Question

Water flows through a vertical tube of variable cross section. The area of cross - section at $A$ and $B$ are $6{\text{mm}}^2$ and $3{\text{mm}}^2$ respectively. If $12{\text{cm}}^3$ of water enters per second through $A$, find the pressure difference $|P_A-P_B|$. The separation between the cross-section at $A$ and $B$ is $100\text{cm}$. (Take density of water as $1000{\text{kg/m}}^3$ and $g=10{\text{m/s}}^2$)

• A. $7000{\text{N/m}}^2$
• B. $3000{\text{N/m}}^2$
• C. $6000{\text{N/m}}^2$
• D. $4000{\text{N/m}}^2$

Answer 1

The correct option is D $4000{\text{N/m}}^2$

Given area $A_A=6{\text{mm}}^2$
area $A_B=3{\text{mm}}^2$
Given, discharge, $Q=A_A{V}_A=A_B{V}_B=12{\text{cm}}^3/\text{sec}$
$V_A=\frac{12\times {10}^{-6}}{A_A}=\frac{12\times {10}^{-6}}{6\times {10}^{-6}}=2\text{m/s}$
$V_B=\frac{12\times {10}^{-6}}{A_B}=\frac{12\times {10}^{-6}}{3\times {10}^{-6}}=4\text{m/s}$
Applying Bernoulli's theorem between $A$ and $B$, we get
$P_A+\rho g{h}_A+\frac{1}{2}\rho V_A^2=P_B+\rho g{h}_B+\frac{1}{2}\rho V_B^2$
$\Rightarrow P_A-P_B=\frac{1}{2}\rho (V_B^2-V_A^2)+\rho g(h_B-h_A)$
$\Rightarrow P_A-P_B=\frac{1}{2}\times 1000\times (4^2-2^2)+{10}^3\times 10\times (-\frac{100}{100})$
$\Rightarrow P_A-P_B=-4000{\text{N/m}}^2$
$\therefore |P_A-P_B|=4000{\text{N/m}}^2$