(C) Let $A = 0.5 \; m^{2}$ be the area of the tank,$a = 1 \; cm^{2} = 10^{-4} \; m^{2}$ be the area of the opening,$M = 25 \; kg$ be the mass applied,

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(C) Let $A = 0.5 \; m^{2}$ be the area of the tank,$a = 1 \; cm^{2} = 10^{-4} \; m^{2}$ be the area of the opening,$M = 25 \; kg$ be the mass applied,and $h = 40 \; cm = 0.4 \; m$ be the height of the water column.Applying Bernoulli's principle at the top surface and the opening:$P_{top} + \rho g h = P_{atm} + \frac{1}{2} \rho v^{2}$Here,$P_{top} = P_{atm} + \frac{Mg}{A}$.Substituting the values:$P_{atm} + \frac{25 \times 10}{0.5} + 1000 \times 10 \times 0.4 = P_{atm} + \frac{1}{2} \times 1000 \times v^{2}$$500 + 4000 = 500 v^{2}$$4500 = 500 v^{2}$$v^{2} = 9$$v = 3 \; m \; s^{-1} = 300 \; cm \; s^{-1}$.

Class 11 Physics Fluid Mechanics and Surface Tension Bernoulli's Theorem and Applications of Bernoulli's Theory

Medium

The area of cross-section of a large tank is $0.5 \; m^{2}$. It has a narrow opening near the bottom having an area of cross-section $1 \; cm^{2}$. $A$ load of $25 \; kg$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank,the velocity of the water coming out of the opening at the time when the height of water level in the tank is $40 \; cm$ above the bottom will be $\dots \; cm \; s^{-1}$. [Take $g = 10 \; m \; s^{-2}$]

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Fluid Mechanics and Surface Tension

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Bernoulli's Theorem and Applications of Bernoulli's Theory

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Rank in order, from highest to lowest, the liquid heights $h_a$ to $h_d$. The air flow is from left to right. The liquid columns are not drawn to scale.

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$A$ fluid is flowing through a horizontal pipe of varying cross-section,with speed $v \ m/s$ at a point where the pressure is $P \ Pa$. At another point where pressure is $\frac{P}{2} \ Pa$,its speed is $V \ m/s$. If the density of the fluid is $\rho \ kg/m^3$ and the flow is streamline,then $V$ is equal to

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Column - $I$	Column - $II$
$(a)$ Velocity head	$(i)$ $\frac{P}{\rho g}$
$(b)$ Pressure head	$(ii)$ $h$
$(c)$ Potential head	$(iii)$ $\frac{v^2}{2g}$

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The speeds of air above and below the wing of an airplane are $120\, m/s$ and $90\, m/s$,respectively. The density of air is $1.3\, kg/m^3$. If the wing has a length of $10\, m$ and a width of $2\, m$,the pressure difference between the top and bottom of the wing is ......... $Pascal$.

Different heads are in Column - $I$ and their formulas are given in Column - $II$. Match them appropriately.Column - $I$Column - $II$$(a)$ Velocity head$(i)$ $\frac{P}{\rho g}$$(b)$ Pressure head$(ii)$ $h$$(c)$ Potential head$(iii)$ $\frac{v^2}{2g}$

In this figure,an ideal liquid flows through a tube of uniform cross-section. The liquid has velocities $v_A$ and $v_B$,and pressures $P_A$ and $P_B$ at points $A$ and $B$ respectively.

Rank in order, from highest to lowest, the liquid heights $h_a$ to $h_d$. The air flow is from left to right. The liquid columns are not drawn to scale.

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Different heads are in Column - $I$ and their formulas are given in Column - $II$. Match them appropriately.
Column - $I$ Column - $II$
$(a)$ Velocity head $(i)$ $\frac{P}{\rho g}$
$(b)$ Pressure head $(ii)$ $h$
$(c)$ Potential head $(iii)$ $\frac{v^2}{2g}$