Four point masses each of value  m  , are placed at the corners of a square  ABCD  of side  l   The moment of inertia of this system about an axis through  A  and  parallel to  BD  is

Option 1)

ml^{2}

Option 2)

2ml^{2}

Option 3)

\sqrt{3}ml^{2}

Option 4)

3ml^{2}

Answers (1)

As we learnt in

Moment of inertia for system of particle -

I= m_{1}r_{1}^{2}+m_{2}r_{2}^{2}+.........m_{n}r_{n}^{2}

\dpi{100} = \sum_{i=1}^{n}\: m_{i}r_{i}^{2}

 

- wherein

Applied when masses are placed discretely.

 

 

AOcos45^{\circ}=\frac{l}{2}\; \; \; \Rightarrow \; \; \; AO=\frac{l}{\sqrt{}2}

 

I=I_{D}+I_{B}+I_{C}\; \; \; or\; \; \; I=\frac{2ml^{2}}{2}+m\left ( \frac{2l}{\sqrt{2}} \right )^{2}

I=\frac{2ml^{2}}{2}+\frac{4ml^{2}}{2}

or\; \; \;I= \frac{6ml^{2}}{2}=3ml^{2}


Option 1)

ml^{2}

This is an incorrect option.

Option 2)

2ml^{2}

This is an incorrect option.

Option 3)

\sqrt{3}ml^{2}

This is an incorrect option.

Option 4)

3ml^{2}

This is the correct option.

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