A hose has been clamped so that the area at the clamp is only one-quarter the area of the rest of the hose. When we ignore the viscosity of water, the ratio of the volume of water delivered per unit time when the clamp is on to the volume of water delivered per unit time without the clamp is:
A) $\frac{1}{16}$ .
B) $\frac{1}{8}$ .
C) $\frac{1}{4}$ .
D) $\frac{1}{2}$ .
E)1.

Question

Quizplus · Accepted Answer

This problem can be solved using the equation of continuity, which states that the product of the cross-sectional area and the fluid velocity must be constant along a flow path. Let A1 and A2 be the cross-sectional areas of the hose before and after the clamp, respectively, and let v1 and v2 be the fluid velocities before and after the clamp, respectively. We know that A2 = A1/4, so v2 = 4v1.

The volume of water delivered per unit time is given by Q = Av, where Q is the volume flow rate, A is the cross-sectional area of the hose, and v is the fluid velocity. Since the hose is assumed to be delivering the same volume of water per unit time before and after the clamp, we have:

A1v1 = A2v2
A1v1 = (1/4)A1(4v1)
A1v1 = A1v1

Therefore, the ratio of the volume flow rate when the clamp is on to the volume flow rate when the clamp is off is:

Q(clamp on)/Q(clamp off) = A2v2/(A1v1) = (1/4)(4v1)/(v1) = 1

So the answer is E, since the volume flow rate is the same with or without the clamp.

A Hose Has Been Clamped So That the Area at the Clamp