Question

Water flows through a horizontal pipe with a cross-sectional area of $2{\text{m}}^2$ at a speed of $5\text{m/s}$ with a pressure of $3\times {10}^5\text{Pa}$ at a point $A$. At point $B$, the cross-sectional area is $1{\text{m}}^2$. Calculate the pressure at point $B$.

• A. $5\times {10}^5\text{Pa}$
• B. $2.625\times {10}^5\text{Pa}$
• C. $3\times {10}^4\text{Pa}$
• D. $1.5\times {10}^5\text{Pa}$

Answer 1

The correct option is B $2.625\times {10}^5\text{Pa}$


Applying equation of continuity at $A\&B$;
$A_1{v}_1=A_2{v}_2...\left(i\right)$
$\Rightarrow 2\times 5=1\times v_2$
$\therefore v_2=10\text{m/s}$
Applying Bernouli's equation at $A\&B$,
$\Rightarrow P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2...\left(ii\right)$
$\Rightarrow 3\times {10}^5+\rho gh+\frac{1}{2}\times {10}^3\times 5^2=P_2+\rho gh+\frac{1}{2}\times {10}^3\times {10}^2$
$\because$Both $A$ and $B$ are at same elevation (level) hence, $h_1=h_2=h$
$\rho ={10}^3{\text{kg/m}}^3$=density of water
$\Rightarrow P_2=3\times {10}^5+\frac{1}{2}\times {10}^3\times 5^2-\frac{1}{2}\times {10}^3\times {10}^2$
$P_2=3\times {10}^5+\frac{1}{2}\times {10}^3\times [25-100]$
$P_2=3\times {10}^5+\frac{1}{2}\times {10}^3\times (-75)$
$\therefore P_2=262500=2.625\times {10}^5\text{Pa}$