Question

Four thin uniform rods each of length $L$ and mass $m$ are joined to form a square. The moment of inertia of square about an axis along its one diagonal is

• A. $\frac{m{L}^2}{6}$
• B. $\frac{2}{3}m{L}^2$
• C. $\frac{3m{L}^2}{4}$
• D. $\frac{4m{L}^2}{3}$

Answer 1

The correct option is B $\frac{2}{3}m{L}^2$

$\begin{array}{c}\text{Moment of inertia about the center of}\\ \text{one rod}{I}_1=\frac{m{l}^2}{12}\text{. Moment of inertia}\\ \text{through the centre of the square, will}\\ \text{be through an axis parallel to moment}\end{array}$

$\begin{array}{c}\text{of inertia popendicular to the rod,}\\ I_2=I_1+m{\left(\frac{l}{2}\right)}^k=\frac{4m{\lambda }^2}{12}\\ \text{For few rods}\\ I_2=\frac{4m{d}^2}{12}\times 4\end{array}$

$\begin{array}{c}\text{From perpendicular axis theorem}\\ I_2=I_3+I_4\text{whose}{I}_3\text{and}{I}_4\\ \text{are moment of inertia through the diagonal}\\ \text{For square both diagonals are equal}\end{array}$

$\begin{array}{c}{I}_3=I_4\\ I_2=2I_3\\ \frac{4m{\lambda }^2}{12}\times 4=2I_3\Rightarrow I_3=\frac{2m{\lambda }^2}{3}\\ \text{Hence option}B\text{is correct}\end{array}$