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Three point masses, each of m, are placed at the corners of an equilateral triangle of side l. Then the moment of inertia of this system about an axis along one side of the triangle is

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a

3ml2

b

ml2

c

34ml2

d

32ml2

answer is C.

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Detailed Solution

Each angle of an equilateral triangle is 60∘⇒d=lsin⁡60∘⇒d=l32We know that the moment of inertia of a system of particles is given byI=∑i miri2⇒I=m(0)2+m(0)2+ml322⇒I=3ml24

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