(A) According to Bernoulli's principle for the top surface of the water and the hole:$P_{top} + \rho gh + \frac{1}{2}\rho v_{top}^{2} = P_{atm} + \fra

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(A) According to Bernoulli's principle for the top surface of the water and the hole:$P_{top} + \rho gh + \frac{1}{2}\rho v_{top}^{2} = P_{atm} + \frac{1}{2}\rho v^{2}$Since the vessel is closed and the top pressure is $0.9 P_{0}$,and assuming $v_{top} \approx 0$:$0.9 P_{0} + \rho gh = P_{0} + \frac{1}{2} \rho v^{2}$$\rho gh - 0.1 P_{0} = \frac{1}{2} \rho v^{2}$Substituting the values: $10^{3} \times 10 \times (9/4) - 0.1 \times 10^{5} = \frac{1}{2} \times 10^{3} \times v^{2}$$22500 - 10000 = 500 v^{2}$$12500 = 500 v^{2}$$v^{2} = 25$$v = 5 \text{ m/s}$

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

Medium

$A$ closed vessel has a small hole in one face near the bottom,as shown in the figure. The velocity of water coming out from the hole at the given instant is ....... $m/s$. (Given: $P_{0} = 10^{5} \text{ N/m}^{2}$,$h = 9/4 \text{ m}$,density of water $\rho = 10^{3} \text{ kg/m}^{3}$,$g = 10 \text{ m/s}^{2}$)

A
$5$
B
$2$
C
$3$
D
$3\sqrt{5}$

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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

$A$ hole is made at the bottom of a tank filled with water (density $1000 \, kg/m^3$). If the total pressure at the bottom of the tank is $3 \, atm$ $(1 \, atm = 10^5 \, N/m^2)$,then the velocity of efflux is:

MediumAIIMS 2000

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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:

MediumAP EAMCET 2024

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$A$ tank is filled with water of density $1 \, g/cm^3$ and oil of density $0.9 \, g/cm^3$. The height of the water layer is $100 \, cm$ and the height of the oil layer is $400 \, cm$. If $g = 980 \, cm/s^2$,then the velocity of efflux from an opening in the bottom of the tank is

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$A$ container is filled with water up to a height $h$. $A$ small hole is made at the bottom. What is the ratio of the time taken for the water level to fall from $h$ to $\frac{h}{2}$ to the time taken for it to fall from $\frac{h}{2}$ to $0$?

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There are two identical small holes of cross-sectional area $a$ on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. The tank is resting on a smooth horizontal surface. The horizontal force that must be applied to the tank to keep it in equilibrium is

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$A$ hole is made at the bottom of a tank filled with water (density $1000 \, kg/m^3$). If the total pressure at the bottom of the tank is $3 \, atm$ $(1 \, atm = 10^5 \, N/m^2)$,then the velocity of efflux is:

$A$ 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:

$A$ tank is filled with water of density $1 \, g/cm^3$ and oil of density $0.9 \, g/cm^3$. The height of the water layer is $100 \, cm$ and the height of the oil layer is $400 \, cm$. If $g = 980 \, cm/s^2$,then the velocity of efflux from an opening in the bottom of the tank is

$A$ container is filled with water up to a height $h$. $A$ small hole is made at the bottom. What is the ratio of the time taken for the water level to fall from $h$ to $\frac{h}{2}$ to the time taken for it to fall from $\frac{h}{2}$ to $0$?

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