Properties of Solids & Fluids
(KVPY – Physics – Historic Edition)
1. Simple Explanation
| Property | Solids | Fluids |
|---|---|---|
| Shape | Maintain a definite shape. | Take the shape of their container. |
| Density | Usually high & nearly constant. | Variable with temperature & pressure, but often taken as constant for liquids. |
| Compressibility | Very low; can be ignored in most problems. | Significant for gases; negligible for liquids. |
| Elasticity / Strength | Can return to original shape after small deformations; characterised by Young’s modulus, yield strength, etc. | No elastic recovery; only viscous flow. |
| Flow/Viscosity | No flow under ordinary conditions. | Flow under shear; governed by viscosity. |
| Buoyancy | Not relevant. | Objects immersed experience an upward buoyant force equal to the weight of displaced fluid. |
2. Key Points
2.1 Solids
| Feature | What to remember |
|---|---|
| Density (ρ) | ρ = m / V (kg/m³). Used for weight, buoyancy calculations. |
| Elasticity | Stress (σ) = E × strain (ε). Hooke’s law: F = kx for springs. |
| Strength | Yield strength, ultimate tensile strength, compressive strength. |
| Plasticity | Permanent deformation when stress > yield strength. |
| Hardness & Ductility | Hardness = resistance to indentation; ductility = ability to deform plastically. |
| Phase | Solids can be crystalline (ordered) or amorphous. |
| Thermal Expansion | ΔL = αL₀ΔT (α = linear expansion coefficient). |
2.2 Fluids
| Feature | What to remember |
|---|---|
| Density (ρ) | Same definition; for liquids often taken as constant. |
| Pressure (P) | P = F / A. Hydrostatic pressure: P = ρgh. |
| Viscosity (μ) | Shear stress τ = μ (du/dy). Low μ → laminar flow; high μ → turbulent. |
| Continuity Equation | A₁v₁ = A₂v₂ (incompressible flow). |
| Reynolds Number | Re = ρvd/μ. Re < 2000 → laminar, Re > 4000 → turbulent. |
| Bernoulli’s Equation | P + ½ρv² + ρgh = constant along a streamline (ideal fluid). |
| Archimedes’ Principle | Buoyant force = ρVg (upward). |
| Surface Tension | γ = force per unit length along the surface. |
| Compressibility | Gases: significant, liquids: negligible. |
3. Worked Example
Problem
A 2‑cm³ steel cube (ρ = 7.8 g/cm³) is completely submerged in water (ρ = 1.0 g/cm³). Find (i) the buoyant force on the cube, (ii) the net downward force, and (iii) the apparent weight of the cube in water.
Step‑by‑step Solution
-
Convert Volume to SI Units
(V = 2 \text{ cm}^3 = 2 \times 10^{-6} \text{ m}^3) -
Weight of the cube in air
(m_{\text{steel}} = ρ_{\text{steel}} V = 7.8,\text{g/cm}^3 \times 2,\text{cm}^3 = 15.6,\text{g} = 0.0156,\text{kg})
(W_{\text{steel}} = m_{\text{steel}} g = 0.0156 \times 9.8 \approx 0.153,\text{N}) -
Buoyant force
(F_b = ρ_{\text{water}} V g = 1.0,\text{g/cm}^3 \times 2,\text{cm}^3 \times 9.8 \approx 0.0196,\text{N}) -
Net downward force
(F_{\text{net}} = W_{\text{steel}} - F_b = 0.153 - 0.0196 \approx 0.133,\text{N}) -
Apparent weight (weight in water)
(W_{\text{app}} = F_{\text{net}} = 0.133,\text{N})
Summary of Results
| Quantity | Value | Units |
|---|---|---|
| Weight in air | 0.153 | N |