Question

A manometer connected to a closed tap reads $3.5\times {10}^{5}N/m^2$. When the value is opened,the reading of manometer fall to $3.0\times {10}^{5}N/m^2$, then velocity of flow of water is

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

Answer 1

The correct option is C $10m/s$

Bernoulli's theorem for unit mass of liquid $\frac{P}{\rho }+\frac{1}{2}{v}^2=cons\tan t$

As the liquid starts flowing . it pressure energy decreases .

$\begin{array}{c}\\ \frac{1}{2}{v}^2=\frac{P_1-P_2}{\rho }\\ \Rightarrow \frac{1}{2}{v}^2=\frac{3.5\times {10}^5-3\times {10}^5}{{10}^3}\\ \Rightarrow v^2=\frac{2\times 0.5\times {10}^5}{{10}^3}\\ \therefore v^2=100\\ \therefore v=10m/s\end{array}$