Question

One end of a horizontal pipe is closed with the help of a valve and the reading of a barometer attached to the pipe is $3\times {10}^5$ pascal. When the valve in the pipe is opened then the reading of the barometer falls to ${10}^5$ pascal. The velocity of water flowing through the pipe will be in m/s.

• A. $0.2$
• B. $2$
• C. $20$
• D. $200$

Answer 1

Answer: (C)

$20$

Applying Bernoulli's principle on a horizontal pipe, we have
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2$
here $P_1=3\times {10}^5$ pascal , $v_1=0$, $P_2={10}^5$ Pascal, $\rho =1000kg/m^3$
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2\Rightarrow 3\times {10}^5={10}^5+\frac{1}{2}\times 1000\times {v_2}^2$
Solving the above equation we have, $v_2=20m/s$