Question

A pump is used to deliver water at a certain rate from a given pipe. To obtain n times water from the same pipe in the same time, by what factor, the force of the motor should be increased?

• A. n times
• B. $n^2$ times
• C. $n^3$ times
• D. $\frac{1}{n}$ times

Answer 1

Answer: (B)

$n^2$ times

Answer 2

To solve this problem, we need to understand the relationship between the mass flow rate, velocity of water, and the force exerted by the motor.

Let:

• $\rho$ be the density of water.
• $A$ be the cross-sectional area of the pipe.
• $v$ be the velocity of water flowing out of the pipe.
• $\frac{dm}{dt}$ be the mass flow rate of water.
• $F$ be the force exerted by the motor to deliver the water.

1. Initial State: The mass flow rate of water is given by: $\frac{dm}{dt} = \rho A v$ The force required to deliver this water is the rate of change of momentum of the water. Assuming the water starts from rest and is ejected with velocity $v$: $F = \frac{d(mv)}{dt} = v \frac{dm}{dt}$ Substituting the expression for $\frac{dm}{dt}$: $F = v (\rho A v) = \rho A v^2$ Let's denote the initial force and velocity as $F_1$ and $v_1$, respectively: $F_1 = \rho A v_1^2 \quad \ldots(1)$

2. Final State: We want to obtain 'n' times water from the same pipe in the same time.

• "n times water" means the mass of water delivered per unit time (mass flow rate) increases by a factor of 'n'. $\left(\frac{dm}{dt}\right)_2 = n \left(\frac{dm}{dt}\right)_1$
• "same pipe" means the cross-sectional area $A$ remains constant.
• "water" means the density $\rho$ remains constant.

From the mass flow rate equation, $\frac{dm}{dt} = \rho A v$: $\rho A v_2 = n (\rho A v_1)$ Since $\rho$ and $A$ are constant, they cancel out: $v_2 = n v_1$ This means the velocity of the water must increase by a factor of 'n' to deliver 'n' times the water in the same time.

Now, let's find the new force $F_2$ required in the final state: $F_2 = \rho A v_2^2$ Substitute $v_2 = n v_1$ into this equation: $F_2 = \rho A (n v_1)^2$ $F_2 = \rho A n^2 v_1^2$ $F_2 = n^2 (\rho A v_1^2)$ From equation (1), we know that $F_1 = \rho A v_1^2$. Therefore: $F_2 = n^2 F_1$

This shows that the force of the motor should be increased by a factor of $n^2$.

Explanation of the solution: The force required by the motor to deliver water is proportional to the square of the water's velocity ($F \propto v^2$). To deliver 'n' times the amount of water in the same time from the same pipe, the velocity of the water must increase 'n' times ($v_2 = n v_1$). Substituting this into the force relation, the new force $F_2$ becomes $F_2 \propto (nv_1)^2 \propto n^2 v_1^2 \propto n^2 F_1$. Thus, the force must be increased by a factor of $n^2$.