# The solid cylinder of length

Question:

The solid cylinder of length $80 \mathrm{~cm}$ and mass $M$ has a radius of $20 \mathrm{~cm}$. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to $\mathrm{AB}$ as shown

in figure is $2.7 \mathrm{~kg} \mathrm{~m}^{2}$.

1. $14.9 \mathrm{~kg} / \mathrm{m}^{3}$

2. $7.5 \times 10^{1} \mathrm{~kg} / \mathrm{m}^{3}$

3. $7.5 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}$

4. $1.49 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}$

Correct Option: , 4

Solution:

Parallel axis theorem

$\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Md}^{2}$

$\mathrm{I}=\frac{\mathrm{Mr}^{2}}{2}+\mathrm{M}\left(\frac{\mathrm{L}}{2}\right)^{2}$

$2.7=\mathrm{M} \frac{(0.2)^{2}}{2}+\mathrm{M}\left(\frac{0.8}{2}\right)^{2}$

$2.7=M\left[\frac{2}{100}+\frac{16}{100}\right]$

$M=15 k g$

$\Rightarrow \rho=\frac{\mathrm{M}}{\pi \mathrm{r}^{2} \mathrm{~L}}=\frac{15}{\pi(0.2)^{2} \times 0.8}$

$=0.1492 \times 10^{3}$

Ans. 4