Question

A manometer connected to a closed tap reads $3.5\times {10}^5{\text{N/m}}^2$. When the valve is opened, the reading of the manometer falls to $3.0\times {10}^5{\text{N/m}}^2$. Then, the velocity of flow of water is:-

• A. $100\text{m/s}$
• B. $10\text{m/s}$
• C. $1\text{m/s}$
• D. $10\sqrt{10}\text{m/s}$

Answer 1

The correct option is B $10\text{m/s}$

Fluid is not at rest. To measure the flow speed of water in a pipe, by applying Bernoulli’s principle in the two situations:
[since the horizontal level of the pipe doesn't change]

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
$\Rightarrow \frac{1}{2}\rho [v_2^2-v_1^2]=P_1-P_2$

From the data given in the question,

$\frac{1}{2}\times {10}^3[v^2-0]=3.5\times {10}^5-3.0\times {10}^5$
[initially, water doesn't flow, so $v_1=0$]
$\Rightarrow v=\sqrt{100}=10\text{m/s}$

Thus, option (b) is the correct answer.