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## Main Question or Discussion Point

A thin wire of length L and uniformly density ρ is bent into a circular loop with center at O.The moment of inertia of it about a tangential axis lying in the plane of loop is.

Ans : Mass M is not given,but ρ is given. So M=ρL3->(1) (L3 means L cube,no idea how to post it in that manner!). For a circular loop, we all know the formula for that condition is (3/2)MR square. Applying this in equation for equation (1) will give (3/2)ρL3 x (L square/4∏ square)

This will give 3ρL raised to 5 / 8∏ square.

But my teacher told the answer is 3ρL cube/8∏square.

This is not a home work.This is just a practice for me.

In this,where is my mistake?

Ans : Mass M is not given,but ρ is given. So M=ρL3->(1) (L3 means L cube,no idea how to post it in that manner!). For a circular loop, we all know the formula for that condition is (3/2)MR square. Applying this in equation for equation (1) will give (3/2)ρL3 x (L square/4∏ square)

This will give 3ρL raised to 5 / 8∏ square.

But my teacher told the answer is 3ρL cube/8∏square.

This is not a home work.This is just a practice for me.

In this,where is my mistake?