Question

Figure shows a thin metallic triangular sheet $ABC$. The mass of the sheet is $M$. The moment of inertia of the sheet about side $AC$ is:

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

Answer 1

The correct option is B $\frac{M{l}^2}{12}$

Lets assume there is a square of length l of mass double that of triangle .So, its mass is '2M' and side 'l'.

Now if we find out the moment of inertia of square about 'AC' , the half it will give the moment of inertia of triangle about 'AC'.

So,Moment of inertia of square about 'AC' is

$=\frac{1}{12}\left(2M\right)L^2$

$=\frac{1}{6}M{l}^2$

So, the moment of inertia of triangle about side AC is

$=\frac{1}{12}\left(\frac{1}{6}M{l}^2\right)$

$=\frac{1}{12}M{l}^2$