Question

A rod $0.5$ m long has two masses each of $20$ gram stuck at its ends. If the masses are treated as point masses and if the mass of the rod is neglected, then the moment of inertia of the system about a transverse axis passing through the centre is

• A. $1.25\times {10}^{-3}kg-m^2$
• B. $2.5\times {10}^{-3}kg-m^2$
• C. $4\times {10}^{-3}kg-m^2$
• D. $5\times {10}^{-3}kg-m^2$

Answer 1

The correct option is B $2.5\times {10}^{-3}kg-m^2$

$\begin{array}{c}I=M{\left(\frac{l}{2}\right)}^2\times 2\\ =\frac{M{l}^2}{2}=\frac{20}{1000}\times \frac{{\left(0.5\right)}^2}{2}\\ =\frac{20}{2\times 1000}\times \frac{25}{100}\\ =2.5\times {10}^{-3}kg/m^2\end{array}$

Hence,

option $\left(B\right)$ is correct answer.