Question

Two discs of same mass and same thickness have densities as $17g/{cm}^3$ and $51g/{cm}^3$. The ratio of their moment of inertia are in the ratio

• A. $1:3$
• B. $3:1$
• C. $\frac{1}{\sqrt[3]{39}}$
• D. $\sqrt[3]{39}$

Answer 1

The correct option is D $\sqrt[3]{39}$

$I=\frac{1}{2}M{R}^2$ ......$\left(1\right)$

Now, $M=\frac{4}{3}\pi R^{3}d\rho$ where $m$ is the mass, $d$ is the thickness and $\rho$ is the density

$\Rightarrow R={\left(\frac{3M}{4\pi d\rho }\right)}^{\frac{1}{3}}$

$\Rightarrow R^2={\left(\frac{3M}{4\pi d\rho }\right)}^{\frac{2}{3}}$

Hence eqn$\left(1\right)$ becomes,

$I=\frac{1}{2}M\times {\left(\frac{3M}{4\pi d\rho }\right)}^{\frac{2}{3}}$ .......$\left(2\right)$

Now if $M$ and $d$ are remaining constant as per the given condition then

$I\propto \frac{I}{{\rho }^{\frac{2}{3}}}$ ......$\left(3\right)$

Given: moment of inertia of first disk of density ${\rho }_1=17$g/${cm}^3$ and

moment of inertia of second disk of density ${\rho }_2=51$g/${cm}^3$

$\frac{I_1}{I_2}={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\frac{2}{3}}$

$={\left(\frac{51}{17}\right)}^{\frac{2}{3}}$

$={\left(\frac{3}{1}\right)}^{\frac{2}{3}}$

$=9^{\frac{1}{3}}$