Question

Question:A water supply maintains a constant rate of flow for water in a hose. You want to change the opening of the nozzle so that water leaving the nozzle will reach a height that is four times the current maximum height the water reaches with the nozzle vertical. To do so, should you (a) decrease the area of the opening by a factor of 16, (b) decrease the area by a factor of 8, (c) decrease the area by a factor of 4, (d) decrease the area by a factor of 2, or (e) give up because it cannot be done?

Short answer

To double the velocity, the area must be reduced by a factor of 2. Hence the correct answer is option (d).

Step-by-step solution

The flow rate

The flow rate (volume flux) through a pipe that varies in cross-sectional area is constant; that is equivalent to stating that the product of the cross-sectional area A and the speed v at any point is a constant. This result is expressed in the equation of continuity for fluids:

A₁V₁ = A₂V₂ = constant

Where

V₁= Initial volume of fluid

V₁=Final volume of fluid

A₁=Initial Area

A₂=Final Area

Find the correct answer

LetsupposetheinitialareaisA,andinitialvolumeisV.

Thus,

A₁ = A₁V₁ = V, and V₂we get from formula V₂=$\sqrt{2g{Y}_2}$

$\frac{V_2}{V_1}=\sqrt{\frac{Y_2}{Y_1}}$

$\frac{V_2}{V_1}=\sqrt{\frac{4Y}{Y}}$

$\frac{V_2}{V_1}=\sqrt{4}=2$

V₂= 2V₁ =2V

Nowbyusingconceptwecanwrite,

A₁V₁ = A₂V₂ = constant

A_.V =A₂.2V

A₂=$\frac{A}{2}$

From V =$\sqrt{2gY}$, the velocity needs to be double if height quadruples. By using concept we have seen that if we need to double the velocity, the area must be reduced by a factor of 2.

Hence the correct answer is option (d).