Question

Water enters a horizontal pipe of non uniform cross section with a velocity of $0.4$ $m{s}^{-1}$ and leaves the other end with a velocity of $0.6$ $m{s}^{-1}$ .pressure of water at the first end is $1500$ $N{m}^{-2}$,then pressure at the other end is

• A. $1000$ $N{m}^{-2}$
• B. $1200$ $N{m}^{-2}$
• C. $1400$ $N{m}^{-2}$
• D. $1600$ $N{m}^{-2}$

Answer 1

The correct option is D $1600$ $N{m}^{-2}$

Bernoulli's equation for a horizontal pipe can be used to find the pressure at the second point of the pipe, which is

$\frac{1}{2}\rho V_1^2+P_1=\frac{1}{2}\rho V_2^2+P_2$

where

$\rho$ is the density of the water $=1000kg/m^3$

$V_1$ is the velocity at first point $=0.6m/s$

$P_1$ is the pressure at first point $=1500N/m^2$

$V_2$ is the velocity of second point $=0.4m/s$

$P_2$ is the pressure at second point $=?$

Putting these values in the Bernoulli's equation,

$\frac{1}{2}\left(1000\right)(0.6{)}^2+1500=\frac{1}{2}(1000\left)\right(0.4{)}^2+P_2$

$P_2=500\left[\right(0.6{)}^2-\left(0.4{)}^2\right]+1500=1600N{m}^{-2}$