The height of the water level in a tank of un | vedclass

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(B) The velocity of efflux $(v)$ from a hole at the bottom of a tank is given by Torricelli's law: $v = \sqrt{2gh}$.Given:Height of water level,$h = 5

Vedclass · Accepted Answer

$120 	imes 10^{-6} \,m^3$

Vedclass · Answer

(B) The velocity of efflux $(v)$ from a hole at the bottom of a tank is given by Torricelli's law: $v = \sqrt{2gh}$.Given:Height of water level,$h = 5 \,m$.Area of the hole,$A = 2.4 \,mm^2 = 2.4 	imes 10^{-6} \,m^2$.Time,$t = 5 \,s$.Acceleration due to gravity,$g = 10 \,ms^{-2}$.The volume of water leaked $(V)$ is given by the product of the area of the hole,the velocity of efflux,and the time:$V = A 	imes v 	imes t = A \sqrt{2gh} 	imes t$.Substituting the values:$V = (2.4 	imes 10^{-6} \,m^2) 	imes \sqrt{2 	imes 10 \,ms^{-2} 	imes 5 \,m} 	imes 5 \,s$.$V = 2.4 	imes 10^{-6} 	imes \sqrt{100} 	imes 5$.$V = 2.4 	imes 10^{-6} 	imes 10 	imes 5$.$V = 2.4 	imes 50 	imes 10^{-6} \,m^3$.$V = 120 	imes 10^{-6} \,m^3$.

The height of the water level in a tank of uniform cross-section is $5 \,m$. The volume of water leaked in $5 \,s$ through a hole of area $2.4 \,mm^2$ made at the bottom of the tank is (Assume the level of the water in the tank remains constant and acceleration due to gravity $= 10 \,ms^{-2}$).

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