| ABC is a right angled triangular plate 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? |
| [AIPMT 2000] |
 |
A) \[{{I}_{1}}={{I}_{2}}={{I}_{3}}\]
B) \[{{I}_{2}}>{{I}_{1}}>{{I}_{3}}\]
C) \[{{I}_{3}}<{{I}_{2}}<{{I}_{1}}\]
D) \[{{I}_{3}}>{{I}_{1}}>{{I}_{2}}\]
Correct Answer: B
Solution :
| 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 |
| \[{{I}_{2}}>{{I}_{1}}>{{I}_{3}}\] |
| Note: In a rotational motion, moment of inertia is also known as rotational inertia. |
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