Question

If a solid sphere of mass $5 kg$ and a disc of mass $4 kg$ have the same radius Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be $\frac{x}{7}$ The the value of $x$ is ___

Answer 1

Correct Answer: 5

The correct answer is 5.

$I_1=\frac{2}{5}{m}_1{R}^2+m_1{R}^2$
$I_1=m_1{R}^2\left(\frac{7}{5}\right)$
$I_1=7R^2$

$I_2=\frac{m_2{R}^2}{4}+m_2{R}^2$
$I_2=\frac{5}{4}{m}_2{R}^2$
$I_2=5R^2$
$\frac{I_2}{I_1}=\frac{5}{7}$
$x=5$

Answer 2

The moment of inertia \( I_1 \) of a solid sphere about a tangent to its surface is calculated as: \[ I_1 = I_{\text{CM}} + mR^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2 \] For a sphere with mass \( m = 5 \, \text{kg} \), we get: \[ I_1 = 7R^2 \] The moment of inertia \( I_2 \) of the disc about a tangent in its plane is: \[ I_2 = \frac{m_2 R^2}{4} + m_2 R^2 = \frac{5}{4} m_2 R^2 \] For the disc with mass \( m = 4 \, \text{kg} \), we get: \[ I_2 = 6R^2 \] The ratio \( \frac{I_2}{I_1} \) is: \[ \frac{I_2}{I_1} = \frac{5R^2}{7R^2} = \frac{5}{7} \] Thus, \( x = 5 \).