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.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions