JEE Main Rotational Motion PYQ

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76. (AIEEE 2003)

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

A. $I_{Y} = 32 I_{X}$

B. $I_{Y} = 16 I_{X}$

C. $I_{Y} = I_{X}$

D. $I_{Y} = 64 I_{X}$

Correct Answer is Option (D)

We know that density $(d) = \frac{m a s s (M)}{v o l u m e (V)}$

$∴$ Mass of disc $M = d \times V = d \times (π R^{2} \times t) .$

The moment of inertia of any disc is $I = \frac{1}{2} M R^{2}$

$∴$ $I = \frac{1}{2} (d \times π R^{2} \times t) R^{2} = \frac{π d}{2} t \times R^{4}$

$∴$ $\frac{I_{X}}{I_{Y}} = \frac{t_{X} R_{X}^{4}}{t_{Y} R_{Y}^{4}}$

$= \frac{t \times R^{4}}{\frac{t}{4} \times {(4 R)}^{4}}$

$= \frac{1}{64}$

77. (AIEEE 2002)

Moment of inertia of a circular wire of mass $M$ and radius $R$ about its diameter is :

A. $\frac{M R^{2}}{2}$

B. $M R^{2}$

C. $2 M R^{2}$

D. $\frac{M R^{2}}{4}$

Correct Answer is Option (A)

Moment of Inertia of a circular wire about an axis $n n^{'}$ passing through the centre of the circle and perpendicular to the plane of the circle $= M R^{2}$
AIEEE 2002 Physics - Rotational Motion Question 166 English Explanation
As shown in the figure, $X$ -axis and $Y$ -axis lies in the plane of the ring. Then by perpendicular axis theorem

$I_{X} + I_{Y} = I_{Z}$

$\Rightarrow 2 I_{X} = M R^{2}$ $[$ as $I_{X} - I_{Y}$ (by symmetry) and $I_{Z} = M R^{2}$ $]$

$∴$ $I_{X} = \frac{1}{2} M R^{2}$