Question

Three identical rods each of mass m and length $\ell$ are joined to form an equilateral triangle. The moment of inertia about an axis passing through one corner and perpendicular to the plane of triangle, is

• A. $\frac{4}{3}m{\ell }^2$
• B. $\frac{9}{8}m{\ell }^2$
• C. $\frac{3}{4}m{\ell }^2$
• D. $\frac{3}{2}m{\ell }^2$

Answer 1

Answer: (C)

$\frac{3}{4}m{\ell }^2$

$\begin{array}{c}\text{Height of equilateral triangle}=\frac{\sqrt{3}}{2}l\\ \text{Moment of inertia at end of}rod=\frac{{ml}^2}{3}\\ \text{and at corner is}=\frac{{ml}^2}{12}\end{array}$

$\begin{array}{cc}{I}_{n0}t& =I_1+I_2+I_3+m\times {\left(\frac{\sqrt{3}}{2}\lambda \right)}^2\\ & =\frac{m{\lambda }^2}{3}+\frac{m{l}^2}{3}+\frac{R{\lambda }^2}{12}+\frac{3m{l}^2}{4}\end{array}$

$\begin{array}{c}=\frac{3}{2}m{d}^2\\ \text{Hence option}C\text{is correct}\end{array}$