2} .For disc Y : I_{Y} = \frac{π ρ ( t /4 ) ( 4 R ) ^{4}}{2} = \frac{π ρt 256 R ^{4}}{8} = 32 \frac{π ρt R ^{4}}{2} \times 2 = 64 I_{X} .Specifically: I_{Y} = \frac{1}{2} π (4 R)^{2} (t /4) ρ (4 R)^{2} = \frac{1}{2} π (16 R^{2}) (t /4) ρ (16 R^{2}) = \frac{1}{2} π (4 R^{2} tρ) (16 R^{2}) = 64 (\frac{1}{2} π R^{2} tρ R^{2}) = 64 I_{X} Wait, checking calculation: m_{Y} = π (4 R)^{2} (t /4) ρ = 4 π R^{2} tρ = 4 m_{X} . I_{Y} = \frac{1}{2} m_{Y} R_{Y}^{2} = \frac{1}{2} (4 m_{X}) (4 R)^{2} = \frac{1}{2} (4 m_{X}) (16 R^{2}) = 64 (\frac{1}{2} m_{X} R^{2}) = 64 I_{X} .

Try it yourself before checking the explanation above.

A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4 R is made from an iron plate of thickness t /4 . The relation between the moment of inertia I_{X} and I_{Y} is:

I_{Y} = 32 I_{X}

I_{Y} = 16 I_{X}

I_{Y} = 32 I_{X}

I_{Y} = 64 I_{X}

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