Question

A horizontal pipe has a cross section of $10{\text{cm}}^2$ in one region and of $5{\text{cm}}^2$ in another. The water velocity at the first is $5{\text{ms}}^{–1}$ and the pressure in the second is $2\times {10}^5{\text{Nm}}^{–2}.$ Find the pressure of water in $1^{st}$ region (in ${10}^2\text{Pa}$).

• A. $2000$
• B. $1500$
• C. $2375$
• D. $1177.5$

Answer 1

The correct option is C $2375$

Given:
$A_1=10{\text{cm}}^2$
$V_1=5\text{m/s}$
$A_2=5{\text{cm}}^2$
$P_2=2\times {10}^5{\text{Nm}}^{-2}$
${\rho }_{water}=1000{\text{kg/m}}^3$

Now, from equation of continuity $Q=AV=constant$
$\Rightarrow A_1{V}_1=A_2{V}_2$
$\Rightarrow 10\times 5=5\times V_2$
$\Rightarrow V_2=10\text{m/s}$

Again, applying Bernoulli's equation

$P_1+\frac{1}{2}{\rho }_{water}{V}_1^2=P_2+\frac{1}{2}{\rho }_{water}{V}_2^2\because h_1=h_2$
$\Rightarrow P_1+\frac{1}{2}\times 1000\times 5^2=2\times {10}^5+\frac{1}{2}\times 1000\times {10}^2$
$\Rightarrow P_1=2375\times {10}^2\text{Pa}$
Hence, $\left(A\right)$ is the correct answer.