Question

The reading of a manometer fitted to a closed tap is $3.5\times {10}^{5}N{m}^2$. If the valve is opened the reading of the manometer falls to $3\times {10}^{5}N{m}^2$. The velocity of water is :

• A. $0.1m{s}^{-1}$
• B. $1m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. $100m{s}^{-1}$

Answer 1

Answer: (C)

$10m{s}^{-1}$

Applying Bernouilli's theorem to the fluid before and after the volume is opened,
$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$

where velocity of water before the tap is opened $v_1=0$

Given : $P_1=3.5\times {10}^{5}N{m}^{-2}$, $P_2=3\times {10}^{5}N{m}^{-2}$ and $\rho =1000kg/m^3$

$3.5\times {10}^5+\frac{1}{2}\left(1000\right)\left(0\right)=3\times {10}^5+\frac{1}{2}\left(1000\right)v_2^2$

Thus we get $v_2^2=1\times {10}^2$

$⟹$ $v_2=10m/s$