Question

The cylinders P and Q are of equal of mas and length but made of metals with densities ${\rho }_P$ and ${\rho }_Q\left({\rho }_P>{\rho }_Q\right).$ If their moment of inertia about an axis passing through centre and normal to the circular face be $I_P$ and $I_Q$, then:

• A. $I_P=I_Q$
• B. $I_P>I_Q$
• C. $I_P<I_Q$
• D. $I_P⩽I_Q$

Answer 1

The correct option is C $I_P<I_Q$

Given: The cylinders P and Q are of equal of mas and length but made of metals with densities ${\rho }_P$ and ${\rho }_Q({\rho }_P>{\rho }_Q)$. If their moment of inertia about an axis passing through centre and normal to the circular face be $I_P$ and $I_Q$,

To find the relation between $I_P$ and $I_Q$

Solution:

Let mass be m,

We know, $m=\rho V$ (as mass=density$\times$volume)

$⟹m=\rho \times \pi r^{2}l⟹r^2=\frac{m}{\pi \rho l}.......\left(i\right)$

Moment of inertia about given axis is,

$I=\frac{1}{2}m{r}^2=\frac{1}{2}m\times \frac{m}{\pi \rho l}⟹I=\frac{m^2}{2\pi \rho l}⟹I\propto \frac{1}{\rho }⟹\frac{I_P}{I_Q}=\frac{{\rho }_Q}{{\rho }_P}$

Given,

${\rho }_P>{\rho }_Q⟹I_P<I_Q$