Overview
This unit studies the macroscopic properties of matter that arise from the collective behaviour of a large number of particles. It examines concepts such as elasticity, stress and strain, mechanical properties of solids and fluids, pressure in fluids, viscosity, surface tension, and heat-related properties like thermal expansion and specific heat. The unit explains how bulk matter responds to forces, how internal forces distribute, and how these responses are measured and used in engineering and everyday life. Understanding these properties helps explain why materials stretch, compress, flow, resist motion, or change volume with temperature. The unit also introduces practical methods to measure these properties and relates microscopic ideas—intermolecular forces and atomic arrangements—with macroscopic observables. This knowledge is important for designing structures, choosing materials, controlling fluid flow in pipes, understanding capillary action in plants, and predicting thermal effects in devices. Overall, the unit connects theoretical definitions and formulas to experiments and real-world applications, preparing students for further study in physics, engineering, and technology.
Learning Objectives
- Describe and distinguish stress, strain, and elastic modulus for solids.
- Apply Hooke's law and calculate elastic potential energy in springs and stretched bodies.
- Explain pressure in fluids and solve problems using Pascal’s law and hydrostatic pressure formula.
- Define and compute bulk modulus and relate it to compressibility.
- Explain viscosity, derive the expression for viscous force in laminar flow (qualitatively), and use Poiseuille’s ideas in simple problems.
- Explain surface tension, capillarity, and calculate rise in capillary tubes.
- Calculate thermal expansion of solids, liquids and gases and understand effects on structures and instruments.
- Determine specific heat capacity and latent heat from calorimetry problems.
Topics in this chapter
15 topics · tap a topic title to jump straight to it.
Introduction to bulk properties and matter phases
Bulk properties are the characteristics of a large amount of matter considered as a whole, not of individual atoms. When we study bulk matter, we focus on measurable quantities such as stiffness, compressibility, viscosity, surface tension and thermal response. These properties arise from microscopic forces between particles and the way particles are arranged, but they appear at the scale of rods, liquids in containers, and gases in balloons.
There are three common phases of matter in everyday experience: solids, liquids and gases. In solids particles occupy fixed average positions in a structure (crystal or amorphous), so solids resist changes of shape and volume. Liquids have no fixed shape but a nearly fixed volume; particles are close but can move relative to each other, allowing flow. Gases have widely separated particles that move freely and fill the container; they are easily compressible. The differences in response to forces and temperature changes stem from these arrangements and from strength of intermolecular forces.
To describe how bulk matter responds to forces we introduce quantities like stress (internal force per unit area) and strain (relative deformation). For fluids we use pressure, which acts equally in all directions at a point. Elastic moduli—Young’s modulus, shear modulus and bulk modulus—connect stress and strain for solids and quantify stiffness. For fluids, viscosity quantifies resistance to flow and surface tension measures energy cost to create surface area. Thermal properties such as specific heat and thermal expansion coefficients describe how materials absorb heat and change size with temperature.
In practical terms bulk properties determine choice of material and design. Engineers choose steel with high Young’s modulus and acceptable ductility for beams; hydraulic systems use incompressible fluids for force transmission; capillarity helps water move in plant tissues and thin tubes; thermal expansion must be allowed for in bridges and railway tracks. Experiments in the laboratory, such as stretching wires, measuring pressure at depth, observing capillary rise, or heating a rod, allow students to measure these properties and connect formulas to real behaviour.
In this unit we will introduce definitions, derive basic relations under simplifying assumptions, solve numerical problems typical for class 11, and point out limits of simple laws. Emphasis is on understanding physical meaning, units and practical implications, preparing you for more advanced study in mechanics, fluid dynamics and thermal physics.
- Compare a steel rod and water in a same-shaped container to see why steel resists change of shape while water takes container shape.
- Compress a sponge and a sealed air balloon to note differences in how volume changes under force.
- Observe capillary rise by dipping a thin glass tube in water and noting the height to which water climbs.
- Stretch a spring within elastic limit and plot force versus extension to verify proportionality.
- No specific formula in this introductory topic
Stress and Strain
Stress and strain are the primary quantities used to describe deformation of solids. Stress is the internal distribution of force per unit area inside a body when external forces are applied. If a force F acts normally on area A, the normal stress is σ = F/A. Normal stress may be tensile (pulling) or compressive (pushing). Shear stress τ acts tangentially and tends to slide layers past each other; τ = F_shear / A where F_shear is tangential force on area A. Stress has SI unit pascal (Pa) which is N/m2.
Strain describes how much a body deforms relative to its original shape or size. For a rod stretched from length L to L + ΔL, the engineering (linear) strain is ε = ΔL/L. Strain is dimensionless and often small for elastic deformations. For shear deformation of a block, shear strain can be taken as the tangent of the small angle change (in radians) between originally perpendicular lines; for small angles shear strain ≈ displacement / original height.
Stress and strain are related through material response. In the elastic regime (small deformations reversible on removing load) many materials show a linear relation between stress and strain. For normal loading, Young’s modulus E is defined by E = σ/ε for this linear region. For shear deformation the shear modulus G relates τ = G γ where γ is shear strain. For volumetric or hydrostatic loading, bulk modulus K relates pressure change to fractional volume change. Each modulus is a measure of stiffness under a particular kind of deformation.
It is important to recognise elastic limit: above a certain stress the material will not return to original shape (plastic deformation starts). Stress has direction and distribution; in complex cases one must consider stress components on different planes. For class 11, focus on basic one-dimensional cases: uniform tensile or compressive loading. Use careful sign convention: tensile stress and extension may be taken positive; compressive negative depending on chosen convention. Always note units and keep consistent significant figures when calculating.
Experimentally, stress-strain data are obtained from tensile tests: specimens of known cross-section are pulled while measuring load and extension. The initial straight portion of the graph yields Young’s modulus E, while yield point, ultimate tensile strength and fracture behaviour appear at higher stresses. Understanding stress and strain is the first step to designing components that carry loads safely.
- A rod of length 2.0 m stretched by 1.0 mm: strain ε = 1.0×10^-3 / 2.0 = 5.0×10^-4.
- A force 1000 N on area 2×10^-4 m2 gives stress σ = 1000 / 2×10^-4 = 5×10^6 Pa (5 MPa).
- Shear example: a block with top face displaced by 2 mm over height 100 mm gives shear strain γ ≈ 2×10^-3/0.1 = 0.02.
- Stress (σ) = Force (F) / Area (A)
- Strain (ε) = Change in length (ΔL) / Original length (L)
- Shear strain ≈ lateral displacement / original perpendicular dimension
Hooke's Law and Elastic Potential Energy
Hooke’s law describes the linear relation between force and displacement for elastic objects within a limited range. For an ideal linear spring, the restoring force F is proportional to displacement x from equilibrium: F = -kx. The constant k indicates stiffness; stiffer springs have larger k. The negative sign shows the force acts opposite to displacement, pulling the system back to equilibrium.
Hooke’s law extends to simple elastic behaviour of materials in tension or compression. For a uniform rod under a tensile force F, stress σ = F/A and strain ε = ΔL/L; within elastic limit σ = E ε where E is Young’s modulus. Combining these gives extension ΔL = (F L)/(A E). This relation is useful to predict extensions of beams, wires and structural members under load and is derived assuming uniform load, constant cross-section, and small deformations.
Elastic potential energy is the energy stored in an object when it is deformed elastically. For a linear spring stretched from x=0 to x, the work done against the restoring force equals the stored energy U = 1/2 k x^2. Graphically this is the area under the force-extension curve. For a stretched rod the strain energy per unit volume (energy density) is (1/2) σ ε; total stored energy is (1/2) σ ε × volume. This energy is recoverable if the material remains in the elastic range; beyond the yield point part of the work is dissipated in plastic deformation and heat.
Verification of Hooke’s law is straightforward in the lab: measure extension x for various forces F, plot F versus x; a straight line through origin indicates linearity and slope gives k. Deviations show the elastic limit or non-linear elasticity. In many real springs and materials Hookean behaviour is an approximation valid only for small displacements. Engineers always check that operating loads remain within elastic limits to avoid permanent deformation.
Energy considerations are practical: springs store energy in clocks, vehicle suspensions and measuring instruments. The formula U = 1/2 k x^2 lets you calculate maximum energy storage and design systems accordingly. For rods under axial loading, use U = (1/2)(F^2 L)/(A E) or equivalently U = (1/2) σ ε × volume when given stresses and strains.
- A spring with k = 200 N/m stretched by 0.05 m stores U = 1/2 × 200 × 0.05^2 = 0.25 J.
- A steel wire (L=2 m, A=1×10^-6 m2, E=2×10^11 Pa) carrying F = 100 N extends by ΔL = FL/(AE) = 1×10^-3 m.
- Area under a force-extension graph linear from (0,0) to (x,F) gives elastic energy = 1/2 Fx.
- Hooke’s law: F = -kx
- Young’s relation for extension: ΔL = FL / (AE)
- Elastic potential energy in spring: U = 1/2 k x^2
- Elastic energy density: u = 1/2 σ ε
Elastic Moduli: Young's Modulus, Shear Modulus and Bulk Modulus
Elastic moduli are material constants that characterise stiffness under different types of deformation. Young’s modulus E quantifies stiffness in tension or compression and is defined by the ratio of normal stress to longitudinal strain in the linear elastic region: E = σ/ε. A large E means the material resists changes in length; for example steel has E of order 2×10^11 Pa while rubber has much smaller E.
Shear modulus G characterises response to shear deformations where one face of a body slides relative to another. If shear stress τ produces shear strain γ (angle in radians for small deformations), then G = τ/γ. Materials with high G resist shape change under tangential forces. Bulk modulus K measures resistance to uniform volumetric compression. For a pressure increase ΔP that produces fractional volume change ΔV/V, the bulk modulus is K = -ΔP/(ΔV/V); negative sign appears because ΔV is negative for positive compressive ΔP. Liquids often have very high K (nearly incompressible) while gases have much lower K.
The three moduli are related for isotropic, linear elastic materials through Poisson’s ratio ν, which measures the ratio of transverse contraction strain to longitudinal extension strain under uniaxial stress. The relations are E = 2 G (1 + ν) and E = 3 K (1 - 2 ν). Knowing any two of E, G, K and ν allows determination of the others. Typical values of ν for metals are around 0.25–0.35; for incompressible materials ν approaches 0.5.
Measurements of elastic moduli use different experiments: tensile test gives E from slope of stress–strain curve; torsion test on cylindrical shafts yields G from torque and angle of twist; hydrostatic compression tests yield K. Units for all three moduli are pascals (Pa). The moduli are strictly valid only in the linear elastic regime and for homogeneous materials; real materials can show anisotropy (different moduli in different directions) and time-dependent behaviour like viscoelasticity.
Applications: selection of materials for structures requires suitable E to limit deflection, adequate G to resist shear distortion, and sufficient K in pressure vessels. In dynamic problems the speed of elastic waves depends on these moduli; for example speed of longitudinal waves in a solid depends on both E and ν. Practically, engineers use tabulated values of E, G, K and ν for design calculations and safety checks.
- Compute E from a tensile test where stress 2×10^8 Pa produces strain 0.001: E = 2×10^11 Pa.
- If K for water is about 2.2×10^9 Pa, a pressure increase of 2.2×10^7 Pa gives fractional volume decrease of 0.01.
- Using ν = 0.3 and E = 2×10^11 Pa, find G = E / [2(1+ν)] = 7.69×10^10 Pa.
- Young’s modulus: E = σ / ε
- Shear modulus: G = shear stress / shear strain
- Bulk modulus: K = -P / (ΔV/V)
- Relations: E = 2G(1 + ν), E = 3K(1 - 2ν)
Stress-Strain Curve and Mechanical Properties of Materials
The stress–strain curve of a material is a primary tool to understand its mechanical properties. It is produced by applying increasing load to a test specimen and plotting engineering stress (load divided by original area) versus engineering strain (extension divided by original length). The initial portion of the curve is often linear, indicating elastic behaviour where stress is proportional to strain. The slope of this linear portion is Young’s modulus E, a measure of stiffness.
Beyond the linear region the curve may show yield behaviour: the elastic limit marks the maximum stress that can be removed without permanent deformation. Some materials show a distinct yield point or a yield plateau where strain increases significantly at nearly constant stress. After yielding, plastic deformation occurs; deformations are permanent. With further loading the curve reaches an ultimate tensile strength (UTS), the maximum engineering stress the specimen can bear. Past UTS necking may occur: localized reduction in cross-sectional area leads to reduction in load-bearing capacity and eventual fracture.
Materials are classified by ductility and brittleness: ductile materials exhibit significant plastic deformation before fracture (large area under the curve), whereas brittle materials fracture with little plastic deformation. Toughness is the total energy absorbed before fracture and equals the area under the stress–strain curve; it is important for impact resistance. Hardness is resistance to localised surface deformation and often correlates with strength but requires separate tests (indentation methods).
Unloading behaviour is also instructive. If unloading occurs within the elastic region, the curve retraces and full recovery occurs. If unloading happens after plastic deformation, a residual strain remains and the unloading path is parallel to the elastic slope, showing elastic recovery superimposed on permanent strain. Distinguish engineering stress–strain (uses original area) from true stress–strain (uses instantaneous area); true stress increases after necking while engineering stress decreases.
Application examples: selecting materials for beams requires adequate yield strength and stiffness; for crash components high toughness is needed; cutting tools require hardness. In class 11, predict modulus from slope, calculate percentage elongation at fracture, estimate elastic energy per unit volume from area under linear portion, and interpret qualitative differences between stress–strain curves of ductile and brittle materials.
- From a stress–strain graph find Young’s modulus from slope of linear portion.
- Calculate percentage elongation if original length 100 mm and fracture length 120 mm: % elongation = 20%.
- Estimate toughness as area under a simple piecewise linear stress–strain curve by summing triangular and rectangular areas.
- Engineering stress = Load / Original area
- Engineering strain = Change in length / Original length
- Toughness per unit volume = ∫ σ dε (area under stress–strain curve)
Elastic Deformation of Solids: Bars, Rods and Springs
Elastic deformation of bars and rods under axial loads is a standard application of Hooke’s law and Young’s modulus. Consider a uniform rod of length L, cross-sectional area A, and Young’s modulus E, subject to an axial tensile force F. Under the assumptions of uniform stress and small strains, longitudinal extension is ΔL = F L / (A E). The derivation uses stress = F/A and strain = ΔL/L with σ = E ε. This formula is widely used to compute elongation of wires, rods and structural members under loads.
When rods are arranged in series or parallel, effective extension and load sharing follow simple rules. For rods in series (end-to-end) carrying the same force, total extension equals sum of individual extensions. For rods in parallel (side-by-side) sharing the same extension, the load splits proportionally to cross-sectional areas, and the effective stiffness sums like springs in parallel. Using the equivalence between a rod and a spring one can define an effective spring constant k for a rod as k = A E / L. Then standard spring combination rules apply: springs in series give 1/k_eq = Σ 1/k_i and in parallel k_eq = Σ k_i.
Springs used in applications are often helical and behave approximately like ideal springs within elastic limits. Energy stored in an elastic rod under axial force can be computed from work done: U = ∫ F d(ΔL) = (1/2) F ΔL for linear elastic behaviour. Using ΔL = FL/(AE) gives U = F^2 L / (2 A E) or U = (1/2) σ ε × volume. These relations are useful to estimate energy stored in stretched wires and to design members to avoid excessive stored energy that might be released dangerously on sudden failure.
Bending and torsion are other elastic deformations that require moment of inertia and shear modulus respectively; these are beyond the basic axial case but worth noting. For composite rods of different materials joined together, thermal and mechanical strain compatibility must be considered: loads may redistribute between materials based on stiffness. In real experiments, end conditions, stress concentration and non-uniformities cause deviations from simple formulas, so engineering design includes safety factors.
- Two identical rods each with k = 1000 N/m in series give k_eq = 500 N/m; in parallel give k_eq = 2000 N/m.
- Convert a rod (A = 1×10^-4 m2, L = 1 m, E = 2×10^11 Pa) to k = AE/L = 2×10^7 N/m.
- A spring of k = 400 N/m stretched by 0.02 m carries force 8 N and stores 0.16 J.
- Extension of rod: ΔL = FL / (AE)
- Spring constant for rod: k = AE / L
- Series springs: 1/k_eq = Σ (1/k_i); Parallel springs: k_eq = Σ k_i
Pressure in Fluids, Pascal's Law and Hydrostatics
Pressure in a fluid at rest is defined as force per unit area acting normal to a surface: p = F/A. A key property of fluids at rest is that pressure at a point acts equally in all directions. Pascal’s law states that any external pressure applied to a confined fluid is transmitted undiminished to every part of the fluid and the walls of the container. This principle is used in hydraulic machines: a small force on a small-area piston produces the same pressure throughout, yielding a larger force on a piston with larger area.
Hydrostatic pressure in a fluid under gravity increases with depth. For a fluid of density ρ, the pressure at depth h below the free surface (where pressure is p0) is p = p0 + ρ g h. This relation follows by balancing forces on a small fluid element: the pressure difference supports the weight of the fluid column above. Hydrostatic pressure depends only on depth, fluid density and gravity, not on the total amount of fluid above or the shape of the container. This explains why pressure on dam walls is greater at the base and why pressure measured at the same depth in connected vessels of the same liquid is equal.
Manometers measure pressure differences using height differences of a liquid column. In a U-tube manometer with two fluids, balance hydrostatic pressures at a common horizontal level to relate applied pressure difference to heights and densities. Gauge pressure is p - p_atm; absolute pressure includes atmospheric pressure. For problems require careful sign convention and consistent units (Pa, m, kg/m3). If fluid densities vary with depth (e.g., layered fluids), apply hydrostatic relation across each layer sequentially.
Applications: hydraulic press magnifies force using area ratio; blood pressure measurement uses hydrostatic concepts; submarines control buoyancy to change depth. Note compressible fluids (gases) need thermodynamic considerations if depth/pressure changes are large; for liquids assume incompressibility unless specified. Class 11 problems typically involve using p = p0 + ρ g h, applying Pascal’s law to forces on pistons, and solving manometer height relations for pressure differences.
- Hydrostatic pressure 10 m below water surface: p = ρ g h ≈ 1000×9.8×10 = 9.8×10^4 Pa above surface pressure.
- Hydraulic lift: small piston area 0.01 m2 with force 100 N produces pressure 100/0.01 = 10000 Pa; larger piston area 0.5 m2 produces force 10000×0.5 = 5000 N.
- U-tube manometer with mercury and water can be solved by matching pressures at same level and using densities.
- Pressure: p = F / A
- Hydrostatic pressure: p = p0 + ρ g h
- Pascal’s principle: p transmitted uniformly in confined fluid
Buoyancy and Archimedes' Principle
Archimedes’ principle is a simple but powerful statement about the buoyant force on bodies in fluids: a body wholly or partially immersed in a fluid experiences an upward force equal in magnitude to the weight of the fluid displaced. This buoyant force acts through the centre of buoyancy, the centroid of the displaced fluid volume. The principle follows from hydrostatic pressure acting on the surfaces of the immersed body: greater pressure at greater depth produces a net upward force.
Whether a body floats or sinks depends on the balance between buoyant force and weight. If buoyant force equals weight, the body floats at equilibrium; if buoyant force is smaller, the body sinks; if larger, the body accelerates upward until equilibrium is reached (often at the surface). For floating objects the fraction of volume submerged equals the ratio of densities: V_sub/V_total = ρ_object/ρ_fluid. This explains how objects denser than water, such as ships made of steel, can float because the overall average density including trapped air is less than water’s density.
Apparent weight of an immersed object is actual weight minus buoyant force. This is the principle behind density measurements using the displacement method: measure weight in air and apparent weight in fluid to find buoyant force and hence displaced fluid volume and object density. Stability of floating bodies involves the positions of centre of gravity G and centre of buoyancy B. If a small tilt produces a restoring moment, the metacentre M (intersection of vertical lines through B for small tilts) lies above G and the equilibrium is stable. If M lies below G the body will capsize; designers ensure metacentric height is positive for ships and boats.
Practical problems include calculating displaced volume, apparent weight, and floating fraction. Keep track of units: use SI units for mass, volume and g. For class 11 qualitative description of metacentric stability is sufficient; detailed hydrostatic stability calculations are advanced. Simple experiments with irregular objects and a beaker of water demonstrate Archimedes’ principle: by measuring rise in water level or weight loss on immersion one can find object volume and density.
- A block of density 600 kg/m3 floats in water: fraction submerged = 600/1000 = 0.6 so 60% submerged.
- Iron object of mass 2 kg immersed in water: if it displaces 0.001 m3 of water buoyant force = 1000×0.001×9.8 = 9.8 N, apparent weight = 19.6 - 9.8 = 9.8 N.
- Using displacement method, find volume of irregular object by measuring rise in water level or weight loss on immersion.
- Buoyant force: F_b = ρ_fluid × V_displaced × g
- Apparent weight = Actual weight - Buoyant force
- Floating equilibrium: ρ_object / ρ_fluid = V_submerged / V_total
Viscosity, Laminar Flow, Poiseuille’s Law and Stoke’s Law
Viscosity is the measure of a fluid’s internal resistance to flow. In a flowing fluid adjacent layers move at different velocities; viscous forces arise from momentum exchange and molecular interactions between these layers. The dynamic viscosity η (Pa·s) appears in Newton’s law of viscosity which in simple shear says the tangential force F necessary to maintain steady relative motion is proportional to area A and velocity gradient dv/dy: F = η A (dv/dy). Fluids that obey this linear relation are called Newtonian (e.g., water, air). Non-Newtonian fluids (e.g., ketchup) have more complex shear-dependent behaviour.
Laminar flow is smooth motion with well-defined streamlines and ordered layers; turbulence is chaotic with eddies and mixing. Reynolds number Re = ρ v L / η is a dimensionless parameter that predicts flow regime: low Re usually means laminar, high Re tends to turbulence. For pipe flow Re below ≈2000 usually indicates laminar flow where Poiseuille’s law applies.
Poiseuille’s law describes steady, incompressible, laminar flow of a Newtonian fluid through a long straight circular pipe. Under no-slip boundary condition (fluid velocity zero at wall) the volumetric flow rate Q is Q = (π r^4 ΔP)/(8 η L), where r pipe radius, ΔP pressure difference across length L. The r^4 dependence shows small changes in radius greatly affect flow; doubling radius increases flow by 16. The velocity profile is parabolic with maximum at centre and zero at walls; average velocity is half the centre velocity.
Stoke’s law gives viscous drag on a small sphere moving slowly through a viscous fluid: F_d = 6 π η r v, valid for very low Reynolds numbers and an isolated sphere in an infinite medium. A sphere falling through a fluid experiences weight downward, buoyant force upward and viscous drag upward; at terminal velocity these balance and one can solve for v_t. Falling-sphere viscometers use this principle to measure η by observing terminal speed. Corrections are needed for wall effects and higher Re. Practical applications of viscosity concepts range from blood flow in capillaries and lubrication to pipe design and syrups’ handling.
In problems ensure units are consistent (η in Pa·s, r and L in metres, ΔP in pascals). Always check that assumptions hold: laminar flow for Poiseuille, low Re for Stoke’s law, incompressibility and steady flow. Understand qualitative temperature dependence: viscosity of liquids usually decreases with temperature; for gases it increases with temperature.
- Velocity profile in a pipe: v(r) = v_max(1 - (r/R)^2) leading to parabolic shape; v_avg = v_max/2.
- Poiseuille example: capillary radius 0.5 mm, length 0.1 m, ΔP = 200 Pa, η = 1×10^-3 Pa·s gives Q ≈ 6.1×10^-8 m3/s.
- Stoke’s law: a small sphere radius 0.001 m falling in glycerin reaches terminal velocity when mg - ρ_fluid V g - 6 π η r v_t = 0.
- Newton’s law of viscosity: F = η A (dv/dy)
- Reynolds number: Re = ρ v L / η
- Poiseuille’s law: Q = (π r^4 ΔP) / (8 η L)
- Stoke’s law: F = 6 π η r v
Surface Tension and Capillarity
Surface tension is the property of a liquid surface that acts like a stretched elastic membrane. It arises because molecules at the surface have unsatisfied attractive forces compared with those in the interior; creating extra surface area requires work. Surface tension γ is the force per unit length along the surface or energy per unit area, with SI unit N/m (dimensionally J/m2). As a result of surface tension, liquid droplets tend to form spherical shapes to minimise surface area for a given volume.
At a solid–liquid–gas contact the contact angle θ describes wettability: cos θ relates interfacial tensions by Young’s equation γ_SG = γ_SL + γ_LG cos θ. If θ is small (liquid wets the solid) the liquid spreads; if θ is large the liquid beads up. Surfactants reduce surface tension by accumulating at the interface and changing molecular interactions; this is why soap helps mixing oil and water and reduces surface tension dramatically.
Capillarity is the phenomenon of liquid rise or depression in narrow tubes due to surface tension and contact angle. A liquid that wets the tube (θ < 90°) rises; if it does not wet (θ > 90°) it is depressed. Balance vertical component of surface tension around the circumference (2 π r γ cos θ) with weight of liquid column (π r^2 ρ g h) to obtain capillary rise formula h = (2 γ cos θ)/(ρ g r). This formula shows inverse relation with radius: narrower tubes cause larger rise. Capillarity explains how water climbs in porous materials and plant xylem; it also affects ink flow in pens and thin-layer wetting in coatings.
Surface tension produces measurable forces: a wire frame with a soap film feels a horizontal force equal to 2 γ L for a movable wire of length L (factor 2 for two surfaces). Methods to measure γ include capillary rise, drop-weight or drop-volume methods, and the Du Noüy ring method. Temperature affects γ, typically reducing it as temperature increases and vanishing at the critical temperature. In engineering and biology, control of surface tension and contact angle is important in printing, painting, detergency and microfluidics.
- Capillary rise: water (γ≈0.072 N/m, θ≈0) in tube radius 1 mm gives h ≈ 14.7 mm using h = 2γ/(ρ g r).
- Force on a wire: for γ = 0.07 N/m and L = 0.1 m, film on both sides gives F = 2γL = 0.014 N.
- Adding soap reduces surface tension so droplets spread and wet surfaces more easily.
- Work to increase area: dW = γ dA
- Force on frame: F = γ × (length of contact) × (number of surfaces)
- Capillary rise: h = (2 γ cos θ) / (ρ g r)
Stoke's Law Applications and Terminal Velocity
Stoke’s law gives the viscous drag on a small sphere moving slowly through a viscous fluid: F_d = 6 π η r v, where η is dynamic viscosity, r sphere radius and v relative speed. This linear dependence on velocity contrasts with higher Reynolds number regimes where drag varies approximately as v^2. Stoke’s law is valid when Reynolds number Re is small (typically Re << 1) so inertial effects are negligible and flow is laminar and steady around the sphere.
Consider a small sphere falling under gravity in a viscous fluid. Three forces act: weight mg downward, buoyant force ρ_f V g upward and viscous drag 6 π η r v upward. As the sphere accelerates, drag increases until net force becomes zero; then it moves at constant terminal velocity v_t. Setting forces in equilibrium yields mg - ρ_f V g - 6 π η r v_t = 0. With V = (4/3) π r^3 and m = ρ_s V we find v_t = (2 r^2 g (ρ_s - ρ_f))/(9 η) for a sphere in an infinite medium. This formula shows terminal velocity increases with r^2 and density difference, and decreases with viscosity.
Stoke’s law underlies falling-sphere viscometry: measure v_t for a sphere of known radius and density to calculate η. Real experiments must correct for wall effects if container size is not large compared to sphere, and for non-spherical particles. In nature, small particles like pollen grains or droplets show slow settling in air due to high relative viscosity effect; sedimentation rates in liquids depend on Stoke’s behaviour for small particles.
Limitations: Stoke’s law breaks down at higher Re where flow separation and wake formation occur; also it assumes smooth rigid spheres and no interactions between particles. Always check Re = ρ v r / η to confirm applicability. For class 11, use the formula for terminal velocity and set up equilibrium of forces; understand qualitative trends and experimental methods to determine viscosity from measured terminal speeds.
- Derive terminal velocity: v_t = (2 r^2 g (ρ_s - ρ_f))/(9 η) using equilibrium mg - ρ_f V g = 6 π η r v_t.
- Compare terminal velocities in water and oil to see effect of viscosity and density difference.
- Use falling-sphere method to estimate viscosity by measuring steady speed of sphere through a column of liquid.
- Stoke’s law: F = 6 π η r v
- Terminal velocity for sphere: v_t = (2 r^2 g (ρ_s - ρ_f))/(9 η) (in infinite medium)
Thermal Expansion: Linear, Area and Volume Expansion
Thermal expansion describes how the dimensions of materials change when temperature changes. For small temperature changes ΔT the linear expansion of a rod or length L is given by ΔL = α L ΔT where α is the coefficient of linear expansion (per K). For thin sheets area expansion approximates ΔA ≈ 2 α A ΔT because the sheet expands in two orthogonal directions. For volume expansion of a solid or liquid, ΔV = β V ΔT where β is the coefficient of volume expansion; for isotropic solids β ≈ 3 α.
Values of α vary widely: metals typically have α ~ 10^-5 to 10^-6 K^-1, glass has smaller α, while polymers and gases may have larger coefficients. For gases ideal behaviour gives large volumetric expansion: for an ideal gas β = 1/T for absolute temperature T in Kelvin under isothermal small perturbations or in idealised linearisation. The practical consequence of different α values is that connected parts made of different materials develop stresses or bend when heated; for this reason engineers include expansion joints in bridges, rails and pipelines.
Bimetallic strips exploit differential expansion: two metals with different α bonded together will bend when temperature changes because one expands more than the other. This bending is used in simple thermostats and thermal switches. When expansion is constrained, thermal stress develops. For a rod clamped at both ends, prevented from expanding by ΔT, the thermal stress is σ = E α ΔT where E is Young’s modulus. This stress can be large and must be considered in design to avoid cracking or buckling.
Measurement of thermal expansion uses dilatometers or simple experimental setups where a rod is heated and the change in length is measured. For long structures even small α leads to significant absolute change; e.g., a 100 m steel rail (α ≈ 1.2×10^-5 K^-1) heated by 30 K would try to expand by ~36 mm. Problems typically require calculating ΔL, ΔA or ΔV, or finding thermal stress under constrained conditions. Units: α in K^-1, lengths in metres, temperature change in kelvin or degrees Celsius (differences are same in magnitude).
- A steel rod 2 m long heated by 50°C with α = 1.2×10^-5 K^-1 expands by ΔL = α L ΔT = 1.2×10^-5×2×50 = 0.0012 m = 1.2 mm.
- Bimetallic strip: when heated the side with larger α bulges outward, causing curvature used in thermostats.
- Volume expansion example: 1 litre of liquid with β = 7×10^-4 K^-1 heated by 20°C increases by ≈14 mL.
- Linear expansion: ΔL = α L ΔT
- Area expansion: ΔA ≈ 2 α A ΔT
- Volume expansion: ΔV = β V ΔT (β ≈ 3 α for isotropic solids)
- Thermal stress when constrained: σ = E α ΔT
Specific Heat Capacity, Latent Heat and Calorimetry
Specific heat capacity c of a substance is the amount of heat required to raise the temperature of unit mass by 1 K. For a mass m undergoing a temperature change ΔT without a phase change the heat absorbed is Q = m c ΔT, assuming no heat loss to surroundings. Different materials have different specific heats; water has a notably high value (~4186 J kg^-1 K^-1) which gives it large thermal inertia and is important in climate and engineering applications.
Latent heat L is the heat required for phase change at constant temperature per unit mass. For melting (fusion) use Q = m Lf, and for vaporisation use Q = m Lv. During phase change temperature remains constant while heat goes into changing internal structure and potential energy of the substance rather than kinetic energy.
Calorimetry uses conservation of energy to measure specific heats and latent heats. In a mixing experiment in an insulated calorimeter, heat lost by hot bodies equals heat gained by cold bodies plus calorimeter: Σ m_i c_i (T_i - T_f) = 0 where T_f is final equilibrium temperature. When a phase change is involved include latent heat terms. Real calorimeters have heat capacity (calorimeter constant) which must be accounted for. Be careful with sign convention: heat gained positive, heat lost negative, or use absolute values and equate lost to gained.
Applications include measuring specific heat of metals by heating in boiling water and then placing in measured water in calorimeter, and finding latent heat of fusion of ice. In practical calculations ensure units are consistent (mass in kg, specific heat in J/kg·K, temperature in K or °C as differences). For gases, specific heats at constant pressure Cp and constant volume Cv differ and relate to gas constants (Cp - Cv = R) for ideal gases; such concepts are introduced but detailed thermodynamic derivations are beyond class 11 scope.
- Mix 0.2 kg of water at 80°C with 0.3 kg at 20°C in insulated container; final temperature T_f satisfies 0.2×4186×(80-T_f) = 0.3×4186×(T_f-20).
- Heat required to raise 0.5 kg copper (c=390 J/kg·K) by 60°C: Q = 0.5×390×60 = 11700 J.
- Melting ice: energy to melt 0.1 kg ice at 0°C with latent heat Lf = 3.34×10^5 J/kg is Q = 33400 J.
- Heat for temperature change: Q = m c ΔT
- Latent heat: Q = m L
- Calorimetry energy balance: Σ heat lost + Σ heat gained = 0
Heat Transfer by Conduction (Qualitative) and Thermal Conductivity
Heat transfer by conduction is the process by which thermal energy moves through a material due to temperature differences, without bulk motion of the material itself. Microscopically, conduction occurs because faster (higher energy) particles collide with slower ones and transfer energy, and in metals conduction is strongly aided by free electrons that carry energy rapidly. The basic idea is simple: heat flows from higher to lower temperature regions until thermal equilibrium is reached.
Fourier’s law gives a quantitative statement for steady one-dimensional conduction: the rate of heat flow Q̇ through a cross-sectional area A is proportional to the temperature gradient dT/dx and the area, and inversely related to the distance over which the temperature changes. In differential form Q̇ = -k A (dT/dx), where k is thermal conductivity. The negative sign indicates heat flows opposite to increasing temperature. Thermal conductivity k is a material property measured in W m^-1 K^-1: metals like copper have high k (fast heat flow), while insulators like wood or foam have low k.
For a uniform slab of thickness L between two faces at temperatures T1 and T2, assuming linear gradient and steady state, the heat current is Q̇ = k A (T1 - T2)/L. This simple relation is useful for estimating heat loss through walls, windows and insulating layers. When several layers are present (composite wall), thermal resistances add: R_th = L/(kA) for each layer, and total heat flow Q̇ = (T_hot - T_cold) / ΣR_th. This analogy with electrical resistance helps design insulation systems and heat exchangers.
Conduction is closely linked with heat capacity and unsteady behaviour. In transient conduction temperature within a body changes with time and spatial position; heat diffuses with a characteristic time depending on thermal diffusivity (α_th = k/(ρ c)), where ρ is density and c specific heat. Materials with high thermal diffusivity respond quickly to temperature changes; those with low diffusivity respond slowly. Although detailed transient analysis requires differential equations (heat equation), class 11 focuses on steady conduction and physical understanding.
Practical aspects include reducing conduction with insulating materials, using thermal bridges carefully in building design, and employing heat sinks to remove heat from electronics by increasing area and using materials with high k. Experimental measurement of k uses steady heat input, measured ΔT across a known thickness and area, and careful minimisation of convective and radiative losses. Always check units (Q̇ in watts, k in W m^-1 K^-1, distances in metres) and remember conduction applies only where there is no bulk flow of the medium; if fluid motion occurs, convection must be considered alongside conduction.
- Heat flow through a 0.05 m thick slab (k = 0.2 W/m·K, A = 1 m2, ΔT = 10 K): Q̇ = 0.2×1×10/0.05 = 40 W.
- Two layers in series: layer1 L1/k1 + layer2 L2/k2 gives total thermal resistance; compute Q̇ = ΔT / R_th.
- Design note: doubling thickness of insulation approximately halves conductive heat flow.
- Fourier’s law (steady 1D): Q̇ = -k A (dT/dx)
- For slab: Q̇ = k A ΔT / L
- Composite walls: R_th = Σ (L_i / (k_i A)); Q̇ = ΔT / R_th
Thermal Stress, Design Considerations and Compressibility
When thermal expansion is prevented, internal stresses develop called thermal stresses. For a rod of length L clamped at both ends and heated by ΔT the free expansion would be ΔL_free = α L ΔT. If expansion is prevented, the mechanical strain that appears is ε = -α ΔT and the thermal stress is σ = E ε = -E α ΔT. The magnitude E α ΔT is compressive if expansion is prevented. These stresses can cause buckling, cracking or failure; engineers use expansion joints, flexible supports and materials with appropriate α to avoid damage.
Bimetallic strips convert differential thermal expansion into bending motion. Two different metals bonded together, with α1 ≠ α2, will curve on heating; the side with larger α expands more and forms the outside of the curve. Such strips are simple and robust thermostatic elements used in switches and gauges. In design, choose materials with low α for precision instruments (e.g., invar alloy) or ensure freedom for thermal movement in long structures like bridges and rails.
Compressibility and bulk modulus describe how volume changes under pressure. Bulk modulus K is defined by K = -ΔP/(ΔV/V). Liquids typically have very large K and are nearly incompressible in many contexts; gases have much lower K and their compressibility depends on thermodynamic process. For an ideal gas under isothermal compression K = P, and under adiabatic small perturbations K = γ P. Bulk modulus affects the speed of sound in a medium via c = sqrt(K/ρ); media with larger K and lower density transmit sound faster. This principle explains why sound travels faster in solids than in gases.
Applications of compressibility include hydraulic systems where slight compressibility of fluid affects response, and geophysics where bulk modulus of rocks influences seismic wave speeds. In many engineering problems treat liquids as incompressible unless high pressures or acoustic phenomena are involved. In class 11 problems you will calculate thermal stress for constrained bodies and use K = -ΔP/(ΔV/V) for simple compressibility problems, and link K to speed of sound when asked.
- Thermal stress: steel rod (E = 2×10^11 Pa, α = 1.2×10^-5 K^-1) heated by 40°C and clamped gives σ = E α ΔT = 9.6×10^7 Pa.
- Water compressibility: with K ≈ 2.2×10^9 Pa, a pressure increase of 2.2×10^7 Pa reduces volume by about 1%.
- Speed of sound in medium with K = 2.2×10^9 Pa and ρ = 1000 kg/m3: c ≈ 1483 m/s.
- Thermal stress for completely constrained body: σ = E α ΔT
- Bulk modulus: K = -ΔP / (ΔV/V)
- Speed of sound: c = sqrt(K/ρ)
Key Concepts
- Stress
- Internal force per unit area acting within a material, measured in pascals.
- Strain
- Dimensionless measure of deformation defined as change in length divided by original length.
- Young’s Modulus
- Ratio of tensile stress to longitudinal strain in the elastic region of a material.
- Shear Modulus
- Ratio of shear stress to shear strain for a material undergoing shear deformation.
- Bulk Modulus
- Measure of a material’s resistance to uniform compression, defined by K = -ΔP/(ΔV/V).
- Hooke’s Law
- Linear relation between restoring force and displacement for elastic deformation: F = -kx.
- Viscosity
- Fluid property quantifying internal resistance to flow, appearing in F = η A (dv/dy).
- Poiseuille’s Law
- Relation for laminar flow in a circular pipe: Q = (π r^4 ΔP)/(8 η L).
- Surface Tension
- Force per unit length or energy per unit area at a liquid surface resulting from molecular forces.
- Capillarity
- Rise or depression of liquid in narrow tubes caused by surface tension and contact angle.
- Specific Heat Capacity
- Heat required per unit mass to raise temperature by one degree: Q = m c ΔT.
- Latent Heat
- Heat required per unit mass for a phase change at constant temperature.
- Thermal Expansion Coefficient
- Proportionality constant α relating fractional linear change to temperature change: ΔL/L = α ΔT.
- Reynolds Number
- Dimensionless quantity Re = ρ v L / η that predicts laminar or turbulent flow regimes.
- Stoke’s Law
- Viscous drag on small sphere at low Reynolds number: F = 6 π η r v.
- Buoyant Force
- Upthrust on an immersed body equal to the weight of fluid displaced, per Archimedes’ principle.
- Thermal Stress
- Stress developed when thermal expansion is constrained: σ = E α ΔT.
- Elastic Potential Energy
- Energy stored in an elastically deformed body, for a spring U = 1/2 k x^2.
Practice Questions
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A steel wire 2.0 m long and cross-sectional area 1.0×10^-6 m2 is stretched by a force of 200 N. If Young’s modulus for steel is 2.0×10^11 Pa, find the extension. / एक स्टील की तार जिसकी लंबाई 2.0 मीटर और क्रॉस-सेक्शनल क्षेत्रफल 1.0×10^-6 m2 है, उस पर 200 N का खिंचाव लागू किया जाता है। यदि स्टील का यंग मॉड्यूल 2.0×10^11 Pa है, तो विस्तार ज्ञात कीजिए।
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Use ΔL = FL/(AE): ΔL = (200×2.0)/(1.0×10^-6×2.0×10^11) = 400 / (2.0×10^5) = 2.0×10^-3 m = 2.0 mm. / सूत्र ΔL = FL/(AE) लगाते हैं: ΔL = (200×2.0)/(1.0×10^-6×2.0×10^11)=2.0×10^-3 m = 2.0 मिमी।
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Define surface tension and give one example where it is important. / सतही तनाव क्या है और इसका एक उदाहरण दीजिए जहाँ यह महत्वपूर्ण होता है।
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Surface tension is the force per unit length at a liquid surface (or energy per unit area) caused by unbalanced molecular forces; it tends to minimise surface area. Example: water droplets form nearly spherical shapes due to surface tension; capillary rise in plant xylem helps draw water upward. / सतही तनाव वह बल प्रति यूनिट लंबाई (या क्षेत्रफल प्रति ऊर्जा) है जो तरल सतह पर अणुओं के असंतुलित आकर्षण से उत्पन्न होता है और सतह क्षेत्र को न्यूनतम करने की प्रवृत्ति देता है। उदाहरण: पानी की बूँदें सतही तनाव के कारण गोलाकार बनती हैं; पौधों में केपिलरी क्रिया पानी को ऊपर खींचने में मदद करती है।
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Water rises 1.5 cm in a capillary tube of radius 1.0 mm. Assuming contact angle is zero and density of water 1000 kg/m3, find surface tension. / पानी एक 1.0 mm त्रिज्या वाले केपिलरी ट्यूब में 1.5 cm तक चढ़ता है। संपर्क कोण शून्य माना जाए और पानी का घनत्व 1000 kg/m3 हो तो सतही तनाव ज्ञात कीजिए।
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Use h = 2γ/(ρ g r) so γ = (ρ g r h)/2. Substitute: γ = (1000×9.8×1.0×10^-3×1.5×10^-2)/2 ≈ 0.0735 N/m ≈ 7.35×10^-2 N/m. / सूत्र h = 2γ/(ρ g r) से γ = (ρ g r h)/2। मान रखकर γ ≈ 0.0735 N/m।
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State Poiseuille’s law and explain the effect of doubling the radius of a pipe on volumetric flow rate for same pressure difference. / पोइसुईल का नियम लिखिए और बताया जाय कि यदि एक पाइप की त्रिज्या दो गुनी कर दी जाए तो उसी दाब के अंतर पर प्रवाह दर पर क्या प्रभाव होगा।
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Poiseuille’s law: Q = (π r^4 ΔP) / (8 η L) for laminar flow of a Newtonian fluid in a circular pipe. Doubling radius increases r^4 by 16, so volumetric flow Q increases by factor 16 for the same ΔP, η and L. / पोइसुईल का नियम: Q = (π r^4 ΔP)/(8 η L)। यदि त्रिज्या दो गुनी हो जाए तो r^4 16 गुना बढ़ेगा, अतः उसी दाब के अंतर पर प्रवाह दर 16 गुना बढ़ेगी।
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A block of density 600 kg/m3 floats in water. What fraction of its volume remains submerged? / घनत्व 600 kg/m3 वाले एक ब्लॉक का पानी में कितना भाग डूबा रहता है?
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For floating block ρ_object/ρ_fluid = V_sub/V_total. Thus fraction submerged = 600/1000 = 0.6, i.e., 60% submerged. / तैरते हुए ब्लॉक के लिए V_sub/V = ρ_object/ρ_fluid = 600/1000 = 0.6 अर्थात् 60% भाग डूबा रहेगा।
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A long rod is rigidly clamped at both ends and is heated through ΔT. Derive expression for the thermal stress developed. / एक लंबी छड़ दोनों सिरों पर कठोरता से जकड़ी हुई है और उसे ΔT ताप बढ़ाया जाता है। विकसित होने वाले थर्मल तनाव का व्यंजक निकालीए।
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If free, expansion would be ΔL_free = α L ΔT. Clamped ends force ΔL_actual = 0, so mechanical strain needed is ε = (ΔL_actual - ΔL_free)/L = -α ΔT. Using Hooke’s law σ = E ε gives σ = -E α ΔT. Magnitude of compressive thermal stress = E α ΔT (compressive sign shown negative). / स्वतंत्र होने पर विस्तार ΔL_free = α L ΔT होता। किन्तु जकड़े होने पर वास्तविक विस्तार 0 होने से यांत्रिक स्ट्रेन ε = -α ΔT होता। Hooke के नियम से σ = E ε = -E α ΔT। तनाव का परिमाण E α ΔT (संपीडनात्मक)।
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Calculate the energy stored in a spring of constant 400 N/m when stretched by 5 cm. / 400 N/m का एक स्प्रिंग 5 cm से खींचा गया है; इसमें संग्रहीत ऊर्जा ज्ञात कीजिए।
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Elastic energy U = 1/2 k x^2 = 0.5×400×0.05^2 = 0.5×400×0.0025 = 0.5×1 = 0.5 J. / U = 1/2 k x^2 = 1/2×400×0.05^2 = 0.5 J।
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Explain why viscosity of liquids generally decreases with increase in temperature while viscosity of gases increases with temperature. / क्यों द्रवों की चिपचिपाहट (विस्कोसिटी) सामान्यतः तापमान बढ़ने पर घटती है जबकि गैसों में यह बढ़ती है?
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In liquids molecules are closely associated; increasing temperature increases molecular motion that overcomes intermolecular attractions, allowing layers to slide more easily, so viscosity decreases. In gases molecular collisions transfer momentum; with higher temperature average molecular speed increases leading to more frequent and more energetic collisions between layers, increasing momentum transfer and hence viscosity increases. / तरल में अणु निकट होते हैं और अधिक ताप पर गतिशीलता बढ़ने से आपसी आकर्षण कम असर डालते हैं, अतः परतें आसानी से सरकती हैं और विस्कोसिटी घटती है। गैसों में ताप बढ़ने पर अणुओं की औसत गति और टकराव बढ़ते हैं जिससे परतों के बीच महत्त्वपूर्ण संवेग आदान-प्रदान बढ़ता है और विस्कोसिटी बढ़ती है।
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An ideal gas is compressed isothermally from volume V to V/2 at pressure P. What is its bulk modulus during this process? / एक आदर्श गैस को समतापीय रूप से आयतन V से V/2 तक संपीड़ित किया जाता है और दबाव P है। इस प्रक्रिया के दौरान इसका बल्क मॉड्यूल क्या होगा?
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For an ideal gas under isothermal conditions K = P (since ΔP/P = -ΔV/V). Thus bulk modulus equals the pressure at that state; so K = P. / समतापीय आदर्श गैस के लिए K = P ही होता है। अतः बल्क मॉड्यूल K = P।
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A 0.1 kg lead shot (density 11340 kg/m3) is dropped in oil of viscosity 0.5 Pa·s and density 900 kg/m3. Assuming Stokes’ law applies and terminal velocity is reached, write the equation to find terminal velocity and indicate how to solve. / घनत्व 11340 kg/m3 वाले 0.1 kg लीड के दाने को तेल में गिराया जाता है; तेल का विस्कोसिटी 0.5 Pa·s और घनत्व 900 kg/m3 है। स्टोक्स के नियम लागू मानते हुए और टर्मिनल वेग पहुँचने पर टर्मिनल वेग खोजने हेतु समीकरण लिखिए और बताइए इसे कैसे हल करेंगे।
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For a sphere, at terminal velocity mg - ρ_fluid V g - 6 π η r v_t = 0. Here mass m and sphere volume V = m/ρ_sphere give radius via V = (4/3) π r^3 so r = [3m/(4 π ρ_sphere)]^(1/3). Substitute r and solve algebraically for v_t: v_t = [ (m g - ρ_fluid V g) ] / (6 π η r). Compute r from m and ρ_sphere, compute V, then evaluate numerator and denominator to get v_t. / टर्मिनल स्थिति में समीकरण: m g - ρ_fluid V g - 6 π η r v_t = 0। यहाँ V = m/ρ_sphere और r = [3m/(4 π ρ_sphere)]^(1/3)। इन्हें प्रतिस्थापित करके v_t = (m g - ρ_fluid V g)/(6 π η r) निकाला जाएगा; पहले r और V निकाल कर पुट करके गणना करें।
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Explain Archimedes’ principle and how it can be used to find the density of an irregular solid. / अर्किमिडीज सिद्धांत क्या कहता है और इसका उपयोग एक अनियमित ठोस के घनत्व को जानने में कैसे किया जा सकता है?
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Archimedes’ principle: a body immersed in a fluid experiences an upward buoyant force equal to the weight of fluid displaced. To find density: measure the mass m of the solid in air, then submerge it fully in a fluid of known density and measure apparent loss of weight equal to buoyant force; displaced fluid mass = weight loss/g. Volume of solid V = mass of displaced fluid / ρ_fluid. Then density of solid ρ_solid = m / V. Practically use a balance and a beaker to measure loss in weight on immersion. / अर्किमिडीज के अनुसार, द्रव में डूबे शरीर पर ऊपर की ओर उत्थापन बल वह होता है जितना द्रव का वजन विस्थापित होता है। अनियमित ठोस का घनत्व मापने के लिए उस वस्तु का द्रव्यमान वायु में नापें, फिर उसे द्रव में पूर्णतया डुबोकर वज़न में कमी नापें; यह कमी विस्थापित द्रव के वजन के बराबर होगी जिससे वस्तु का आयतन और अन्ततः घनत्व ρ = m/V निकाला जा सकता है।
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Calculate heat required to raise temperature of 0.5 kg of water from 20°C to 80°C. / 0.5 kg पानी का तापमान 20°C से 80°C तक बढ़ाने हेतु आवश्यक ऊष्मा की गणना कीजिए।
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Use Q = m c ΔT with c_water = 4186 J kg^-1 K^-1. ΔT = 60 K. Q = 0.5×4186×60 = 125580 J ≈ 1.26×10^5 J. / Q = m c ΔT = 0.5×4186×60 = 125580 J ≈ 1.26×10^5 J।
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