Question

A uniform triangular plate of mass $M$ whose vertices are ABC has lengths $l,\frac{l}{\sqrt{2}}$ and $\frac{l}{\sqrt{2}}$ as shown in figure. The moment of inertia of this plate about an axis passing through point A and perpendicular to the plane of the plate is

• A. $\frac{m{l}^2}{6}$
• B. $\frac{m{l}^2}{3}$
• C. $\frac{2m{l}^2}{3}$
• D. $\frac{m{l}^2}{12}$

Answer 1

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

Consider a square plate having side length $\frac{l}{\sqrt{2}}$
MI of the square plate is 2 times that of the triangular plate.
$I_{square}=\frac{M{a}^2}{6}$ where $a$ is the length of any side and $M$ is the mass of the square plate, MI is about an axis passing through the center of mass of the square.
$I_{squareaboutcornerA}=I_A=\frac{M{a}^2}{6}+M{\left(\frac{\sqrt{a}}{2}\right)}^2=\frac{2}{3}M{a}^2$
$I_{triangleaboutA}=\frac{1}{2}{I}_A=\frac{1}{2}\times \frac{2}{3}\times M{a}^2=\frac{1}{3}M{a}^2$
$M_{triangle}=\frac{1}{2}M$
Given, $a=\frac{l}{\sqrt{2}}$
$\therefore I_{tirangleaboutA}=\frac{1}{3}M{a}^2=\frac{2}{3}{M}_{triangle}{a}^2=\frac{2}{3}{M}_{triangle}{\left(\frac{l}{\sqrt{2}}\right)}^2=\frac{1}{3}{M}_{triangle}{l}^2$