Question

Water coming out of a horizontal tube at a speed $$v$$ strikes normally vertically was close to the mouth of the tube and falls down vertically after impact. When the speed of water is increased to $$2v$$.
A) The thrust exerted by the water on the wall will be doubled.
B) The thrust exerted by the water on the wall will be four times
C) The energy lost per second by water strike up the wall will also be four times
D) The energy lost per second by water striking the wall will be increased eight times

Answer 1

Hint: Here, initial velocity of water jet is given as $$v$$. Thrust exerted and energy loss per second is mentioned in the options. So, we need to calculate both the values. We know that, thrust or force exerted by water is related to change in linear momentum and the velocity. Similarly, energy loss per second is proportional to mass and velocity of water. Using these two formulas we can find out the changes when velocity of water is doubled.

 Formula used:
 $$F=\rho A{v}^2$$
 $$E={\displaystyle \frac{1}{2}}{\displaystyle \frac{dm}{dt}}{v}^2$$

Complete step by step answer:
 Here, water is coming out of a horizontal tube at a speed $$v$$.
 We know that,
 Force exerted by the jet on the wall, $$F=\rho A{v}^2$$
 Where,
$$v$$ is the velocity of water
$$\rho$$ is the density of the substance
$$m$$ is the mass striking per second
$$A$$ is the area of the cross section of the tube.
Here, initial thrust, $$F=\rho A{v}^2$$
When the velocity is doubled, i.e.,$$v=2v$$,
 Thrust, $$F^{\prime }=\rho A{\left(2v\right)}^2=4\rho A{v}^2=4F$$
 If velocity is doubled then thrust will increase 4 times.
 Energy loss per second, $$E={\displaystyle \frac{1}{2}}{\displaystyle \frac{dm}{dt}}{v}^2$$ ------- 1
We have,
 Mass flow rate,
 $${\displaystyle \frac{dm}{dt}}=\rho Av$$
 Substitute the above formula in equation 1, we get,
 Energy loss per second, $$E={\displaystyle \frac{1}{2}}\rho Av\times v^2={\displaystyle \frac{1}{2}}\rho A{v}^3$$
 When velocity is doubled,, i.e.,$$v=2v$$,
 Energy lost per second, $$E^{\prime }={\displaystyle \frac{1}{2}}\rho A\left(2v^3\right)=8\times {\displaystyle \frac{1}{2}}\rho A{v}^3=8E$$
Thus, when velocity is doubled the energy loss per second increases 8 times.

Therefore, options B and D are correct.

Note:
Thrust is a push or a force or the reaction force described quantitatively by Newton's third law is known as thrust. When a system accelerates or expels mass in a direction, the expelled mass will exert a force of equal magnitude but opposite direction on that system. The force applied on a surface in a direction normal or perpendicular to the surface is also called thrust. Force and thrust is measured in Newton.