Question

When water $(\text{viscosity,}{\eta }_w=0.01\text{poise})$ and benzene $(\text{viscosity,}{\eta }_b=0.0065\text{poise})$ are allowed to flow through a pipe, it was found that the same amount of liquids are collected in the same time but the pressure drop that caused the flow are different. If the pressure drop in water is $0.015\text{atm}$, then the pressure drop in benzene will be

• A. $2.3\times {10}^{-3}\text{atm}$
• B. $975\times {10}^{-5}\text{atm}$
• C. $4.33\times {10}^{-3}\text{atm}$
• D. $2.3\times {10}^{-5}\text{atm}$

Answer 1

The correct option is B $975\times {10}^{-5}\text{atm}$

Now, as the rate of flow of both the water and the benzene are same, so we can say from Poiseuille's equation,
$Q_{water}=Q_{benzene}$
$\Rightarrow \frac{\pi P_w{r}^4}{8{\eta }_{w}l}=\frac{\pi P_b{r}^4}{8{\eta }_{b}l}$
$\Rightarrow \frac{\pi \times 0.015\times r^4}{8\times 0.01\times l}=\frac{\pi \times P_b\times r^4}{8\times 0.0065\times l}$
$\Rightarrow P_b=\frac{0.0065\times 0.015}{0.01}\text{atm}$
$\Rightarrow P_b=9.75\times {10}^{-3}\text{atm}$
$\Rightarrow P_b=975\times {10}^{-5}\text{atm}$