A circular disc X of radius R is made from an iron  plate of thickness t  and another disc Y of radius 4R  is made from an iron plate of thickness t/4. Then the relation between the moment of inertia. I_{X} and I_{Y} is  

Option 1)

I_{Y}= 32I_{X}

Option 2)

I_{Y}= 16I_{X}

Option 3)

I_{Y}= I_{X}

Option 4)

I_{Y}= 64I_{X}

Answers (1)

As we learnt in

Moment of inertia for disc -

I=\frac{1}{2} MR^{2}

 

- wherein

About an axis perpendicular to the plane of disc & passing through its centre .

 

 Mass of Disc X=\left ( \pi R^{2} t \sigma \right )\ \; \Rightarrow\ \; I_{X}=\frac{MR^{2}}{2}=\frac{\left ( \pi R^{2}t\sigma \right )R^{2}}{2}=\frac{\pi R^{2}\sigma t}{2}

Mass of Disc Y=I_{Y}=\frac{M\left ( {4R} \right )^{2}}{2}=\frac{\pi\left ( 4R \right )^{2}}{2}\frac{t}{4}\sigma 16R^{2}=32{\pi}R^{4}t \sigma

\frac{I_{X}}{I_{Y}}=\frac{\pi R^{4}\sigma t}{2}\times\frac{1}{32{\pi}R^{4}\sigma t}=\frac{1}{64}

I_{Y}=64 \; I_{X}

 


Option 1)

I_{Y}= 32I_{X}

This is an incorrect option.

Option 2)

I_{Y}= 16I_{X}

This is an incorrect option.

Option 3)

I_{Y}= I_{X}

This is an incorrect option.

Option 4)

I_{Y}= 64I_{X}

This is the correct option.

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