ABC is the right-angled triangular plane of uniform thickness. The sides are such that AB>BC as shown in the figure. I₁, I₂, I₃ are moments of inertia about AB, BC and AC, respectively. Then which of the following relations is correct?

I₁ = I₂ = I₃

I₂ > I₁> I₃

I₃<I₂<1₁

I₃>I₁>I₂

Solution

I₂ > I₁> I₃

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 the same, so it will depend only on the distribution of mass about the axis. The mass is farthest from axis BC, so I₂ is maximum mass is nearest to axis AC, so I₃ is minimum.

Hence, the correct sequence will be I₂>I₁>I₃

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ABC is the right-angled triangular plane of uniform thickness. The sides are such that AB>BC as shown in the figure. I1, I2, I3 are moments of inertia about AB, BC and AC, respectively. Then which of the following relations is correct? I1 = I2 = I3 I2 > I1> I3 I3<I2<11 I3>I1>I2

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Four point masses, each of value m, are placed at the corners of a square ABCD of side A. The moment of inertia through A and parallel to BD is

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