Question

The pressure of water in a water pipe when tap is opened and closed is respectively $3\times {10}^{5}N/m^2$ and $3.5\times {10}^{5}N/m^2$. With open tap, the velocity of water flowing is

• A. 10 m/s
• B. 5 m/s
• C. 20 m/s
• D. 15 m/s

Answer 1

Answer: (C)

20 m/s

$P_2=3.5\times {10}^{5}N/m^2$

$P_1=3\cdot 0\times {10}^{5}N/m^2$, but due to atmospheric pressure

$P_1^{\prime }={10}^{5}N/m^2$

$\begin{array}{cc}\text{now:}\left(\Delta P\right)A& =\frac{1}{2}\rho A{V}^2\\ 2\times {10}^4\times 2=\rho V^2& \\ \rho & ={10}^{3}kg/m^3\\ V& =20m/s\end{array}$

So, correct option is $\left(C\right)$.