Mechanical Properties of Fluids

(a) Blood pressure at feet vs brain:
The height of the blood column is greater at the feet compared to the brain. According to hydrostatic pressure $P = h\rho g$, pressure increases with depth ($h$). Thus, BP at feet > BP at brain.

(b) Atmospheric pressure vs height:
Density of air $\rho$ is not constant; it decreases rapidly with height. Most of the atmosphere’s mass is concentrated near the surface. Hence, pressure drops significantly (exponentially) even within 6 km.

(c) Hydrostatic pressure is scalar:
Pressure acts normally to any area element placed in the fluid. Since it has no unique direction (acts in all directions equally at a point) and is defined as the magnitude of force per unit area, it is a scalar.

(a) Contact Angles:
For mercury-glass, Cohesive force (Hg-Hg) > Adhesive force (Hg-Glass). Resultant force is inward $\Rightarrow$ Obtuse angle.
For water-glass, Adhesive force (Water-Glass) > Cohesive force (Water-Water). Resultant is outward $\Rightarrow$ Acute angle.

(b) Spreading vs Drops:
Strong adhesion makes water spread (wet) the glass. Strong cohesion makes mercury “ball up” to minimize surface area.

(c) Surface Tension vs Area:
Surface Tension is a property of the liquid interface material, defined as force per unit length. It does not depend on the total surface area.

(d) Detergents:
Detergents lower surface tension, increasing adhesion with dirt/fabric. This reduces the contact angle, helping water penetrate pores better.

(e) Spherical Drop:
In the absence of external forces (like gravity), surface tension tends to minimize surface area. For a given volume, a sphere has the minimum surface area.

(a) Blowing over paper:

Blowing over the paper increases air speed. By Bernoulli’s principle, higher speed means lower pressure. The pressure under the paper (atmospheric) is higher, pushing it up.

(b) Closing tap with fingers:
According to Equation of Continuity ($Av = \text{constant}$), reducing the area ($A$) of the outlet forces the velocity ($v$) to increase dramatically, creating fast jets.

(c) Syringe needle size:
Flow rate $Q \propto r^4$ (Poiseuille’s Law). A small change in radius ($r$) affects flow rate much more than pressure changes ($P$). Hence, needle size controls flow better.

(d) Backward thrust from hole:
Fluid exiting carries momentum forward. By conservation of momentum (Newton’s 3rd Law), the vessel receives an equal and opposite backward reaction force.

(e) Spinning cricket ball (Magnus Effect):

Spinning drags air on one side (increasing speed, lowering pressure) and opposes air on the other (decreasing speed, increasing pressure). The pressure difference creates a side force (swing), deviating it from a parabola.

Given: $m = 50 \text{ kg}$, Diameter $D = 1.0 \text{ cm} = 0.01 \text{ m}$.
Radius $r = 0.005 \text{ m}$. Area $A = \pi r^2 = \pi (25 \times 10^{-6}) \approx 7.85 \times 10^{-5} \text{ m}^2$.
Force $F = mg = 50 \times 9.8 = 490 \text{ N}$.

Normal atmospheric pressure $P_a = 1.01 \times 10^5 \text{ Pa}$.
Density of wine $\rho = 984 \text{ kg m}^{-3}$. $g = 9.8 \text{ m s}^{-2}$.

Formula: $P_a = h \rho g \Rightarrow h = \frac{P_a}{\rho g}$.
$h = \frac{1.01 \times 10^5}{984 \times 9.8} = \frac{101000}{9643.2} \approx \mathbf{10.5 \text{ m}}$.

Mass to lift $M = 3000 \text{ kg}$. Area $A = 425 \text{ cm}^2$ (irrelevant for pressure calculation on small piston).
According to Pascal’s Law, pressure is transmitted equally.

Pressure required to lift load: $P = \frac{F}{A} = \frac{Mg}{A}$.
$P = \frac{3000 \times 9.8}{425 \times 10^{-4}} = \frac{29400}{0.0425} \approx \mathbf{6.92 \times 10^5 \text{ Pa}}$.
This is the maximum pressure the smaller piston must bear.

Equilibrium Condition: Pressure at the same level in mercury is equal.
$P_{water\_col} = P_{spirit\_col}$ (relative to interface).
$h_w \rho_w g = h_s \rho_s g \Rightarrow h_w \rho_w = h_s \rho_s$.

Given: $h_w = 10.0 \text{ cm}$, $h_s = 12.5 \text{ cm}$.
Specific gravity of spirit $S_s = \frac{\rho_s}{\rho_w} = \frac{h_w}{h_s}$.
$S_s = \frac{10.0}{12.5} = \mathbf{0.8}$.

Add 15.0 cm of water and spirit. Total $h_w = 25.0 \text{ cm}$, $h_s = 27.5 \text{ cm}$.
Pressure difference caused by columns: $\Delta P = P_w – P_s = g(\rho_w h_w – \rho_s h_s)$.
This difference is balanced by the mercury level difference $h_{Hg}$.
$\rho_{Hg} g h_{Hg} = g(\rho_w h_w – \rho_s h_s)$.
$13.6 \rho_w h_{Hg} = \rho_w (25.0) – (0.8 \rho_w)(27.5)$.
$13.6 h_{Hg} = 25.0 – 22.0 = 3.0 \text{ cm}$.
$h_{Hg} = \frac{3.0}{13.6} \approx \mathbf{0.22 \text{ cm}}$.

Poiseuille’s Equation: $Q = \frac{\pi P r^4}{8 \eta L}$.
Given: Mass rate $= 4.0 \times 10^{-3} \text{ kg/s}$. Density $\rho = 1.3 \times 10^3$.
Volume rate $Q = \frac{\text{Mass rate}}{\rho} = \frac{4 \times 10^{-3}}{1.3 \times 10^3} \approx 3.08 \times 10^{-6} \text{ m}^3/\text{s}$.
$r = 0.01 \text{ m}, L = 1.5 \text{ m}, \eta = 0.83 \text{ Pa s}$.

Rearranging for $P$:
$P = \frac{8 \eta L Q}{\pi r^4} = \frac{8 \times 0.83 \times 1.5 \times 3.08 \times 10^{-6}}{3.1416 \times (10^{-2})^4}$
$P = \frac{3.067 \times 10^{-5}}{3.1416 \times 10^{-8}} \approx \mathbf{976 \text{ Pa}}$.

Bernoulli’s Principle: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$.
Pressure diff $\Delta P = P_2 – P_1 = \frac{1}{2}\rho (v_1^2 – v_2^2)$.
$v_1 = 70 \text{ m/s}$ (Upper), $v_2 = 63 \text{ m/s}$ (Lower). $\rho = 1.3$.
$\Delta P = 0.5 \times 1.3 \times (70^2 – 63^2) = 0.65 \times (4900 – 3969) = 0.65 \times 931 = 605.15 \text{ Pa}$.

Lift Force: $F = \Delta P \times A = 605.15 \times 2.5 \approx \mathbf{1.51 \times 10^3 \text{ N}}$.

Equation of Continuity: $A_1 v_1 = n A_2 v_2$.
Tube Area $A_1 = 8.0 \text{ cm}^2 = 8 \times 10^{-4} \text{ m}^2$.
Speed $v_1 = 1.5 \text{ m/min} = 1.5/60 = 0.025 \text{ m/s}$.
Holes $n = 40$. Hole diameter $d = 1 \text{ mm} \Rightarrow r = 0.5 \text{ mm}$.
Hole Area $A_2 = \pi (0.5 \times 10^{-3})^2 \approx 7.85 \times 10^{-7} \text{ m}^2$.

Velocity $v_2 = \frac{A_1 v_1}{n A_2} = \frac{8 \times 10^{-4} \times 0.025}{40 \times 7.85 \times 10^{-7}}$
$v_2 = \frac{2 \times 10^{-5}}{3.14 \times 10^{-5}} \approx \mathbf{0.64 \text{ m/s}}$.

Radius $R = 3.00 \text{ mm} = 3 \times 10^{-3} \text{ m}$. $S = 0.465 \text{ N/m}$.
Atm Pressure $P_0 = 1.01 \times 10^5 \text{ Pa}$.

Excess Pressure: $P_{ex} = \frac{2S}{R}$ (One surface for drop).
$P_{ex} = \frac{2 \times 0.465}{3 \times 10^{-3}} = \frac{0.93}{0.003} = \mathbf{310 \text{ Pa}}$.

Total Inside Pressure: $P_{in} = P_0 + P_{ex}$
$P_{in} = 1.01 \times 10^5 + 310 = 101000 + 310 = \mathbf{1.0131 \times 10^5 \text{ Pa}}$.

(a) Excess pressure in soap bubble (in air):
Bubble has 2 surfaces. $P_{ex} = \frac{4S}{R}$.
$S = 2.50 \times 10^{-2}$. $R = 5 \times 10^{-3}$.
$P_{ex} = \frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}} = \mathbf{20 \text{ Pa}}$.

(b) Air bubble inside soap solution (depth 40 cm):
Air bubble in liquid has only 1 surface. $P_{ex} = \frac{2S}{R} = 10 \text{ Pa}$.
Total Pressure $P_{in} = P_{atm} + P_{hydrostatic} + P_{ex}$.
$P_{hydro} = h \rho g = 0.4 \times 1200 \times 9.8 = 4704 \text{ Pa}$.
$P_{in} = 1.01 \times 10^5 + 4704 + 10 = 105714 \text{ Pa} \approx \mathbf{1.06 \times 10^5 \text{ Pa}}$.