Physics 9702 / 44 — May/June 2025

(b)

The gravitational field strength $g$ at a distance $x$ from the centre of a uniform spherical planet of mass $M$ is given by the expression

g = \frac{GM}{x^2}

where $G$ is the gravitational constant and distance $x$ is greater than the radius of the planet.

4M

(i)

Describe the pattern of the field lines outside the planet that represent the gravitational field due to the planet.

DifficultyMedium-Easy

Worked solution

diagram-b-i-field-lines

Answer

Field lines are radial and directed towards the centre of the planet. They are closer together near the planet and become more widely spaced with increasing distance.

Final answer

Field lines are radial and directed towards the centre of the planet, with line density decreasing with distance.

Detailed explanation

Background Concept

Field lines are a representation of a field. The rules are:

The arrow on a field line shows the direction of the force on a small test mass.
The spacing (density) of field lines indicates the magnitude of the field (closer lines = stronger field).

For a spherically symmetric mass, the gravitational field is the same in all directions at a given radius, so the field must be radial.

Understanding the Question

You are told the planet is a uniform sphere and you are considering points outside it ( $x$ greater than the radius). You must describe what the gravitational field lines look like in space around the planet.

Approach

Use symmetry + the “field line rules”:

direction: gravity is attractive, so lines point towards the mass.
pattern: spherical symmetry means radial lines.
spacing: because $g \propto 1/x^2$ , the field weakens with distance, so lines spread out.

Step-by-Step Reasoning

diagram-b-i-field-lines

Since the field only depends on distance from the centre ( $g = GM/x^2$ ), the pattern must be the same at every angle: radial lines.
A test mass is pulled towards the planet, so arrows point inward.
As $x$ increases, $g$ decreases (inverse-square), so the number of lines crossing a given area effectively decreases: the lines are more widely spaced further from the planet.

Key Takeaways

Outside a spherical mass: field lines are radial and point towards the centre.
Field strength decreases with distance, shown by decreasing line density.

Common Mistakes

Drawing circular (tangential) lines instead of radial lines.
Having arrows pointing away from the planet (would represent repulsion).
Keeping the spacing constant even though $g$ changes with $x$ .

Things to Be Careful About

This question is specifically for outside the planet, so you should not discuss the inside-field variation.
Include both: (i) direction and (ii) change of spacing, to access full marks.

Techniques used

use field-line rules to describe direction and spacingrelate radial symmetry to a spherical mass distributionlink field strength variation to field-line density

(ii)

Explain why, for small changes in vertical height near the surface of the planet, $g$ may be assumed to be constant.

DifficultyMedium-Easy

Worked solution

Answer

Near the surface, $x \approx R$ and a small change in height $\Delta x$ gives only a very small fractional change in $x$ .

Since

g = \frac{GM}{x^2}

a small fractional change in $x$ produces an even smaller change in $g$ , so $g$ may be taken as constant for small height changes.

Final answer

Because near the surface a small height change makes a negligible fractional change in distance from the centre, and since $g \propto 1/x^2$ , the change in $g$ is negligible.

Detailed explanation

Background Concept

Outside a spherical planet,

g = \frac{GM}{x^2}

So $g$ depends on the distance $x$ from the centre, not on height above ground directly. The inverse-square form means $g$ decreases as you move away.

For small changes, it is useful to think in terms of fractional (percentage) change:

If $x$ changes by a tiny fraction, $g$ changes by about twice that fraction (because of the power 2).

Understanding the Question

“Near the surface” means $x$ is about the planet radius $R$ (which is very large). “Small changes in vertical height” means $\Delta x$ is tiny compared with $R$ .

You must explain why, in that situation, treating $g$ as constant is a good approximation.

Approach

Compare $x$ and $x + \Delta x$ :

when $\Delta x \ll R$ , the ratio $\frac{x+\Delta x}{x}$ is almost 1.
since $g \propto 1/x^2$ , the corresponding change in $g$ is negligible.

Step-by-Step Reasoning

Let $x = R$ at the surface. A small height change gives $x = R + h$ where $h \ll R$ .

Because

g \propto \frac{1}{x^2}

we can compare:

\frac{g(R+h)}{g(R)} = \frac{R^2}{(R+h)^2}

If $h$ is very small compared with $R$ , then $(R+h)^2$ is only very slightly bigger than $R^2$ , so the ratio is very close to 1, meaning $g(R+h) \approx g(R)$ .

So for everyday heights compared with the Earth’s radius, the change in $g$ is too small to matter, and we can treat $g$ as constant.

Key Takeaways

The reason is geometric: the Earth’s radius is huge, so small heights barely change distance to the centre.
Inverse-square dependence makes the change in $g$ small for small fractional changes in $x$ .

Common Mistakes

Saying “ $g$ is constant because gravity is constant” without referencing $g = GM/x^2$ .
Confusing height above ground with distance from the centre (the formula uses $x$ ).

Things to Be Careful About

Always relate “small height” to “ $\Delta x \ll R$ ”. That comparison is the heart of the argument.
Do not claim $g$ is exactly constant; it is approximately constant over small height ranges.

Techniques used

use the inverse-square relationship to compare values at nearby radiiapply the small-change approximation near a large radiusjustify a constant-field approximation qualitatively

(c)

Assume that the Earth is a uniform sphere. For the Earth, the product $GM$ is equal to $3.99 \times 10^{14}\ \text{m}^3\ \text{s}^{-2}$ .

4M

(i)

Determine a value, to three significant figures, for the radius $R$ of the Earth.

$R$ = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

At the surface,

g = \frac{GM}{R^2}

so

R = \sqrt{\frac{GM}{g}}

R = \sqrt{\frac{3.99 \times 10^{14}}{9.81}} = 6.38 \times 10^{6}\ \text{m}

Answer

$R = 6.38 \times 10^{6}\ \text{m}$

Final answer

6.38 × 10^6 m

Detailed explanation

Background Concept

For a spherically symmetric mass, the gravitational field strength at distance $r$ from the centre is

g = \frac{GM}{r^2}

At the planet’s surface, $r = R$ (the radius), so

g_{\text{surface}} = \frac{GM}{R^2}

Understanding the Question

You are given $GM = 3.99 \times 10^{14}\ \text{m}^3\ \text{s}^{-2}$ . To find the Earth’s radius $R$ , you must use the surface value of $g$ (commonly taken as $g = 9.81\ \text{m s}^{-2}$ ).

Unknown: $R$ .

Approach

Use

g = \frac{GM}{R^2}

and rearrange to make $R$ the subject, then substitute the values and quote the result to 3 s.f.

Step-by-Step Reasoning

Starting point:

g = \frac{GM}{R^2}

Rearrange:

R^2 = \frac{GM}{g}

so

R = \sqrt{\frac{GM}{g}}

Substitute:

R = \sqrt{\frac{3.99 \times 10^{14}}{9.81}}

Compute inside the root:

\frac{3.99 \times 10^{14}}{9.81} \approx 4.07 \times 10^{13}

Then

R \approx \sqrt{4.07 \times 10^{13}} \approx 6.38 \times 10^{6}\ \text{m}

The unit is metres because the expression inside the square root has units:

\frac{\text{m}^3\ \text{s}^{-2}}{\text{m s}^{-2}} = \text{m}^2

and the square root gives $\text{m}$ .

Key Takeaways

At the surface of a spherical planet, $g$ and $R$ are related by $g = GM/R^2$ .
Rearranging an inverse-square law often introduces a square root.

Common Mistakes

Forgetting the square root and using $R = GM/g$ .
Using $g = 9.8$ without matching the required significant figures (usually fine, but check expectations).
Giving the answer in km without converting or without stating the unit.

Things to Be Careful About

Use consistent SI units throughout.
Quote to three significant figures as asked.
Make sure the final unit is $\text{m}$ , not $\text{m}^2$ .

Techniques used

rearrange an inverse-square law to make the unknown the subjectsubstitute numerical values consistently in SI unitsuse square-root evaluation and significant figures

(ii)

Calculate the gravitational potential at the Earth’s surface. Give a unit with your answer.

gravitational potential = ______ unit ______

DifficultyMedium-Easy

Worked solution

Working

Gravitational potential at distance $r$ from the centre:

V = -\frac{GM}{r}

At the Earth’s surface $r = R = 6.38 \times 10^{6}\ \text{m}$ :

V = -\frac{3.99 \times 10^{14}}{6.38 \times 10^{6}} = -6.25 \times 10^{7}\ \text{J kg}^{-1}

Answer

$V = -6.25 \times 10^{7}\ \text{J kg}^{-1}$

Final answer

−6.25 × 10^7 J kg−1

Detailed explanation

Background Concept

Gravitational potential $V$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point (with no change in kinetic energy). For a spherical mass (or point mass) of mass $M$ , outside the sphere:

V = -\frac{GM}{r}

The negative sign appears because gravity is attractive and we take $V = 0$ at infinity.

Unit: since it is energy per unit mass, $V$ has units $\text{J kg}^{-1}$ .

Understanding the Question

You have already found the Earth’s radius $R$ in part (i). You are asked for the gravitational potential at the Earth’s surface, so use $r = R$ in $V = -GM/r$ , with $GM$ given.

Approach

Write $V = -GM/R$ .
Substitute $GM$ and $R$ .
Keep the negative sign and quote the correct unit.

Step-by-Step Reasoning

Use the formula:

V = -\frac{GM}{r}

At the surface, $r = R = 6.38 \times 10^{6}\ \text{m}$ and $GM = 3.99 \times 10^{14}\ \text{m}^3\ \text{s}^{-2}$ :

V = -\frac{3.99 \times 10^{14}}{6.38 \times 10^{6}}

Calculate:

\frac{3.99}{6.38} \approx 0.625, \quad 10^{14}/10^{6} = 10^{8}

So

V \approx -(0.625 \times 10^{8}) = -6.25 \times 10^{7}\ \text{J kg}^{-1}

Check units:

\frac{\text{m}^3\ \text{s}^{-2}}{\text{m}} = \text{m}^2\ \text{s}^{-2} = \text{J kg}^{-1}

Key Takeaways

Outside a spherical body, gravitational potential is $V = -GM/r$ .
Gravitational potential is negative if zero is chosen at infinity.

Common Mistakes

Missing the negative sign.
Giving units as $\text{J}$ instead of $\text{J kg}^{-1}$ .
Using $V = -GM/r^2$ (confusing potential with field strength).

Things to Be Careful About

Use the radius from part (i) consistently.
Do not round too early; keep enough significant figures to give a 3 s.f. final value if appropriate.
Ensure your final power of ten is correct when dividing standard form numbers.

Techniques used

use the formula for gravitational potential of a point/spherical masssubstitute values and handle the negative sign correctlystate an appropriate unit for potential

(d)

Explain why the gravitational potential energy of two point masses is always negative.

DifficultyMedium-Easy

Worked solution

Answer

Gravitational potential energy is defined to be zero when the masses are infinitely far apart. Since gravity is attractive, energy must be supplied (work done against the force) to separate them to infinity, so at any finite separation the potential energy is less than zero:

U = -\frac{Gm_1m_2}{r}

so $U$ is always negative.

Final answer

Because zero is taken at infinity and the gravitational force is attractive, the potential energy at any finite separation is below zero (e.g. $U=-Gm_1m_2/r$ ).

Detailed explanation

Background Concept

Gravitational potential energy $U$ of a pair of masses is defined via the work done to move them. A standard convention is:

Take $U = 0$ when the two masses are infinitely separated.

For two point masses $m_1$ and $m_2$ separated by distance $r$ :

U = -\frac{Gm_1m_2}{r}

The sign comes from the fact that gravity is an attractive force.

Understanding the Question

You must explain why gravitational potential energy is always negative for two point masses (with the usual convention $U=0$ at infinity). This is a conceptual sign question.

Approach

Use one (or both) of these arguments:

The formula $U = -Gm_1m_2/r$ directly shows it is negative.
Work/energy argument: to separate the masses to infinity you must do positive work against attraction, so the initial energy at finite separation must be lower than the zero level at infinity.

Step-by-Step Reasoning

If the masses are at separation $r$ , they attract each other.
If you slowly pull them apart to infinity (so kinetic energy doesn’t change), you must supply energy because you are doing work against the attractive gravitational force.
That means the potential energy increases from its value at $r$ up to the value at infinity.
Since we define the value at infinity as $U(\infty)=0$ , the value at finite $r$ must be below this:

U(r) < 0

This matches the mathematical expression:

U = -\frac{Gm_1m_2}{r}

where $G$ , $m_1$ , $m_2$ , and $r$ are all positive, so the result is always negative.

Key Takeaways

The negativity comes from (i) the choice of zero at infinity and (ii) the attractive nature of gravity.
At finite separation, the system is “bound”; you must add energy to separate it completely.

Common Mistakes

Saying “it’s negative because of the minus sign” without connecting to the physical meaning (zero at infinity and attraction).
Choosing the zero level at the Earth’s surface (a different convention) and then getting confused about signs.

Things to Be Careful About

The statement “always negative” assumes the standard convention $U=0$ at infinite separation, which is what A-Level uses.
Distinguish gravitational potential $V$ (per unit mass) from gravitational potential energy $U$ (energy for a particular pair of masses).

Techniques used

use the reference level of potential energy at infinite separationlink attractive force to negative work done by the fielduse the expression for gravitational potential energy to justify the sign

Q2Medium-EasyTemperature

(a)

State what is meant by two objects being in thermal equilibrium.

DifficultyEasy

Worked solution

Answer

Two objects are in thermal equilibrium when they are at the same temperature and there is no net transfer of thermal energy between them.

Final answer

Same temperature and no net heat transfer between them.

Detailed explanation

Background Concept

Thermal equilibrium is a state related to temperature and heat transfer. Temperature indicates the direction of net thermal energy transfer: if two bodies at different temperatures are placed in thermal contact, thermal energy flows from the higher temperature body to the lower temperature body.

When no net thermal energy transfer occurs between two bodies in contact, their temperatures are equal and they are said to be in thermal equilibrium.

Understanding the Question

You are asked to state what it means for two objects to be in thermal equilibrium. This is a definition-type question worth 2 marks, so two distinct points are expected.

Approach

Give the two standard marking points:

same temperature, and
no net heat flow (no net transfer of thermal energy) between them.

Step-by-Step Reasoning

If object A is hotter than object B, there will be a net transfer of thermal energy from A to B.
This transfer continues until both objects reach the same temperature.
At that point, energy transfers may still occur microscopically in both directions, but they balance, so the net transfer is zero.
Therefore, the definition requires both equal temperature and zero net heat transfer.

Key Takeaways

Thermal equilibrium implies equal temperature.
Equal temperature implies no net heat transfer between objects in contact.

Common Mistakes

Saying only "same temperature" and not mentioning "no net heat transfer" (often loses 1 mark).
Saying "no heat transfer" without "net" (some transfer occurs both ways microscopically; examiners usually want the phrase "no net transfer").

Things to Be Careful About

Use the wording "no net transfer of thermal energy" (or "no net heat flow").
Make it clear the statement refers to the situation when the objects are in thermal contact (or able to exchange thermal energy).

Techniques used

state the condition for thermal equilibriumlink thermal equilibrium to net heat transfer

(b)

A mass $X$ of ice at $0\,^{\circ}\text{C}$ is placed in a beaker containing a mass $M$ of water at Celsius temperature $t$ . The beaker is perfectly insulated and has negligible heat capacity. After some time, the ice that was added reaches thermal equilibrium with the original water in the beaker.

The specific latent heat of fusion of water is $L$ . The specific heat capacity of water is $c$ . The final Celsius temperature of the system is $\theta$ .

Give expressions, in terms of some or all of $X$ , $M$ , $t$ , $\theta$ , $L$ and $c$ , for the thermal energy:

3M

(i)

$E_1$ , gained by the ice as it melts to become water at $0\,^{\circ}\text{C}$

$E_1$ = ______

(ii)

$E_2$ , lost by the water as its Celsius temperature decreases from $t$ to $\theta$

$E_2$ = ______

(iii)

$E_3$ , gained by the melted ice as its Celsius temperature increases from $0\,^{\circ}\text{C}$ to $\theta$ .

$E_3$ = ______

(c)

Use your answers in (b) to show that the final Celsius temperature $\theta$ of the system is given by

\theta = \frac{Mct - XL}{c(M + X)}.

DifficultyMedium-Easy

Worked solution

Working

Perfect insulation and negligible beaker heat capacity give

E_2 = E_1 + E_3.

So

Mc(t-\theta)=XL+Xc\theta.

Expand and collect $\theta$ terms:

Mct-Mc\theta=XL+Xc\theta

Mct-XL=c\theta(M+X)

Hence

\theta = \frac{Mct-XL}{c(M+X)}.

Answer

\theta = \frac{Mct-XL}{c(M+X)}.

Final answer

θ = (Mct − XL) / (c(M + X))

Detailed explanation

Background Concept

For calorimetry problems (mixing, melting, heating), the key idea is conservation of energy.

In an insulated system, there is no thermal energy transfer to the surroundings.
If the container’s heat capacity is negligible, it does not absorb a significant amount of energy.

Therefore, thermal energy lost by the hotter part equals thermal energy gained by the colder part(s).

The relevant equations are:

E = mc\Delta T \quad \text{(temperature change, no phase change)}

and

E = mL \quad \text{(phase change at constant temperature)}.

Understanding the Question

Ice of mass $X$ at $0\,^{\circ}\text{C}$ is added to water of mass $M$ at temperature $t$ . Eventually, everything reaches a common final temperature $\theta$ .

During the process:

the ice melts at $0\,^{\circ}\text{C}$ (energy $E_1$ gained),
the original water cools from $t$ to $\theta$ (energy $E_2$ lost),
the melted ice (now water) warms from $0$ to $\theta$ (energy $E_3$ gained).

You must use these to show the provided formula for $\theta$ .

Approach

Write an energy balance using insulation:
heat lost by original water = heat gained in melting + heat gained in warming melted water.
Substitute the expressions from (b):
$E_2 = Mc(t-\theta)$ , $E_1 = XL$ , $E_3 = Xc\theta$ .
Rearrange to make $\theta$ the subject and match the required expression.

Step-by-Step Reasoning

Because the beaker is perfectly insulated and has negligible heat capacity:

E_2 = E_1 + E_3.

Substitute each expression:

Mc(t-\theta) = XL + Xc\theta.

Now expand the left side:

Mct - Mc\theta = XL + Xc\theta.

Gather the terms without $\theta$ on one side and the $\theta$ terms on the other. Add $Mc\theta$ to both sides:

Mct = XL + Xc\theta + Mc\theta.

Factor out $c\theta$ on the right (since $Xc\theta + Mc\theta = c\theta(X+M)$ ):

Mct = XL + c\theta(M+X).

Subtract $XL$ from both sides:

Mct - XL = c\theta(M+X).

Finally divide by $c(M+X)$ :

\theta = \frac{Mct - XL}{c(M+X)}.

This is exactly the expression required.

Key Takeaways

In an insulated mixing problem: heat lost = heat gained.
Separate phase change energy ( $mL$ ) from temperature change energy ( $mc\Delta T$ ).
When multiple bodies gain energy, add their gains and equate to the loss.

Common Mistakes

Forgetting one of the gain terms (often missing the warming of the melted ice, $Xc\theta$ ).
Using $Mc(\theta-t)$ instead of $Mc(t-\theta)$ when treating energy lost as a positive quantity.
Including the beaker’s heat capacity even though it is stated negligible.
Algebra slip when collecting $\theta$ terms (e.g. writing $c(M-X)$ instead of $c(M+X)$ ).

Things to Be Careful About

The final temperature is $\theta$ for both portions of water (original and melted ice).
The melted ice starts warming only after it has melted, so the heating term uses mass $X$ and temperature rise $0 \to \theta$ .
Keep signs consistent: treat $E_1$ , $E_2$ , $E_3$ as positive magnitudes and use the balance $E_2 = E_1 + E_3$ .

Techniques used

apply conservation of energy in an insulated systemequate heat lost by one part to heat gained by anothersubstitute expressions for latent heat and specific heat capacityrearrange algebra to isolate the final temperature

Q3MediumIdeal GasesThermodynamics

(a)

State what is meant by an ideal gas.

DifficultyEasy

Worked solution

Answer

An ideal gas is a gas in which:

molecules have negligible volume and exert no intermolecular forces (except during collisions), and
collisions are perfectly elastic (so it obeys $pV = nRT$ ).

Final answer

An ideal gas has molecules of negligible volume with no intermolecular forces (except during collisions), undergoing perfectly elastic collisions so that it obeys pV = nRT.

Detailed explanation

Background Concept

An ideal gas is a model used to simplify real gas behaviour. In the kinetic theory model, gas molecules are treated as point particles moving randomly.

The key assumptions are:

the molecules occupy negligible volume compared with the container volume,
there are no intermolecular forces except during collisions,
collisions between molecules and with the container walls are perfectly elastic (kinetic energy is conserved in each collision),
molecules move in random directions with a wide range of speeds.

Under these assumptions, the gas obeys the equation of state

pV = nRT

(or equivalently $pV = NkT$ ).

Understanding the Question

The question asks you to state what is meant by an ideal gas (2 marks). That means giving a concise definition with the essential assumptions (typically any two correct points are enough).

Approach

Provide two clear, standard points from the ideal-gas assumptions, and (optionally) link them to the fact that an ideal gas obeys $pV = nRT$ .

Step-by-Step Reasoning

State a microscopic assumption: e.g. negligible molecular volume and/or no intermolecular forces.
State another: e.g. collisions are perfectly elastic.
Optionally mention the macroscopic consequence: the gas obeys $pV = nRT$ .

Any two of these core points usually secure full credit.

Key Takeaways

“Ideal gas” is defined by assumptions about molecules and collisions, not by a particular substance.
Ideal gases obey $pV = nRT$ because of those assumptions.

Common Mistakes

Saying only “it obeys $pV = nRT$ ” with no assumptions (often not enough on its own).
Confusing with “incompressible” (gases are compressible).
Mentioning temperature/pressure conditions without stating the model assumptions.

Things to Be Careful About

Use the phrasing “no intermolecular forces except during collisions” (forces act during collisions to change momentum).
If you mention $pV = nRT$ , make sure it’s in addition to at least one microscopic assumption to secure the definition marks.

Techniques used

state the defining assumptions of the ideal gas modellink microscopic assumptions to macroscopic equation of state

(b)

An ideal gas at a pressure of $1.6 \times 10^5\ \text{Pa}$ has a density of $1.9\ \text{kg m}^{-3}$ .

5M

(i)

Show that the root-mean-square (r.m.s.) speed of molecules of this gas is approximately $500\ \text{m s}^{-1}$ .

DifficultyMedium-Easy

Worked solution

Working

For an ideal gas,

p = \frac{1}{3}\rho c_{\text{rms}}^2

c_{\text{rms}} = \sqrt{\frac{3p}{\rho}}

c_{\text{rms}} = \sqrt{\frac{3(1.6 \times 10^5)}{1.9}} = \sqrt{2.53 \times 10^5} = 5.0 \times 10^2\ \text{m s}^{-1}

Answer

$c_{\text{rms}} \approx 5.0 \times 10^2\ \text{m s}^{-1}$ (about $500\ \text{m s}^{-1}$ ).

Final answer

5.0 × 10^2 m s^-1

Detailed explanation

Background Concept

Kinetic theory links macroscopic pressure to microscopic molecular motion. For a gas of density $\rho$ in which molecules have a range of speeds, the pressure is related to the root-mean-square speed $c_{\text{rms}}$ by

p = \frac{1}{3}\rho c_{\text{rms}}^2

where:

$p$ is the gas pressure,
$\rho$ is the mass density of the gas,
$c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ is the square root of the mean of the squared speeds.

This formula comes from molecules repeatedly colliding elastically with the container walls; the factor $1/3$ arises because only one-third of the random motion contributes, on average, to any given direction.

Understanding the Question

You are given:

$p = 1.6 \times 10^5\ \text{Pa}$
$\rho = 1.9\ \text{kg m}^{-3}$

and asked to show that $c_{\text{rms}}$ is approximately $500\ \text{m s}^{-1}$ . So the method is implied: use the kinetic theory relation between $p$ , $\rho$ and $c_{\text{rms}}$ .

Approach

Start from

p = \frac{1}{3}\rho c_{\text{rms}}^2

Rearrange for $c_{\text{rms}}$ .
Substitute the given values.
Evaluate and round to an appropriate significant figure to demonstrate “approximately $500$ ”.

Step-by-Step Reasoning

Start with the kinetic theory equation:

p = \frac{1}{3}\rho c_{\text{rms}}^2

Rearrange:

c_{\text{rms}}^2 = \frac{3p}{\rho} \qquad \Rightarrow \qquad c_{\text{rms}} = \sqrt{\frac{3p}{\rho}}

Now substitute:

\frac{3p}{\rho} = \frac{3(1.6 \times 10^5)}{1.9} = \frac{4.8 \times 10^5}{1.9} \approx 2.53 \times 10^5

Take the square root:

c_{\text{rms}} = \sqrt{2.53 \times 10^5} \approx 5.03 \times 10^2\ \text{m s}^{-1} \approx 5.0 \times 10^2\ \text{m s}^{-1}

So $c_{\text{rms}}$ is indeed about $500\ \text{m s}^{-1}$ .

Key Takeaways

Use the kinetic theory link: $p = \frac{1}{3}\rho c_{\text{rms}}^2$ .
Rearranging gives $c_{\text{rms}} = \sqrt{\frac{3p}{\rho}}$ .
Always keep track of units: $\text{Pa} = \text{N m}^{-2}$ works consistently with $\rho$ in $\text{kg m}^{-3}$ to give speed in $\text{m s}^{-1}$ .

Common Mistakes

Forgetting the factor of $\frac{1}{3}$ (giving a speed too small by a factor of $\sqrt{3}$ ).
Using $\rho/p$ instead of $p/\rho$ .
Rounding too early (though here “approximately 500” allows some rounding).

Things to Be Careful About

$c_{\text{rms}}$ is a speed (scalar), so no direction is needed.
Don’t confuse $c_{\text{rms}}$ with mean speed; the kinetic theory relation specifically uses mean square speed.

Techniques used

use the kinetic theory relation between pressure, density and mean square speedrearrange to obtain the root-mean-square speedsubstitute values and take a square root

(ii)

One molecule of the gas has a mass of $4.7 \times 10^{-26}\ \text{kg}$ .

Determine the thermodynamic temperature of the gas.

temperature = ______ $\text{K}$

DifficultyMedium-Easy

Worked solution

Working

Average translational kinetic energy:

\frac{1}{2}mc_{\text{rms}}^2 = \frac{3}{2}kT

T = \frac{mc_{\text{rms}}^2}{3k}

Using $m = 4.7 \times 10^{-26}\ \text{kg}$ , $c_{\text{rms}} = 5.0 \times 10^2\ \text{m s}^{-1}$ and $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ :

T = \frac{(4.7 \times 10^{-26})(5.0 \times 10^2)^2}{3(1.38 \times 10^{-23})} \approx 2.8 \times 10^2\ \text{K}

Answer

temperature $\approx 2.8 \times 10^2\ \text{K}$

Final answer

2.8 × 10^2 K

Detailed explanation

Background Concept

For an ideal gas, temperature is linked to the average translational kinetic energy of molecules:

\langle E_k \rangle = \frac{3}{2}kT

Also,

\langle E_k \rangle = \frac{1}{2}m\langle c^2 \rangle = \frac{1}{2}mc_{\text{rms}}^2

Combining these gives

\frac{1}{2}mc_{\text{rms}}^2 = \frac{3}{2}kT

where:

$m$ is the mass of one molecule,
$k$ is the Boltzmann constant ( $1.38 \times 10^{-23}\ \text{J K}^{-1}$ ),
$T$ is thermodynamic temperature in kelvin.

Understanding the Question

You are told one molecule has mass $m = 4.7 \times 10^{-26}\ \text{kg}$ and, from part (i), the gas has $c_{\text{rms}} \approx 5.0 \times 10^2\ \text{m s}^{-1}$ . You must find $T$ .

Approach

Use

\frac{1}{2}mc_{\text{rms}}^2 = \frac{3}{2}kT

Rearrange for $T$ .
Substitute values and evaluate.

Step-by-Step Reasoning

Start from the energy–temperature relation:

\frac{1}{2}mc_{\text{rms}}^2 = \frac{3}{2}kT

Cancel the factor of $\frac{1}{2}$ :

mc_{\text{rms}}^2 = 3kT

So

T = \frac{mc_{\text{rms}}^2}{3k}

Substitute the given values:

$c_{\text{rms}}^2 = (5.0 \times 10^2)^2 = 2.5 \times 10^5\ \text{m}^2\text{s}^{-2}$
$mc_{\text{rms}}^2 = (4.7 \times 10^{-26})(2.5 \times 10^5) = 1.175 \times 10^{-20}\ \text{J}$
$3k = 3(1.38 \times 10^{-23}) = 4.14 \times 10^{-23}\ \text{J K}^{-1}$

Therefore:

T = \frac{1.175 \times 10^{-20}}{4.14 \times 10^{-23}} \approx 2.84 \times 10^2\ \text{K}

So the temperature is about $2.8 \times 10^2\ \text{K}$ .

Key Takeaways

Temperature measures average translational kinetic energy: $\langle E_k \rangle = \frac{3}{2}kT$ .
Using $c_{\text{rms}}$ is natural because $c_{\text{rms}}^2 = \langle c^2 \rangle$ .

Common Mistakes

Using $\frac{1}{2}mc_{\text{rms}}$ instead of $\frac{1}{2}mc_{\text{rms}}^2$ .
Forgetting the factor of 3 (using $T = mc_{\text{rms}}^2/k$ ).
Mixing up $R$ and $k$ (here you are working per molecule, so use $k$ ).

Things to Be Careful About

Use kelvin for thermodynamic temperature (not degrees Celsius).
The final value should be sensible: $\sim 300\ \text{K}$ corresponds to “room temperature”, consistent with molecular speeds of a few hundred $\text{m s}^{-1}$ .

Techniques used

apply the kinetic theory relation between average kinetic energy and temperaturesubstitute the r.m.s. speed into the kinetic energy expressionrearrange to solve for temperature

(c)

Calculate the internal energy $U$ of $6.0\ \text{mol}$ of the gas in (b). Explain your reasoning.

$U$ = ______ $\text{J}$

DifficultyMedium

Worked solution

Working

For an ideal gas, intermolecular potential energy is negligible, so internal energy is the total random kinetic energy.

Average translational kinetic energy per molecule is $\frac{3}{2}kT$ , hence for $n$ moles:

U = \frac{3}{2}NkT = \frac{3}{2}nRT

With $n = 6.0\ \text{mol}$ , $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ and $T \approx 2.8 \times 10^2\ \text{K}$ :

U = \frac{3}{2}(6.0)(8.31)(2.8 \times 10^2) \approx 2.1 \times 10^4\ \text{J}

Answer

$U \approx 2.1 \times 10^4\ \text{J}$

Final answer

2.1 × 10^4 J

Detailed explanation

Background Concept

The internal energy $U$ of a substance is the sum of the random microscopic energies of its particles:

random kinetic energy (translational, rotational, vibrational),
plus any microscopic potential energy due to intermolecular forces.

For an ideal gas, one key assumption is that there are no intermolecular forces (except during collisions). That means the intermolecular potential energy is negligible, so $U$ depends only on the random kinetic energy.

From kinetic theory, the average translational kinetic energy per molecule is

\langle E_k \rangle = \frac{3}{2}kT

For $N$ molecules,

U = N\left(\frac{3}{2}kT\right) = \frac{3}{2}NkT

Using $N = nN_A$ and $R = N_Ak$ gives the molar form:

U = \frac{3}{2}nRT

(This is the form typically expected at this level when modelling the gas as having only translational degrees of freedom.)

Understanding the Question

You must calculate the internal energy of $6.0\ \text{mol}$ of the same gas as in part (b), so you use the temperature found in (b)(ii), about $2.8 \times 10^2\ \text{K}$ . The question also asks you to explain your reasoning, i.e. why $U$ can be written in terms of $T$ .

Approach

State the ideal-gas reasoning: negligible intermolecular potential energy, so $U$ is kinetic energy only.
Use the kinetic theory result for average kinetic energy, leading to

U = \frac{3}{2}nRT.

Substitute $n = 6.0$ and your value of $T$ .

Step-by-Step Reasoning

Because the gas is ideal, intermolecular potential energy is negligible. Internal energy is therefore just the total random kinetic energy.

Average translational kinetic energy per molecule:

\langle E_k \rangle = \frac{3}{2}kT

Total for $N$ molecules:

U = N\langle E_k \rangle = \frac{3}{2}NkT

Convert to moles using $Nk = (nN_A)k = n(N_Ak) = nR$ :

U = \frac{3}{2}nRT

Now substitute $n = 6.0\ \text{mol}$ , $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ , and $T \approx 2.84 \times 10^2\ \text{K}$ :

U = \frac{3}{2}(6.0)(8.31)(2.84 \times 10^2)

Compute stepwise:

$\frac{3}{2} \times 6.0 = 9.0$
$9.0 \times 8.31 = 74.79$
$74.79 \times 2.84 \times 10^2 \approx 2.12 \times 10^4$

So

U \approx 2.1 \times 10^4\ \text{J}

Key Takeaways

For an ideal gas, internal energy depends only on temperature because potential energy is negligible.
The standard model used here gives $U = \frac{3}{2}nRT$ .

Common Mistakes

Using $U = nRT$ (missing the factor $\frac{3}{2}$ ).
Forgetting to use kelvin.
Using $k$ with $n$ moles (if you use $k$ , you must use number of molecules $N$ , not moles).

Things to Be Careful About

Quote $U$ in joules and to sensible significant figures (typically 2 or 3).
Ensure consistency: the temperature used should be the same as in part (b)(ii), or error-carried-forward should be clear.
The justification must explicitly link “ideal gas” to “negligible potential energy”, which is why $U$ depends on $T$ only.

Techniques used

use the expression for internal energy of an ideal gas in terms of temperaturejustify neglect of intermolecular potential energy for an ideal gassubstitute amount of substance and temperature into U = 3/2 nRT

Q4MediumOscillations

(a)

State what is meant by simple harmonic motion.

DifficultyEasy

Worked solution

Answer

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from the equilibrium position and is directed towards the equilibrium position:

a \propto -x \quad \text{(i.e. } a = -\omega^2 x\text{)}.

Final answer

Acceleration is proportional to displacement from equilibrium and directed towards equilibrium: a = -ω^2 x.

Detailed explanation

Background Concept

In oscillatory motion, an object moves back and forth about an equilibrium position. Simple harmonic motion (SHM) is a special type of oscillation where the restoring effect becomes stronger the further you are from equilibrium.

The defining condition for SHM is that the acceleration (and hence resultant force) is:

directly proportional to the displacement $x$ from equilibrium, and
opposite in direction to the displacement (i.e. always towards equilibrium).

Mathematically this is written as

a = -\omega^2 x

where:

$a$ is acceleration,
$x$ is displacement from equilibrium,
$\omega$ is the angular frequency (a constant for the motion).

Understanding the Question

You are not asked to do any calculation: you must state what SHM means. For full marks, you must include both:

proportionality to displacement, and 2) direction towards equilibrium (negative sign).

Approach

Write the defining equation or its worded equivalent. Make sure the "towards equilibrium" part is clear (this is what the minus sign represents).

Step-by-Step Reasoning

"Acceleration proportional to displacement" means $a \propto x$ .
"Directed towards equilibrium" means if $x$ is positive, $a$ must be negative, and vice versa, so $a \propto -x$ .
The constant of proportionality is written as $\omega^2$ , giving $a = -\omega^2 x$ .

Key Takeaways

SHM is defined by the relationship between $a$ and $x$ , not by the object being a pendulum or having a sine graph.
The negative sign (towards equilibrium) is essential.

Common Mistakes

Saying only "it moves in a sine wave" (not a definition).
Missing the direction (writing $a \propto x$ instead of $a \propto -x$ ).
Stating "velocity proportional to displacement" (wrong: velocity is largest at equilibrium where $x=0$ ).

Things to Be Careful About

Use the word equilibrium explicitly.
Make clear it is the acceleration (or restoring force) that is proportional to displacement.

Techniques used

state the defining proportionality between acceleration and displacementspecify the direction of acceleration relative to displacement

(b)

A small sphere is suspended from a fixed point P by a string of negligible mass, as shown in Fig. 4.1.

Fig. 4.1

The sphere is given a small horizontal displacement and is then released.

The variation with time of the horizontal velocity $v$ of the sphere is shown in Fig. 4.2.

Fig. 4.2

3M

(i)

State two times at which the sphere is passing in the same direction through the equilibrium position.

time ______ and time ______

DifficultyMedium-Easy

Worked solution

Answer

At the equilibrium position the speed is maximum, so this corresponds to peaks/troughs of the $v$ – $t$ graph.

Two suitable times are $t_1$ and $t_5$ (both have negative maximum velocity, so the sphere passes equilibrium in the same direction).

Final answer

t1 and t5

Detailed explanation

Background Concept

In SHM, the relationship between displacement $x$ and velocity $v$ is such that:

At equilibrium ( $x=0$ ), the acceleration is zero and the speed is maximum.
At the extremes ( $x=\pm x_0$ ), the speed is zero.

So, on a velocity–time graph:

Crossing $v=0$ indicates turning points (maximum displacement).
Maximum positive or negative $v$ indicates passing through equilibrium.

The direction of motion is given by the sign of $v$ (positive or negative).

Understanding the Question

You are given a graph of horizontal velocity against time. The question asks for two times when the sphere is:

at the equilibrium position (so horizontal displacement is zero), and
moving through it in the same direction (so velocity must have the same sign at those two times).

Approach

Find times of maximum magnitude of velocity (peaks/troughs) because those correspond to equilibrium.
Choose two times where the velocity has the same sign (both positive peaks or both negative troughs).

Step-by-Step Reasoning

From the sinusoidal $v$ – $t$ curve, the trough at $t_1$ is the most negative value of $v$ , so this is an equilibrium crossing with motion in the negative direction.
The next trough occurs at $t_5$ , also most negative, so this is again an equilibrium crossing in the same (negative) direction.
Therefore, $t_1$ and $t_5$ satisfy the requirement.

(Equally acceptable would be $t_3$ and $t_7$ , which are both positive peaks.)

Key Takeaways

In SHM, equilibrium corresponds to maximum speed, not zero speed.
The sign of $v$ tells you the direction of travel.

Common Mistakes

Choosing $t_2, t_4, t_6$ (these are where $v=0$ , i.e. turning points, not equilibrium).
Choosing one maximum and one minimum (directions opposite).

Things to Be Careful About

The question says "passing ... through the equilibrium position" which implies it is moving (so $v \neq 0$ ).
Use the velocity sign to ensure "same direction".

Techniques used

interpret a velocity–time sinusoidal graphidentify equilibrium crossings from maximum speeduse the sign of velocity to determine direction of motion

(ii)

The time interval between $t_1$ and $t_6$ is $2.2\ \text{s}$ .

Calculate the frequency of oscillation of the sphere.

frequency = ______ $\text{Hz}$

DifficultyMedium

Worked solution

Working

From the $v$ – $t$ graph: $t_1$ is a minimum and $t_6$ is the next-but-one zero crossing, so

t_1 \to t_5 = T, \quad t_5 \to t_6 = \frac{1}{4}T \;\Rightarrow\; t_1 \to t_6 = \frac{5}{4}T.

Given $t_6 - t_1 = 2.2\ \text{s}$ ,

\frac{5}{4}T = 2.2 \Rightarrow T = 2.2\times \frac{4}{5} = 1.76\ \text{s}.

f = \frac{1}{T} = \frac{1}{1.76} = 0.568\ \text{Hz} \approx 0.57\ \text{Hz}.

Answer

f \approx 0.57\ \text{Hz}

Final answer

0.57 Hz

Detailed explanation

Background Concept

Oscillations repeat in time with a period $T$ (time for one full cycle). The frequency is

f = \frac{1}{T}.

For a sinusoidal velocity–time graph, particular features repeat every fixed fraction of a period:

Minimum $\to$ next zero crossing: $\frac{1}{4}T$
Minimum $\to$ maximum: $\frac{1}{2}T$
Minimum $\to$ next minimum: $T$

Understanding the Question

You are told that the time between $t_1$ and $t_6$ is $2.2\ \text{s}$ . From the graph:

$t_1$ is a minimum of velocity.
$t_6$ is a zero crossing after the next minimum ( $t_5$ ).

You must convert this time gap into the period $T$ , then find $f$ .

Approach

Decide what fraction of a full cycle corresponds to the interval $t_1 \to t_6$ .
Use that to calculate $T$ .
Use $f = 1/T$ .

Step-by-Step Reasoning

From the graph sequence: minimum at $t_1$ , then later another minimum at $t_5$ . So

t_5 - t_1 = T.

From a minimum to the next time the curve crosses zero is a quarter period, so

t_6 - t_5 = \frac{1}{4}T.

Therefore the total from $t_1$ to $t_6$ is

(t_6-t_1) = (t_5-t_1) + (t_6-t_5) = T + \frac{1}{4}T = \frac{5}{4}T.

Substitute the given $t_6-t_1 = 2.2\ \text{s}$ :

\frac{5}{4}T = 2.2 \Rightarrow T = 2.2\times \frac{4}{5} = 1.76\ \text{s}.

Then the frequency is

f = \frac{1}{1.76} = 0.568\ \text{Hz} \approx 0.57\ \text{Hz}.

Key Takeaways

A sinusoidal graph lets you find $T$ even if you are not given two identical points directly, by using fractions of the cycle.
Use $f = 1/T$ .

Common Mistakes

Assuming $t_1 \to t_6$ is exactly one period.
Mixing up quarter-period and half-period steps between maxima/minima/zero crossings.
Writing $f=T$ instead of $f=1/T$ .

Things to Be Careful About

Identify the features correctly: $t_1$ is a trough, $t_6$ is a zero crossing.
Give frequency in $\text{Hz}$ and to a sensible number of significant figures (typically 2 s.f. here).

Techniques used

relate time intervals on a sinusoidal graph to fractions of a perioddetermine the period from multiple cyclescalculate frequency using f = 1/T

(c)

The sphere in (b) is undergoing simple harmonic motion.

Use your answer in (b)(ii) and data from Fig. 4.2 to determine the maximum displacement of the sphere from its equilibrium position.

maximum displacement = ______ $\text{m}$

DifficultyMedium

Worked solution

Working

From Fig. 4.2, maximum horizontal speed $v_{\max} \approx 0.12\ \text{m s}^{-1}$ .

Using $\omega = 2\pi f$ with $f = 0.57\ \text{Hz}$ ,

\omega = 2\pi(0.57) = 3.58\ \text{rad s}^{-1}.

For SHM,

v_{\max} = \omega x_0 \Rightarrow x_0 = \frac{v_{\max}}{\omega} = \frac{0.12}{3.58} = 3.35\times 10^{-2}\ \text{m}.

Answer

x_0 \approx 3.4\times 10^{-2}\ \text{m}

Final answer

3.4 × 10^-2 m

Detailed explanation

Background Concept

In SHM, displacement and velocity vary sinusoidally. If we write

x = x_0 \sin(\omega t),

then differentiating gives velocity

v = \frac{\mathrm{d}x}{\mathrm{d}t} = \omega x_0 \cos(\omega t).

So the maximum speed is

v_{\max} = \omega x_0.

Also, angular frequency $\omega$ is related to frequency $f$ by

\omega = 2\pi f.

Understanding the Question

You have already found the frequency (from part (b)(ii)). You are now told the motion is SHM and must use:

your value of $f$ to find $\omega$ ,
the velocity–time graph to read $v_{\max}$ ,
then calculate the maximum displacement (the amplitude) $x_0$ .

Approach

Read the velocity amplitude $v_{\max}$ from the graph (the peak value).
Convert $f$ into $\omega$ using $\omega = 2\pi f$ .
Rearrange $v_{\max} = \omega x_0$ to get $x_0 = v_{\max}/\omega$ .

Step-by-Step Reasoning

From Fig. 4.2, the peaks of the sinusoidal velocity curve give $v_{\max}$ . Reading from the vertical scale gives approximately $0.12\ \text{m s}^{-1}$ .
With $f \approx 0.57\ \text{Hz}$ ,

\omega = 2\pi f = 2\pi(0.57) = 3.58\ \text{rad s}^{-1}.

Use the SHM amplitude relation:

x_0 = \frac{v_{\max}}{\omega} = \frac{0.12}{3.58} = 0.0335\ \text{m} = 3.35\times 10^{-2}\ \text{m}.

Rounding appropriately gives $x_0 \approx 3.4\times 10^{-2}\ \text{m}$ .

Key Takeaways

For SHM, the link between velocity amplitude and displacement amplitude is $v_{\max} = \omega x_0$ .
You often use a graph to obtain an amplitude (here, $v_{\max}$ ).

Common Mistakes

Using $v_{\max} = 2\pi f x_0$ but forgetting the $2\pi$ .
Reading $v_{\max}$ from a zero crossing (wrong: $v=0$ at turning points, not maximum speed).
Mixing up $f$ and $\omega$ (units: $f$ in $\text{Hz}$ , $\omega$ in $\text{rad s}^{-1}$ ).

Things to Be Careful About

Ensure you read the peak of the curve, not the axis limit.
Use consistent units: $v$ in $\text{m s}^{-1}$ gives $x_0$ in $\text{m}$ .
Quote the final answer to a sensible number of significant figures based on the graph reading.

Techniques used

use v_max = omega x0 for SHMconvert frequency to angular frequency using omega = 2 pi fread an amplitude from a graph

Q5MediumElectric FieldsCapacitance

(a)

Define electric potential at a point.

DifficultyMedium-Easy

Worked solution

Answer

Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point (without change in kinetic energy).

Final answer

Work done per unit positive charge to bring a test charge from infinity to the point.

Detailed explanation

Background Concept

Electric potential $V$ is a way of describing the "electrical energy situation" at a point in an electric field.

It is defined using the work done $W$ when moving charge $q$ :

V = \frac{W}{q}

So the unit of potential is $\text{J C}^{-1}$ , which is called the volt (V).

A key idea in Cambridge definitions is that the reference point is usually infinity for isolated charges, so that potential is measured relative to $V=0$ at infinity.

Understanding the Question

The question asks for a definition of electric potential at a point. For full credit, you must:

mention work done (or potential energy change),
state “per unit positive charge”,
specify “from infinity to the point” (the reference).

Approach

Write the standard definition in words, making sure to include all the marking points (work done, per unit charge, from infinity).

Step-by-Step Reasoning

Electric potential is work done per unit charge: $V = W/q$ .
For a point in an electric field due to isolated charges, we define the zero of potential at infinity.
Therefore the definition must say: work done per unit positive charge in bringing a test charge from infinity to the point.
To be safe, add “without change in kinetic energy” (i.e. moved slowly), since otherwise work done could go into kinetic energy.

Key Takeaways

$V$ is energy per unit charge.
For isolated charge distributions, the reference is infinity.

Common Mistakes

Defining potential difference instead of potential (missing “from infinity”).
Saying “force per unit charge” (that is electric field strength $E$ ).
Missing “per unit charge” or not stating that the charge is a small test charge.

Things to Be Careful About

Use the wording “work done per unit positive charge” (or “potential energy per unit charge”).
Include the reference point “infinity” for full marks in this context.

Techniques used

state the definition of electric potential in terms of work done per unit chargereference bringing a test charge from infinity to the point

(b)

An isolated solid metal sphere of radius $r$ is given a positive charge.

The potential at the surface of the sphere is $9.0 \times 10^4\ \text{V}$ . At a distance of $3r$ from the centre of the sphere, the electric field strength is $2.0 \times 10^5\ \text{N C}^{-1}$ .

8M

(i)

Determine the electric field strength at the surface of the sphere.

electric field strength = ______ $\text{N C}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

For $r$ outside the sphere, the field is that of a point charge, so $E \propto \dfrac{1}{R^2}$ .

At $R = 3r$ , $E = 2.0 \times 10^5\ \text{N C}^{-1}$ .

E(r) = E(3r)\left(\frac{3r}{r}\right)^2 = 2.0 \times 10^5 \times 9

E(r) = 1.8 \times 10^6\ \text{N C}^{-1}

Answer

$1.8 \times 10^6\ \text{N C}^{-1}$

Final answer

1.8 \times 10^6 N C^{-1}

Detailed explanation

Background Concept

For a spherically symmetric charge distribution (such as a charged isolated conducting sphere), the electric field outside the sphere behaves as if all the charge were concentrated at the centre.

So at distance $R$ from the centre:

E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}

This shows $E \propto 1/R^2$ .

Understanding the Question

You are told that at distance $3r$ from the centre, the field strength is $2.0 \times 10^5\ \text{N C}^{-1}$ . You must find the field strength at the surface, i.e. at $R=r$ .

Approach

Use the inverse-square relationship:

\frac{E(r)}{E(3r)} = \left(\frac{3r}{r}\right)^2

Then multiply the known field by $9$ .

Step-by-Step Reasoning

Starting from

E \propto \frac{1}{R^2}

we compare two radii:

\frac{E(r)}{E(3r)} = \frac{(1/r^2)}{(1/(3r)^2)} = \frac{(1/r^2)}{(1/9r^2)} = 9

Therefore

E(r)=9E(3r)=9\times 2.0\times 10^5 = 1.8\times 10^6\ \text{N C}^{-1}

Key Takeaways

Outside a charged conducting sphere, treat it like a point charge at the centre.
Field strength scales as $1/R^2$ .

Common Mistakes

Using $E \propto 1/R$ (that is for potential $V$ , not field).
Dividing by $9$ instead of multiplying by $9$ (field is stronger closer in).

Things to Be Careful About

The distance given is from the centre, so use $R=3r$ not $R=2r$ .
Keep units as $\text{N C}^{-1}$ .

Techniques used

use the inverse-square dependence of electric field strength with distancescale the field strength between two radial distances

(ii)

Show that the radius of the sphere is $5.0\ \text{cm}$ .

DifficultyMedium

Worked solution

Working

From (i):

E(r)=1.8\times 10^6\ \text{N C}^{-1}

For a charged sphere (outside):

E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2},\qquad V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

So

V(r)=E(r)\,r

9.0\times 10^4 = (1.8\times 10^6)r

r=\frac{9.0\times 10^4}{1.8\times 10^6}=5.0\times 10^{-2}\ \text{m}=5.0\ \text{cm}

Answer

$r=5.0\ \text{cm}$

Final answer

5.0 cm

Detailed explanation

Background Concept

For a spherically symmetric charge distribution, outside the sphere:

E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}

and the electric potential (taking $V=0$ at infinity) is:

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}

These are linked: dividing $V$ by $E$ gives

\frac{V}{E} = \frac{\left(\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}\right)}{\left(\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}\right)} = R

So for a point charge field,

V = ER

Understanding the Question

At the surface, $R=r$ and the potential is given as $V=9.0\times 10^4\ \text{V}$ . From part (i), the field at the surface is $E(r)=1.8\times 10^6\ \text{N C}^{-1}$ . You must show that this implies $r=5.0\ \text{cm}$ .

Approach

Use the point-charge relationships at the surface and eliminate $Q$ . The quickest way is to use $V=Er$ (valid for a spherical field outside the sphere).

Step-by-Step Reasoning

At the surface:

V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

and

E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}

Divide the expressions or recognise the known identity for a spherical field:

V(r)=E(r)\,r

Substitute values:

9.0\times 10^4 = 1.8\times 10^6\, r

So

r = \frac{9.0\times 10^4}{1.8\times 10^6} = 5.0\times 10^{-2}\ \text{m}

Convert to cm:

5.0\times 10^{-2}\ \text{m} = 5.0\ \text{cm}

Key Takeaways

For spherical fields: $E \propto 1/R^2$ and $V \propto 1/R$ .
A useful link is $V=ER$ (with $R$ measured from the centre).

Common Mistakes

Using $V=Ed$ for a uniform field; here the field is not uniform.
Forgetting to convert $0.050\ \text{m}$ into $5.0\ \text{cm}$ .
Using $E$ at $3r$ instead of the surface value.

Things to Be Careful About

The relation $V=ER$ is specific to a $1/R^2$ field (point charge / spherical symmetry), not general.
Keep consistent significant figures: the data are mostly 2 s.f., so $r=5.0\ \text{cm}$ is appropriate.

Techniques used

relate electric field strength and potential for a point chargeeliminate charge to solve for radiususe algebraic rearrangement and unit conversion

(iii)

Calculate the charge on the sphere.

charge = ______ $\text{C}$

DifficultyMedium-Easy

Worked solution

Working

At the surface:

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

So

Q = 4\pi\varepsilon_0 V r = \frac{Vr}{8.99\times 10^9}

Using $V=9.0\times 10^4\ \text{V}$ and $r=5.0\times 10^{-2}\ \text{m}$ :

Q = \frac{(9.0\times 10^4)(5.0\times 10^{-2})}{8.99\times 10^9} = 5.0\times 10^{-7}\ \text{C}

Answer

$5.0 \times 10^{-7}\ \text{C}$

Final answer

5.0 \times 10^{-7} C

Detailed explanation

Background Concept

The potential due to an isolated charge (or an isolated conducting sphere, measured outside it) is

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}

At the surface of a conducting sphere, $R=r$ .

You can rearrange to find charge:

Q = 4\pi\varepsilon_0 V r

Since

\frac{1}{4\pi\varepsilon_0} = 8.99\times 10^9\ \text{N m}^2\ \text{C}^{-2}

you may also write $Q = Vr/(8.99\times 10^9)$ .

Understanding the Question

You already found the radius $r=5.0\ \text{cm}$ (i.e. $5.0\times 10^{-2}\ \text{m}$ ). The potential at the surface is $9.0\times 10^4\ \text{V}$ . Use these to calculate the charge $Q$ on the sphere.

Approach

Use the spherical potential formula at the surface, substitute $V$ and $r$ , and solve for $Q$ .

Step-by-Step Reasoning

Start with

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

Rearrange:

Q = 4\pi\varepsilon_0 V r

Convert $r$ to metres:

r = 5.0\ \text{cm} = 5.0\times 10^{-2}\ \text{m}

Now calculate using $1/(4\pi\varepsilon_0)=8.99\times 10^9$ :

Q = \frac{Vr}{8.99\times 10^9} = \frac{(9.0\times 10^4)(5.0\times 10^{-2})}{8.99\times 10^9}

Compute numerator:

(9.0\times 10^4)(5.0\times 10^{-2}) = 4.5\times 10^3

So

Q = \frac{4.5\times 10^3}{8.99\times 10^9} \approx 5.0\times 10^{-7}\ \text{C}

Key Takeaways

For a conducting sphere, outside behaviour matches a point charge at the centre.
Potential at the surface is $V = (1/4\pi\varepsilon_0)(Q/r)$ .

Common Mistakes

Leaving $r$ in cm rather than converting to m (gives a factor of 100 error).
Using the field formula $E \propto 1/r^2$ instead of the potential formula $V \propto 1/r$ .

Things to Be Careful About

Quote $Q$ to appropriate significant figures (here 2 s.f.).
Make sure you use $r$ (surface radius), not $3r$ .

Techniques used

use the potential of a charged sphere treated as a point chargerearrange to solve for chargesubstitute numerical values with correct units

(iv)

Use your answer in (b)(iii) to determine the capacitance of the sphere.

capacitance = ______ $\text{F}$

DifficultyMedium-Easy

Worked solution

Working

Capacitance:

C = \frac{Q}{V}

C = \frac{5.0\times 10^{-7}}{9.0\times 10^4} = 5.6\times 10^{-12}\ \text{F}

Answer

$5.6 \times 10^{-12}\ \text{F}$

Final answer

5.6 \times 10^{-12} F

Detailed explanation

Background Concept

Capacitance $C$ is defined as the ratio of charge stored to potential (potential difference):

C = \frac{Q}{V}

For an isolated conducting sphere, the potential $V$ is the potential of the sphere relative to infinity, and $Q$ is the charge on it. The unit is the farad (F), where $1\ \text{F} = 1\ \text{C V}^{-1}$ .

Understanding the Question

You are told to use your answer to (b)(iii) (the charge on the sphere) together with the given potential at the surface ( $9.0\times 10^4\ \text{V}$ ) to find the sphere’s capacitance.

Approach

Use the definition $C=Q/V$ and substitute the known $Q$ and $V$ .

Step-by-Step Reasoning

From (b)(iii):

Q = 5.0\times 10^{-7}\ \text{C}

Given:

V = 9.0\times 10^4\ \text{V}

So

C = \frac{Q}{V} = \frac{5.0\times 10^{-7}}{9.0\times 10^4}

Combine powers of ten:

\frac{10^{-7}}{10^{4}} = 10^{-11}

And numeric part:

\frac{5.0}{9.0} = 0.556

Thus

C \approx 0.556\times 10^{-11} = 5.56\times 10^{-12}\ \text{F} \approx 5.6\times 10^{-12}\ \text{F}

Key Takeaways

Capacitance is not just for parallel plates: any conductor has a capacitance.
For an isolated conductor, $C=Q/V$ where $V$ is relative to infinity.

Common Mistakes

Using $C=V/Q$ (inverting the definition).
Forgetting that the given $V$ is already the correct potential for the isolated sphere.

Things to Be Careful About

Keep powers of ten under control: dividing by $10^4$ shifts the exponent by $-4$ .
Final unit must be $\text{F}$ .

Techniques used

apply the definition of capacitancesubstitute charge and potential values to compute capacitancecheck units for consistency

Q6Medium-EasyMagnetic Fields

A rectangular coil PQRS of wire is free to rotate about its axis XY, as shown in Fig. 6.1.

Fig. 6.1

The coil has length QR of $5.4\ \text{cm}$ , width PQ of $2.5\ \text{cm}$ and has 190 turns of wire.
The plane of the coil is at an angle $\theta$ to a uniform magnetic field of flux density $5.2 \times 10^{-3}\ \text{T}$ .
The axis XY of the coil is normal to the field.
The current in the coil is $1.2\ \text{A}$ .

(a)

8M

(i)

Calculate the magnitude of the force on side QR of the coil.

force = ______ $\text{N}$

DifficultyMedium-Easy

Worked solution

Working

Force on one side per turn:

F_1 = BIL

Total force on side $QR$ for $N$ turns:

F = NBIL

F = 190 \times \left(5.2 \times 10^{-3}\right) \times 1.2 \times 0.054

F = 6.4 \times 10^{-2}\ \text{N}

Answer

$6.4 \times 10^{-2}\ \text{N}$

Final answer

6.4 × 10^-2 N

Detailed explanation

Background Concept

A straight current-carrying conductor of length $L$ in a magnetic field of flux density $B$ experiences a magnetic force

F = BIL\sin\phi

where $I$ is the current and $\phi$ is the angle between the direction of the current and the magnetic field. The force is maximum when the conductor is perpendicular to the field ( $\sin\phi = 1$ ).

For a coil with $N$ turns, each turn experiences the same force (in a uniform field), so the total force on one side of the coil is multiplied by $N$ .

Understanding the Question

We are asked for the magnitude of the force on side $QR$ of a rectangular coil.

Given:

$B = 5.2 \times 10^{-3}\ \text{T}$
$I = 1.2\ \text{A}$
length $QR = 5.4\ \text{cm} = 0.054\ \text{m}$
number of turns $N = 190$

From the diagram/setup, side $QR$ is perpendicular to the magnetic field, so $\sin\phi = 1$ .

Approach

Convert the side length into metres.
Use $F = BIL$ for one turn.
Multiply by $N$ turns to get the force on that side of the whole coil.

Step-by-Step Reasoning

Convert length:

L = 5.4\ \text{cm} = 5.4 \times 10^{-2}\ \text{m} = 0.054\ \text{m}

For one turn on side $QR$ (with $\sin\phi = 1$ ):

F_1 = BIL = (5.2\times 10^{-3})(1.2)(0.054)

Compute:

(5.2\times 10^{-3})(1.2) = 6.24\times 10^{-3}

F_1 = (6.24\times 10^{-3})(0.054) = 3.37\times 10^{-4}\ \text{N}

Now multiply by $N=190$ turns:

F = NF_1 = 190(3.37\times 10^{-4}) = 6.4\times 10^{-2}\ \text{N}

Key Takeaways

Use $F = BIL\sin\phi$ for a straight conductor in a magnetic field.
For a multi-turn coil in a uniform field, total force on a side scales as $N$ .
Always convert centimetres to metres before substituting.

Common Mistakes

Forgetting to multiply by $N = 190$ turns.
Using the wrong side length (e.g. $2.5\ \text{cm}$ instead of $5.4\ \text{cm}$ ).
Leaving $L$ in cm, causing an answer off by a factor of $100$ .

Things to Be Careful About

The $\sin\phi$ factor: check whether the conductor is perpendicular to the field. Here it is, so $\sin\phi=1$ .
Significant figures: the data are mostly 2 s.f., so $6.4\times 10^{-2}\ \text{N}$ is appropriate.

Techniques used

apply the magnetic force on a current-carrying conductor equationconvert dimensions into SI unitsaccount for the effect of multiple turns on the total force

(ii)

Use your answer in (a)(i) to show that the torque $\tau$ on the coil is given by

\tau = 1.6 \times 10^{-3} \cos \theta\ \text{N m}.

DifficultyMedium-Easy

Worked solution

Working

The forces on sides $QR$ and $PS$ form a couple.

Perpendicular separation of the forces:

d = PQ\cos\theta = 0.025\cos\theta

Torque of the couple:

\tau = Fd = F\,(0.025\cos\theta)

Using $F = 6.4 \times 10^{-2}\ \text{N}$ from (a)(i):

\tau = (6.4 \times 10^{-2})(0.025)\cos\theta

\tau = 1.6 \times 10^{-3}\cos\theta\ \text{N m}

Answer

$\tau = 1.6 \times 10^{-3}\cos\theta\ \text{N m}$

Final answer

τ = 1.6 × 10^-3 cosθ N m

Detailed explanation

Background Concept

A coil in a uniform magnetic field experiences forces on its opposite sides. If these forces are equal in magnitude, opposite in direction, and act along different parallel lines, they form a couple.

The moment (torque) of a couple is

\tau = F \times d

where $F$ is the magnitude of one of the forces, and $d$ is the perpendicular distance between the lines of action of the two forces.

For a rectangular coil, the general result is

\tau = NBI A\cos\theta

if $\theta$ is defined as the angle between the plane of the coil and the magnetic field (because the usual formula $\tau = NBI A\sin\alpha$ uses $\alpha$ as the angle between the normal to the plane and $\vec{B}$ , and $\sin\alpha = \cos\theta$ ).

Understanding the Question

You have already found the force $F$ on side $QR$ in part (a)(i). Now you must use that to show the torque is

\tau = 1.6 \times 10^{-3} \cos\theta\ \text{N m}.

The coil width is $PQ = 2.5\ \text{cm} = 0.025\ \text{m}$ . The two forces act on the long sides ( $QR$ and $PS$ ) and produce a turning effect about axis $XY$ .

Approach

Treat the two magnetic forces as a couple.
Find the perpendicular separation $d$ between the lines of action. Because the coil plane is tilted by $\theta$ , the separation perpendicular to the force is reduced by a factor $\cos\theta$ .
Use $\tau = Fd$ and substitute the value of $F$ from (a)(i).

Step-by-Step Reasoning

diagram-a-ii-couple

The forces on $QR$ and $PS$ are equal ( $F$ ) and opposite, forming a couple.

The geometric separation between the sides is the coil width $PQ = 0.025\ \text{m}$ , but only the component perpendicular to the force contributes to the couple. From the end view, the perpendicular distance is

d = PQ\cos\theta = 0.025\cos\theta

Torque is then

\tau = Fd = F(0.025\cos\theta)

Using $F = 6.4\times 10^{-2}\ \text{N}$ from (a)(i):

\tau = (6.4\times 10^{-2})(0.025)\cos\theta

Multiply the numbers:

(6.4\times 10^{-2})(2.5\times 10^{-2}) = 16\times 10^{-4} = 1.6\times 10^{-3}

So

\tau = 1.6\times 10^{-3}\cos\theta\ \text{N m}

which is the required result.

Key Takeaways

Two equal and opposite forces on different lines of action produce a couple.
Torque of a couple is $\tau = Fd$ where $d$ is the perpendicular separation.
The $\cos\theta$ comes from geometry because $\theta$ is the angle between the plane of the coil and the field.

Common Mistakes

Using $\sin\theta$ instead of $\cos\theta$ (mixing up whether $\theta$ is to the plane or to the normal).
Using $d = PQ$ instead of $d = PQ\cos\theta$ .
Forgetting to convert $2.5\ \text{cm}$ to $0.025\ \text{m}$ .

Things to Be Careful About

Always use the perpendicular distance between the forces, not the direct distance between the sides.
Keep units consistent: $F$ in N, distance in m, so $\tau$ in N m.

Techniques used

model the forces on opposite sides of the coil as a coupleuse torque equals force times perpendicular separationsubstitute a previously calculated force value

(iii)

Using the expression in (a)(ii) sketch, on the axes of Fig. 6.2, a graph to show the variation of the torque $\tau$ with angle $\theta$ for values of $\theta$ between $0$ and $360^{\circ}$ . Label the $\tau$ axis with an appropriate scale.

Fig. 6.2

DifficultyMedium-Easy

Worked solution

Answer

Using

\tau = 1.6 \times 10^{-3} \cos\theta\ \text{N m},

the graph is a cosine curve with amplitude $1.6 \times 10^{-3}\ \text{N m}$ :

$\theta = 0^\circ$ : $\tau = +1.6 \times 10^{-3}\ \text{N m}$ (maximum)
$\theta = 90^\circ$ : $\tau = 0$
$\theta = 180^\circ$ : $\tau = -1.6 \times 10^{-3}\ \text{N m}$ (minimum)
$\theta = 270^\circ$ : $\tau = 0$
$\theta = 360^\circ$ : $\tau = +1.6 \times 10^{-3}\ \text{N m}$

Label the vertical axis (in units of $10^{-3}\ \text{N m}$ ) to at least $\pm 1.6$ .

Final answer

Cosine curve from +1.6×10^-3 at 0° to 0 at 90°, −1.6×10^-3 at 180°, 0 at 270°, back to +1.6×10^-3 at 360° (τ-axis scaled to ±1.6×10^-3 N m).

Detailed explanation

Background Concept

When a quantity depends on $\cos\theta$ , its graph against $\theta$ (in degrees) has the cosine shape:

maximum at $0^\circ$ ,
crosses zero at $90^\circ$ ,
minimum at $180^\circ$ ,
crosses zero at $270^\circ$ ,
returns to maximum at $360^\circ$ .

The amplitude is the coefficient in front of $\cos\theta$ .

Understanding the Question

You are told the torque is

\tau = 1.6 \times 10^{-3}\cos\theta\ \text{N m}

and must sketch $\tau$ against $\theta$ for $0^\circ$ to $360^\circ$ . The axes already show the key angles, and the $y$ -axis label is $\tau / 10^{-3}\ \text{N m}$ , so values should be plotted in multiples of $10^{-3}\ \text{N m}$ .

Approach

Identify the amplitude: $1.6\times 10^{-3}\ \text{N m}$ .
Calculate/recall cosine values at $0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ$ .
Plot those points and draw a smooth cosine curve through them.
Choose a vertical scale that clearly includes $+1.6$ and $-1.6$ on the provided axis units.

Step-by-Step Reasoning

Evaluate at standard angles:

\tau(0^\circ) = 1.6\times 10^{-3}(\cos 0^\circ) = 1.6\times 10^{-3}

\tau(90^\circ) = 1.6\times 10^{-3}(\cos 90^\circ) = 0

\tau(180^\circ) = 1.6\times 10^{-3}(\cos 180^\circ) = -1.6\times 10^{-3}

\tau(270^\circ) = 0, \qquad \tau(360^\circ)=+1.6\times 10^{-3}

On the axis labelled $\tau/10^{-3}\ \text{N m}$ these correspond to $+1.6, 0, -1.6, 0, +1.6$ .

Then sketch a smooth cosine curve through these points.

Key Takeaways

A cosine relationship produces one full cycle between $0^\circ$ and $360^\circ$ .
The coefficient sets the amplitude (maximum magnitude).
Use the axis label carefully: $\tau/10^{-3}\ \text{N m}$ means you plot numbers like $1.6$ , not $1.6\times 10^{-3}$ .

Common Mistakes

Drawing a sine curve (starting at zero) instead of a cosine curve (starting at maximum).
Getting the sign wrong at $180^\circ$ .
Using an unsuitable $y$ -axis scale that does not reach $\pm 1.6$ .

Things to Be Careful About

Ensure the curve is smooth and symmetric about $180^\circ$ .
The axis label implies a scaling; mark ticks clearly so the peak is at $1.6$ on that scaled axis.

Techniques used

interpret a trigonometric torque-angle relationshipidentify key angles where cosine is maximum, zero, or minimumchoose an appropriate axis scale for a sketch graph

(b)

The coil is now replaced by an identical coil wound on a ferrous core.

Suggest, with a reason, how the torque on this coil compares with the torque on the original coil.

DifficultyMedium-Easy

Worked solution

Answer

The torque is larger with the ferrous core.

Reason: the core increases the magnetic flux density $B$ (higher permeability concentrates the field), and

\tau \propto B.

Final answer

Torque increases because the ferrous core increases flux density B (higher permeability), and τ ∝ B.

Detailed explanation

Background Concept

For a current-carrying coil in a magnetic field, the maximum torque is

\tau = NBI A\cos\theta

For a given coil ( $N$ , $I$ , $A$ fixed) and a given angle $\theta$ , the torque is directly proportional to the magnetic flux density $B$ .

A ferrous (iron/steel) core has high relative permeability. It provides an easier path for magnetic field lines and can concentrate the magnetic field inside/through the coil, increasing the flux density experienced by the conductors.

Understanding the Question

The original coil is air-cored. It is replaced by an identical coil, but wound on a ferrous core. Nothing else is stated to change (same dimensions, turns, current, external field arrangement). You must compare torques and give a reason.

Approach

Use the proportionality $\tau \propto B$ and consider what a ferrous core does to the magnetic field/flux density threading the coil.

Step-by-Step Reasoning

In the torque expression, $N$ , $I$ , and $A$ are unchanged (identical coil).
A ferrous core increases the effective magnetic flux density within the coil region because its high permeability concentrates magnetic field lines.
Therefore, with a larger $B$ , the torque is larger for the same $\theta$ .

Key Takeaways

Torque on a coil depends linearly on $B$ .
Ferrous cores increase flux density due to high permeability (field lines preferentially pass through the core).

Common Mistakes

Saying the torque decreases because the core is heavier (mass is not in the torque formula here).
Claiming torque is unchanged because the external field is “uniform” (the core can still increase local $B$ in the coil).

Things to Be Careful About

In real devices, cores can saturate at high fields; exam questions usually assume it increases $B$ without saturation unless stated otherwise.
The comparison should be qualitative (“greater torque”) plus a physics reason tied to $B$ and permeability.

Techniques used

use proportionality of torque to magnetic flux densityrelate ferrous cores to increased permeability and flux density

Q7MediumOscillationsMagnetic Fields

A bar magnet is suspended from a spring. One pole of the magnet oscillates freely in a coil of wire, as shown in Fig. 7.1.

Fig. 7.1

The switch S is initially open.

(a)

The switch S is now closed. As a result, the oscillations of the magnet are lightly damped.

6M

(i)

State what is meant by damping.

DifficultyEasy

Worked solution

Answer

Damping is the loss of energy from an oscillating system to the surroundings, causing the amplitude of oscillation to decrease with time.

Final answer

Loss of energy from the oscillator to the surroundings so the amplitude decreases with time.

Detailed explanation

Background Concept

In oscillations, the system has mechanical energy that continually swaps between kinetic energy and potential energy. In an ideal (undamped) oscillator there is no energy loss, so the total energy stays constant and the amplitude stays constant.

Damping means there is a resistive effect (often a force opposite to the motion) that transfers energy out of the oscillating system into the surroundings. Because the total oscillation energy decreases, the amplitude decreases.

Understanding the Question

A bar magnet is hanging on a spring and oscillating up and down. The question asks for the meaning of “damping” in this context: what changes in the motion and what causes that change (in general terms).

Approach

State damping in terms of:

energy being dissipated from the oscillating system, and
a consequence that is easy to observe: amplitude reduces with time.

Step-by-Step Reasoning

If energy is removed from the oscillation each cycle, the maximum kinetic energy and maximum potential energy become smaller.
Maximum kinetic/potential energy in SHM is proportional to amplitude squared, so reduced energy means reduced amplitude.
Therefore damping is described as energy loss that causes amplitude to decrease with time.

Key Takeaways

Damping = energy dissipation from an oscillator.
Observable effect = amplitude decreases with time.

Common Mistakes

Saying only “amplitude decreases” without mentioning energy loss (often loses a mark).
Confusing damping with a change in frequency (frequency may change slightly, but that is not the definition).

Things to Be Careful About

The definition is about energy transfer out of the system; “friction” is just one possible cause.
Damping does not necessarily stop oscillations immediately; it depends on whether damping is light/heavy.

Techniques used

define damping in terms of energy loss from an oscillatorrelate amplitude decrease to dissipative forces

(ii)

Describe what is observed to indicate that the damping is light.

(iii)

By reference to electromagnetic induction and to conservation of energy, explain why the oscillations are damped.

DifficultyMedium

Worked solution

Answer

As the magnet moves in the coil, the magnetic flux linkage through the coil changes, so an emf is induced and a current flows in the closed circuit.

The induced current produces a magnetic field that opposes the change causing it (Lenz’s law), giving a force opposing the motion of the magnet.

Work is done against this opposing force; the magnet’s mechanical energy is transferred to electrical energy and is dissipated as thermal energy in the resistor, so the oscillation energy (and hence amplitude) decreases with time.

Final answer

Changing flux induces an emf and current; the induced field opposes the motion (Lenz), doing negative work, so mechanical energy is dissipated as heat in the resistor and the amplitude decreases.

Detailed explanation

Background Concept

Electromagnetic induction occurs when the magnetic flux linkage $N\Phi$ through a coil changes:

\varepsilon = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

$\varepsilon$ is the induced emf.
The minus sign is Lenz’s law: the induced effect opposes the change that produces it.

If the circuit is closed, the induced emf drives a current $I$ . That current in the coil produces its own magnetic field, which can exert a force on the moving magnet. This force typically acts like a resistive (opposing) force, so it removes mechanical energy from the oscillation.

By conservation of energy, the “lost” mechanical energy is not destroyed: it becomes electrical energy in the circuit and is finally dissipated as thermal energy (mainly in the resistor) at a rate $P = I^2 R$ .

Understanding the Question

With the switch open, the coil circuit is incomplete, so any induced emf cannot drive a sustained current (and the electromagnetic braking effect is very small).

When the switch is closed, the magnet continues to oscillate through the coil, but now the oscillations are damped. The question asks you to explain this damping specifically using:

electromagnetic induction (why current flows and what it does), and
conservation of energy (where the oscillation energy goes).

Approach

Argue that the moving magnet changes the magnetic flux linkage through the coil, so an induced emf appears (Faraday’s law).
Since the circuit is now closed, an induced current flows.
Use Lenz’s law to state that the induced current’s magnetic field opposes the motion / opposes the change in flux, producing a force opposite to the magnet’s velocity (magnetic braking).
Conclude that work is done against this opposing force, so mechanical energy is converted into electrical energy and then into heat in the resistor, reducing amplitude.

Step-by-Step Reasoning

diagram-7a-iii-1

As the magnet oscillates, it moves into and out of the coil. The magnetic field pattern from the magnet threads the turns of the coil.
When the magnet’s position changes, the flux $\Phi$ through the coil changes, so $N\Phi$ changes.
Therefore an emf is induced in the coil.
With switch $S$ closed, the circuit is complete, so the induced emf drives a current $I$ .
The current produces a magnetic field of the coil. By Lenz’s law, this induced field is in the direction that opposes the change in flux:
- if the magnet is moving further into the coil (flux increasing), the coil produces a field that opposes the magnet approaching,
- if the magnet is moving out (flux decreasing), the coil produces a field that tends to keep the flux from decreasing (again opposing the motion).
In both cases, the net effect is a force opposite to the magnet’s velocity, like a damping force.
Because there is a force opposing motion, the magnet must do work against it each cycle. That work comes from the oscillation’s mechanical energy.
The energy is transferred into the circuit as electrical energy and is dissipated as thermal energy in the resistor (and coil) due to Joule heating.
Hence the total mechanical energy decreases, so the amplitude of oscillation decreases with time (damping).

Key Takeaways

Motion of magnet in coil $\Rightarrow$ changing flux $\Rightarrow$ induced emf.
Closed circuit $\Rightarrow$ induced current.
Lenz’s law $\Rightarrow$ induced magnetic effect opposes motion (magnetic braking).
Energy conservation $\Rightarrow$ mechanical energy becomes heat in the resistor, so amplitude decays.

Common Mistakes

Mentioning “emf induced” but not stating that a current flows only because the switch is closed.
Forgetting Lenz’s law (must state the induced effect opposes the change / opposes the motion).
Saying energy is “lost” without stating it is dissipated as thermal energy in the resistor.
Claiming the induced current “helps” the motion (wrong direction).

Things to Be Careful About

The key is opposition to the change in flux; in an oscillation the direction reverses, so the induced current reverses too, but the force remains opposing the motion.
Do not confuse the induced current with a permanent current source; it only exists while flux is changing.
Energy transfer must be consistent with conservation of energy: mechanical energy decreases, thermal energy increases.

Techniques used

apply Faraday’s law to a changing magnetic flux linkageuse Lenz’s law to determine the direction of the induced effectlink electrical power dissipation to loss of mechanical energy via conservation of energyrelate a resistive force to reduction in amplitude of oscillations

(b)

The procedure in (a) is repeated after replacing the resistor with one of greater resistance.

Suggest, with a reason, the effect of this change on the oscillations.

DifficultyMedium-Easy

Worked solution

Answer

Greater resistance gives a smaller induced current (for the same induced emf), so the induced magnetic field and opposing force are smaller.

Therefore the damping is reduced: the magnet oscillates for longer and the amplitude decreases more slowly.

Final answer

Damping decreases; amplitude falls more slowly / oscillations last longer because higher $R$ gives smaller induced current and weaker opposing magnetic force.

Detailed explanation

Background Concept

When a magnet oscillates in a coil, an induced emf $\varepsilon$ is produced by the changing flux linkage. In a closed circuit, the induced current is approximately

I = \frac{\varepsilon}{R}

where $R$ is the total resistance of the circuit.

The induced current produces a magnetic field that opposes the motion (Lenz’s law). The strength of this opposing effect depends on how large the induced current is. Also, the rate at which energy is dissipated as heat is

P = I^2 R

So changing $R$ changes the current and therefore the damping strength.

Understanding the Question

The only change from part (a) is that the resistor has been replaced by one with a greater resistance. You are asked to suggest what happens to the oscillations and give a reason.

Approach

Use $I = \varepsilon/R$ to decide how current changes when $R$ increases.
Link smaller $I$ to a weaker induced magnetic field and hence a weaker opposing (damping) force.
Conclude how the observable oscillations change (amplitude decay rate).

Step-by-Step Reasoning

For the same magnet motion, the change of flux linkage with time is similar, so the induced emf $\varepsilon$ is of similar size.
Increasing the resistance increases $R$ , so the induced current decreases because $I = \varepsilon/R$ .
A smaller current produces a weaker magnetic field from the coil, so the magnetic braking force opposing the magnet’s motion is smaller.
With a smaller opposing force, less mechanical energy is transferred out of the oscillation per cycle.
Therefore the oscillations are less damped: the amplitude decreases more slowly and the magnet continues oscillating for more cycles.

Key Takeaways

Larger $R$ in the external circuit generally reduces induced current.
Reduced induced current reduces the Lenz-law opposing effect, so damping is smaller.

Common Mistakes

Stating “greater resistance means greater damping” without considering that the current becomes smaller.
Mixing up the ideas of emf and current: emf depends mainly on rate of change of flux, not on resistance.

Things to Be Careful About

The question asks for a qualitative effect (more/less damping) and a reason; a full numerical treatment is not needed.
Don’t claim the oscillation necessarily stops instantly; the key change is the rate of amplitude decay.

Techniques used

apply the idea that induced current depends on circuit resistancelink current magnitude to the strength of the induced opposing forcepredict qualitative changes to damping from energy dissipation rate

Q8MediumAlternating CurrentsCapacitance

An incomplete circuit diagram of a bridge rectifier is shown in Fig. 8.1.

Fig. 8.1

(a)

Complete Fig. 8.1 for the bridge rectifier such that the point A is at a positive potential with respect to point B.

DifficultyMedium-Easy

Worked solution

Answer

Arrange the diodes so that:

the two diodes connected to point $C$ have their cathodes at $C$ (anodes at the two a.c. input terminals),
the two diodes connected to point $D$ have their anodes at $D$ (cathodes at the two a.c. input terminals),
so that current through the load is from $A(\equiv C)$ to $B(\equiv D)$ for both half-cycles.

Final answer

Diodes oriented with both cathodes meeting at C (A) and both anodes meeting at D (B).

Detailed explanation

Background Concept

A bridge rectifier uses four diodes to make the current through a load flow in the same direction for both halves of an a.c. input cycle.

A diode conducts only when forward biased (anode at higher potential than cathode). In a bridge, two diodes conduct on each half-cycle, steering current so that the same output terminal is always at higher potential.

Understanding the Question

You are given a diamond-shaped bridge with two opposite corners connected to the a.c. supply. The other two corners are labelled $C$ and $D$ , and the load resistor is connected between $A$ and $B$ where $A$ is connected to $C$ and $B$ is connected to $D$ .

You must add diode symbols (with correct orientation) so that $A$ is always at a higher potential than $B$ , i.e. $C$ is the positive output terminal and $D$ is the negative output terminal.

Approach

Check each half-cycle separately:

Assume the left a.c. input terminal is positive relative to the right. Choose two diodes that will conduct so current goes through the load from $C$ to $D$ .
Then swap polarity (right positive relative to left). Choose the other pair of diodes that will conduct, again making current through the load go from $C$ to $D$ .
The diode orientations must satisfy both conduction paths.

Step-by-Step Reasoning

diagram-8a-1

Half-cycle 1: left a.c. terminal positive.

We want current to go: left a.c. terminal $\rightarrow C \rightarrow$ load $\rightarrow D \rightarrow$ right a.c. terminal.
So the diode from left a.c. terminal to $C$ must conduct when left is positive: anode at left, cathode at $C$ .
The diode from $D$ to right a.c. terminal must conduct for current to leave $D$ toward the right terminal: anode at $D$ , cathode at right.

Half-cycle 2: right a.c. terminal positive.

We want current to go: right a.c. terminal $\rightarrow C \rightarrow$ load $\rightarrow D \rightarrow$ left a.c. terminal.
So the diode from right a.c. terminal to $C$ must have anode at right, cathode at $C$ .
The diode from $D$ to left a.c. terminal must have anode at $D$ , cathode at left.

Combining these requirements:

Both diodes connected to $C$ must have their cathodes at $C$ .
Both diodes connected to $D$ must have their anodes at $D$ .
This ensures $C$ (and hence $A$ ) is always the positive output.

Key Takeaways

In a bridge rectifier, two diodes conduct per half-cycle.
To make $C$ the positive output, arrange the diodes so current is always delivered into $C$ and always leaves from $D$ .

Common Mistakes

Reversing one diode so that one half-cycle gives the correct polarity but the other half-cycle gives $B$ positive.
Placing both anodes at $C$ (this makes $C$ tend to become the negative output instead).

Things to Be Careful About

Always label anode/cathode consistently: current enters the diode at the anode and leaves at the cathode (when forward biased).
The required condition is about potential polarity ( $A$ positive relative to $B$ ), equivalently current direction through the load.

Techniques used

use diode forward-bias direction to determine current pathsanalyse current direction through the load for both half-cyclesorient diodes so the load current is unidirectional

(b)

The variation with time $t$ of the potential difference (p.d.) $V$ across the load resistor is shown in Fig. 8.2.

Fig. 8.2

A capacitor is now connected between points C and D of the bridge rectifier. This results in smoothing of the p.d. across the load resistor. The difference between the maximum and minimum values of the smoothed p.d. is $33\%$ of the peak p.d. $V_0$ .

8M

(i)

On Fig. 8.2, draw a line to show the variation of the potential difference $V$ across the load resistor with time $t$ . Your line should extend from $t = 0.5T$ to $t = 2.0T$ .

DifficultyMedium

Worked solution

diagram-8b1-1

Answer

From $t=0.5T$ to $2.0T$ , draw a smoothing curve that:

rises quickly to $V_0$ at each peak ( $t=0.5T,\ 1.0T,\ 1.5T,\ 2.0T$ ),
then decays approximately exponentially between peaks,
reaching a minimum of about $0.67V_0$ just before the next peak (since $V_{\max}-V_{\min}=0.33V_0$ ).

Final answer

Smoothed ripple: quick charge to V0 at each peak, exponential decay to about 0.67V0 before the next peak, repeating every 0.5T.

Detailed explanation

Background Concept

With a capacitor connected across the output of a rectifier, the capacitor charges up when the rectified p.d. increases (diodes conduct) and then discharges through the load resistor when the rectified p.d. falls (diodes stop conducting). This reduces the variation (ripple) of the output.

Charging: fast (small effective resistance through conducting diodes / source), so the voltage rises quickly to near the peak.
Discharging: slower, through the load resistor $R$ , following an exponential decay:

V = V_0 e^{-t/\tau}

where $\tau = RC$ .

For a full-wave rectified output, peaks occur every $T/2$ (twice the mains frequency), so the discharge time between recharge events is about $T/2$ .

Understanding the Question

Fig. 8.2 shows the full-wave rectified waveform across the load (no smoothing): repeated positive half-sine pulses with peak $V_0$ .

A capacitor is added across points $C$ and $D$ (i.e. across the load). You are told the ripple size is:

V_{\max} - V_{\min} = 0.33V_0

You must sketch the new (smoothed) output from $t=0.5T$ to $2.0T$ .

Approach

Identify the times of the peaks on the existing rectified waveform (every $0.5T$ ).
At each peak, show the capacitor charging quickly to approximately $V_0$ .
Between peaks, show exponential discharge through the load down to $V_{\min}$ .
Use the given ripple fraction to place $V_{\min}$ relative to $V_0$ .

Step-by-Step Reasoning

diagram-8b1-1

Take $V_{\max} \approx V_0$ (the capacitor charges to the peak).
Then

V_{\min} = V_{\max} - 0.33V_0 \approx V_0 - 0.33V_0 = 0.67V_0.

On the sketch:
- At $t=0.5T$ the curve should be at $V_0$ .
- From $0.5T$ to $1.0T$ draw a smooth exponential-like decay down to about $0.67V_0$ .
- At $t=1.0T$ show a rapid rise back up to $V_0$ .
- Repeat the same pattern from $1.0T$ to $1.5T$ and from $1.5T$ to $2.0T$ .

Key Takeaways

A smoothing capacitor charges near the peak and discharges between peaks.
Full-wave rectification halves the time between peaks to $T/2$ , improving smoothing.
Ripple size determines how far the discharge curve falls before the next recharge.

Common Mistakes

Drawing a straight-line discharge instead of a curved exponential decay.
Using a discharge interval of $T$ instead of $T/2$ for a full-wave rectified output.
Making the smoothed waveform drop below zero (it remains positive here).

Things to Be Careful About

The ripple is given as a fraction of $V_0$ , not as a fraction of the mean output.
The curve should touch (or almost touch) $V_0$ at each peak time and only fall to $0.67V_0$ before the next peak.

Techniques used

use capacitor charge-discharge behaviour to sketch a smoothed waveformidentify that peaks in full-wave rectification occur every half-periodshow exponential discharge between successive peaks

(ii)

Use your line in (b)(i) to determine, in terms of $T$ , the time constant of the smoothing circuit.

time constant = ______ $T$

DifficultyMedium

Worked solution

Working

For full-wave rectification, time between peaks is $\Delta t = 0.5T$ .

Ripple: $V_{\max}-V_{\min}=0.33V_0$ , with $V_{\max}\approx V_0$ so $V_{\min}\approx 0.67V_0$ .

Capacitor discharge:

V_{\min} = V_{\max} e^{-\Delta t/\tau}

0.67 = e^{-0.5T/\tau}

\tau = \frac{0.5T}{-\ln(0.67)} = 1.25T

Answer

$\tau \approx 1.25T$

Final answer

1.25T

Detailed explanation

Background Concept

When the rectified input drops below the capacitor voltage, the diodes stop conducting and the capacitor discharges through the load resistance $R$ . The voltage across a discharging capacitor follows

V = V_0 e^{-t/\tau}

where $\tau = RC$ is the time constant (time to fall to $1/e$ of its initial value).

In a full-wave rectifier, the output peaks occur twice per input cycle, so the time between charging peaks is $T/2$ .

Understanding the Question

You have drawn (or are meant to draw) a smoothed waveform where the capacitor charges to about the peak value $V_0$ and then discharges to a minimum before the next peak.

You are told that the ripple amplitude (difference between maximum and minimum smoothed p.d.) is $33\%$ of $V_0$ . You must use this to find the time constant in units of $T$ .

Approach

Convert “ $33\%$ of $V_0$ ” into a ratio $V_{\min}/V_{\max}$ .
Use the exponential discharge over the time between peaks ( $T/2$ ) to relate that ratio to $\tau$ .
Rearrange using a natural logarithm.

Step-by-Step Reasoning

If the capacitor charges to the peak, then $V_{\max} \approx V_0$ .

Given

V_{\max} - V_{\min} = 0.33V_0,

so

V_{\min} \approx V_0 - 0.33V_0 = 0.67V_0.

Hence

\frac{V_{\min}}{V_{\max}} \approx \frac{0.67V_0}{V_0} = 0.67.

For full-wave rectification, the capacitor is recharged every half-cycle, so the discharge time is

\Delta t = \frac{T}{2} = 0.5T.

Apply the discharge equation from $V_{\max}$ down to $V_{\min}$ :

V_{\min} = V_{\max} e^{-\Delta t/\tau}.

Divide by $V_{\max}$ :

0.67 = e^{-0.5T/\tau}.

Take natural logs:

\ln(0.67) = -\frac{0.5T}{\tau}.

So

\tau = \frac{0.5T}{-\ln(0.67)} \approx \frac{0.5T}{0.4005} \approx 1.25T.

Key Takeaways

Smoothing uses exponential discharge between successive peaks.
For a bridge (full-wave) rectifier, the relevant discharge time is $T/2$ .
A ripple percentage lets you find $V_{\min}/V_{\max}$ , which directly gives $\tau$ via logs.

Common Mistakes

Using $\Delta t = T$ instead of $T/2$ (this would roughly double the value of $\tau$ ).
Using $0.33$ as $V_{\min}/V_{\max}$ instead of recognising it is the difference.
Forgetting the negative sign when taking $\ln$ of a number less than 1.

Things to Be Careful About

The approximation $V_{\max} \approx V_0$ is standard for these questions (diode drop ignored).
Use $\ln$ (natural logarithm), not $\lg$ .
Keep $T$ symbolic until the end so the final answer is directly in multiples of $T$ .

Techniques used

read maximum and minimum values from a ripple waveformapply exponential discharge equation for a capacitorsolve for the time constant using natural logarithmsuse the time between successive peaks for full-wave rectification

(iii)

The resistance of the load resistor is now increased. The capacitance of the capacitor is unchanged.

State and explain the effect of this change on the smoothed output p.d.

DifficultyMedium-Easy

Worked solution

Answer

Increasing the load resistance increases the time constant $\tau = RC$ .

So the capacitor discharges more slowly between peaks, giving less ripple (minimum p.d. closer to $V_0$ ) and a smoother / larger mean output p.d.

Final answer

Ripple decreases; output is smoother (Vmin closer to V0) because increasing R increases \tau = RC so the capacitor discharges more slowly.

Detailed explanation

Background Concept

In a smoothing circuit, the capacitor discharges through the load resistance $R$ when the rectified input falls below the capacitor voltage. The discharge rate is set by the time constant

\tau = RC.

A larger $\tau$ means a slower exponential decay, so the voltage falls less in a given time.

Understanding the Question

Only the load resistance is changed (increased). The capacitance is unchanged. You are asked for the effect on the smoothed output p.d. and an explanation.

Approach

Use $\tau = RC$ to see how the time constant changes.
Relate time constant to how much the capacitor voltage drops between peaks.
Translate that into ripple size and smoothness of the output.

Step-by-Step Reasoning

If $R$ increases and $C$ is constant, then

\tau = RC \text{ increases.}

Between successive peaks (time roughly $T/2$ for a full-wave rectifier), the capacitor voltage decays like

V = V_0 e^{-t/\tau}.

With a larger $\tau$ , the factor $e^{-t/\tau}$ is closer to 1 for the same $t$ , so the voltage drops by a smaller amount.

Therefore:

$V_{\min}$ increases (it does not fall as far before the next recharge),
the ripple amplitude $V_{\max}-V_{\min}$ decreases,
the output is smoother and its mean value is higher.

Key Takeaways

Increasing $R$ increases $\tau$ .
Larger $\tau$ means slower discharge, so the capacitor maintains the output voltage better between peaks.

Common Mistakes

Stating that ripple increases when $R$ increases (it is the opposite).
Talking about the capacitor charging slower: in smoothing, the key change is the discharge through the load.

Things to Be Careful About

The smoothing capacitor is across the load, so the relevant resistance for discharge is the load (or effective load) resistance.
Make sure you describe a physical effect (smaller ripple / higher minimum voltage), not just “time constant increases”.

Techniques used

use the relation \tau = RClink larger time constant to slower exponential dischargeinfer effect on ripple magnitude and minimum output voltage

Q9MediumQuantum Physics

(a)

4M

(i)

Describe what is meant by wave–particle duality.

DifficultyMedium-Easy

Worked solution

Answer

Wave–particle duality is the idea that the same entity (e.g. an electron or photon) can show wave behaviour (e.g. interference/diffraction) and also particle behaviour (e.g. localised impacts / quantised energy transfer), depending on the experiment.

Final answer

The same entity can exhibit both wave properties (e.g. diffraction/interference) and particle properties (localised impacts/quantised transfer), depending on the experiment.

Detailed explanation

Background Concept

In classical physics, waves and particles are distinct:

Waves spread out and can superpose, producing interference and diffraction patterns.
Particles are localised objects that transfer energy and momentum in discrete interactions and arrive as individual “hits”.

Quantum physics shows that microscopic objects (photons, electrons, neutrons, atoms) do not fit purely into one category.

Understanding the Question

You are asked to describe what is meant by wave–particle duality. For 2 marks, you typically need:

a statement that the same entity can behave as a wave, and
a statement that it can behave as a particle (often with an example of each).

Approach

Give a clear definition and support it with examples of the types of behaviour that are considered wave-like and particle-like.

Step-by-Step Reasoning

State that a single quantum object (such as an electron) can show wave effects.
- Suitable wave effects are diffraction and interference.
State that the same object can also show particle effects.
- Suitable particle effects are localised detection events (individual spots on a screen) and quantised interactions (e.g. photons transferring energy in packets).
Mention that which behaviour is observed depends on the measurement/experiment.

Key Takeaways

Wave–particle duality: one object, two kinds of behaviour.
Wave behaviour: diffraction/interference.
Particle behaviour: localised impacts/quantised exchanges.

Common Mistakes

Saying only “electrons are waves” or only “electrons are particles” (needs both).
Referring to two different objects (e.g. “light is a wave and electrons are particles”) instead of the same entity.

Things to Be Careful About

Use examples that are unambiguously wave-like (diffraction/interference) and particle-like (localised hits/quantised transfer).
Keep the statement general: “depending on the experiment/observation.”

Techniques used

state a definition using contrasting observable propertieslink behaviour to the type of experiment or observation

(ii)

State the relationship between the de Broglie wavelength $\lambda$ of a particle and its momentum $p$ . State the meaning of any other symbols that you use.

(b)

A narrow beam of electrons, all with the same speed, is incident normally on a carbon film. The electrons then move on to a fluorescent screen, as illustrated in Fig. 9.1.

Fig. 9.1

The apparatus is in a vacuum. The pattern produced on the screen is shown in Fig. 9.2.

Fig. 9.2

4M

(i)

Explain why the pattern in Fig. 9.2 provides experimental evidence to indicate a wave nature for the electrons.

DifficultyMedium-Easy

Worked solution

Answer

The concentric bright and dark rings are a diffraction / interference pattern. Diffraction (and the resulting constructive/destructive interference) is a wave effect, so the electrons must have a wave nature.

Final answer

The rings are a diffraction/interference pattern (constructive and destructive interference), which is a wave effect, so electrons show wave nature.

Detailed explanation

Background Concept

A hallmark of waves is superposition: when waves overlap they interfere.

Constructive interference gives maxima (bright regions).
Destructive interference gives minima (dark regions).

When a wave passes through a thin crystal (or polycrystalline film), it can be scattered by regularly spaced атомs. The scattered waves can interfere, producing a pattern of maxima and minima. This is the basis of diffraction patterns (e.g. X-ray diffraction). If electrons produce the same kind of pattern, they must be behaving as waves.

Understanding the Question

Electrons of one speed strike a carbon film and then hit a fluorescent screen. The observed result is concentric bright and dark rings.
The question asks why this is evidence of a wave nature for electrons.

Approach

Identify what physical phenomenon creates rings (diffraction/interference), then state why that phenomenon is wave-specific.

Step-by-Step Reasoning

A polycrystalline carbon film has many tiny crystallites with different orientations.
Electron waves are scattered by atomic planes in many directions.
For certain scattering angles, the path difference between waves scattered from successive planes satisfies a constructive condition (often described by Bragg-type reasoning), producing intensity maxima.
Because many crystallite orientations are present, these maxima occur on cones of angle; when they hit a flat screen, they appear as rings.
Since diffraction and interference require wave superposition, the ring pattern is direct experimental evidence that electrons have an associated wavelength.

Key Takeaways

A pattern of alternating maxima/minima (bright/dark) indicates interference.
Interference/diffraction is a defining property of waves.
Therefore, electrons exhibit wave behaviour.

Common Mistakes

Saying “electrons are waves because they make rings” without mentioning diffraction/interference.
Claiming the pattern is due to electrostatic or magnetic deflection (those typically shift the whole beam rather than create multiple maxima/minima).

Things to Be Careful About

Use the correct wave terminology: diffraction/interference, constructive/destructive.
The vacuum is important experimentally (reduces collisions), but the main evidence comes from the diffraction pattern itself, not from the vacuum.

Techniques used

identify diffraction/interference features in an observed patternlink a ring pattern to constructive and destructive interferenceargue that diffraction is evidence of wave behaviour

(ii)

The speed of the electrons is increased.

Suggest, with a reason, how this change affects the pattern observed on the screen.

DifficultyMedium

Worked solution

Working

Increasing speed increases momentum $p$ .

\lambda = \frac{h}{p}

So $p$ increases $\Rightarrow \lambda$ decreases.

Smaller $\lambda$ gives smaller diffraction angles, so the ring radii (and ring spacing) on the screen decrease.

Answer

The rings move closer to the centre / become smaller (reduced spacing) because the de Broglie wavelength decreases when the electron speed increases.

Final answer

Ring radii (and spacing) decrease: higher speed → larger momentum → smaller de Broglie wavelength → smaller diffraction angle.

Detailed explanation

Background Concept

For electrons,

\lambda = \frac{h}{p}

So when momentum $p$ increases, wavelength $\lambda$ decreases.

Diffraction angles depend on wavelength: for a fixed spacing in the scattering structure (atomic spacing in the carbon), a smaller wavelength produces smaller diffraction angles. Geometrically, on a screen at distance $L$ , the ring radius $R$ is related to the scattering angle $\theta$ approximately by $R \approx L\tan\theta \approx L\theta$ for small angles. So smaller $\theta$ means smaller $R$ .

Understanding the Question

The electron beam and carbon film produce a ring diffraction pattern on a fluorescent screen. You increase the speed of the electrons and must predict what happens to the pattern and give a reason.

Approach

Link speed increase to momentum increase.
Use de Broglie to deduce wavelength decrease.
Use the idea “smaller wavelength → less diffraction (smaller angles)” to predict the ring pattern change.

Step-by-Step Reasoning

Increasing electron speed $v$ increases momentum $p$ (non-relativistically $p = mv$ ).
From de Broglie:

\lambda = \frac{h}{p}

so increasing $p$ makes $\lambda$ smaller.

Diffraction/interference conditions in the carbon film depend on wavelength; smaller $\lambda$ corresponds to smaller scattering angle $\theta$ for maxima.
On the screen, each diffraction angle maps to a ring radius.

diagram-bii-1

Therefore the rings contract: smaller ring diameters (and typically rings are closer together).

Key Takeaways

Speed up electrons → momentum increases → de Broglie wavelength decreases.
Smaller wavelength → reduced diffraction angles → smaller ring radii on the screen.

Common Mistakes

Predicting the rings spread out when speed increases (it is the opposite).
Forgetting the reason must explicitly mention de Broglie wavelength or momentum.

Things to Be Careful About

State the chain clearly: $v\uparrow \Rightarrow p\uparrow \Rightarrow \lambda\downarrow \Rightarrow \theta\downarrow \Rightarrow R\downarrow$ .
If you mention “less diffraction”, translate it into what is seen: rings move inward / radii decrease.

Techniques used

relate change in speed to change in momentumapply the de Broglie relationship to infer wavelength changelink smaller wavelength to reduced diffraction angle and ring size

Q10Medium-EasyMedical Physics

(a)

Describe how the piezoelectric crystal in a transducer generates ultrasound waves for use in medical diagnosis.

DifficultyMedium-Easy

Worked solution

Answer

An alternating p.d. is applied across the piezoelectric crystal.
The crystal repeatedly expands and contracts (changes thickness) at the same frequency.
This makes the face of the transducer vibrate and produce longitudinal pressure variations in the adjacent medium, i.e. ultrasound waves.

Final answer

An a.c. p.d. across the piezoelectric crystal makes it alternately expand/contract, so the transducer face vibrates at the driving frequency and generates longitudinal pressure waves (ultrasound) in the medium.

Detailed explanation

Background Concept

The piezoelectric effect is the coupling between электрical and mechanical behaviour in certain crystals.

Direct effect: mechanical stress on the crystal produces a charge separation and hence a p.d.
Inverse effect: an applied p.d. produces a change in the crystal dimensions.

In an ultrasound transducer used for imaging, we use the inverse piezoelectric effect to generate ultrasound. If the crystal face vibrates, it pushes and pulls on the material in contact with it, creating longitudinal pressure variations that propagate as an ultrasound wave.

Understanding the Question

The question asks specifically how the piezoelectric crystal generates ultrasound for diagnosis (so: electrical drive (\rightarrow) crystal vibration (\rightarrow) ultrasound in tissue/coupling gel). The key ideas needed are:

apply an alternating voltage,
the crystal changes shape periodically,
those vibrations launch a longitudinal wave.

Approach

State the cause-and-effect chain that creates a sound wave:

apply an a.c. (or rapid pulses) p.d. to the crystal,
inverse piezoelectric effect makes the crystal expand/contract,
the oscillating face drives the medium, creating pressure oscillations (ultrasound).

Step-by-Step Reasoning

Alternating voltage applied: The transducer electronics apply an alternating p.d. (often as short pulses) across electrodes on the crystal.
Crystal deforms: Because it is piezoelectric, the applied electric field causes the crystal lattice to distort. When the p.d. reverses, the distortion reverses, so the crystal repeatedly increases and decreases in thickness.
Vibrations produce ultrasound: The front face of the crystal is in contact with a coupling medium (gel) and then tissue. As the face moves back and forth, it creates alternating compressions and rarefactions in the medium. These propagate away as a longitudinal wave at ultrasound frequency.

(Extra context: the transducer is often designed so the driving frequency matches a mechanical resonance of the crystal to increase amplitude, but the question only requires the basic generation mechanism.)

Key Takeaways

Ultrasound is produced by mechanical vibration of a surface.
A piezoelectric crystal converts an alternating electrical signal into alternating mechanical deformation.
The oscillation launches longitudinal pressure waves in the contacting medium.

Common Mistakes

Describing the direct piezoelectric effect (pressure (\rightarrow) voltage) instead of generation (voltage (\rightarrow) vibration).
Saying the wave is transverse (ultrasound in tissue is longitudinal).
Forgetting that it is the alternating nature of the applied p.d. that produces continuous oscillation.

Things to Be Careful About

Use correct language: “expand/contract”, “vibrate/oscillate”, “compressions/rarefactions”.
The ultrasound frequency is the same as the driving frequency (or pulse repetition determines bursts), so make that link explicitly if needed.

Techniques used

link an alternating potential difference to periodic mechanical deformation via the piezoelectric effectrelate the vibration frequency of a source to the frequency of the emitted sounddescribe production of longitudinal ultrasound waves from an oscillating surface

(b)

A parallel ultrasound beam is incident on the boundary between two media, as illustrated in Fig. 10.1.

Fig. 10.1

The media have specific acoustic impedances $Z_1$ and $Z_2$ .

At the boundary, a fraction $\alpha$ of the incident intensity of the ultrasound beam is reflected. The remainder is transmitted.

4M

(i)

State what is meant by specific acoustic impedance.

DifficultyMedium-Easy

Worked solution

Answer

Specific acoustic impedance $Z$ of a medium is the ratio of acoustic pressure to particle velocity in the wave:

Z = \frac{p}{u}

and for a plane wave

Z = \rho c

Final answer

Specific acoustic impedance is the ratio of acoustic pressure to particle velocity (for a plane wave, Z = ρc).

Detailed explanation

Background Concept

For a travelling plane sound wave, particles of the medium oscillate back and forth while the wave carries energy forward. Two useful wave quantities are:

acoustic pressure (pressure variation) (p) (units (\text{Pa})),
particle velocity (u) (units (\text{m s}^{-1})).

The specific acoustic impedance (Z) describes how “hard” it is for the wave to make the particles move:

Z = \frac{p}{u}

For a plane wave in a uniform medium, it can also be shown that

Z = \rho c

where (\rho) is density and (c) is wave speed in that medium. Unit check: (\rho c) has units (\text{kg m}^{-3} \times \text{m s}^{-1} = \text{kg m}^{-2} \text{s}^{-1}), which is equivalent to (\text{Pa s m}^{-1}).

Understanding the Question

You are asked to state what is meant by specific acoustic impedance, so you need a clear definition (usually ratio form), and an accepted equivalent expression (often (\rho c)).

Approach

Give the definition and, if possible, the common plane-wave relation:

definition: (Z = p/u),
relation to material properties: (Z = \rho c).

Step-by-Step Reasoning

Start from the definition: impedance is a ratio of a “driving” quantity to a “response” quantity.
In acoustics, the “drive” is the pressure variation (p) and the response is particle speed (u), hence (Z = p/u).
For plane waves in a medium, pressure and particle speed are linked through the medium’s density and sound speed, giving (Z = \rho c).

Key Takeaways

(Z) measures opposition of a medium to particle motion in a sound wave.
(Z) can be defined by (Z = p/u) and (for plane waves) equals (\rho c).

Common Mistakes

Confusing impedance with attenuation (they are different: impedance affects reflection/transmission at boundaries).
Writing (Z = \rho / c) (wrong: it is (\rho c)).
Missing the idea of it being a ratio of pressure to particle velocity.

Things to Be Careful About

Use particle velocity (u), not wave speed (c), in the ratio definition.
State clearly it is a property of the medium for plane waves (not of the source).

Techniques used

state a physics definition in terms of a ratio of two wave quantitiesrelate specific acoustic impedance to material properties

(ii)

Describe how $\alpha$ depends on the relative values of $Z_1$ and $Z_2$ .

DifficultyMedium-Easy

Worked solution

Answer

If $Z_1 = Z_2$ then $\alpha = 0$ (no reflection).

As the difference between $Z_1$ and $Z_2$ increases, $\alpha$ increases (greater fraction reflected), tending towards $1$ for a very large mismatch.

(For normal incidence:)

\alpha = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2

Final answer

α is zero when Z1 = Z2 and increases as the mismatch |Z2 − Z1| increases, approaching 1 for a very large mismatch (α = ((Z2 − Z1)/(Z2 + Z1))^2 for normal incidence).

Detailed explanation

Background Concept

When a wave meets a boundary between two media, part of it can be reflected and part transmitted. The amount reflected depends on how well the two media “match” in terms of specific acoustic impedance (Z).

For normal (perpendicular) incidence, the intensity reflection coefficient (\alpha) is

\alpha = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2

This is a squared ratio, so (\alpha) is always between 0 and 1.

Understanding the Question

A parallel beam hits the boundary between media with impedances (Z_1) and (Z_2). You are asked to describe how the reflected fraction (\alpha) depends on the relative values of (Z_1) and (Z_2), so the key is “impedance mismatch”.

diagram-bii-1

Approach

Use limiting cases and/or the standard formula:

if impedances are equal, expect no reflection,
larger mismatch gives more reflection,
very large mismatch gives almost total reflection.

Step-by-Step Reasoning

Start with the formula for normal incidence:

\alpha = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2

Case (Z_1 = Z_2): numerator ((Z_2 - Z_1) = 0), so

\alpha = 0

meaning all intensity is transmitted (ideal matching).

As (|Z_2 - Z_1|) increases: the numerator grows relative to the denominator, so the fraction increases and hence (\alpha) increases.
Very large mismatch: if (Z_2 \gg Z_1), then

\frac{Z_2 - Z_1}{Z_2 + Z_1} \approx \frac{Z_2}{Z_2} = 1

so (\alpha \to 1) (almost all intensity reflected). Similarly if (Z_1 \gg Z_2).

Key Takeaways

Reflection at a boundary is controlled by impedance mismatch.
(Z_1 = Z_2) gives no reflection.
The larger the mismatch, the larger the reflected fraction, approaching 1 for an extreme mismatch.

Common Mistakes

Saying reflection depends on the absolute size of (Z) rather than the difference between (Z_1) and (Z_2).
Claiming (\alpha) could be greater than 1 (it cannot, it is a fraction of intensity).
Forgetting that (\alpha) is the fraction of intensity, not amplitude.

Things to Be Careful About

The question states a fraction of intensity, so the dependence is squared compared with many amplitude reflection formulas.
Ensure you mention the special case (Z_1 = Z_2) explicitly; it is a common marking point.

Techniques used

use qualitative reasoning about impedance mismatch at a boundaryapply the intensity reflection coefficient dependence on impedanceidentify limiting cases for equal impedances and large mismatch

(c)

A parallel ultrasound beam of intensity $I_0$ enters a region of soft tissue. After passing a distance of $2.1\ \text{cm}$ through this tissue, the intensity of the ultrasound is $0.62I_0$ .

Calculate the linear attenuation coefficient $\mu$ of ultrasound in the soft tissue. Give a unit with your answer.

$\mu$ = ______ unit ______

DifficultyMedium-Easy

Worked solution

Working

Use

I = I_0 e^{-\mu x}

Given $I = 0.62 I_0$ and $x = 2.1\ \text{cm} = 0.021\ \text{m}$ :

0.62 = e^{-\mu (0.021)}

\mu = -\frac{\ln(0.62)}{0.021} = 2.28 \times 10^{1}\ \text{m}^{-1}

Answer

$\mu = 23\ \text{m}^{-1}$

Final answer

23 m^-1

Detailed explanation

Background Concept

As ultrasound travels through tissue, its intensity decreases because energy is absorbed and scattered. This is modelled by exponential attenuation:

I = I_0 e^{-\mu x}

where

(I_0) is the intensity at the start of the material,
(I) is the intensity after travelling distance (x),
(\mu) is the linear attenuation coefficient.

The unit of (\mu) must be the inverse of distance (e.g. (\text{m}^{-1}) if (x) is in metres).

Understanding the Question

You are told that after travelling (2.1\ \text{cm}) in soft tissue, the intensity is (0.62 I_0). You must calculate (\mu) and state a suitable unit.

Known:

(I/I_0 = 0.62)
(x = 2.1\ \text{cm})

Unknown:

(\mu)

Approach

Start from (I = I_0 e^{-\mu x}).
Divide by (I_0) to use the ratio (I/I_0).
Take (\ln) to bring (\mu) out of the exponent.
Convert (x) to SI units and compute (\mu).

Step-by-Step Reasoning

Write the attenuation law and form a ratio:

\frac{I}{I_0} = e^{-\mu x}

Substitute (I/I_0 = 0.62) and convert distance:

0.62 = e^{-\mu (2.1\ \text{cm})}

Using SI:

2.1\ \text{cm} = 2.1 \times 10^{-2}\ \text{m} = 0.021\ \text{m}

So

0.62 = e^{-\mu (0.021)}

Take natural logs:

\ln(0.62) = -\mu (0.021)

Rearrange for (\mu):

\mu = -\frac{\ln(0.62)}{0.021}

Calculate: (\ln(0.62) \approx -0.478), so

\mu \approx \frac{0.478}{0.021} \approx 22.8\ \text{m}^{-1}

Rounded appropriately:

\mu \approx 23\ \text{m}^{-1}

Key Takeaways

Intensity attenuation in tissue is exponential: (I = I_0 e^{-\mu x}).
Taking (\ln) is the standard way to rearrange exponential decay.
The unit of (\mu) is the inverse of the distance unit used for (x).

Common Mistakes

Using base-10 log instead of natural log (must use (\ln) for (e^{-\mu x})).
Forgetting to convert (2.1\ \text{cm}) to metres when giving (\mu) in (\text{m}^{-1}).
Dropping the minus sign: since (\ln(0.62)) is negative, (\mu) must come out positive.

Things to Be Careful About

Be consistent: if you keep (x) in cm, you must give (\mu) in (\text{cm}^{-1}). If you convert to metres, use (\text{m}^{-1}).
Quote a sensible number of significant figures (matching the given data, typically 2 s.f. here).

Techniques used

apply exponential attenuation of intensity with distancerearrange an exponential equation using natural logarithmsconvert lengths to SI units and assign the correct unit to a coefficient

Q11MediumNuclear PhysicsAstronomy and Cosmology

The deuterium nucleus ( $^2_1\text{H}$ ) has a mass defect of $0.002388\ \text{u}$ . The helium-4 nucleus ( $^4_2\text{He}$ ) has a mass defect of $0.030377\ \text{u}$ . Helium-4 is formed from deuterium in a nuclear reaction that can be represented by the equation

3\ ^2_1\text{H} \longrightarrow\ ^4_2\text{He} + \ ^1_1\text{p} + \ ^1_0\text{n}.

(a)

4M

(i)

State the name of this type of nuclear reaction.

(ii)

Show that the energy released when one nucleus of helium-4 is formed from deuterium is $3.47 \times 10^{-12}\ \text{J}$ .

DifficultyMedium-Easy

Worked solution

Working

Binding energy $E_b = (\Delta m)c^2$ .

Initial total mass defect:

\Delta m_i = 3(0.002388\ \text{u}) = 0.007164\ \text{u}

Final total mass defect (only the helium nucleus is bound):

\Delta m_f = 0.030377\ \text{u}

Increase in mass defect:

\Delta m = \Delta m_f - \Delta m_i = 0.030377 - 0.007164 = 0.023213\ \text{u}

Convert to energy ( $1\ \text{u} = 1.66\times 10^{-27}\ \text{kg}$ , $c = 3.00\times 10^8\ \text{m s}^{-1}$ ):

E = (0.023213)(1.66\times 10^{-27})(3.00\times 10^8)^2

E = 3.47\times 10^{-12}\ \text{J}

Answer

$3.47\times 10^{-12}\ \text{J}$ .

Final answer

3.47 × 10^−12 J

Detailed explanation

Background Concept

A mass defect for a nucleus is the difference between:

the total mass of its separated nucleons, and
the actual mass of the nucleus.

This “missing” mass corresponds to the binding energy of the nucleus:

E_b = (\Delta m)c^2.

If a reaction produces nuclei with a greater total binding energy than the reactants, the difference is released as energy.

Understanding the Question

You are given the mass defects:

Deuterium nucleus: $0.002388\ \text{u}$
Helium-4 nucleus: $0.030377\ \text{u}$

The reaction is:

3\ ^2_1\text{H} \longrightarrow\ ^4_2\text{He} + \ ^1_1\text{p} + \ ^1_0\text{n}.

A proton and a neutron on the product side are free nucleons, so they have no nuclear binding energy to include. The energy released is the increase in total binding energy from reactants to products.

Approach

Convert each nucleus’s mass defect into a binding energy conceptually (you can keep everything in u until the end).
Find the total mass defect (i.e. total binding) initially and finally.
The energy released is:

E = (\Delta m_f - \Delta m_i)c^2.

Convert u to kg and apply $E = mc^2$ .

Step-by-Step Reasoning

1) Total initial binding (in u):
There are 3 deuterium nuclei, each with mass defect $0.002388\ \text{u}$ :

\Delta m_i = 3\times 0.002388 = 0.007164\ \text{u}.

2) Total final binding (in u):
Only the helium-4 nucleus is bound:

\Delta m_f = 0.030377\ \text{u}.

3) Increase in binding (in u):

\Delta m = \Delta m_f - \Delta m_i = 0.030377 - 0.007164 = 0.023213\ \text{u}.

This increase corresponds to energy released.

4) Convert to joules:
Using $1\ \text{u} = 1.66\times10^{-27}\ \text{kg}$ ,

\Delta m = 0.023213\times 1.66\times 10^{-27} = 3.85\times 10^{-29}\ \text{kg} \ \text{(approx)}

Then

E = \Delta m c^2 = (3.85\times 10^{-29})(3.00\times 10^8)^2 \approx 3.47\times 10^{-12}\ \text{J}.

Key Takeaways

Mass defect corresponds to nuclear binding energy via $E = mc^2$ .
Energy released in a fusion reaction is the increase in total binding energy.
Free protons/neutrons contribute no nuclear binding energy.

Common Mistakes

Subtracting the wrong way round (using $\Delta m_i-\Delta m_f$ gives a negative value).
Adding mass defects of the free proton and neutron (they are not bound nuclei).
Forgetting to convert u to kg before using $E = mc^2$ .

Things to Be Careful About

Keep track of what the given “mass defect” represents: it is already the “missing mass” for that nucleus.
Use consistent significant figures; the final value is typically quoted to 3 s.f. here.
Ensure you use $c = 3.00\times 10^8\ \text{m s}^{-1}$ and square it correctly.

Techniques used

use mass defect to determine nuclear binding energycalculate change in total binding energy between reactants and productsconvert atomic mass unit to kilograms and apply E = mc^2

(b)

A star has a radius of $6.96 \times 10^8\ \text{m}$ . Helium-4 is produced in this star, from deuterium, at a mass rate of $7.34 \times 10^{11}\ \text{kg s}^{-1}$ . All the energy released from this process is radiated away from the star. All the energy that is radiated from the star is released by this process.

5M

(i)

Calculate the luminosity of the star.

luminosity = ______ $\text{W}$

DifficultyMedium

Worked solution

Working

Energy released per helium nucleus formed (from (a)(ii)):

E = 3.47\times 10^{-12}\ \text{J}

Mass of one helium-4 nucleus:

m_{\text{He}} \approx 4\ \text{u} = 4(1.66\times 10^{-27}) = 6.64\times 10^{-27}\ \text{kg}

Number of helium nuclei formed per second:

\dot{N} = \frac{7.34\times 10^{11}}{6.64\times 10^{-27}} = 1.11\times 10^{38}\ \text{s}^{-1}

Luminosity $L$ (energy per second):

L = \dot{N}E = (1.11\times 10^{38})(3.47\times 10^{-12}) = 3.83\times 10^{26}\ \text{W}

Answer

$3.83\times 10^{26}\ \text{W}$ .

Final answer

3.83 × 10^26 W

Detailed explanation

Background Concept

Luminosity $L$ is the total energy emitted per unit time (power) by a star:

L = \frac{\Delta E}{\Delta t}.

If each nuclear reaction releases energy $E$ and happens at a rate of $\dot{N}$ reactions per second, then:

L = \dot{N}E.

Here the reaction rate is not given directly, but the mass production rate of helium is given. If helium is produced at a mass rate $\dot{m}$ , then

\dot{N} = \frac{\dot{m}}{m_{\text{one nucleus}}}.

Understanding the Question

Given:

Helium-4 is produced at mass rate $\dot{m} = 7.34\times 10^{11}\ \text{kg s}^{-1}$ .
Energy released per helium-4 nucleus formed (from part a(ii)) is $E = 3.47\times 10^{-12}\ \text{J}$ .
All this energy is radiated away, and all radiated energy comes from this process, so luminosity equals nuclear power output.

We must find $L$ .

Approach

Find the mass of one helium-4 nucleus (use $4\ \text{u}$ as its mass).
Convert the mass production rate into number of helium nuclei produced per second.
Multiply by energy released per helium nucleus.

Step-by-Step Reasoning

1) Mass of one helium nucleus
Take approximately $4\ \text{u}$ :

m_{\text{He}} = 4(1.66\times 10^{-27}) = 6.64\times 10^{-27}\ \text{kg}.

2) Number produced per second

\dot{N} = \frac{\dot{m}}{m_{\text{He}}} = \frac{7.34\times 10^{11}}{6.64\times 10^{-27}} \approx 1.11\times 10^{38}\ \text{s}^{-1}.

3) Convert to luminosity (power)

L = \dot{N}E = (1.11\times 10^{38})(3.47\times 10^{-12}) \approx 3.83\times 10^{26}\ \text{W}.

The unit check is helpful: $\text{s}^{-1}\times\text{J} = \text{J s}^{-1} = \text{W}$ .

Key Takeaways

Luminosity is power output.
Use $L = \dot{N}E$ if you know the energy per reaction and the reaction rate.
Convert a mass rate to a particle rate using $\dot{N} = \dot{m}/m$ .

Common Mistakes

Using $m = 4\times 10^{-27}\ \text{kg}$ instead of $4\times 1.66\times 10^{-27}\ \text{kg}$ .
Forgetting that the mass rate refers to helium produced, not deuterium consumed.
Using the mass defect (in u) directly as a mass of the nucleus.

Things to Be Careful About

Keep powers of ten under control when dividing by $10^{-27}$ .
Use the energy per one helium nucleus formed, not per deuterium nucleus.
Quote the final luminosity to appropriate significant figures (typically 3 s.f.).

Techniques used

convert a mass production rate to a particle production ratecalculate power from energy per reaction and number of reactions per seconduse unit analysis to check consistency of the final luminosity

(ii)

Use your answer in (b)(i) to determine the surface temperature of the star.

temperature = ______ $\text{K}$