Problem Statement
Solve the fluid mechanics problem: Find the dimensions of coefficient of viscosity $\eta$. Newton’s viscosity law: $F = \eta A (dv/dy)$ Step 1: $\eta = F / (A \cdot dv/dy)$. Step 2: $[dv/dy] = (LT^{-1})/L = T^{-1}$. Step 3: $[\eta] = MLT^{-2}/(L^2 \cdot T^{-1}) = ML^{-1}T^{-1}$. $$\boxed{[\eta] = ML^{-1}T^{-1},\quad \text{SI unit} =
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{[\eta] = ML^{-1}T^{-1},\quad \text{SI unit} =
Given Information
- Fluid density $\rho$, velocities, cross-sections, and heights as given
- Atmospheric pressure $P_0 = 1.013\times10^5\,\text{Pa}$
- $g = 9.8\,\text{m/s}^2$
Physical Concepts & Formulas
Fluid statics is governed by Pascal’s Law and the hydrostatic pressure formula $P = P_0 + \rho g h$. Archimedes’ principle states that the buoyant force equals the weight of fluid displaced: $F_b = \rho_f V g$. Fluid dynamics for ideal (incompressible, non-viscous, steady) flow uses two key results: the continuity equation $A_1 v_1 = A_2 v_2$ (mass conservation) and Bernoulli’s equation $P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$ (energy conservation per unit volume).
- $P = P_0 + \rho g h$ — hydrostatic pressure
- $F_b = \rho_f V g$ — Archimedes buoyancy
- $A_1 v_1 = A_2 v_2$ — continuity equation
- $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const}$ — Bernoulli’s equation
- $v_{\text{efflux}} = \sqrt{2gh}$ — Torricelli’s theorem
Step-by-Step Solution
Step 1 — Identify the situation: Static (use $P = P_0+\rho gh$ and Archimedes) or dynamic (use continuity + Bernoulli).
Step 2 — Apply Bernoulli between two points at the same streamline:
$$
$$
Step 3 — Use continuity to relate $v_1$ and $v_2$: $v_2 = v_1 A_1/A_2$.
Step 4 — Solve for the unknown (pressure, velocity, or flow rate).
Worked Calculation
Substituting all values with units:
Torricelli: Tank depth $h = 2\,\text{m}$, hole at bottom:
$$
$$
Answer
$$
Answer
$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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