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(D) The thrust exerted by the water on the wall is equal to the rate of change of linear momentum.Thrust $F = \frac{dp}{dt} = \dot{m}v$,where $\dot{m}

Vedclass · Accepted Answer

$B$ and $C$ both.

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(D) The thrust exerted by the water on the wall is equal to the rate of change of linear momentum.Thrust $F = \frac{dp}{dt} = \dot{m}v$,where $\dot{m}$ is the mass flow rate.Since $\dot{m} = \rho A v$,we have $F = (\rho A v) v = \rho A v^2$.Thus,$F \propto v^2$. When speed is increased from $v$ to $2v$,the thrust becomes $(2)^2 = 4$ times the original value.Energy lost per second $P = \frac{1}{2} \dot{m} v^2 = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} \rho A v^3$.Thus,$P \propto v^3$. When speed is increased from $v$ to $2v$,the energy lost per second becomes $(2)^3 = 8$ times the original value.Therefore,both options $B$ and $C$ are correct.

Column $I$	Column $II$
$A$. Stoke's law	$I$. Pressure and energy
$B$. Turbulence	$II$. Hydraulic lift
$C$. Bernoulli's Principle	$III$. Viscous drag
$D$. Pascal's law	$IV$. Reynold's number

Water coming out of a horizontal tube at a speed $v$ strikes a vertical wall normally,close to the mouth of the tube,and falls down vertically after impact. When the speed of water is increased to $2v$:

Similar Questions

$A$ large number of droplets,each of radius $r$,coalesce to form a bigger drop of radius $R$. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. The velocity of the drop is ($T=$ surface tension,$\rho =$ density)

Match the columns $I$ and $II$:

Column $I$ Column $II$

$A$. Stoke's law $I$. Pressure and energy

$B$. Turbulence $II$. Hydraulic lift

$C$. Bernoulli's Principle $III$. Viscous drag

$D$. Pascal's law $IV$. Reynold's number

The correct match is:

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