(A) Let $a$ be the area of the hole,$v_e$ be the velocity of efflux,and $h$ be the height of the liquid above the hole. Let $v$ be the speed with whic

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(A) Let $a$ be the area of the hole,$v_e$ be the velocity of efflux,and $h$ be the height of the liquid above the hole. Let $v$ be the speed with which the level decreases in the container.From the equation of continuity,$a v_e = A v \Rightarrow v = \frac{a v_e}{A}$.Using Bernoulli's theorem at the top surface and the orifice:$p_0 + h \rho g + \frac{1}{2} \rho v^2 = p_0 + \frac{1}{2} \rho v_e^2$$h \rho g + \frac{1}{2} \rho \left(\frac{a v_e}{A}\right)^2 = \frac{1}{2} \rho v_e^2$$v_e^2 = \frac{2gh}{1 - (a/A)^2}$Given $h = 3 \,m - 0.525 \,m = 2.475 \,m$,$a/A = 0.1$,and $g = 10 \,ms^{-2}$:$v_e^2 = \frac{2 \times 10 \times 2.475}{1 - (0.1)^2} = \frac{49.5}{1 - 0.01} = \frac{49.5}{0.99} = 50 \,m^2 \,s^{-2}$.

Class 11 Physics Fluid Mechanics and Surface Tension Velocity of Efflux and Torricelli's law

Medium

$A$ cylindrical tank is filled with water to a level of $3 \,m$. $A$ hole is opened at a height of $52.5 \,cm$ from the bottom. The ratio of the area of the hole to that of the cross-sectional area of the tank is $0.1$. The square of the speed with which water will be coming out from the orifice is $(g=10 \,ms^{-2})$.

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A
$50 \,m^2 \,s^{-2}$
B
$40 \,m^2 \,s^{-2}$
C
$51.5 \,m^2 \,s^{-2}$
D
$50.5 \,m^2 \,s^{-2}$

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Class 11

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Fluid Mechanics and Surface Tension

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Velocity of Efflux and Torricelli's law

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$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of the time taken for the level to fall from $h$ to $h/2$ and that taken for the level to fall from $h/2$ to $0$ is:

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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