Question

In a metallic triangular sheet ABC, $AB=BC=l$. If M is mass of sheet, what is the moment of inertia about AC?

• A. $\frac{M{l}^2}{4}$
• B. $\frac{M{l}^2}{12}$
• C. $\frac{M{l}^2}{6}$
• D. $\frac{M{l}^2}{18}$

Answer 1

Answer: (B)

$\frac{M{l}^2}{12}$

The Moment of inertia of complete square ABCD about AC, is $\left(2M\right)l^2/12=M{l}^2/6$

As ABC and ADC are symmetric about AC,

Moment of Inertia of triangle ABC $=$ Moment of Inertia of triangle ADC

Thus, Moment of inertia of complete square ABCD = Moment of Inertia of triangle ABC + Moment of Inertia of triangle ADC = 2 x Moment of Inertia of triangle ABC

So, Moment of Inertia of triangle ABC about AC is $\frac{1}{2}\times M{l}^2/6=M{l}^2/12$

Option B is correct.