At what speed,the velocity head of water is e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The velocity head of water is given by $\frac{V^2}{2g}$.The pressure head of mercury is given as $h_{Hg} = 40 \, cm = 0.4 \, m$.The pressure exert

Vedclass · Accepted Answer

$10.32$

Vedclass · Answer

(B) The velocity head of water is given by $\frac{V^2}{2g}$.The pressure head of mercury is given as $h_{Hg} = 40 \, cm = 0.4 \, m$.The pressure exerted by this mercury column is $P = ho_{Hg} g h_{Hg}$.Equating the pressure head of water to the pressure head of mercury:$\frac{P}{ho_w g} = \frac{ho_{Hg} g h_{Hg}}{ho_w g} = \frac{ho_{Hg}}{ho_w} h_{Hg}$.Given that the velocity head $\frac{V^2}{2g}$ is equal to this pressure head:$\frac{V^2}{2g} = \frac{ho_{Hg}}{ho_w} h_{Hg}$.$V^2 = 2g 	imes \frac{ho_{Hg}}{ho_w} 	imes h_{Hg}$.Substituting the values: $V^2 = 2 	imes 9.8 	imes 13.6 	imes 0.4$.$V^2 = 106.5984$.$V = \sqrt{106.5984} \approx 10.32 \, m/s$.

At what speed,the velocity head of water is equal to the pressure head of $40 \, cm$ of mercury? (Given: Density of mercury = $13.6 \times 10^3 \, kg/m^3$,Density of water = $10^3 \, kg/m^3$,$g = 9.8 \, m/s^2$)

Similar Questions

$A$ barometer is constructed using a liquid (density $= 760 \; kg/m^3$). What would be the height (in $m$) of the liquid column,when a mercury barometer reads $76 \; cm$? (density of mercury $= 13600 \; kg/m^3$)

Explain how an open-ended tube manometer measures pressure.

What would be the absolute pressure at a depth of $1 \,km$ below the ocean surface? [Given: density of water $\rho = 10^{3} \,kg/m^{3}$,$g = 10 \,m/s^{2}$,$1 \,atm = 1.01 \times 10^{5} \,N/m^{2}$]

On what factors does the pressure at any place depend? Explain.

Explain why you cannot make a decent cup of tea on a mountain.

Vedclass Test Series

Exam Paper Generator

Online Exam Module