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A circular disc $\mathrm{x}$ 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 $t / 4$. Then the relation between the moment of inertia $I_x$ and $I_Y$ is
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The correct answer is:
$I_Y=64 I_X$
Given,
The radius of the disc $=R$
The thickness of the disc $=t$
The radius of the second disc $=4 R$
The thickness of the second $\operatorname{disc}=\frac{t}{4}$
Now,
Moment of Inertia of $\operatorname{disc}^{\prime} X^{\prime}, I_x=\frac{M_x R_x^2}{2}$
Where $M_x=\rho\left(\pi R_x^2\right) t_x=\rho \pi R^2 t$
Moment of Inertia of $\operatorname{disc}^{\prime} Y^{\prime}, I_y=\frac{M_y R_y^2}{2}$
Where $M_y=\rho \pi(4 R)^2 \frac{t}{4}=4 \rho \pi R^2 t$
Ratio of the moment of inertia of both the discs
$\therefore \frac{I_x}{I_y}=\frac{\left(\rho \pi R^2 t\right) R^2}{2} \times \frac{2}{\left(4 \rho \pi R^2 t\right) 16 R^2}$
After solving
$\Rightarrow \frac{\mathrm{I}_x}{I_y}=\frac{1}{64} \Rightarrow I_y=64 I_x$
Hence, the moment of inertia of $Y$ is 64 times the moment of inertia of $X$
The radius of the disc $=R$
The thickness of the disc $=t$
The radius of the second disc $=4 R$
The thickness of the second $\operatorname{disc}=\frac{t}{4}$
Now,
Moment of Inertia of $\operatorname{disc}^{\prime} X^{\prime}, I_x=\frac{M_x R_x^2}{2}$
Where $M_x=\rho\left(\pi R_x^2\right) t_x=\rho \pi R^2 t$
Moment of Inertia of $\operatorname{disc}^{\prime} Y^{\prime}, I_y=\frac{M_y R_y^2}{2}$
Where $M_y=\rho \pi(4 R)^2 \frac{t}{4}=4 \rho \pi R^2 t$
Ratio of the moment of inertia of both the discs
$\therefore \frac{I_x}{I_y}=\frac{\left(\rho \pi R^2 t\right) R^2}{2} \times \frac{2}{\left(4 \rho \pi R^2 t\right) 16 R^2}$
After solving
$\Rightarrow \frac{\mathrm{I}_x}{I_y}=\frac{1}{64} \Rightarrow I_y=64 I_x$
Hence, the moment of inertia of $Y$ is 64 times the moment of inertia of $X$
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