Question

ABC is right angled triangular plane of uniform thickness. The sides are such that $AB>BC$ as shown in figure. $I_1,I_2,I_3$ are moments of inertia about AB, BC and AC, respectively. Then which of the following relations is correct?

• A. $I_1=I_2=I_3$
• B. $I_2>I_1>I_3$
• C. $I_3<I_2<I_1$
• D. $I_3>I_1>I_2$

Answer 1

The correct option is B $I_2>I_1>I_3$

The moment of inertia of a body about an axis depends not only on the mass of the body but also on the distribution of mass about the axis. For a given body mass is same, so it will depend only on the distribution of mass about the axis. The mass is farthest from axis BC, so $I_2$ is maximum mass is nearest to axis AC, so $I_3$ is minimum.
Hence, the correct sequence will be $I_2>I_1>I_3$.