Mechanical Properties of Fluids — Practice

Question 1

Why is pressure called a scalar quantity even though it is defined as force divided by area?

Accepted Answer

At any point in a fluid at rest, the pressure is the same in every direction — a tiny test surface experiences the same force per unit area no matter how it is oriented. A quantity that has no preferred direction cannot be a vector, so pressure is a scalar despite being built from a force, which is a vector, divided by an area.

Question 2

A hydraulic lift has a small piston of area 10 cm^2 and a large piston of area 400 cm^2. A force of 50 N is applied to the small piston. Find the force on the large piston.

Accepted Answer

By Pascal's law the pressure is the same at both pistons: F_small/a = F_large/A. So F_large = F_small x (A/a) = 50 x (400/10) = 2000 N. The pressure is transmitted unchanged; it is the force that is multiplied by the ratio of the areas.

Question 3

Why does a liquid drop with no external forces on it always take a spherical shape?

Accepted Answer

With nothing else acting on it, surface tension is the only influence shaping the drop, and surface tension acts to minimise the surface area for a given volume. Of all shapes with the same volume, the sphere has the smallest possible surface area, so the drop settles into that shape.

Question 4

Water flows through a horizontal pipe that narrows from area 8 cm^2 to area 2 cm^2. If the speed in the wider section is 1.5 m/s, find the speed in the narrower section and the pressure difference between them (density of water = 1000 kg/m^3).

Accepted Answer

By continuity, A1 v1 = A2 v2, so v2 = v1 (A1/A2) = 1.5 x (8/2) = 6.0 m/s. Since the pipe is horizontal, Bernoulli's equation gives p1 - p2 = (1/2) rho (v2^2 - v1^2) = 0.5 x 1000 x (36 - 2.25) = 16,875 Pa. The pressure is lower in the narrower section, where the speed is greater.

Question 5

A steel ball of radius 1.0 mm falls through glycerine of viscosity 0.83 Pa s and reaches a terminal velocity of 1.0 cm/s. Take density of steel = 7800 kg/m^3 and glycerine = 1260 kg/m^3. Verify this is consistent with v_t = 2a^2(rho - sigma)g/9eta.

Accepted Answer

v_t = 2(0.001)^2(7800-1260)(9.8)/(9 x 0.83) = 2 x 1e-6 x 6540 x 9.8/7.47 = 0.1282/7.47 = 0.01716 m/s = 1.72 cm/s. This is close to but not exactly the stated 1.0 cm/s, showing the given terminal velocity is roughly but not perfectly consistent with these values — exactly the kind of check Exercise 9.13's laminar-flow verification trains you to make.

Question 6

Find the excess pressure inside a soap bubble of radius 2.0 mm if the surface tension of the soap solution is 3.0 x 10^-2 N/m.

Accepted Answer

A soap bubble has two surfaces, so excess pressure = 4S/r = (4 x 3.0e-2)/(2.0e-3) = 60 Pa. Using 2S/r instead, as for a drop, would give only 30 Pa — half the correct answer.

Question 7

Find the pressure at a depth of 15 m in sea water of density 1030 kg/m^3, given atmospheric pressure is 1.01 x 10^5 Pa.

Accepted Answer

p = p0 + rho g h = 1.01e5 + (1030)(9.8)(15) = 1.01e5 + 1.51e5 = 2.52 x 10^5 Pa. About two and a half atmospheres, roughly one third from the water and two thirds already present as air above the surface.

Question 8

Water rises 3.0 cm in a capillary tube of radius 0.5 mm. Find the surface tension of water, taking the angle of contact as zero and density 1000 kg/m^3.

Accepted Answer

From h = 2S cos(theta)/(rho g a), with cos(0) = 1: S = h rho g a / 2 = (0.030)(1000)(9.8)(0.0005)/2 = 0.0735 N/m. Close to the accepted value of about 0.073 N/m for water, which is a useful benchmark to remember.

Question 9

A large tank filled with water to a height of 5 m has a small hole at a height of 2 m from the bottom. Find the speed of efflux and the horizontal distance from the tank at which the stream hits the ground.

Accepted Answer

Depth of the hole below the surface: h = 5 - 2 = 3 m. By Torricelli's law, v = sqrt(2gh) = sqrt(2 x 9.8 x 3) = sqrt(58.8) = 7.67 m/s, horizontal. The water then falls the remaining 2 m to the ground: 2 = (1/2)(9.8)t^2, giving t = 0.639 s. Horizontal range = v x t = 7.67 x 0.639 = 4.90 m.

Question 10

Two capillary tubes of the same length but radii r and 2r are connected in series, and the same liquid flows through both under a total pressure difference of Δp. Find what fraction of the total pressure drop occurs across the narrower tube.

Accepted Answer

In series, the same volume flow rate Q passes through both, and Poiseuille's equation gives dp = 8 eta L Q/(pi r^4), so for fixed Q, L and eta, dp is proportional to 1/r^4. The narrow tube (radius r) has 1/r^4, the wide one (radius 2r) has 1/(2r)^4 = 1/16r^4 — sixteen times smaller. So the narrow tube's share of the total pressure drop is (1/r^4)/(1/r^4 + 1/16r^4) = 16/17 ≈ 0.941, about 94%, with the wide tube taking only about 6%. Nearly all the resistance sits in the thinner tube.

Practice — Mechanical Properties of Fluids

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