PHYSICS FORM FIVE – devinevisiontech.com

Form Six Advanced Physics — Complete Text Notes with Examples

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Mechanics: measurement, dimensions, derivation, gravitation, viscosity and fluid flow.
Thermal physics: thermodynamics, heat capacities, ideal gases, heat transfer and radiation.
Waves and optics: resonance, Doppler effect, interference, polarization.
Electricity and fluids: electric flux, fields, potential, capacitors, surface tension and capillarity.

Quick equation reference

Area	Key equations	Use
Dimensions	$[A]=L^2$, $[V]=L^3$, $[v]=LT^{-1}$	Checking formula correctness and deriving relationships.
Gravitation	$F=GM_1M_2/r^2$, $g’=g(1-d/R)$, $g’=gR^2/(R+h)^2$	Weight change with distance, depth and altitude.
Fluids	$Q=V/t=Av$, $P_2=\frac{P_1r_1^4l_2+P_3r_2^4l_1}{r_1^4l_2+r_2^4l_1}$	Flow rate and pressure at tube junctions.
Thermal	$\Delta Q=\Delta U+W$, $PV=nRT$, $PV^\gamma=constant$, $Q/t=kA\Delta T/L$	Gas processes, work, internal energy and conduction.
Waves/Optics	$T=2\pi\sqrt{l/g}$, $f_0=\frac{V}{V-U_s}f$, $ heta=\lambda/a$	Pendulum, Doppler effect and interference fringes.
Electricity/Capillarity	$\Phi=Q/\varepsilon_0$, $C_p=C_1+C_2+…$, $1/C_s=1/C_1+…$, $h=2\gamma/( ho gr)$	Flux, capacitor combinations and capillary rise.

Formula writing — Important equations are boxed and rendered using MathJax style.

Question practice — Problem banks are grouped in expandable panels for revision.

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Measurement & Dimensional Analysis

Mechanics

Measurement is the process of assigning numbers to a given physical quantity. To describe the behaviour of objects we must consider matter, space and time — every measurable property in mechanics reduces to combinations of these.

Physical quantities

Fundamental (basic) quantities — independent physical quantities such as mass, length and time. Each has both a dimension and a standard unit. Dimensions are represented as M, L and T respectively.

Derived quantities — obtained by combining fundamental quantities, e.g. area, volume, density, speed and momentum.

Common derived dimensions

\[ [A] = L\times L = L^2 \qquad [V] = L\times L\times L = L^3 \] \[ [\rho] = \frac{M}{V} = ML^{-3} \qquad [v] = \frac{\text{Distance}}{\text{Time}} = LT^{-1} \]

Dimension & dimensional analysis

Dimension is the way in which a physical quantity is related to the fundamental physical quantities. Dimensional analysis shows how physical quantities relate to each other. A quantity whose dimension involves more than one fundamental unit is written generally as $K(M)^x(L)^y(T)^z$, where $K$ is a pure number and $x,y,z$ are the powers (“dimensions”) of the derived unit.

Example — the area of a square of side 1 m: $(L)\times(L)=L^2$; so area has the dimension of length squared. Velocity is the rate of change of displacement, $V = L/T = LT^{-1}$.

Uses of dimensions: deriving formulae, and checking the homogeneity of an equation (both sides must carry the same dimensions).

Worked derivation — relating Work, mass and velocity solved

Suppose observation suggests work $W$ is proportional to mass $m$ and velocity $v$:

\[ W = k\,m^x v^y \qquad (i) \]

Since $W = F\times s$ and $F=ma$:

\[ [W] = [F][s] = MLT^{-2}\cdot L = ML^2T^{-2} \] \[ [v] = LT^{-1} \]

Substituting the dimensions into (i):

\[ M^1L^2T^{-2} = k\,M^xL^yT^{-z} \]

Equating indices: for M, $x=1$; for L, $y=2$; for T, $-y=-2 \Rightarrow y=2$.

Result: $ W = k\,m\,v^2 $ (with $k=\tfrac12$ from full mechanics, giving the familiar kinetic-energy form).

Additional worked example for revision

Extra example — test dimensional correctness

Check whether $s=ut+\frac12at^2$ is dimensionally correct.

Solution

\[ [s]=L \] \[ [ut]=(LT^{-1})(T)=L \] \[ [at^2]=(LT^{-2})(T^2)=L \]

Every term has dimension $L$, so the equation is dimensionally homogeneous.

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Complete text-only source transcript for this chapter (pages 1-4)

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MECHANICS 1.0 MEASUREMENTMeasurement is the process of assigning numbers to a given physical quantity. 1.0 Physical Quantity In describing the behavior of objects around us we have to consider to matter, space and time. A moving body covers distance with time and for an object to move energy is required. For the motion to take place, force must be applied. When an object is in the course of motion changes its speed within a given time interval we said that it is undergoing acceleration. In all this we have physical quantities which are measurable and whose values can be used in the mathematical expressions to give numerical description about the object in a question. The physical quantities are divided into two categories which are fundamental / basic quantitics and derived quantities. (a) Fundamental quantities These are independent physical quantities such as mass, length and time. These quantities have both dimensions and standard units which can be expressed dimensionally.The dimensions ofmass, length and time are represented as M, L and T respectively. The term dimension is used to denote the nature of physical quantity.(b) Derived quantities The physical quantities which are obtained from fundamental quantities are called derived quantities. An example of derived quantities are such as area, volume, density, speed, and momentum, These quantities can be obtained by combining the fundamental quantities in one way or the other. The following are the few examples:(i) Area =Length Length[AJ=L x L=L? (ii) Volume =Length x Length x Length[V]=LxLxL

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(iii) Density = Mass/Volume [p] = M/V (iv)Speed = Distance/Time(VJ=UT=_LT”DIMENSION Dimension is the way in which the physical quantities are related to fundamental physical quantities.DIMENSIONAL ANALYSIS Dimensional analysis is the way of showing how physical quantities are related to each other. The alphabets used to represent particular unit may be called a symbol. There are various systemsin use for the same unit , however the symbol MI. and I are dimensionally used for mass, lengthand time tespectively. The dimensions of a physical quantity refers to a fundamental units contained in it. Any quantity which can be measured in mass unit only, may said to have the dimension of mass. The derived units are based on the fundamental quantities and in many cases it involves more thanone fundamentals in such case the dimension of such quantity is expressed in generalas K(M)*(I.)*(T)”where K is the pure numeral of x, y and z which indicate how many times a particular unit is involved.’The power to which the fundamental units are raised can be obtained and are called the dimensionof the derived unit.For example the area of a square whose sides are in m_ each.lmxlm=1m? The dimension of the area of a square 1m? is; (L) x(L)=L?; then area has the dimension of length.The dimension of velocity can be obtained from the definition of velocity which is; Velocity is the rate of change in displacement. Its unit is meter per second.Thedimension of the velocity V=L/T=LT”USES OF DIMENSIONDimensions of physical quantities can be used in the derivation of formula, checking ofhomogeneity of the formula etc

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Derivation of FormulaDimensions are sometimes used as a tool in establishing relationship between physical quantities. For example through observation one would like to establish the connection between mass (m), its velocity (v) and the work done(w) on it.The following are steps to follow:Form a statement that: Work is proportional to mass and velocityWhere k is the proportionality constant Dimensions Work = Force x Distance W=FxSWhere F = ma [W] = [F] [S] = [] [a] [s]= MLT?L= MI2T2[V]= LT! Substitute the dimension in equation (i) M’L?T? = kM*LYT?Compare and equate the indices of corresponding dimensionsFor M: x= 1 Li:y=2 T: -y =-2 or =2Substitute for x and y in equation (i)

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W = km¹v² or W = kmv²

Gravitation

Mechanics

Variation of g

Effect of rotation — at the poles

\[ g’ = g – \omega^2 R\cos^2\lambda \]

At the poles $\lambda = 90^\circ$, so $\cos^2 90^\circ = 0$ and $g_p = g$: rotation has no effect on $g$ at the poles.

Newton’s law of gravitation

\[ F = \frac{GM_1M_2}{r^2} \]

Doubling the separation ($r\rightarrow 2r$) reduces the force to one quarter: $F_2 = F_1/4$, since $Fr^2=\text{constant}$.

Variation of g with depth

\[ g’ = g\left[1-\frac{d}{R}\right] \]

At the centre of the Earth ($d=R$) the weight of a body is zero.

Variation of g with altitude

\[ g’ = \frac{gR^2}{(R+h)^2} \]

Worked example — % decrease in weight 16 km below the surface solved

Given $d = 16\text{ km}$, $R = 6400\text{ km}$:

\[ \frac{mg-mg’}{mg} = \frac{d}{R} = \frac{16}{6400}\times100\% = 0.25\% \]

So the weight decreases by 0.25 % at that depth. At the centre of the Earth the weight becomes exactly zero.

Worked example — altitude at which g is half its surface value solved

\[ \frac{g}{2} = \frac{gR^2}{(R+h)^2} \;\Rightarrow\; (R+h)^2 = 2R^2 \] \[ h = (\sqrt{2}-1)R \]

Worked example — angular velocity for which g at the equator is zero solved

At the equator: $g_e = g – R\omega^2$. Setting $g_e=0$:

\[ g – R\omega^2 = 0 \;\Rightarrow\; \omega = \sqrt{\frac{g}{R}} \]

This imaginary angular velocity is the spin rate at which objects at the equator would become weightless — used to find the corresponding (much shorter) length of the day.

Jumping & pendulums off-Earth

Since $g_{\text{moon}} \approx \tfrac16\,g_{\text{earth}}$, and $v^2 = 2gh$ for a given take-off speed, a smaller $g$ gives a larger jump height $h$ — hence one can jump higher on the Moon.

Simple pendulum period

\[ T = 2\pi\sqrt{\frac{l}{g}} \]

Since $T \propto 1/\sqrt{g}$ and $g$ is smaller on a mountain than on a plain, a pendulum’s period is longer at altitude.

Additional worked example for revision

Extra example — force when distance is tripled

If gravitational force is $F$ at distance $r$, find the force at distance $3r$.

Solution

\[ F\propto \frac{1}{r^2} \Rightarrow F_2 = F\left(\frac{r}{3r} ight)^2=\frac{F}{9} \]

Tripling distance makes the force one ninth.

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Complete text-only source transcript for this chapter (pages 5-8)

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d) Find the percentage decrease in weight of a body when taken 16km below the surface of the earth. What happens to the weight of the body center of the earth’s/ radius of earth = 6400kmSolution a) At poles A 90° sothat 9 =9— @R?cos?AIp = 9 — WR’ cos?90 In =9Hence there is no effect of rotational motion of carth on the value of 9 at the poles.b) According to Newton’s law of gravitation_ MyM:Tf the distance between the objects is double gravitational force between the objects decrease to onc —fourth MyM. oe F,=Gt “ory tas ces ees eee cee eee see sae ove (di) Also, 2 = 472 Then, F,r*° = 4r°F, Ff, 2 Fy =”1/, 4 c) Solution The gravitational force between two bodies is «

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Fr = constant Fyr? = Fyrf 1x 1? = F; x (27)? 1×1? =F, x 472 F,= 0.25 d) Solution d=16km R= 6400kmThe acceleration due to gravity at a depth d isg’- al “e 1-d mg’™o| 1™9′ Ing = [: – 4,-™9 /ing =4pmg a ae = dp ™I~™4 Ing = 18/64q9 x 100%I~ ™9′ Ing = 0.25%At the center of the earth,

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ad=R=mg’ =mg|{1-4/p Weight of the body mg’ = mg[1—¥/p] Weight of body =0 2.a) Explain, why one can jump higher on the surface of the moon than on the earth? b) What will be the effect on the time period of a simple pendulum on taking it to a mountain?c) Assuming that the carth is a sphere of radius R at what altitude will be the value of acceleration due to gravity be half its value on the earth’s surface.d) Calculate that imaginary angular velocity of the earth for which the effective acceleration due togravity at the equator becomes zero. In this condition what will be the length (in hours) of the day? RE = 6400 kmI =10mis? a) SolutionLet us be the speed of the man while taking a jump and h the height of the jump. Thenv”—w=2gh 0? —UZ =1(g)h —UZ =—2ghSince acceleration due to gravity 4 on the moon is 1/6″ of that on the earth, one can jump higheron the surfaceb) Time period of simple pendulum TT=2

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Therefore, TX 1/ Vg . Since the value of g is less at mountain than at plain the time period of simple pendulum will be more at mountain than at plain.©) Solution g’=9 F – “al Also, g’= oF lg +h)?Now,19; ~gR? 9’= 9a = hee + in?(+n)? IRR+K= 2rR Wh= 2-1)R n-02- Dpd) Solution Support g is the acceleration due to gravity in the absence of rotational motion of earth. The acceleration due to gravity at the equator in the presence of earth’s rotation is given by Ie = 9 – Ru? Here w is the angular velocity of the earth for g = 0, we haveg-Rw? =0

Viscosity & Fluid Dynamics

Mechanics of Fluids

Pressure at a junction of tubes

Since $V/t = \text{constant}$ along a connected system, for Tube 1 (radius $r_1$, pressure drop $P_1-P_2$, length $l_1$) and Tube 2 (radius $r_2$, pressure drop $P_2-P_3$, length $l_2$) joined at a common junction:

Setting the two Poiseuille flow rates equal

\[ \frac{\pi r_1^4(P_1-P_2)}{8\eta\, l_1} = \frac{\pi r_2^4(P_2-P_3)}{8\eta\, l_2} \]

Cancelling $\pi/8\eta$ and cross-multiplying:

\[ r_1^4 l_2 (P_1-P_2) = r_2^4 l_1 (P_2-P_3) \] \[ P_1 r_1^4 l_2 – P_2 r_1^4 l_2 = P_2 r_2^4 l_1 – P_3 r_2^4 l_1 \]

Collecting all $P_2$ terms on one side:

\[ P_2\left(r_1^4 l_2 + r_2^4 l_1\right) = P_1 r_1^4 l_2 + P_3 r_2^4 l_1 \]

Junction pressure:

\[ P_2 = \frac{P_1 r_1^4 l_2 + P_3 r_2^4 l_1}{r_1^4 l_2 + r_2^4 l_1} \]

Applications of viscosity

Ink quality is judged by its coefficient of viscosity.
Variation of viscosity with temperature helps select the best lubricant for a machine.
Blood-flow studies: viscosity changes affect pressure and circulatory efficiency.
Damping in instruments, e.g. a car’s shock-absorber suspension.
Production and transport of oils.
High-viscosity liquids are used as buffers at railway stations.

Fluid dynamics — governing assumptions

The fluid is non-viscous (offers no internal frictional resistance).
The fluid is incompressible (density is constant).
The fluid motion is steady — velocity, density and pressure at each point don’t change with time.

Types of flow

Streamline (laminar) flow — every fluid particle passing a given point follows the same path at the same speed; also called orderly or uniform flow. A streamline is a curve whose tangent at any point gives the velocity direction there; streamlines never cross.

Turbulent flow — speed and direction of particles vary with time at a point; also called disorderly flow.

A tube of flow is a tabular region of a flowing fluid bounded by a set of streamlines; no fluid crosses the sides of the tube.

Rotational flow — fluid elements have net angular velocity about an axis. Irrotational flow — fluid elements have zero net angular velocity.

Rate of flow (discharge)

Definition & derivation

\[ Q = \frac{V}{t} \]

For a pipe of cross-section $A$ with average flow speed $v$, the volume passing in time $t$ is $V = A\,v\,t$, so:

\[ Q = Av \]

This is the discharge / continuity equation; $Av$ is called the flow rate or volume flux. SI unit: m³ s⁻¹.

Additional worked example for revision

Extra example — discharge through a pipe

A liquid flows through a pipe of cross-sectional area $2.0\times10^{-4}\,m^2$ at speed $3.0\,m/s$. Find the rate of flow.

Solution

\[ Q=Av=(2.0\times10^{-4})(3.0)=6.0\times10^{-4}\,m^3s^{-1} \]

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Complete text-only source transcript for this chapter (pages 9-13)

Source page 9

Expression of the pressure at the junctionSince, v = Constant t For Tubell yv wy (AR) A(AB) t & I,an(R-B) _*3(P,-P) 87 1; 87 12n@-B)_2@-8) ] 1,nil, (R-P)=r$1,P,-P) Prth, -Path, =Prih-Breh PBA +P {bh = Btn +PrhPGE +Hh)=Rrih + PrkPrih +PyrthR= nih +rhApplications of Viscosity1. The quality of ink is decided by the coefficient of viscosity of ink. 2. The study of variation of viscosity with temperature helps us to pick up the best lubricant of a certain machine. 3. Applied in the study of circulation of blood. The variation in the coefficient of viscosity of blood affects the pressure and that in turn affects the efficiency of our bloody.

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4. For damping the motion of certain instruments such as, shock absorber in car’s suspension system.5. Used in production and transportation of oils.6. Liquids having higher values of coefficient of viscosity are used as buffers at railway station.FLUID DYNAMICSFluid dynamics is the study of fluids in motion.While discussing fluid flow, we generally make the following assumptions. (i) The fluid is non-viscous (ii) The fluid is incompressible (iii) The fluid motion is steady Non viscous fluid. Non viscous fluid is the fluid which docs not offer a resistance to the motion through it of any solid body. There is no internal friction between the adjacent layers of the fluid,Tncompressible fluid.An incompressible fluid is the fluid in which changes in pressure produce no change in the density of the fluidThis means that density of the fluid is constant. Steady flow of a fluidThis means that the velocity, density and pressure at each point in the fluid do not change with time.TYPES OF LIQUID FLOWThe liquid flow is of two main types(i) Streamline flow or steady flow (ii) Turbulent flowStream line flow.Streamline flow is the flow of a fluid when all the fluid particles that pass any given point follow the same path at the same speed.The fluid particles have the same velocity. This flow is also called orderly flow or uniform flow.

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Characteristics of streamline. 1. The velocity of a particle at any point is a constant and is independent of time. 2. The liquid layer in contact with the solid surface will be at rest 3. The motion of the fluid (liquid) follows Newton’s law of viscous forceFor example, consider a liquid flowing through a pipe as shown in figure belowThe flowing liquid will have a certain velocity v: at a, a velocity v2 at b and so on.As time goes, the velocity of whatever liquid particle happens pas to pass be at a is still vi, that at b is still V2, then the flow is said to be steady or streamline flow.Every particle starting at a will follow the same path abc. The line abc is called streamline. Streamline.A streamline is a curve whose tangent at any point is along the direction of the velocity of the liquid particle at that point.Streamlines never cross each other otherwise, particles reaching the intersection would not have a unique velocity at that point in space.Tube of flow.

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A tube of flow is a tabular region of a flowing fluid whose boundaries are defined by a set of streamlines.JahSince the streamlines represent the path of particles, we see that no liquid conflation or out of the sidesof a tube of flow.Ina steady flow, the velocity, density and pressure at each point in the fluid do not change with time. Laminar flow.Laminar flow is a special case of steady flow in which the velocities of all particles on any given streamline are the same, though the particles of different streamlines may move at different speeds.Turbulent flow.Turbulent flow is the flow of fluid when the speed and direction of fluid particles passing any point vary with time,It is also known as disorderly flow.Line of flow.A line of flow is the path followed by a particle of the fluid.Rotational flowThis is when the element of fluid at each point the angular velocity is equal to zero. Irrotation flow.Irrotation flow is the type of fluid flow by which the element of fluid at each irrotation no net angular velocity about that axis.RATE OF FLOWA rate of flow is the volume of a liquid that passes the cross-section of a vessel (pipe) in one second

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It is denoted by the symbol Qo-” tThe SI unit of rate of flow of liquid is m’/s Consider a pipe of uniform cross-sectional area A as shown below.Prt, +Pyril,3 nL, +rfL,Tf the liquid is flowing at an average velocity of v, then distance Ltrrougn which the liquid moves in time tisThis may be regarded as the length of an imaginary cylinder of the liquid that has passed the section $ in time t.Then the volume of liquid that has passed section S$ in time t is VRate of flowQ=Av This is called discharge equation.The quantity Av is called the flow rate or volume flux.

Thermodynamics

Heat & Gases

A large problem set covers isobaric, isothermal and adiabatic processes for ideal gases, using the first law of thermodynamics, molar heat capacities, and latent heat. Key relations used throughout:

First law of thermodynamics

\[ \Delta Q = \Delta U + \Delta W \]

Heat supplied to a system equals the increase in internal energy plus the external work done by the system.

Ideal-gas process equations

\[ PV = nRT \quad\text{(ideal gas equation)} \] \[ W = P\Delta V \quad\text{(work in a small expansion at constant pressure)} \] \[ PV^\gamma = \text{constant} \quad\text{(reversible adiabatic process)}, \quad \gamma = \frac{C_P}{C_V} \]

Adiabatic change — a process in which no heat enters or leaves the system, yet the temperature can still change because all the energy exchange is via work done on/by the gas. A gas cools on adiabatic expansion and heats on adiabatic compression, because work is done at the expense of (or adding to) internal energy with no heat flowing in to compensate.

Isothermal change — temperature stays constant while heat is exchanged; internal energy of an ideal gas is unchanged since $U$ depends only on $T$.

Molar / specific heat capacities

Difference between principal specific heats

\[ C_P – C_V = R \quad\text{(per mole)} \]

An ideal gas can, in principle, have infinitely many heat capacities depending on the process path — but only $C_P$ (constant pressure) and $C_V$ (constant volume) are the two “principal” values.

Heated at constant volume: all the energy added increases internal energy (no work is done, no volume change).
Heated at constant pressure: energy added increases internal energy and does external work as the gas expands — hence $C_P > C_V$.

Work & heat in isothermal expansion

For an ideal gas in thermal contact with a large constant-temperature reservoir, expanding isothermally from $V_1$ to $V_2$:

\[ W = nRT\ln\!\left(\frac{V_2}{V_1}\right) \]

Latent heat & vaporization

Why heat is needed to vaporize a liquid: energy must be supplied to overcome intermolecular attractive forces as molecules separate from the liquid into the vapour phase, and to do external work as the vapour expands against atmospheric pressure.

Heat, work and internal-energy change on vaporization

\[ Q = mL \qquad W = P\Delta V \qquad \Delta U = Q – W \]

For water at 100 °C and 101 kPa, volume increases by a factor of ~1670 on turning to steam — most of the latent heat goes into internal energy (separating molecules), with a smaller share doing work of expansion.

Worked outline — heat-engine cycle (constant volume → adiabatic → constant pressure) method

A classic exam cycle: process 1→2 at constant volume, 2→3 adiabatic, 3→1 at constant pressure (e.g. 1 atm), with $\gamma = 5/3$ for a monatomic gas. Method:

Use $PV=nRT$ at each labelled state to find the unknown P, V from given temperatures.
Apply $PV^\gamma=\text{const}$ across the adiabatic leg.
Net work of the cycle = area enclosed on the P–V diagram = sum of $W$ for each leg (using $W=P\Delta V$ for isobaric legs, $W=0$ for constant-volume legs, and the adiabatic work formula $W=\dfrac{P_iV_i-P_fV_f}{\gamma-1}$ for the adiabatic leg).

Piston-cylinder P–V cycle (precedes Problem 81)

A cylinder fitted with a piston contains gas at pressure $P_1$ and temperature $T_1$. Its state changes are plotted on a Pressure–Volume diagram with points A ($P_1,V_1$), B ($P_1,V_2$) and C ($P_2,V_1$), where A→B is a horizontal (constant-pressure) line and A→C, B→C trace a curve back through C.

(i) A → B — the air is heated to 373 K at constant pressure. Calculate the new volume.
(ii) B → C — the air is compressed isothermally to volume $V_1$. Calculate the new pressure $P_2$.
(iii) Calculate the root-mean-square speed of nitrogen molecules at a temperature of 27 °C.

P–V cycle: heat engine (Problem 87)

A heat engine carries 1 mole of an ideal gas around a cycle shown on a P–V diagram: process 1→2 is at constant volume, process 2→3 is adiabatic, and process 3→1 is at a constant pressure of 1 a.t.m. The value of $\gamma$ for this gas is $5/3$. Given $T_1=300\,\text{K}$, $T_2=453\,\text{K}$, $T_3=600\,\text{K}$. Find: (i) the pressure and volume at points 1, 2 and 3; (ii) the net work done by the gas in the cycle.

Full problem bank — verbatim statements

Every problem from the thermodynamics problem set (Problems 81–140), reproduced as given in the source notes:

Problems 81 – 88

Problem 81. a) State the 1^st law of thermodynamics and write its equation. b) A litre of air initially at 25 °C and 760 mmHg is heated at constant pressure until the volume is doubled. Determine: (i) the final temperature; (ii) the external work done by the air in expanding it; (iii) the quantity of heat supplied.

Problem 82. 0.15 mol of an ideal mono-atomic gas is enclosed in a cylinder at a pressure of 250 kPa and a temperature of 320 K. The gas is allowed to expand adiabatically and reversibly until its pressure is 100 kPa. (a) Sketch a P–V curve for the process. (b) Calculate the final temperature and the amount of work done by the gas.

Problem 83. i) Define the bulk modulus of a gas. ii) Find the ratio of the adiabatic bulk modulus of a gas to that of its isothermal bulk modulus in terms of the specific heat capacities of the gas.

Problem 84. (a) A gas expands adiabatically and its temperature falls, while the same gas when compressed adiabatically its temperature rises. Explain, giving reasons, why this happens. (b) A mole of oxygen at 280 K is insulated in an infinitely flexible container at $5\times10^5\,\text{Nm}^{-2}$. When 580 J of heat is supplied to the oxygen the temperature increases to 300 K and the volume of the container increases by $3.32\times10^{-4}\,\text{m}^3$. Calculate the values of the principal molar heat capacities and the specific universal gas constant. Given that molar mass of oxygen = $32\times10^{-3}\,\text{kg}$.

Problem 85. (a)(i) Why is heat needed to change liquid water into vapour? What amount of energy is needed? (ii) The molar heat capacity of hydrogen at constant volume is $20.2\,\text{J mol}^{-1}\text{K}^{-1}$. What is the molar heat capacity at constant pressure? (b) In an industrial refrigerator ammonia is vaporized in the cooling unit to produce a low temperature. Why should the evaporation of ammonia reduce the temperature in the refrigerator? How much energy is needed to convert 150 g of water at 20 °C into steam at 100 °C?

Problem 86. An ideal gas is kept in thermal contact with a very large body of constant temperature T and undergoes an isothermal expansion in which its volume changes from $V_1$ to $V_2$. Derive an equation for the work done by the gas.

Problem 87. A heat engine carries 1 mole of an ideal gas around a cycle as shown in the figure. Process 1–2 is at constant volume, process 2–3 is adiabatic, and process 3–1 is at a constant pressure of 1 a.t.m. The value of $\gamma$ for this gas is 5/3 (with $T_1=300\,\text{K}$, $T_2=453\,\text{K}$, $T_3=600\,\text{K}$). Find: (i) the pressure and volume at points 1, 2 and 3; (ii) the net work done by the gas in the cycle.

Problem 88. [Given $C_P/C_V = \gamma = 1.40$, continues into Problem 120.]

Problems 120 – 128

Problem 120. In a diesel engine, fuel oil is injected into a cylinder in which air has been heated by adiabatic compression to above the ignition temperature of the oil. The ignition temperature of a certain fuel is 630 °C, and the air enters the cylinder, which has an initial volume of $5.0\times10^{-4}\,\text{m}^3$ at a pressure of $1.0\times10^{5}\,\text{Pa}$ and a temperature of 28 °C. (a) What minimum compression ratio (the ratio of the initial to the final volume of the cylinder) is required to heat the air to the fuel ignition temperature? (b) How much work is done in compressing the air? Given that for air $\gamma = 1.40$.

Problem 121. (a) A cylinder fitted with a piston which can move without friction contains 0.05 mole of a mono-atomic ideal gas at a temperature of 27 °C and a pressure of $1.0\times10^{5}\,\text{Pa}$. Calculate: (i) the volume of the gas; (ii) the internal energy of the gas. (b) The temperature of the gas in (a) above is raised to 77 °C, the pressure remaining constant. Calculate: (i) the change in internal energy; (ii) the external work done; (iii) the total heat energy supplied. Given that molar gas constant = $8.3\,\text{J mol}^{-1}\text{K}^{-1}$.

Problem 122. (a) Give one practical example of each of the following: (i) A process in which heat is supplied to a system without causing an increase in temperature. (ii) A process in which no heat enters or leaves a system but the temperature changes. (b) What happens to the energy added to an ideal gas when it is heated: (i) At constant volume? (ii) At constant pressure? (c) Deduce an expression for the difference between the specific heat capacities of a gas at constant pressure and at constant volume. (d) If the ratio of the principal specific heat capacities of a certain gas is 1.40 and its density at S.T.P is $0.09\,\text{kg m}^{-3}$, calculate the values of the specific heat capacity at constant pressure and at constant volume. Standard atmospheric pressure = $1.01\times10^{5}\,\text{Nm}^{-2}$.

Problem 123. A steel pressure vessel of volume $2.2\times10^{-2}\,\text{m}^3$ contains $4.0\times10^{5}\,\text{Pa}$ and temperature 300 K. An explosion suddenly releases $6.48\times10^{4}\,\text{J}$ of energy, which raises the pressure instantaneously to $1.0\times10^{6}\,\text{Pa}$. Assuming no loss of heat to the vessel, and ideal gas behaviour, calculate: (a) the maximum temperature attained; (b) the two principal specific heat capacities of the gas. What is the velocity of sound in this gas at a temperature of 300 K?

Problem 124. (a) Explain why an ideal gas can have an infinite number of molar heat capacities and define the principal values. (b) A thermally-insulated tube through which a gas may be passed at constant pressure contains an electric heater and thermometers for measuring the temperature of the gas as it enters and as it leaves the tube. $3.0\times10^{-3}\,\text{m}^3$ of gas of density $1.8\,\text{kg m}^{-3}$ flows into the tube in 90 seconds and, when electrical power is supplied to the heater at a rate of 0.16 W, the temperature difference between the outlet and inlet is 2.5 K. Calculate a value for the specific heat capacity of the gas at constant pressure.

Problem 125. (a) Explain clearly and concisely why, for a fixed mass of a perfect gas: (i) the internal energy remains constant when the gas expands isothermally; (ii) the heat capacity at constant pressure is greater than the heat capacity at constant volume. (b) A vessel of volume $1.0\times10^{-2}\,\text{m}^3$ contains an ideal gas at a temperature of 300 K and pressure $1.5\times10^{5}\,\text{Pa}$. Calculate the mass of gas, given that the density of the gas at temperature 285 K and pressure $1.0\times10^{5}\,\text{Pa}$ is $1.2\,\text{kg m}^{-3}$. (c) 750 J of heat is suddenly released in the gas, causing an instantaneous rise of pressure to $1.8\times10^{5}\,\text{Pa}$. Assuming ideal gas behaviour, and no loss of heat to the containing vessel, calculate the temperature rise, and hence the specific heat capacity at constant volume of the gas.

Problem 126. (a) What is an adiabatic change? A vessel of volume $8.00\times10^{-3}\,\text{m}^3$ contains an ideal gas at a pressure of $1.14\times10^{5}\,\text{Pa}$. A stopcock in the vessel is opened and the gas expands adiabatically, expelling some of its original mass, until its pressure equals that outside the vessel, $1.01\times10^{5}\,\text{Pa}$. The stopcock is then closed and the vessel is allowed to stand until the temperature returns to its original value; in this equilibrium state the pressure is $1.06\times10^{5}\,\text{Pa}$. (i) Explain why there was a temperature change as a result of the adiabatic expansion. (ii) Find the volume which the mass of gas finally left in the vessel occupied under the original conditions. (iii) Sketch a graph showing the way in which the pressure and volume of the mass of gas left in the vessel changed during the operations described above. (iv) What is the value of $\gamma$, the ratio of the principal heat capacities of the gas? (v) What can you deduce about the molecules of the gas? Give your reasons.

Problem 127. The diagram represents an energy cycle whereby a mole of an ideal gas is firstly cooled at constant pressure (A → B), then heated at constant volume (B → C), and returned to its original state (C → A). Pressure/volume values: at A, $P=1\times10^{5}\,\text{Nm}^{-2}$, $V=0.025\,\text{m}^3$; at B, $P=1\times10^{5}\,\text{Nm}^{-2}$, $V=0.125\,\text{m}^3$; at C, $P=2\times10^{5}\,\text{Nm}^{-2}$, $V=0.125\,\text{m}^3$. (a) Calculate the temperature of the gas at A, at B and at C. (b) Calculate the heat given out by the gas in the process A → B. (c) Calculate the heat absorbed in the process B → C. (d) Calculate the net amount of heat transferred in the cycle. Given that $R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}$ and $C_V = \tfrac{5}{2}R$.

Problem 128. The specific latent heat of vaporization of a particular liquid at 130 °C and a pressure of $2.60\times10^{5}\,\text{Pa}$ is $1.84\times10^{6}\,\text{J kg}^{-1}$. The specific volume of the liquid under these conditions is $2.00\times10^{-3}\,\text{m}^3\text{kg}^{-1}$, and that of the vapour is $5.66\times10^{-1}\,\text{m}^3\text{kg}^{-1}$.

Problems 134 – 140

Problem 134. (a) By considering the expansion of an ideal gas contained in a cylinder and enclosed by a piston, show that the work done in a small expansion is equal to the pressure times the volume change. (b) An ideal gas, at a temperature of 290 K and a pressure of $1.0\times10^{5}\,\text{Nm}^{-2}$, occupies a volume of $1.0\times10^{-3}\,\text{m}^3$. Its density under these conditions is $0.30\,\text{kg m}^{-3}$. It expands at constant pressure to a volume of $1.5\times10^{-3}\,\text{m}^3$. Calculate the energy added. (c) The gas is now compressed isothermally to its original volume. Calculate: (i) its final pressure and temperature; (ii) the difference between its final and initial internal energies. Given that the specific heat capacity at constant volume of this gas $= 7.1\times10^{2}\,\text{J kg}^{-1}\text{K}^{-1}$.

Problem 135. A litre of air, initially at 20 °C and at 760 mmHg pressure, is heated at constant pressure until its volume is doubled. Find: (a) the final temperature; (b) the external work done by the air in expanding; (c) the quantity of heat supplied. Assume that the density of air at S.T.P is $1.293\,\text{kg m}^{-3}$ and that the specific heat capacity of air at constant volume is $714\,\text{J kg}^{-1}\text{K}^{-1}$.

Problem 136. a) Deduce an expression for the difference between the specific heat capacities of an ideal gas. (b) If the specific heat capacity of air at constant pressure is $1013\,\text{J kg}^{-1}\text{K}^{-1}$ and the density at S.T.P is $1.29\,\text{kg m}^{-3}$, estimate a value for the specific heat capacity of air at constant volume.

Problem 137. (a) What is the importance of the ratio of the specific heat capacities of an ideal gas? (b) A mass of air occupying initially a volume $2\times10^{-3}\,\text{m}^3$ at a pressure of 760 mmHg and a temperature 20 °C is expanded adiabatically and reversibly to twice its volume, and then compressed isothermally and reversibly to a volume of $3\times10^{-3}\,\text{m}^3$. Find the final temperature and pressure, assuming the ratio of the specific heat capacities of air to be 1.40.

Problem 138. Air initially at 27 °C and at 750 mmHg pressure is compressed isothermally until its volume is halved. It is then expanded adiabatically until its original volume is recovered. Assuming the changes to be reversible, find the final pressure and temperature; take $\gamma = 1.40$.

Problem 139. When water at 100 °C and pressure of 101 kPa changes to steam under the same conditions, its volume increases by a factor of 1670, given the density of water is $960\,\text{kg m}^{-3}$ at 100 °C and 101 kPa, and its specific latent heat of vaporization is $2.26\times10^{6}\,\text{J kg}^{-1}$. Calculate: (a) the heat supplied to convert 1 kg of water at 100 °C to steam at the same temperature; (b) the work done when 1 kg of water turns to steam at 101 kPa pressure; (c) the increase of internal energy.

Problem 140. A fixed mass of ideal gas is contained in a cylinder. The cylinder volume can be varied by moving a piston in or out. The gas has an initial volume of $0.01\,\text{m}^3$ at 100 kPa pressure and its temperature is initially [statement continues in source with the volume/pressure change to be evaluated].

Additional worked example for revision

Extra example — using the first law

A gas receives $500\,J$ of heat and does $180\,J$ of work. Find the increase in internal energy.

Solution

\[ \Delta Q=\Delta U+W \Rightarrow \Delta U=500-180=320\,J \]

No scanned page is embedded here. The full source content is preserved as searchable text below, while the study notes above are rewritten for easier understanding.

Complete text-only source transcript for this chapter (pages 14-25)

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PistonPlTlPressure> Volumei. (i) AB ~ the air heated to 373 K at constant pressure. Calculate the new volume.i. (ii) BC ~ the air is compressed isothermally to volume1 Calculate the new pressure P,i. (iii) Calculate the root mean square speed of nitrogen molecules at a temperature of27 °C Problem 81a. () State the 1* law of thermodynamics and write its equation.b. () A liter of air initially at 28¢ and 760mmHg is heated at constant pressure until the volume is doubled. Determine:(i) The final temperature(ii) The external work done by the air in expanding it.

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(iii) | The quantity of heat supplied Problem 820.15 mol of an ideal mono atomic gas is enclosed in a cylinder at a pressure of 250 KPa and a temperature of 320K. The gas is allowed to expand adiabatically and reversibly until its pressures is 100KPa(a) Sketch a P — V curve for the process. (b) Calculate the final temperature and the amount of work done by the gas. Problem 83i. () Define the bulk modulus of a gasi. (1) Find the ratio of the adiabatic bulk modulus of a gas to that of its isothermal bulk modulus in terms of the specific heat capacities of the gas.Problem 84 (a) A gas expands adiabatically and its temperature falls while the same gas whencompressed adiabatically its temperature rises, Explain giving reasons why this happens.(b) A mole of oxygen at 280K is insulated in an infinitely flexible container is 5 5 m2 x 10°Nm™” When 5804 of heat is supplied to the oxygen the temperature increasesto 300K and the volume of the container increases by 3.32 * 10™”m” ‘Calculate the values of the principal molar heat capacities and the specific universal gas constant.-3 Given that molar mass of oxygen = 32 x 10-°kgProblem85 (a) (i) Why is heat needed to change liquid water into vapour? What amount of energy is needed (ii) The molar heat capacity of hydrogen at constant volumeis 20,2/mol*K™

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What is the molar heat capacity at constant pressure? (b) In an industrial refrigerator ammonia is vaporized in the cooling unit to producea low temperature. Why should the evaporation of ammonia reduce the temperature in the refrigerator?How much energy is needed to convert 150g of water at 20°C into steam at °CProblem 86An ideal gas is kept in thermal contact with a very large body of constant temperature T andundergoes an isothermal expansion in which its volume changes from Vito Vo. Derive an equation for the work done by the gas. Problem 87A heat engine carries | mole of an ideal gas around a cycle as shown in the figure below. Process 1 — 2 is at constant volume, process 2 — 3 is adiabatic and process 3 — 1 is at a constant5 1 pressure of | a.t.m. The value of ” for this gas isFind: i. ) The pressure and volume at points 1, 2 and 3ii ) The net work done by the gas in the cycle.Problem 88

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C, “lo, Given that =Y=140Problem 120In a diesel engine, fuel oil is injected into a cylinder in which air has been heated by adiabatic compression to above the ignition temperature of the oil. The ignition temperature of a certainfuel is 630°C, and the air enters the cylinder, which has an initial volume of 5.0 X10™%m?* ata Ss pressure of 1.0 * 10° pa and a temperature of 28 °c(a) What minimum compression ratio (the ratio of the initial to the final volume of the cylinder) is required to heat the air to the fuel ignition temperature?(b) How much work is done in compressing the air?Given that for air ¥ = 1.40 Problem 121(a) A cylinder fitted with a piston which can move without friction contains 0.05 mole of a monoatomic ideal gas at a temperature of 27°C anda pressure of 1.0 X10 py,Calculate:(i) The volume of the gas.(ii) The internal energy of the gas (b) The temperature of the gas in (a) above is raised to Ee the pressure remaining constant. Calculate:(i) The change in internal energy(ii)The external work done(iii) The total heat et energy supplied Given that molar gas constant = 33/’ mol”*k : Problem 122 (a) Give one __ practical example of each of the following:

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(i) A process in which heat is supplied to a system without causing an increase in

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temperature. (ii) A process in which no heat enters or leaves a system but the temperature changes.(b) What happens to the energy added to an ideal gas when it is heated: (i) At constant volume? (ii) At constant pressure?(c) Deduce an expression for the difference between the specific heat capacities of a gas at constant pressure and at constant volume. (d) Ifthe ratio of the principal specific heat capacities of’a certain gas is 1.40 and its density <3 at S.T.P is oogkgm , calculate the values of the specific heat capacity at constant5 Nen-2 pressure and at constant volume. Standard atmospheric pressure = 1.01 x 10° NmProblem 123-2 3 5 A steel pressure vessel of volume 2.2 * 10°” m contains 4.0 * 10° pa and temperature 300 K. An explosion suddenly releases 6.48 x 10 J of energy, which taises the pressure instantaneously to 1.0 x10 Pa, Assuming no loss of heat to the vessel, and ideal gas behaviour, Calculate: (a) The maximum temperature attained (b) The two principal specific heat capacities of the gas. What is the velocity of sound in this gas at a temperature of 300 K?Problem 124(a) Explain why an ideal gas can have infinity number of molar heat capacities and define the principal values.(b)A thermally — insulated tube through which a gas may be passed at constant pressure contains an electric heater and thermometers for measuring the temperature of thegas as it enters and as it leaves the tube. 3.0 * 10™” m of gas of density 1.8 kgm flows into the tube in 90 seconds and, when electrical power is supplied tothe heater at a rate of 0.16W, the temperature difference between the out let and inlet is 2.5 K. Calculate a value for the specific heat capacity of the gas at constant

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pressure.

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Problem 125 (a) Explain clearly and concisely why, for a fixed mass of a perfect gas: (i) The internal energy remains constant when the gas expands isothermally. (ii) The heat capacity at constant pressure is greater than the heat capacity at constant volume.x 10-?m? ae(b) A vessel of volume 1.0 contain an ideal gas at a temperature of 300 K and pressure15% 10 Pa. Calculate the mass of gas, given that the density of the gas at temperature 285 K and-3 pressure 1.0 x 10°p, is 1.2 kgm(c) 750 J of heat is suddenly releases in the gas, causing an instantaneous rise of pressure to 1.8 x 10° pa. Assuming ideal gas behavior, and no loss of heat to the containing vessel, Calculate the temperature rise, and hence the specific heat capacity at constant volume of the gas.Problem 126 (a) What is an adiabatic change?-343 5 09% 10° “m 4 X10°p,A vessel of volume 8. contains an ideal gas at a pressure of 1.1A stopcock in the vessel is opened and the gas expands adiabatically, expelling some of its originalmass, until its pressure is equal to that outside the vessel 1.01 *10°pa The stopcock is then closed and the vessel is allowed to stand until the temperature returns to its original valuc; in this equilibrium. 5 state, the pressure is 1.06 * 10 Pa, (i) Explain why there was a temperature change as a result of the adiabatic expansion.(ii)Find the volume which the mass of gas finally left in the vessel occupied under the original conditions.(iii)Sketch a graph showing the way in which the pressure and volume of the massof gas left in the vessel changed during the operations described above:(iv) What is the value of Y the ratio of the principal heat capacities of the gas. (v) What can you deduce about the molecules of the gas? Give your reasons.Problem 127

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P/Nm22X10?0.125 0.025 V/m? The diagram above represents an energy cycle whereby a mole of an ideal gas is firstly cooled at constant pressure (A md B) then heated a constant volume @ ° c) and retumed to its original state (c > A)(a) Calculate the temperature of the gas at A, at B and at C (b) Calculate the heat given out by the gas in the process A B(c) Calculate the heat absorbed in the process B 7c (d) Calculate the net amount of heat transferred in the cycle.3/mol-*k*Given that R = 8. andCy==RProblem 128s The specific latent heat of vaporization of particular liquid at 130 °C and a pressure of 2.60 x10 Pais 1.84 % 10° Jkg™*-3 9-3 g-1 The specific volume of the liquid under these conditions is 2.00 x 107° m™”kg . And that of the vapor is 5.66 x 107? mekg™*

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(b) The volume is kept constant (Given that relative molecular mass of oxygen = 32) Problem 134 (a) By considering the expansion of an ideal gas contained in a cylinder and enclosed by a piston, show that the work done in a small expansion is equal to the pressure times the volume change. 5 Nm-2 (b) An ideal gas, at a temperature of 290 K and a pressure of 1.0 x 10° Nm ; A 1 X 1073 m3 . . kgm-? occupies a volume of 1.0 . Its density conditions is 0.30 :-3 It expands at constant pressure to a volume of 1.5 x 10 – Calculate the energy added.(c) The gas is now compressed isothermally to its original volume. Calculate.(i) Its final pressure and temperature (ii) The difference between its final and initial internal energies. Given that specific heat capacity at constant volume of this gas = w= 107JKg*K7*Problem 135A litre of air, initially at 29 and at 760mmHg pressure, is heated at constant pressure until its volume is doubled. Find(a) The final temperature(b) The external work done by the air in expanding (c) The quantity of heat supplied.-3 Assume that the density of air at S.T_P is L293K gm and that the specific heat capacity of air at-1n-1 constant volume is 714/ Kg” *K~”Problem 136

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a) Deduce an expression for the difference between the specific heat capacities of an ideal gas.JKg”tK-1(b) If the specific heat capacity of air at constant pressure is 1013 and the density at-3 S.T.P is 129kgm , estimate a value for the specific heat capacity of air at constant volume.Problem 137(a) What is the importance of the ratio of the specific heat capacities of an ideal gas? (b) A mass of air occupying initially a volume 2 x 10°°m at a pressure of 760mmHg and a temperature 29°F is expanded adiabatically and reversibly to twice its volume,and then compressed isothermally and reversibly to a volume of 3 x 10m . Find the final temperature and pressure, assuming the ratio of the specific heat capacities of air to be 1.40.Problem 138Air initially at 27 Gna at 750mmHg pressure is compressed isothermally until its volume is halved. It is then expanded adiabatically until its original volume is recovered. Assuming the changes to be reversible find the final pressure andtemperature take Y= 1.40Problem 139When water at 100°C and pressure of 101 kPa changes to steam under the same conditions, its volume increases by a factor of 1670 given the density of water is 960-3 Kgm at 100°C and 101 kPa, and its specific latent heat of vaporization is 2.26 x 10°9JKgtCalculate(a) The heat supplied to convert | kg of water at 100°C to steam at the same temperature.(b) The work done when 1 kg of water tums to steam at 101kPa pressure. (c) The increase of internal energy.Problem 140 A fixed mass of ideal gas is contained in a cylinder. The cylinder volume can be varied by moving m3?

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apiston in or out. The gas has an initial volume 0.01 —_at 100 kPa pressure and its temperature is initially

Heat Transfer

Conduction · Radiation

Conduction

Rate of heat conduction (steady state, rod/slab)

\[ \frac{Q}{t} = \frac{kA\,\Delta T}{L} \]

where $k$ = thermal conductivity, $A$ = cross-sectional area, $\Delta T$ = temperature difference across length $L$.

Applied in the notes to: heat conducted along a steel rod between two fixed temperatures, heat lost through single vs double-glazed windows (air gap reduces conduction loss), and Searle’s method for measuring the thermal conductivity of a metal bar — using a flow of cooling water, two thermometers a known separation apart, and the temperature rise of the outflowing water.

Radiation

Black-body radiation — the thermal radiation emitted by an idealised perfect absorber/emitter, whose spectrum depends only on temperature.

Stefan–Boltzmann law

\[ P = \sigma A T^4 \quad (\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}) \]

Used to estimate: the operating temperature of a radiant wall heater from its rated power, the maximum temperature of a blackened sphere at the focus of a solar concave mirror, and the equilibrium temperature of a thin blackened plate exposed to solar radiation while radiating to surroundings at 300 K.

Earth’s equilibrium-temperature relation

\[ T^4 = \frac{E}{16\sigma\pi R^2} \]

where $E$ is the sun’s total radiant power output and $R$ is the Earth’s orbital radius — derived by treating the Earth as a black body in radiative equilibrium.

Also treated: power loss per unit length from an unlagged copper pipe carrying hot water, radiating into surroundings at a lower temperature (via Stefan’s law applied to the temperature difference).

Full problem bank — verbatim statements

Problems 173 – 183

Problem 173. (a) Outline an experiment to measure the thermal conductivity of a solid which is a poor conductor, showing how the result is calculated from the measurements. (b) Calculate the theoretical percentage change in heat lost by conduction achieved by replacing a single glass window by a double window consisting of two sheets of glass separated by 10 mm of air.

Problem 174. The silica cylinder of a radiant wall heater is 0.6 m long and has a radius of 5 mm. If it is rated at 1.5 kW estimate its temperature when operating. State two assumptions you have made in making your estimate. (The Stefan constant, $\sigma = 6\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$.)

Problem 175. (a) Explain what is meant by black body radiation. (b) A blackened metal sphere of diameter 10 mm is placed at the focus of a concave mirror of diameter 0.5 m directed towards the sun. If the solar power incident on the mirror is $1600\,\text{W m}^{-2}$, calculate the maximum temperature which the sphere can attain. State the assumptions you have made. (The Stefan’s constant $= 6\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$.)

Problem 176. If the mean equilibrium temperature of the Earth’s surface is T and the total rate of energy emission by the sun is E, show that $T^4 = \dfrac{E}{16\sigma\pi R^2}$, where $\sigma$ is the Stefan constant and R is the radius of the Earth’s orbit around the sun. (Assume that the Earth behaves like a black body.)

Problem 177. An unlagged thin-walled copper pipe of diameter 2.0 cm carries water at a temperature of 40 K above that of the surrounding air. Estimate the power loss per unit length of the pipe if the temperature of the surroundings is 300 K and the Stefan constant, $\sigma$, is $5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$. State two important assumptions you have made.

Problem 178. The solar radiation falling normally on the surface of the Earth has an intensity of $1.40\,\text{kW m}^{-2}$. If this radiation fell normally on one side of a thin, freely suspended blackened metal plate and the temperature of the surroundings was 300 K, calculate the equilibrium temperature of the plate. Assume that all heat interchange is by radiation. (The Stefan constant $= 5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$.)

Problem 179. A steel rod has length 1.5 m and radius 1 cm. One end of the rod is maintained at 100 °C and the other end is at 0 °C. Find the quantity of heat conducted through the rod in 2 minutes. The thermal conductivity of steel is 50.4 W/m K.

Problem 180. A glass window pane of a room has dimensions $2\,\text{m}\times0.5\,\text{m}\times0.002\,\text{m}$. The temperatures on its two sides are 300 K and 295 K respectively. Find the quantity of heat conducted out of the room in 10 minutes if the room has two windows, each having two such panes. ($K_{glass}=0.84\,\text{W/m K}$)

Problem 181. In Searle’s method, a metal rod of length 50 cm and area of cross-section $8\,\text{cm}^2$ is used. The flow of water through the tube is adjusted at 20 grams per minute. Steady temperatures of 65 °C and 55 °C respectively are shown by the two thermometers inserted in the rod. The separation between the thermometers is 4 cm. The outflowing water shows a rise of 6 °C. Find the thermal conductivity of the metal.

Problem 182. In Searle’s experiment for the measurement of thermal conductivity of a metal, a rod having a cross-sectional area of $10\,\text{cm}^2$ is used. The flow of water through the cooling tube is adjusted at 150 gm/minute. When a steady state is reached, two thermometers, inserted in the rod at a distance of 5 cm from each other, record temperatures of 60 °C and 50 °C respectively. If the rise in temperature of the water flowing through the cooling tube is 5 °C, find the thermal conductivity of the metal.

Problem 183. [Statement not captured in the source scan — page contains only the problem heading.]

Additional worked example for revision

Extra example — heat conducted through a slab

A slab has $k=0.8\,Wm^{-1}K^{-1}$, area $2m^2$, thickness $0.02m$ and temperature difference $10K$. Find heat flow rate.

Solution

\[ \frac{Q}{t}=\frac{kA\Delta T}{L}=\frac{(0.8)(2)(10)}{0.02}=800\,W \]

No scanned page is embedded here. The full source content is preserved as searchable text below, while the study notes above are rewritten for easier understanding.

Complete text-only source transcript for this chapter (pages 26-29)

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(a) Outline an experiment to measure the thermal conductivity of a solid which is a poor conductor, showing how the result is calculated from the measurements,(b) Calculate the theoretical percentage change in heat loose by conduction achieved by replacing a single glass window by a double window consisting of two sheets of glass separated by 10mm of air,Problem 174The silica cylinder of a radiant wall heater is 0.6m long and has a radius of Smm. If it is rated at 1.5 kW estimates its temperature when operating. State two assumptions you have made in making your estimate.-8 -2K-4 (The Stefan constant, 7 = 6 x10 w™ K ).Problem 175 (a) Explain what is meant by black body radiation (b) A blackened metal sphere of diameter 10mm is placed at the focus of a concave mirror of -2 diameter 0.5m directed towards the sun. If the solar power incident on the mirror is 1600 W m . Calculate the maximum temperature in which the sphere can attain. State the assumptions you have estimated. -8 -2R-4 (The Stcfan’s constant, = 6 * 10° wm” RK)Problem 176If the mean equilibrium temperature of the Earth’s surface is ‘I and the total rate of energy emission by the sun is K Show that4 —~_ T* _ YeanR?Where 9 is the Stephan constant and R is the radius of the Earth’s orbit around the sun. (Assume that the Earth behaves like a black body) Problem 177An unlagged thin-walled copper pipe of diameter 2.0 cm carries water at a temperature of 40 K above that the surrounding air. Estimate the power loss per unit length of the pipe if the temperature of the surroundings is 300K and the Stefan constant, o is 5.67-8 —2 1-4 x 10 wm K ).

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State two important assumptions you have made.

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Problem 178The solar radiation falling normally on the surface of the Earth has an intensity 1.40 k W ™ifthis radiation fell normally on one side of a thin, freely suspended blackened metal plate and the temperature of the surroundings was 300 K, calculate the equilibrium temperature of the plate. Assume that all heat interchange is by radiation.-2 2-4 (The Stefan constant = 5.67 x 10 w™ K ).Problem 179A steel rod has length 1.5m and radius 1 cm. One end of the rod is maintained at 100° Cand theother end is at 0CFind the quantity of heat conducted through the rod in 2 minutes. The thermal conductivity of steel is 50.4 W/m K.Problem 180A glass window pane of a room has dimensions 2nf0.5 m x 0.002m. The temperature on its two sides are 300 K and 295 K respectively. Find the quantity of heat conducted out of the room in 10minutes if the room has two windows, each having two such panes. (Kotass = 0.84w/m K)Problem 181In Searle’s method, A metal rod of length 50cm and area of cross-section 8 ©” is used. ‘The flowof water through the tube is adjusted at 20 grams per minute. The stead temperature of 65 °C and 55°C respectively are shown by the two thermometers instead in the rod. The separation between the thermometers is 4 cm. The out flowing water shows a rise of 6°. Find the thermal conductivity of the metal.Problem 182In Searle’s experiment for the measurement of thermal conductivity of a metal, a road having across-sectional area of 10° jis used. The flow of water through the cooling tube is adjusted at 150gm/minute. When a steady state is reached, two thermometers, inserted in the road at a distanceCcof Sem from each other, record temperature of 60°C and 50″ respectively. If the rise intemperature of the water flowing through the cooling tube is se find the thermal conductivity of metal.

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Problem 183

Stationary Waves & Resonance in Pipes

Waves

Harmonics of an open pipe

\[ f_0,\ 2f_0,\ 3f_0,\ 4f_0,\ \dots \]

An open pipe supports all harmonics of its fundamental $f_0$; e.g. $f_2 = \tfrac32\cdot\tfrac{v}{l} = 3f_0$ is the third harmonic.

End correction

In practice, the air just outside the open end of a pipe vibrates too, so the displacement antinode sits a small distance $C$ — the end correction — beyond the physical opening. The effective air-column length therefore slightly exceeds the pipe’s length.

Closed pipe

\[ \tfrac14\lambda_0 = l + C \;\Rightarrow\; \lambda_0 = 4l + 4C \]

Open pipe

\[ \tfrac12\lambda_0 = l + 2C \;\Rightarrow\; \lambda_0 = 2l + 4C \]

Resonance-tube experiment (closed pipe)

Apparatus: a glass tube standing in a tall jar of water (a resonance jar); the length of the enclosed air column is varied by raising/lowering the tube. A vibrating tuning fork is held over the mouth of the tube while the column length is gradually increased.

First resonance at length $l_1$: the air vibrates at its fundamental (first harmonic). $\lambda = 4l_1 + 4C$.
Second resonance at $l_2 \approx 3l_1$: the air vibrates at its third harmonic (first overtone). $3\lambda = 4l_2 + 4C$.

Subtracting the two relations eliminates the unknown end correction $C$, allowing the wavelength (and hence the speed of sound) to be found from $l_1$ and $l_2$ alone.

Additional worked example for revision

Extra example — closed pipe first resonance

A closed pipe resonates at $l_1=0.17\,m$ for a tuning fork. Ignoring end correction, find wavelength.

Solution

\[ \frac{\lambda}{4}=l_1 \Rightarrow \lambda=4l_1=4(0.17)=0.68\,m \]

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Complete text-only source transcript for this chapter (pages 30-33)

Source page 30

f₂ = v / (2l/3)

Source page 31

3 V h = 3 * T v -=2 – From equation * ¢ fo h= 3 *2fo = d [Po = BPO | eeaneenneennnennneee which is the 3″ harmonic– Thus, for an open pipe the frequency of harmonics are:For for Bho» Aflgy vx ov sv se se ne as nesEND CORRECTION OF A PIPE– In practice the air just outside the open end of a pipe is set into vibration and the displacement antinode of a stationary wave occurs a distance “C” called end correction beyond the open end.– The effective length of the air column is therefore slight greater than the length of the pipe. Example (a) Fora closed pipe > 740 =14+C Ay = 414+ 4C(b ) For an open pipe> FA = 1426 Ag = 214+ 4C€ Definition-The end correction C of a pipe is that small length of a stationary wave which protrudes just outside the open end of a pipe instrument where the air inside it is set into vibration.

Source page 32

RESONANCE IN A CLOSED PIPEApparatus A resonance jar is used.It consists of a glass tube which stands in a tall jar full of water.Air column { Glass tubeNeaTall jarWaterAThe length of the air column is varied by raising or lowering the glass tube. WORKINGStarting with a very short air column, a vibrating fork is held over the mouth of the tube and the length of the column is then gradually increased.Strong resonance occurs when the column reaches a certain critical length hy (say). This is called the first position of resonance.At this position the air in the tube vibrates at its fundamental note/first harmonic

Source page 33

N A < > Se 4 L a Where> A= Wavelength of sound produced C=end correction9 A= +e A=4L,+4C ———-~—-—~-– If the length of the air column is now increased still further a second position of resonance is obtained when the column is approximately three times as long as (2 say).rd – At this position the air in the tube vibrates with its 3″ narmonic/first overtone. Third harmoniche43L f= 3v/4L> 3=1,4C @ 32 =41,4+4C ——

The Doppler Effect

Waves · Sound

The apparent frequency of a wave changes when the source and/or observer are in relative motion. Let $V$ = speed of sound, $f$ = true (source) frequency, $f_0$ = apparent (observed) frequency.

Case 1 — Source moving

Source moving toward a stationary observer

\[ \lambda_0 = \frac{V-U_s}{f} \qquad f_0 = \left[\frac{V}{V-U_s}\right]f \]

The apparent wavelength shortens ahead of the moving source, so the observer measures a higher frequency.

Source moving away from a stationary observer

\[ \lambda_0 = \frac{V+U_s}{f} \qquad f_0 = \left[\frac{V}{V+U_s}\right]f \]

Here the wavelength stretches out, giving a lower observed frequency.

Case 2 — Observer moving

When the observer moves, the wavelength emitted by the (stationary) source is unaffected — only the relative velocity of waves reaching the observer changes: $\lambda = V/f$ unchanged, while $f_0 = \dfrac{\text{relative velocity of waves}}{\lambda}$.

Observer moving toward a stationary source

Relative wave speed at the observer $= V + U_0$, so:

\[ f_0 = \left[\frac{V+U_0}{V}\right]f \]

(By symmetry, an observer moving away from a stationary source measures relative wave speed $V-U_0$, giving $f_0 = \left[\frac{V-U_0}{V}\right]f$.)

Additional worked example for revision

Extra example — source moving towards observer

A source of frequency $500Hz$ moves towards a stationary observer at $20m/s$. Take sound speed $340m/s$. Find apparent frequency.

Solution

\[ f_0=\frac{V}{V-U_s}f=\frac{340}{340-20}\times500=531.25\,Hz \]

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Complete text-only source transcript for this chapter (pages 34-37)

Source page 34

λ₀ = (V – Uₛ) / f …………………… (1)

Source page 35

– Where f = true frequency of the source – Let fo be apparent frequency– From V = 4ofo– Substitute equation (1) in equation (2)v— Ww h=>* Vv f= [alt (ii)– Consider a source $ ofsound wave to be moving with velocity Us away from a stationary observer O—— ©u s | | | e Observer -Where V = Velocity of sound wave – Velocity of wave relative to observer at 0= V + Us– The apparent wavelength Yo reaching the observer at 0 is :V+U,A, = a (4)

Source page 36

vJa eeeeeecceeececeecnoee (5)– The apparent frequency 2– Substitute equation (4) in equation (5) v Ffo = lag lf 6CASE2 OBSERVER MOVING – The motion of the observer affects the velocity of the waves he receives.– In this case the wavelength is unchanged and is given by:aa=l7)– The apparent frequency fo is given byRelative velocity of waves wavelengthh=(@) OBSERVER MOVING TOWARDS 4 STATIONARY SOURCEv* (|||.– Where Uy = Velocity of observer V = Velocity of wave– Velocity of wave relative to observer = V + Uo

Source page 37

– The apparent frequency f₀ of the wave is

Wave Optics — Interference

Light

Young’s double-slit experiment — key observations

If the source slit S is made wider, the fringes on the screen disappear (loss of spatial coherence).
With white light (instead of monochromatic light), the central fringe at O is white, but bright fringes either side are coloured — violet nearest O, red furthest from O.

Angular width of a fringe

Let $a$ = slit separation, $D$ = slit-to-screen distance, $\omega$ = fringe width, $\theta$ = angular fringe width.

\[ \tan\theta = \frac{\omega}{D} \approx \theta \ \text{(small-angle, in radians)} \]

Since the fringe width is $\omega = \dfrac{\lambda D}{a}$:

\[ \theta = \frac{\omega}{D} = \frac{\lambda}{a} \]

Air-wedge fringes (thin-film interference)

Two glass slides are clamped together at one edge and separated by a thin spacer (e.g. a piece of paper) at the other, forming a thin wedge of air. Monochromatic light from an extended source reflects partially down onto the wedge via a glass plate G; a microscope focused on the wedge shows bright and dark fringes parallel to the line of contact.

Dark fringe condition

Path difference between the two reflected rays at thickness $l$: $2l = n\lambda$

\[ l = \frac{n\lambda}{2}, \qquad n = 0,1,2,3,\dots \]

Wedge angle from two dark fringes at positions $S_1, S_2$

For a dark fringe at P: $2S_1\theta = n\lambda$. For the $(n+k)^{th}$ dark fringe at Q: $2S_2\theta = (n+k)\lambda$. Subtracting:

\[ \theta = \frac{k\lambda}{2(S_2-S_1)} \]

Worked example — fringe wavelength from slit geometry method

“Two slits are 0.2 mm apart and the screen is 1 m away; the third bright fringe is displaced 7.5 mm from the centre. Find the wavelength.”

Using $x_n = \dfrac{n\lambda D}{a}$ with $n=3$, $a=0.2\times10^{-3}\,\text{m}$, $D=1\,\text{m}$, $x_3 = 7.5\times10^{-3}\,\text{m}$:

\[ \lambda = \frac{x_n\, a}{nD} = \frac{(7.5\times10^{-3})(0.2\times10^{-3})}{3\times1} \approx 5.0\times10^{-7}\ \text{m} = 500\ \text{nm} \]

Full problem bank — verbatim statements

Problems 37 – 43

Problem 37. Two slits are at a distance of 0.2 mm apart and the screen is at a distance of 1 m. The third bright fringe is found to be displaced 7.5 mm from the central fringe. Find the wavelength of the light used.

Problem 38. A yellow light from a sodium vapour lamp of wavelength 5893 Å is directed upon two narrow slits 0.1 cm apart. Find the position on the screen 100 cm away from the slits.

Problem 39. In Young’s experiment, the distance of the screen from the two slits is 1.0 m. When light of wavelength 6000 Å is allowed to fall on the slits the width of the fringes obtained on a screen is 2.0 mm. Determine: (i) the distance between the two slits; (ii) the width of the fringes if the wavelength of the incident light is 4800 Å.

Problem 40. In a double slit experiment, light has a frequency of $6\times10^{14}\,\text{s}^{-1}$. The distance between the centres of adjacent bright fringes is 0.75 mm. What is the distance between the slits if the screen is 1.5 m away? Given that the speed of light in free space is $3\times10^{8}\,\text{m s}^{-1}$.

Problem 41. In a Young’s double slit experiment, two narrow slits 0.8 mm apart are illuminated by the same source of yellow light ($\lambda=5893\,\text{Å}$). How far apart are the adjacent bright bands in the interference pattern observed on a screen 2 m away?

Problem 42. In Young’s double slit experiment the angular width of a fringe formed on a distant screen is 0.1°. The wavelength of light used is 6000 Å. What is the spacing between the slits?

Problem 43. A beam of light consisting of two wavelengths, 6500 Å and 5200 Å, is used to obtain interference fringes in a Young’s double slit experiment. [Statement continues in source with the requested comparison of fringe positions for the two wavelengths.]

Additional worked example for revision

Extra example — fringe width in Young’s experiment

Find fringe width if $\lambda=6.0\times10^{-7}m$, screen distance $D=1.5m$, and slit separation $a=3.0\times10^{-4}m$.

Solution

\[ \omega=\frac{\lambda D}{a}=\frac{(6.0\times10^{-7})(1.5)}{3.0\times10^{-4}}=3.0\times10^{-3}m=3.0mm \]

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Complete text-only source transcript for this chapter (pages 38-44)

Source page 38

(4) If the source slits S is made wider, then the fringes on the screen disappears.

Source page 39

(5) If white light is used instead of monochromatic light source then the central fringe at O is white but the other bright fringe on either sides of the central fringe are colored with violet near “O” and red far away from “O”.ANGULAR WIDTH OF A FRINGEIf P is the position of the first bright fringe then OP = o= fringe width and 6. angular width ofa fringe e tan@ = D Since 9 is very small, therefore tan 9 * 9 (in radian) @=2 DA (1) From the expression of fringe width o a 2)

Source page 40

Problem 37

Source page 41

Two slits are at a distance of 0.2mm apart and the screen is at a distance of Im. The third bright fringe is found to be displaced 7.5mm from the central fringe. Find the wavelength of the light used.Problem 38A yellow light from a sodium vapour lamp of wavelength 5893 is directed upon two narrow slits of 0.lem apart, Find the position on the screen 100cm away from the slits,Problem 39In Young’s experiment, the distance of the screen from the two slits is 1.0m. When light of wavelength6000s allowed to fall on the slits the width of the fringes obtained on a screen is 2.0mm. Determine:(i) The distance between the two slits and(ii) The width of the fringes if the wavelength of the incident light is 4800 Problem 40140-1 In double slit experiment, light has a frequency of 6 x 10 S “The distance between the centres ofadjacent bright fringes is 0.75mm. What is the distance between the slits if the screen is 1.5m away?8 mns-t Given that the speed of light in free space 3 xy o’ms : Problem 41In a Young’s double slit experiment, two narrow slits 0.8mm apart are illuminated by the same source ofyellow light ¢ = 5893A) How far apart are the adjacent bright bands in the interference pattern observed on a screen 2m away?Problem 42° In young’s double slit experimen} the angular width of a fringe formed on a distant sereen is 0.1 . The wavelength of light used is 6000 . What is the spacing between the slits?Problem 43A beam of light consisting of two wavelength 6s004 and s004 is used to obtain interference fringes in a Young’s double slit experiment.

Source page 42

Monocromatic Source of lightPiece ofPaper Bottom of ise of hottom top slide WeMonochromatic light from an extended source is partially reflected vertically downwards by the glass plate G.When the microscope is focused on the wedge, bright and dark equally spaced fringes are seen parallel to the edge of contact of the wedge.Some of the light falling on the wedge is reflected upwards from the bottom surface of the top slide and the rest which is transmitted through the wedge is reflected upwards from the top surface of the bottom slide.Let als be thickness of the air wedge at P.Ifa dark fringe is formed at P then: Path difference between the rays atna

Source page 43

– Where n= 0, 1, 2, 3, ……2eeceeeeeeeeee – First bright fringe, n = 0– Second bright fringe, n= 1DETERMINATION OF ANGLE OF THE WEDGEIn the figure above, Ict * be the angle between the slides/plates (=wedge angle).For a dark fringe at P we have:é 2l=na If 8 is in radians then we have: l= S,6 a 2G8 = mA @)– For the (n+ K™ dark fringe at Q, we have:

Source page 44

28,0 =(n+Ka > eqn (4) — eqn(3) > 25,6 — 28,0 =(n+KA—na > 20(S,—S,) = nA+kA—nd > 20(S,-—S,)=KAkA6==>— 2(S2 = >)

Polarization of Light

Light

Methods of producing polarized light

By selective absorption (Polaroids)
By reflection
By double refraction
By using a Nicol prism

Polaroids

A Polaroid is an artificial crystalline material made in thin sheets that allows light vibrations of only one particular polarization direction to pass through.

Used in sunglasses to reduce light intensity and eliminate glare.
Used to control the intensity of light entering trains and aeroplanes.
Used in automobile windshields.

Polarization by reflection

The reflecting surface of a transparent medium (e.g. glass) can produce plane-polarized light: when unpolarized light strikes the surface at the polarizing (Brewster) angle $\theta_B$, the reflected ray is strongly plane-polarized, while the refracted ray is only partially (slightly) polarized.

Additional worked example for revision

Extra example — identifying polarized light

If a Polaroid is rotated and the transmitted light intensity becomes zero at one position, the incident light is plane polarized. If intensity only changes slightly and never becomes zero, it is partially polarized.

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Complete text-only source transcript for this chapter (pages 45-47)

Source page 45

(2) By reflection

Source page 46

(3) By double refraction (4)By using Nicol prism POLARIZATION BY POLAROIDSPolaroid is an artificial crystalline material which can be made in thin sheets. It has the property of allowing light vibrations only of a particular polarization to pass through.Uses of Polaroid (i) They are used in sunglasses to reduce the intensity of light and to eliminate glare.(ii) They are used to control the intensity of light entering trains and aeroplanes. (iii) They are used in wind shields of automobiles. POLARIZATION BY REFLECTION The reflecting surface of a transparent medium can be able to produce plane polarized light.This happens when unpolarized light is incident to any transparent medium e.g. glass.Reflected ray ( polarized)Incident ray ( unpolarized)H c i Refracted ray (slightly polarized )Where, AO – Is an incident natural light

Source page 47

OB – is a strongly plane-polarized reflected ray

Electrostatics

Fields & Potential

Lines of force

Field-line patterns illustrated in the notes: for two equal positive charges, lines emerge from each charge and curve outward, repelling one another with a neutral point midway between the charges; for two equal but opposite charges, lines leave the positive charge and curve directly into the negative charge, showing mutual attraction along connecting field lines.

Electric flux

The electric flux $\Phi_E$ through an area perpendicular to the field lines is the product of the electric field intensity $E$ and that area.

Flux through a sphere concentric with a point charge Q

\[ \Phi_E = E \times 4\pi r^2 = \frac{Q}{4\pi\varepsilon_0 r^2}\times 4\pi r^2 = \frac{Q}{\varepsilon_0} \]

In a medium of permittivity $\varepsilon$: $\Phi_E = Q/\varepsilon$. The total flux through any concentric sphere outside a point charge is constant — independent of the sphere’s radius.

Field of a charged sphere

Outside the sphere

\[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \]

Outside a charged sphere the field behaves exactly as if all the charge were concentrated at the centre.

Inside a charged, empty (hollow) sphere

\[ E = 0 \]

Since no enclosed charge lies within the hollow interior, Gauss’s law gives zero field there.

Motion of a charge in a uniform field

A particle of mass $M$ and charge $q$, released from rest in a uniform field $E$ between two charged plates, undergoes motion analogous to free fall in a gravitational field.

\[ a = \frac{F}{m} = \frac{qE}{m} \] \[ v = at = \frac{qEt}{m} \qquad y = \tfrac12 at^2 = \frac{1}{2}\cdot\frac{qEt^2}{m} \] \[ v^2 = 2ay = \frac{2qEy}{m} \qquad \Rightarrow \qquad K.E. = \tfrac12 mv^2 = qEy \]

Electric potential

The electric potential difference between points A and B is the work done per unit positive test charge in moving that charge from A to B: $V_B – V_A = \dfrac{W_{AB}}{q_0}$. Its SI unit, joule per coulomb, is called the volt (1 JC⁻¹ = 1 V).

Taking the reference point A at infinity ($V_A = 0$):

\[ V = \frac{W}{q} \]

Definition: the electric potential at a point is the work done by the field in bringing a unit positive charge from infinity to that point.

Work done moving a test charge $Q_0$ near a source charge $Q$

\[ F = \frac{QQ_0}{4\pi\varepsilon_0 x^2} \qquad dW = \frac{Q_0Q}{4\pi\varepsilon_0 r^2}\,dr \]

Additional worked example for revision

Extra example — kinetic energy gained by a charge

A charge $q=2.0\times10^{-6}C$ moves through potential difference $50V$. Find work done.

Solution

\[ W=qV=(2.0\times10^{-6})(50)=1.0\times10^{-4}J \]

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Complete text-only source transcript for this chapter (pages 48-53)

Source page 48

SE! ANNLine of force for the equal positive charges Line of forces for equal but oppositeELECTRIC FLUX (Oz. )The electric flux Og. through an arca perpendicular total lines of force is the product of. Ex areca where F. is the electric intensity at that place,Consider a sphere of radius drawn in a space concentric with a point charge.

Source page 49

Total flux through the sphere is given by Doe =E x Arca of the sphere =Ex4de >’?=4n£0r’x 460 ‘Peie Qe=— ® €0If the charge is placed at any other medium apart from air or vacuum thenQ DoeHeThe above equation shows that the total flux crossing any point at drawn sphere concentrically outsidethe point charge is constant.1. (OUT SIDE THE CHARGED SPHERE)

Source page 50

ELECTRIC FIELD INTENSITY DUE TO A CHARGED SPHEREThe flux across a spherical surface of radius or concentric with a small sphere carrying charge Q is givena by flux= € Ex4anre = 2 €0 ___Q ~ 4n€ 0 r?This result shows that the outside of a charged sphere the field behaves as if all charges on the sphereare concentrated at the centre.2. INSIDE A CHARGED EMPTY SPHERE.(CHARGED SPHEREOUTSIDE THE CHARGE SPHERE WITH RADIUS, R.Inside the empty charged sphere there are no charges so the electric field strength E = O therefore;sinceEX4nr = 2 £0 Q=0,then the value of inside”the sphere is also 0

Source page 51

The intensity of in the field must be perpendicular to the surface and the charges will produce this field are those in projection of the area P on the surface S i.e those within the shaded region A.Question.A particle of mass M and charge of q is placed at rest in a uniform electric field see the fig below and released. Describe its motion.t+ + + + + + $+ 4+ HHHThe motion reassemble that of the falling body in the earth’s gravitational field. The constant to Facecleration is given by mThe equation of uniform acceleration to apply therefore with with Vo=0 we haveV=at put eqn(1) in (2) we have ai)By putting eqn(1) in (3) we get1 actee (4)

Source page 52

From the third equation of motion we havev = 2Ay Putting eqn(1) into eqn(5)The kinetic energy attached at the moving a distance y is formed from :-– Ke= —substitute eqn(6) into (7) we get– 1 ‘2ae Ke= -m2q£y a ie Ke = atyELECTRIC POTENTIAL (V)The electric field around a charged and can be described not only by a vector electric field strength E but also by a scalar quantity i.¢ the electric potential,v.To find the electric potential difference between two points A and B in an electric field we move a test charge q from A to B and we measure the workWAB that must be done by agent moving the charge.Electric potential difference ,v ca be expressed in the form ofVB —VA = “AB qoThe unit of the potential difference is obtained for equation (i) that is JC. However volts is also used.

Source page 53

1JC’=1VoltsIf point A is chosen to be at very far (say at infinity) then the electric potential at infinity distance is arbitrarily taken et zero.Therefore then putting VA=0 1 in equation (i) and dropping the subscripts we get– (ii) Definition:The electric potential at the point is the work done by the foree in taking the unit chart from infinity to that point.CALCULATION OF WORK DONE ‘Consider a positive charge Q to be at Ra distance as indicate in the figure the work done in taking the charge from A to B is cqual to the work donc in taking the same distance from B to A. If QO is moved by the fo….. from A to B then the foree acting on it is= 20 4ne0xa If the charge has moved a distance 6x the work done is_ 209 dw 4m&r?

Capacitors

Circuits

Parallel arrangement

All left-hand plates are joined together, and all right-hand plates are joined together. Each capacitor then shares the same potential difference $V$ across it.

\[ Q_1=C_1V,\quad Q_2=C_2V,\quad Q_3=C_3V \] \[ Q = Q_1+Q_2+Q_3 = V(C_1+C_2+C_3) \] \[ \boxed{C_{parallel} = C_1+C_2+C_3} \]

Series arrangement

The right-hand plate of one capacitor connects to the left-hand plate of the next, and so on. Each capacitor in the chain carries the same charge $Q$.

\[ V_{AB}=\frac{Q}{C_1},\quad V_{DF}=\frac{Q}{C_2},\quad V_{GH}=\frac{Q}{C_3} \] \[ V = V_{AB}+V_{DF}+V_{GH} \] \[ \boxed{\frac{1}{C_{series}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}} \]

Worked example — 2 µF and 8 µF in series across 300 V solved

\[ \frac{1}{C}=\frac{1}{2}+\frac{1}{8}\ (\mu F^{-1}) \;\Rightarrow\; \frac{300}{Q}=\frac{1}{C_1}+\frac{1}{C_2} \] \[ Q = 4.8\times10^{-4}\ \text{C} \] \[ V_1 = \frac{Q}{C_1} = \frac{4.8\times10^{-4}}{2\times10^{-6}} = 240\ \text{V}, \qquad V_2=\frac{4.8\times10^{-4}}{8\times10^{-6}}=60\ \text{V} \]

Check: $V_1+V_2 = 240+60 = 300\ \text{V}$ ✓ — matches the supply voltage.

Worked example — 5 µF and 25 µF in series across 90 V solved

\[ V = Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right) \;\Rightarrow\; 90 = Q\times10^{6}\left(\frac{1}{5}+\frac{1}{25}\right) \] \[ Q \approx 3.75\times10^{-4}\ \text{C} \] \[ V_1 = \frac{Q}{C_1} = 75\ \text{V}, \qquad V_2 = \frac{Q}{C_2} = 15\ \text{V} \]

The larger capacitor (25 µF) takes the smaller share of the voltage, since $V \propto 1/C$ for a fixed series charge.

Energy stored in a capacitor

Moving a small charge $dQ$ onto a capacitor already at potential $V=Q/C$ requires work $dw = V\,dQ = \dfrac{Q}{C}dQ$. Integrating from 0 to the final charge $Q_1$:

\[ W = \int_0^{Q_1}\frac{Q}{C}\,dQ = \frac{1}{2}\frac{Q_1^2}{C} \]

Equivalent forms (using $Q=CV$):

\[ \boxed{W = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV} \]

Worked example — energy stored, before and after joining two charged capacitors solved

$C_1 = 3\,\mu F$ charged to $V_1=200\ \text{V}$; $C_2 = 2\,\mu F$ charged to $V_2=100\ \text{V}$.

\[ E_1 = \tfrac12 C_1V_1^2 = \tfrac12(3\times10^{-6})(200)^2 = 0.06\ \text{J} \] \[ E_2 = \tfrac12 C_2V_2^2 = \tfrac12(2\times10^{-6})(100)^2 = 0.01\ \text{J} \]

When the capacitors are connected together (like plates joined), the combination behaves as a parallel pair, $C = C_1+C_2$, and the total energy after connection can be found from the common final voltage — generally less than $E_1+E_2$, the difference being dissipated as heat/radiation during the redistribution of charge.

Additional worked example for revision

Extra example — two capacitors in parallel

Find equivalent capacitance of $4\mu F$ and $6\mu F$ connected in parallel.

Solution

\[ C_p=C_1+C_2=4+6=10\mu F \]

No scanned page is embedded here. The full source content is preserved as searchable text below, while the study notes above are rewritten for easier understanding.

Complete text-only source transcript for this chapter (pages 54-60)

Source page 54

1) Parallel arrangement of capacitorV = is constantAll the left hand plates are connected together and all the right hand plates are connected together and in the case of parallel arrangement of capacitors (see figure above)When a cell is connected across these capacitor is parallel they have the same potential difference (v)So @MQ=civ (ii) Q2 = Q2V (ii) Qs =C3VLet the total charge be Q then Q= Qi + Q2 +Q3.. (4)

Source page 55

Put ® @ din)=C,V+C,u+ Cv Q= VG, +0, +6, Divide equation (V) by U both side we getc= g= C,+C,+C; Or C=C +C,+C,Where “C” is the eqhivalent capacitance.(il) SERIES ARRANGEMENT OF CAPACITOR.cyQ is constantWhen the right hand plates of one capacitor is connected to the left hand of the next and so on thenthese capacitors are said to be connected in series,When the cell is connected across of the end of the system a charge is transferred from the plates H to A, A charge —Q being left on it. This charge induce a charge +Q on plate Q. This process is repeated with other plate.

Source page 56

Now Vas=2 voF=2 vcH=2 Cs csCyBut. VAB + UDF + VOHvary 242 C, Cz Cs Divide both sides by Qa= Vet Ve, + Uc,ay 1 1 1 Le Yo = “/¢,+ “1c, + “Ic, Where C is the equivalent capacitance. QuestionTwo capacitors of capacitance C\=2pF and C:= 8yF are connected in series and the resulting combination is connected across 300volt. Calculate the charge and potential differenceSolutionC, = uF C, = 8uF,300V

Source page 57

@ InF = 10-*FGi) IpF = 10-FGii) InF = 107°F]Y =2 1Vp 2 ButV = V, + V, where V = 300voltsQ = 300 = ceases)—+ 300 = aos 8×10-¢=Q as = os = = = 4.8 x 10-*¢ =Q=48×10*C(b) Potential difference;v, =*280F = 24 x 107 = 240V, = 42807 = 6 x 10″ = 60V =V, +V, = 240 + 60 = 300 proued The charge across the capacitor is 4.8 x 10~*C and the pd Across the 1*capacitor = 240vpd Across the 2″4 capacitor =60VQn. 1998 Qn 13Capacitor of 5uF and 251F are connected in series and the combination is connected to the battery of 90 volt. Calculatea) Charge on each capacitorb) The p.d across each capacitor

Source page 58

C C V; vs20M | |Where C, = SpFEy = 25uF =2 =2 1G ~ Cy«V=V, +¥, Butv=90 voltsSo V=Q(Z+2) =90=Q(F+ 5) x 10°F90 = Qx 10°F (=2Q= RYE =15 x25 x 10-6cQ=357 x 10°C From above formular V=3.75×10-* Yao =705 x10 =75 VoltYe = 32 x 10-6 = 15 volt.. The charges across the capacitor is 3.75 x 107*C and the pd of the 1* capacitor is 75 volts and the second capacitor = 15 volts.ENERGY STORED IN A CAPACITORConsider a capacitor of capacitance C to have been charged to a potential difference V and let a small charge dQ be transferred from the negative plate to positive plate. Then the work don’t in moving a charge dQ will be

Source page 59

dw = VdQ but V=Q/C Hence dw= 2 dQSuppose a capacitor is at first discharge d and then charged until the final charge on the plate is Q The work done in charging it is given byW = fan fp dQ1/0 =i@ai rah Q0Q =~ 210191? 2CBut Q is equal to CV then (i) becomes,A380 C®….cessereeseesseeeseeeee (iii) put into (i) we getw= v..Equation (i), (ii) and (iv) give the energy stored in a capacitor. Question1998 P2BQn

Source page 60

li.A capacitor of a capacitance 30° ce}F is charged until a potential difference of 200v is developed across. Its plat…..another capacitor of capacitance 26+ ce{F developed apd of 100v across its plates on being charged.What is the energy stored on each capacitor?The capacitors…them connected by a wire of negligible resistance so that the plates carrying like charges are connected together. What is the total energy stored in the combined capacitors?Solution C1=3×10° *FVI= 200v C2=2×10 °F V2=100vFormulaF1=c1v?=1/2x3x10%x(200) =6x 10-2 JoulesE1=0.06JoulesE2= C2V2 =1/2x2x10°x10* =10-24oubes(iiy¢ £1402

Surface Tension & Capillarity

Mechanics of Fluids

Capillary rise (wetting liquid, e.g. water)

For a capillary tube dipped in water the angle of contact is practically zero. If $H$ is atmospheric pressure, $h$ the height risen, $\rho$ the liquid density and $\gamma$ the surface tension:

\[ P_2 – P_1 = \frac{2\gamma}{r} \qquad\Rightarrow\qquad h = \frac{2\gamma}{\rho g r} \]

The height $h$ increases as the tube radius $r$ decreases — narrower tubes draw water higher.

If the tube is too short (height above the liquid $<\,h$), the meniscus at the top adjusts its own radius of curvature $R > r$, meeting the walls at an acute angle of contact $\theta$, so that: $\cos\theta = r/R$.

Capillary depression (non-wetting liquid, e.g. mercury)

For mercury, the (obtuse) angle of contact makes $\cos\theta$ effectively negative, producing a depression rather than a rise:

\[ P_2 – P_1 = \frac{2\gamma\cos\theta}{r} \qquad\Rightarrow\qquad h = \frac{2\gamma\cos\theta}{\rho g r} \]

The depth of depression $h$ increases as the tube radius decreases, exactly analogous to capillary rise but directed downward.

Two different capillary radii, same liquid

Comparing menisci in two connected limbs of radii $r_1$ and $r_2$:

\[ h = \frac{2\gamma}{\rho g}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \]

Worked example — inverted U-tube with two capillary limbs method

“A glass U-tube is inverted with open ends of diameters 0.5 mm and 1.0 mm below the surface of water in a beaker; the air pressure in the upper part is increased until the water level in one limb is level with the water outside. Find the water level in the other limb.”

Method: apply the excess-pressure (capillary) formula $P_1-P_2 = 2\gamma/r$ at each meniscus with radii $r_1 = 0.25\ \text{mm}$, $r_2 = 0.5\ \text{mm}$, and combine with the hydrostatic pressure difference $\rho g h$ to solve for the unknown level $h$:

\[ h = \frac{2\gamma}{\rho g}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \]

Full problem bank — verbatim statements

Questions 1 – 3 (closing problems in the notes)

Question 1. Water rises to a height of 5 cm in a certain capillary tube. In the same tube, mercury is depressed by 1.71 cm. Compare the surface tension of water and mercury; specific gravity of mercury is 13.6, angle of contact for water is zero and that of mercury is 135°.

Question 2. A liquid of surface tension $\gamma$ is used to form a film between a horizontal rod of length L and another shorter rod of mass m, supported from the two light, extensible strings of equal length, joining adjacent ends of each rod. The film fills the vertical plane within the rods and strings. What is the shape of each string? Show that the tension in each is $T = \dfrac{mg-2\gamma L}{2\sin\theta}$, where $\theta$ is the angle which the tangent to each string makes with the upper rod.

Question 3. A soap bubble in vacuum has a radius of 3 cm, and another soap bubble in the vacuum has a radius of 6 cm. If the two bubbles coalesce under isothermal conditions, calculate the radius of the bubble formed (under isothermal conditions, $PV$ is constant).

Additional worked example for revision

Extra example — capillary rise comparison

If the radius of a capillary tube is halved, what happens to the capillary rise of water?

Solution

\[ h=\frac{2\gamma}{ ho gr} \Rightarrow h\propto \frac{1}{r} \]

Halving the radius doubles the height of rise.

No scanned page is embedded here. The full source content is preserved as searchable text below, while the study notes above are rewritten for easier understanding.

Complete text-only source transcript for this chapter (pages 61-65)

Source page 61

If the capillary tube is dipped into water the angle of contact is practically zero. Fig (i) Thisis the atom ospheric pressure and P1 is the pressure in the liquid we haveP; — Pj =—_r If His the atmospheric pressure his the height of the liquid the liquid in the table andits density p P2=H and P; = H—h pEj. . 26 “HH – hpEj)= 47inereascs as r decrease ic the narrow the tube the greater the height to which the waterraised. Suppose tube is pulled down until the top of heightif the height | of the tube is above the water than the calculated value of h in the above formula the watersurface at the top of the tube now meet at an angle of contact. This angleis an acute one. The radius R of the meniscus is greater than the capillary tube of radius r.“TRFrom the excess formula for the meniscusFrom figure we have COSY =2eP-Pi=

Source page 62

H-(H-hpé= R=LP9. 28“L=RPg Tah,Cos AY = =Let the depression of the mercury inside the tube of radius r be h in figure below, thepressure P2 below the curved surface of mercury is the atmospheric pressure ROut side the curved surfaceP2— P; = 2E5 CosEYWhen AY is the supplement of an obtuse angle of contact of mercury with the glass.

Source page 63

ie AY is an acute angle and its cosine is positiveBut P; =H and P) = H + hE;. . 2xCos®“°(H + hpi) -H=_ 2x%cos®hpE; = ¥2x¥cosO prgThe height of —— depression h inercased as the radius decrease.“neExampleA glass U-tube is inverted with an open end of straight limbs of diameters respectively0.5mm and 1,00min below the surface of water in the beaker.The air pressure in the upper part is immersed until the maximum in one limb is level withthewater outside. Find the level of — water in the other limb. Solution H h —_ — —_ | | P2 Le. == ® _—, _ ~ Given

Source page 64

E¥%a-= 0.025, E%s,,=0.05H-(H -AjEjh) = 7 Pi-p2= TtBut P: =H – pEjh_ 2 P-H+pkjh=).2r 2yot phEj = v1_ 20 20 2y1 1 *= par. “72)Fo=7,5 x 107Questions1, Water riscs to a height of Sem in a certain capillary tube in the same tube. Mercury is depressed by 1.71 em, Compare the surface tension of water and mercury specific gravity of mercury is 13.6, angle of contact for water is zero and that of mercury is 135°2. A liquid of surface tension y is used to form a film between a horizontal rod of length Land another shorter of rod of mass m supported from by the two light mentensible stringsof egual length. Joining adjacent end of each rod, The film fells the verticle place within

Source page 65

rods and strings. What is the shape of each string? mg—2sL Show how that the tension in each is 2SimeWhere AY is the angle which the tangent to each string make with the upper net.3. A soup bubbles in Vacuum has a radius of 3cm and another soap bubbles in the vaccum has a radius of 6cm. If the two bubble coalesce under isothermal conditions. Calculate the radius of the bubble formed under isothermal conditions Ec is constant.