(B) According to Bernoulli's theorem,the kinetic energy of the water stream is converted into potential energy at the top of the jet.Let $v = 2.45 \ m

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(B) According to Bernoulli's theorem,the kinetic energy of the water stream is converted into potential energy at the top of the jet.Let $v = 2.45 \ m/s$ be the velocity of the water stream.The maximum height $H$ to which the water can rise above the level of the stream is given by $H = \frac{v^2}{2g}$.Substituting the values,$H = \frac{(2.45)^2}{2 \times 9.8} = \frac{6.0025}{19.6} = 0.30625 \ m = 30.625 \ cm$.The height of the jet above the water surface is $h_{jet} = H - h_{above}$,where $h_{above} = 10.6 \ cm$.Therefore,$h_{jet} = 30.625 \ cm - 10.6 \ cm = 20.025 \ cm \approx 20.0 \ cm$.

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

Medium

An $L$-shaped tube with a small orifice is held in a water stream as shown in the figure. The upper end of the tube is $10.6 \ cm$ above the surface of the water. What will be the height (in $cm$) of the jet of water coming from the orifice? The velocity of the water stream is $2.45 \ m/s$. (Take $g = 9.8 \ m/s^2$)

A
$0$
B
$20.0$
C
$10.6$
D
$40.0$

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Fluid Mechanics and Surface Tension

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Bernoulli's Theorem and Applications of Bernoulli's Theory

$A$ horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity $1 \ m/s$ when the pressure is $50 \ kPa$ at a point. If the velocity of flow has to be $2 \ m/s$ at some other point, the pressure at that point should be: (in $kPa$)

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An aeroplane of mass $4.5 \times 10^4 \,kg$ and total wing area of $600 \,m^2$ is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

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The flow speeds of air on the lower and upper surfaces of the wing of an aeroplane are $v$ and $2v$ respectively. The density of air is $\rho$ and the surface area of the wing is $A$. The dynamic lift on the wing is:

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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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$Assertion :$ The pressure of water increases when it flows from a narrow pipe to a wider pipe.
$Reason :$ According to the equation of continuity,$A_1v_1 = A_2v_2$,and by Bernoulli's principle,where velocity is high,pressure is low.

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$A$ horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity $1 \ m/s$ when the pressure is $50 \ kPa$ at a point. If the velocity of flow has to be $2 \ m/s$ at some other point, the pressure at that point should be: (in $kPa$)

An aeroplane of mass $4.5 \times 10^4 \,kg$ and total wing area of $600 \,m^2$ is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

The flow speeds of air on the lower and upper surfaces of the wing of an aeroplane are $v$ and $2v$ respectively. The density of air is $\rho$ and the surface area of the wing is $A$. The dynamic lift on the wing is:

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.

$Assertion :$ The pressure of water increases when it flows from a narrow pipe to a wider pipe.$Reason :$ According to the equation of continuity,$A_1v_1 = A_2v_2$,and by Bernoulli's principle,where velocity is high,pressure is low.

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$Assertion :$ The pressure of water increases when it flows from a narrow pipe to a wider pipe.
$Reason :$ According to the equation of continuity,$A_1v_1 = A_2v_2$,and by Bernoulli's principle,where velocity is high,pressure is low.