Question

The height of a fountain of water, measured from the open end of the output pipe of a pump is H. The original power is P. What should be the power of the pump so that the total height of the fountain is unchanged when a vertical pipe of the same diameter and of height h<H is attached to the output pipe ?

• A. P
• B. $P(1-\frac{h}{H})^{1/2}$
• C. $P(1-\frac{h}{H})$
• D. $P(1-\frac{h}{H})^{3/2}$

Answer 1

Answer: (C)

P(1-$\frac{h}{H}$)

Answer 2

Let $m'$ be the mass flow rate. In the original setup, power $P$ is delivered to achieve height $H$. Assuming power is proportional to kinetic energy, $P \propto \frac{1}{2}mv_1^2$, and to reach height $H$, $\frac{1}{2}v_1^2 = gH$. Thus, $P = m'gH$ (if power is energy per unit time, and $m'$ is mass flow rate, $P = m' \times \frac{1}{2}v_1^2 = m'gH$).

When a pipe of height $h$ is attached, the water must reach a total height $H$. Let the velocity at the exit of the pipe (at height $h$) be $v_2$. The total energy per unit mass required to reach $H$ is $gH$. This energy is provided by the pump as kinetic energy and potential energy gained up to height $h$: $\frac{1}{2}v_2^2 + gh = gH$. This means the required kinetic energy per unit mass is $\frac{1}{2}v_2^2 = g(H-h)$.

The new power, $P_{new}$, is assumed to be proportional to this new kinetic energy: $P_{new} = m' \frac{1}{2}v_2^2$. Substituting the value of $\frac{1}{2}v_2^2$: $P_{new} = m' g(H-h)$.

Comparing with the original power $P = m'gH$, we get: $P_{new} = P \frac{H-h}{H} = P(1 - \frac{h}{H})$.