Question

Two discs of the same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio 1:3. The moments of inertia of these discs about the respective axes passing through their centres and perpendicular to their planes will be in the ratio of

• A. 1:3
• B. 3:1
• C. 1:9
• D. 9:1

Answer 1

The correct option is B 3:1

Given,

Thickness $=t_1=t_2$

Masses $=m_1=m_2$

Density, $\left(\frac{{\rho }_1}{{\rho }_2}­\right)=\left(\frac{1}{3}\right)$

$I=MI$ of disc $=\left[\frac{(MR{)}^2}{2}\right]$ .......(1)

Also

$\rho =\left[\frac{Mass}{Volume}\right]=\left[\frac{Mass}{Area\times Thickness)}\right]$

$=\left[\frac{m}{\left(\pi R_{2}t\right)}\right]$

$R_2=\left[\frac{m}{\left(\pi t\rho \right)}\right]$

From equation (1)

$I=\left(\frac{M}{2}\right)R_2=\left(\frac{M}{2}\right)\times \left[\frac{M}{\left(\pi \pi t\rho \right)}\right]=\left[\frac{M_2}{\left(2\pi T\rho \right)}\right]$

Given,

Thickness & masses same.

Hence $l\propto \left(\frac{1}{\rho }\right)$

$\left(\frac{I_1}{I_2}\right)=\left(\frac{{\rho }_2}{{\rho }_1}\right)$ as $t_1=t_2$ and $m_1=m_2$

$\left(\frac{I_1}{I_2}\right)=\left(\frac{3}{1}\right)$

So, $I_1:I_2=3:1$