Question

The ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. I_AB, I_BC and I_CA are the moments of inertia of the plate about AB, BC and CA respectively. Which one of the following relations is correct ?

• A. $I_{AB}>I_{BC}$
• B. $I_{BC}>I_{AB}$
• C. $I_{AB}+I_{BC}=I_{CA}$
• D. $I_{CA}ismaximum$

Answer 1

The correct option is B

$I_{BC}>I_{AB}$

Step-1: Definition of moment of inertia

The formula for moment of inertia is the “sum of the product of mass” of each particle with the “square of its distance from the axis of the rotation”.

The formula of Moment of Inertia is expressed as $I=\Sigma m_i{r_i}^2$

Step-2:distance if center of mass from sides of triangle

The intersection of medians is the center of mass of the triangle. Since the distances of center of mass from the sides is related as –

$X_{BC}>X_{AB}>X_{AC}$

Therefore, $I_{BC}>I_{AB}>I_{AC}$

Hence option (b) is the correct answer