Vardaan Institute - Detailed Solutions: Master Sheet (Mechanical Properties of Fluids)

Section A: Pascal's Law Solutions

A hydraulic lift has piston diameters $5\text{ cm}$ and $30\text{ cm}$. Find force on smaller piston to lift $1500\text{ kg}$.

Sol: $\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_1 = F_2 \left(\frac{d_1}{d_2}\right)^2 = (1500 \times 9.8) \left(\frac{5}{30}\right)^2 = 14700 \times \frac{1}{36} = 408.33\text{ N}$.

Derive pressure variation with depth $P = P_0 + \rho g h$.

Sol: Balancing forces on vertical fluid column of area $A$ and height $h$: $P A = P_0 A + m g = P_0 A + \rho (A h) g \implies P = P_0 + \rho g h$.

Section B: Bernoulli's Principle Solutions

Derive Torricelli's law for speed of efflux $v = \sqrt{2gh}$.

Sol: Applying Bernoulli's equation at top surface and hole: $P_0 + \rho g h + 0 = P_0 + \frac{1}{2}\rho v^2 \implies \rho g h = \frac{1}{2}\rho v^2 \implies v = \sqrt{2gh}$.

Venturimeter flow rate calculation ($A_1=20\text{ cm}^2, A_2=10\text{ cm}^2, h=20\text{ cm}$).

Sol: $v_1 = \sqrt{\frac{2gh}{(A_1/A_2)^2 - 1}} = \sqrt{\frac{2 \times 980 \times 20}{4 - 1}} = \sqrt{\frac{39200}{3}} \approx 114.3\text{ cm/s}$. $Q = A_1 v_1 = 20 \times 114.3 \approx 2286\text{ cm}^3/\text{s} = 2.286\text{ L/s}$.

Section C: Viscosity & Surface Tension Solutions

State Stokes' Law and derive terminal velocity formula.

Sol: Viscous drag $F_v = 6\pi\eta r v$. At terminal velocity, $6\pi\eta r v_t + \frac{4}{3}\pi r^3 \sigma g = \frac{4}{3}\pi r^3 \rho g \implies v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$.

Calculate excess pressure inside a soap bubble of radius $r = 5\text{ mm}$ ($S = 0.03\text{ N/m}$).

Sol: A soap bubble has 2 surfaces: $\Delta P = \frac{4S}{r} = \frac{4 \times 0.03}{0.005} = 24\text{ Pa}$.

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