A cylindrical vessel of cross-sectional area | vedclass

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(B) The pressure at the hole is due to the atmospheric pressure,the weight of the piston,and the height of the water column.Let $A = 1000 \ cm^2 = 0.1

Vedclass · Accepted Answer

$3.4$

Vedclass · Answer

(B) The pressure at the hole is due to the atmospheric pressure,the weight of the piston,and the height of the water column.Let $A = 1000 \ cm^2 = 0.1 \ m^2$ be the area of the piston and $m = 10 \ kg$ be its mass.The pressure exerted by the piston is $P_p = \dfrac{mg}{A} = \dfrac{10 	imes 9.8}{0.1} = 980 \ Pa$.The equivalent height of water column $h_p$ that produces this pressure is $h_p = \dfrac{P_p}{ho g} = \dfrac{980}{1000 	imes 9.8} = 0.1 \ m = 10 \ cm$.The total effective height of the water column above the hole is $H_{eff} = h_p + h = 10 \ cm + 50 \ cm = 60 \ cm = 0.6 \ m$.Using Torricelli's law,the velocity of efflux is $v = \sqrt{2gH_{eff}}$.$v = \sqrt{2 	imes 9.8 	imes 0.6} = \sqrt{11.76} \approx 3.43 \ m/s$.Thus,the velocity is approximately $3.4 \ m/s$.

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