Question

A solid metal cylinder A and a hollow metal cylinder B have same mass but their radii are in the ratio 2: 1. Then the ratio of their respective moments of inertia about their own axis is

• A. 1:1
• B. 2:1
• C. 4:1
• D. 1:4
• E. 1:2

Answer 1

The Correct Option is B

The moment of inertia \( I \) of a solid cylinder about its own axis is given by: \[ I_{\text{solid}} = \frac{1}{2} M r^2 \] Where \( M \) is the mass and \( r \) is the radius of the cylinder. For a hollow cylinder, the moment of inertia is: \[ I_{\text{hollow}} = M r^2 \] Given that the radii of the solid and hollow cylinders are in the ratio 2:1, the ratio of their moments of inertia will be: \[ \frac{I_{\text{solid}}}{I_{\text{hollow}}} = \frac{\frac{1}{2} M (2r)^2}{M r^2} = \frac{\frac{1}{2} M \cdot 4r^2}{M r^2} = \frac{2}{1} \] Thus, the ratio of their moments of inertia is \(2:1\).

The correct option is (B) : \(2:1\)

Answer 2

Let the mass of both cylinders be $M$. 
Let the radius of solid cylinder A be $R$ and that of hollow cylinder B be $\dfrac{R}{2}$ (since the ratio is 2:1).

For a solid cylinder (A), moment of inertia about its own axis is:
$I_A = \dfrac{1}{2} M R^2$

For a hollow cylinder (B), moment of inertia about its own axis is:
$I_B = M r^2 = M \left(\dfrac{R}{2}\right)^2 = \dfrac{1}{4} M R^2$

Now, the ratio:
$\dfrac{I_A}{I_B} = \dfrac{\dfrac{1}{2} M R^2}{\dfrac{1}{4} M R^2} = \dfrac{1/2}{1/4} = 2$

Answer: 2:1