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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.

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 $2v$:

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