Question

Water flows through a pipe of radius 1.0 cm . The viscosity of water is 10${}^{-3}$ kg $m^{-1}{s}^{-1}$. If the velocity of the flow at the centre is 10 cm s${}^{-1}$, the pressure drop along a 2m section of the pipe due to viscosity is (in Nm${}^{-2}$):

• A. 0.008
• B. 0.08
• C. 0.8
• D. 8

Answer 1

The correct option is D 8

(Poiseulli's theorem)
As the Q is constant along the pipe
$\Rightarrow \frac{\pi P{r}^4}{8\eta L}=\frac{\pi P_1{r}^4}{8\eta (L-2)}$
$\Rightarrow \frac{P}{L}=\frac{P_1}{L-2}$
$\Rightarrow PL-2P=P_{1}L$
$\Rightarrow (P-P_1)L=2P$
$\Rightarrow P-P_1=\frac{2P}{L}$-------------------(1)
$Q=\pi \frac{({10}^{-2}{)}^2}{10}=\pi \times {10}^{-5}{m}^3/s$ [from poiseulli's equation and equation]
$\Rightarrow \pi \times {10}^{-5}=\frac{\pi \times P\times ({10}^{-2}{)}^4}{8\times {10}^{-3}\times L}$
$\Rightarrow 8\times {10}^{-8}=2\times {10}^{-8}\times \frac{P}{L}\Rightarrow \frac{2P}{L}=8$---------------(2)
From (1) and (2),
$P-P_1=8$