Physics 9702 / 42 — October/November 2024

DifficultyMedium-Easy

Worked solution

Working

v = r\omega

v = 0.85 \times 140 = 1.19 \times 10^{2}\ \text{m s}^{-1}

Answer

v = 1.2 \times 10^{2}\ \text{m s}^{-1}

Final answer

1.2 × 10^2 m s^-1

Detailed explanation

Background Concept

For motion in a circle, a point at radius $r$ has tangential (linear) speed $v$ related to angular speed $\omega$ by

v = r\omega

Here:

$\omega$ is in $\text{rad s}^{-1}$ ,
$r$ is in $\text{m}$ ,
$v$ comes out in $\text{m s}^{-1}$ .

Understanding the Question

Point X is on the rim, so its distance from the axle is the wheel radius $r = 0.85\ \text{m}$ . The wheel rotates with angular speed $\omega = 140\ \text{rad s}^{-1}$ . The question asks for the speed (magnitude of tangential velocity) of X.

Approach

Use the direct circular-motion link between angular and linear speed:

Identify $r$ for point X.
Calculate $v = r\omega$ .
Round sensibly based on given data (both $0.85$ and $140$ are 2 s.f.).

Step-by-Step Reasoning

Write the relationship:

v = r\omega

Substitute $r = 0.85\ \text{m}$ and $\omega = 140\ \text{rad s}^{-1}$ :

v = 0.85 \times 140 = 119\ \text{m s}^{-1}

Quote to 2 s.f. (limited by inputs):

v = 1.2 \times 10^{2}\ \text{m s}^{-1}

Key Takeaways

For uniform circular motion, linear speed is proportional to radius: $v = r\omega$ .
Use the radius of the point in question (rim point uses the wheel radius).

Common Mistakes

Using diameter instead of radius.
Using $\omega = 2\pi f$ or $\omega = 2\pi/T$ unnecessarily and then introducing an arithmetic error.
Rounding to too many significant figures compared with the data.

Things to Be Careful About

Ensure $r$ is in metres and $\omega$ in $\text{rad s}^{-1}$ .
"Speed" is a scalar; no direction is required here.

Techniques used

use the tangential speed relation v = r\omegasubstitute values with consistent SI unitsround to an appropriate number of significant figures

(ii)

the centripetal acceleration.

acceleration = ______ $\text{m s}^{-2}$

DifficultyMedium-Easy

Worked solution

Working

a = r\omega^2

a = 0.85 \times (140)^2 = 0.85 \times 19600 = 1.666 \times 10^{4}\ \text{m s}^{-2}

Answer

a = 1.7 \times 10^{4}\ \text{m s}^{-2}

Final answer

1.7 × 10^4 m s^-2

Detailed explanation

Background Concept

In uniform circular motion, the velocity direction changes continuously, requiring an inward (centre-seeking) acceleration called the centripetal acceleration:

a = r\omega^2 = \frac{v^2}{r}

This acceleration always points towards the centre of the circle.

Understanding the Question

Point X is on the rim, so $r = 0.85\ \text{m}$ . The angular speed is $\omega = 140\ \text{rad s}^{-1}$ . We must calculate the magnitude of centripetal acceleration of X.

Approach

Use the form that matches the data given. Since $\omega$ and $r$ are given directly, use

a = r\omega^2.

Step-by-Step Reasoning

Start with

a = r\omega^2

Substitute:

a = 0.85 \times (140)^2

Calculate $140^2 = 19600$ :

a = 0.85 \times 19600 = 16660\ \text{m s}^{-2}

Round to 2 s.f.:

a = 1.7 \times 10^{4}\ \text{m s}^{-2}

Key Takeaways

Centripetal acceleration depends on both angular speed and radius: doubling $\omega$ makes $a$ four times bigger.
Use $a = r\omega^2$ when $\omega$ is given.

Common Mistakes

Using $a = r\omega$ (missing the square).
Using $r = 0.85\ \text{m}$ incorrectly as $1.7\ \text{m}$ (diameter).
Forgetting units or writing incorrect units.

Things to Be Careful About

Even though the wheel rotates clockwise, the question asks only for the magnitude; direction (towards the centre) is not required unless asked.

Techniques used

apply the centripetal acceleration formula a = r\omega^2square the angular speed correctly and multiply by radiusround to an appropriate number of significant figures

(b)

There is a uniform magnetic field of flux density $0.18\text{ T}$ into the plane of the page.

9M

(i)

State Lenz’s law of electromagnetic induction.

DifficultyMedium-Easy

Worked solution

Answer

The induced e.m.f. (and current) is in a direction such that the magnetic field it produces opposes the change in magnetic flux (flux linkage) that causes it.

Final answer

Induced e.m.f./current acts to oppose the change in magnetic flux (linkage) producing it.

Detailed explanation

Background Concept

Electromagnetic induction occurs when the magnetic flux (or flux linkage) through a circuit changes, producing an induced e.m.f. Faraday’s law gives the magnitude:

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

The negative sign is Lenz’s law: it sets the direction of the induced e.m.f./current.

Understanding the Question

You are asked to state Lenz’s law (no numbers needed). The wheel later rotates in a magnetic field, so the law will be used to decide the direction of induced effects.

Approach

Give a clear one-sentence statement connecting:

induced e.m.f./current direction
opposition to the change in flux (or flux linkage) that produces it.

Step-by-Step Reasoning

A correct statement must include two ideas:

A changing flux causes an induced e.m.f./current.
The induced current produces a magnetic field that opposes the change in flux.

So, you state that the induced e.m.f./current acts to oppose the change in magnetic flux (linkage).

Key Takeaways

Lenz’s law is about direction: induced effects oppose the change that created them.
It is effectively an energy-conservation idea: you cannot get energy “for free” from induction.

Common Mistakes

Saying the induced current opposes the magnetic field itself (it opposes the change in flux).
Only stating “the induced e.m.f. is negative” without explaining what is opposed.

Things to Be Careful About

Either “flux” or “flux linkage” is acceptable if used correctly.
Mentioning “opposes the change” is essential; leaving out “change” loses the key meaning.

Techniques used

state Lenz's law in terms of opposing change in flux linkagelink induced e.m.f. direction to the magnetic effect of the induced current

(ii)

Show that the time taken for point X to complete one revolution is $45\text{ ms}$ .

DifficultyEasy

Worked solution

Working

\omega = \frac{2\pi}{T} \Rightarrow T = \frac{2\pi}{\omega}

T = \frac{2\pi}{140} = 4.49 \times 10^{-2}\ \text{s} = 44.9\ \text{ms}

Answer

T \approx 45\ \text{ms}

Final answer

45 ms

Detailed explanation

Background Concept

For uniform circular motion, angular speed $\omega$ and period $T$ are related by

\omega = \frac{2\pi}{T}

because one full revolution corresponds to an angular displacement of $2\pi$ radians.

Understanding the Question

You are given $\omega = 140\ \text{rad s}^{-1}$ and asked to show the time for one revolution is $45\ \text{ms}$ . That time is the period $T$ .

Approach

Use $\omega = 2\pi/T$ .
Rearrange to $T = 2\pi/\omega$ .
Calculate in seconds then convert to milliseconds.

Step-by-Step Reasoning

Rearrange:

T = \frac{2\pi}{\omega}

Substitute $\omega = 140\ \text{rad s}^{-1}$ :

T = \frac{2\pi}{140} = 4.49 \times 10^{-2}\ \text{s}

Convert using $1\ \text{s} = 1000\ \text{ms}$ :

T = 4.49 \times 10^{-2} \times 1000 = 44.9\ \text{ms} \approx 45\ \text{ms}

Key Takeaways

One revolution is always $2\pi$ radians.
$T = 2\pi/\omega$ is often the quickest route.

Common Mistakes

Using $T = 1/\omega$ (missing the $2\pi$ ).
Forgetting to convert seconds to milliseconds.

Things to Be Careful About

Radians are dimensionless, so $\omega$ in $\text{rad s}^{-1}$ is consistent with $2\pi$ .
Rounding: $44.9\ \text{ms}$ is correctly written as $45\ \text{ms}$ .

Techniques used

use \omega = 2\pi/Trearrange to find the periodconvert seconds to milliseconds

(iii)

Calculate the magnetic flux cut by spoke AX during one revolution of the wheel.
Give a unit with your answer.

magnetic flux = ______ unit ______

DifficultyMedium

Worked solution

Working

Area swept by spoke in one revolution:

A = \pi r^2 = \pi(0.85)^2 = 2.27\ \text{m}^2

Magnetic flux cut:

\Phi = BA = 0.18 \times 2.27 = 4.09 \times 10^{-1}\ \text{Wb}

Answer

\Phi \approx 0.41\ \text{Wb}

Final answer

0.41 Wb

Detailed explanation

Background Concept

Magnetic flux through an area is

\Phi = BA\cos\theta

where:

$B$ is magnetic flux density,
$A$ is the area,
$\theta$ is the angle between $\vec{B}$ and the normal to the area.

Here the field is into the page and the wheel is in the plane of the page, so $\vec{B}$ is perpendicular to the wheel. That makes $\theta = 0$ and $\cos\theta = 1$ , so $\Phi = BA$ .

The phrase “flux cut” in this context refers to the amount of flux associated with the area swept out by the conductor as it rotates.

Understanding the Question

Spoke AX has length equal to the wheel radius $r = 0.85\ \text{m}$ . In one full revolution, the spoke sweeps out the entire circular area of radius $r$ . With uniform $B = 0.18\ \text{T}$ into the page, the flux corresponding to that swept area is required.

Approach

Work out the swept area in one revolution: a full circle of radius $r$ .
Use $\Phi = BA$ (since the field is normal to the area).
Include the correct unit: webers (Wb), because $1\ \text{Wb} = 1\ \text{T m}^2$ .

Step-by-Step Reasoning

Swept area in one revolution:

A = \pi r^2 = \pi(0.85)^2

Calculate:

(0.85)^2 = 0.7225

so

A = \pi \times 0.7225 = 2.27\ \text{m}^2

Flux through that area (field perpendicular):

\Phi = BA = 0.18 \times 2.27 = 0.409\ \text{T m}^2

Convert unit $\text{T m}^2$ to Wb:

\Phi = 0.409\ \text{Wb} \approx 0.41\ \text{Wb}

Key Takeaways

If $\vec{B}$ is perpendicular to the plane, $\Phi = BA$ .
A rotating radius sweeps out a full circle in one revolution, so $A = \pi r^2$ .
Flux unit: $\text{T m}^2 = \text{Wb}$ .

Common Mistakes

Using $A = 2\pi r^2$ (confusing area with circumference).
Using $A = \frac{1}{2}\pi r^2$ (only a semicircle).
Forgetting that $\theta = 0$ here and incorrectly including a cosine less than 1.
Missing the unit or giving $\text{T m}^2$ without recognising it as Wb.

Things to Be Careful About

The radius is $0.85\ \text{m}$ , not the diameter.
Keep enough precision in intermediate steps to avoid rounding error; round at the end.

Techniques used

identify the area swept out by a rotating spokecalculate magnetic flux using \Phi = BA for a uniform field normal to the areause consistent units and quote an appropriate unit for flux

(iv)

Determine the magnitude of the electromotive force (e.m.f.) induced across spoke AX.

induced e.m.f. = ______ $\text{V}$

DifficultyMedium-Easy

Worked solution

Working

From (iii), flux cut in one revolution:

\Delta\Phi = 0.409\ \text{Wb}

Time for one revolution:

\Delta t = T = 4.49 \times 10^{-2}\ \text{s}

Average induced e.m.f.:

\mathcal{E} = \frac{\Delta\Phi}{\Delta t} = \frac{0.409}{4.49 \times 10^{-2}} = 9.11\ \text{V}

Answer

\mathcal{E} \approx 9.1\ \text{V}

Final answer

9.1 V

Detailed explanation

Background Concept

Faraday’s law links induced e.m.f. to the rate of change of magnetic flux linkage. For one conductor cutting flux, an average e.m.f. over a time interval can be found from

\mathcal{E}_{\text{avg}} = \frac{\Delta\Phi}{\Delta t}

(The sign is set by Lenz’s law, but this question asks for the magnitude.)

Understanding the Question

Over one full revolution, spoke AX sweeps out a full circular area, so it “cuts” a certain amount of magnetic flux (found in part (iii)). The time for one revolution is the period $T$ (found in part (ii)). The question asks for the magnitude of induced e.m.f. across the spoke.

Approach

Take $\Delta\Phi$ as the flux cut in one revolution.
Take $\Delta t = T$ as the time for one revolution.
Use $\mathcal{E} = \Delta\Phi/\Delta t$ .

Step-by-Step Reasoning

Use the value from (iii):

\Delta\Phi = 0.409\ \text{Wb}

Use the period from (ii):

T = 44.9\ \text{ms} = 4.49 \times 10^{-2}\ \text{s}

Apply Faraday’s law (average over the revolution):

\mathcal{E} = \frac{\Delta\Phi}{T} = \frac{0.409}{4.49 \times 10^{-2}} = 9.11\ \text{V}

So

\mathcal{E} \approx 9.1\ \text{V}

(You may also see the equivalent generator result for a rotating spoke: $\mathcal{E} = \frac{1}{2}B\omega r^2$ , which gives the same number.)

Key Takeaways

Induced e.m.f. magnitude is proportional to how quickly flux is cut.
Using one-revolution quantities: $\mathcal{E} = (B\pi r^2)/T$ .

Common Mistakes

Using $\mathcal{E} = \Phi/T$ but taking $\Phi$ as $Br$ or $B(2\pi r)$ (wrong geometry).
Forgetting to convert $45\ \text{ms}$ to seconds before dividing.
Including a negative sign when the question asks for magnitude.

Things to Be Careful About

Use consistent units: Wb and s gives volts directly.
Keep at least 2–3 significant figures through the division; round at the end.

Techniques used

apply Faraday's law in the average form \mathcal{E} = \Delta\Phi/\Delta tuse the period to represent the time for one revolutioncalculate and round the induced e.m.f.

(v)

Use Lenz’s law to explain whether the potential is higher at end A or end X of the spoke.

DifficultyMedium

Worked solution

Answer

To oppose the clockwise rotation (Lenz’s law), the magnetic force on the spoke must be opposite to the motion, so the induced conventional current in AX is from X to A. Hence end X is at higher potential than end A.

Final answer

Potential is higher at X.

Detailed explanation

Background Concept

When a conductor moves through a magnetic field, charges in it experience a magnetic force $q\vec{v} \times \vec{B}$ , causing a separation of charge and an induced e.m.f.

If the conductor forms part of a complete circuit, an induced current flows. By Lenz’s law, the induced current produces a magnetic effect that opposes the change producing it. In a generator situation (mechanical motion producing e.m.f.), this usually shows up as a magnetic force that opposes the motion (a resisting torque), ensuring energy conservation.

The direction of force on a current-carrying conductor is given by

\vec{F} = I\,\vec{L} \times \vec{B}

where $\vec{L}$ points in the direction of conventional current.

Understanding the Question

The wheel rotates clockwise in a uniform magnetic field directed into the page. The spoke AX is a conductor from the axle (centre) to the rim (point X). You must decide whether end A or end X is at higher potential, using Lenz’s law.

Approach

Lenz’s law tells us the induced effects must oppose the cause (the rotation cutting flux). That means the magnetic force on the spoke must oppose the motion (produce a counter-torque).
Choose a current direction in the spoke and use $\vec{F} = I\vec{L} \times \vec{B}$ to see whether the force would be in the direction of motion or against it.
The correct current direction gives the higher-potential end (conventional current flows from higher to lower potential through the conductor).

Step-by-Step Reasoning

diagram-bv-1

The wheel is rotating clockwise. For the induced current to oppose this, the magnetic force on the spoke must act tangentially anticlockwise (opposite the motion).
The magnetic field is into the page.
Test current direction:

If conventional current were from A to X (radially outward), then $\vec{L}$ is outward. With $\vec{B}$ into the page, $\vec{L} \times \vec{B}$ is tangential clockwise, i.e. it would help the rotation. That contradicts Lenz’s law.
Therefore the induced conventional current must be from X to A (radially inward). Then $\vec{L}$ is inward and $\vec{L} \times \vec{B}$ is tangential anticlockwise, opposing the motion, consistent with Lenz’s law.

Since conventional current in the spoke is from X to A, end X must be at higher potential than end A.

Key Takeaways

Lenz’s law often means: induced current causes a magnetic force that resists the motion producing it.
Use $\vec{F} = I\vec{L} \times \vec{B}$ to check which current direction gives an opposing force.
Conventional current direction tells you which end is at higher potential.

Common Mistakes

Claiming the induced current aids the motion (violates Lenz’s law / energy conservation).
Mixing up electron flow and conventional current: electrons move opposite to conventional current.
Using Fleming’s rules with inconsistent directions for field and motion.

Things to Be Careful About

The conclusion (which end is higher potential) must match the requirement that the induced magnetic effect opposes the clockwise rotation.
You are asked about potential, so state clearly which end is at higher potential (not just the current direction).

Techniques used

apply Lenz’s law to decide the direction of the induced current that opposes motionuse the motor effect direction (force on a current in a magnetic field) to link current direction to opposing torquededuce which end must be at higher potential from conventional current direction

Q2MediumGravitational FieldsAstronomy and Cosmology

The Sun may be considered as a uniform sphere with a mass of $1.99 \times 10^{30}\text{ kg}$ and a surface temperature of $5780\text{ K}$ .

A probe with a mass of $2.63\text{ kg}$ moves in a straight line towards the Sun.
When it is at a distance $x$ from the centre of the Sun, the probe measures the gravitational field strength $g$ due to the Sun and the radiant flux intensity $F$ of radiation from the Sun.

(a)

Define gravitational field.

(b)

For the position of the probe where $x = 1.47 \times 10^{11}\text{ m}$ :

4M

(i)

calculate $g$

$g$ = ______ $\text{N kg}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

For the Sun,

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

g = \frac{(6.67\times 10^{-11})(1.99\times 10^{30})}{(1.47\times 10^{11})^2} = 6.14\times 10^{-3}

Answer

$g = 6.1 \times 10^{-3}\ \text{N kg}^{-1}$

Final answer

6.1 × 10^-3 N kg^-1

Detailed explanation

Background Concept

For a spherically symmetric mass (like the Sun treated as a uniform sphere), the gravitational field outside it is the same as if all the mass were concentrated at the centre. The gravitational field strength at distance $r$ from the centre is

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

where $G$ is the gravitational constant and $M$ is the mass producing the field.

Understanding the Question

Given:

Mass of Sun $M = 1.99 \times 10^{30}\ \text{kg}$
Distance of probe from Sun’s centre $x = 1.47 \times 10^{11}\ \text{m}$

Find $g$ due to the Sun at that position.

Approach

Use the inverse-square gravitational field formula $g = GM/x^2$ and substitute the given values.

Step-by-Step Reasoning

Write the equation:

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

Substitute:

GM = (6.67\times 10^{-11})(1.99\times 10^{30}) \approx 1.33\times 10^{20}

Square the distance:

(1.47\times 10^{11})^2 = (1.47)^2 \times 10^{22} \approx 2.16\times 10^{22}

Divide:

g \approx \frac{1.33\times 10^{20}}{2.16\times 10^{22}} \approx 6.1\times 10^{-3}\ \text{N kg}^{-1}

This is much smaller than Earth’s $g$ (about $9.8\ \text{N kg}^{-1}$ ) because the probe is very far from the Sun.

Key Takeaways

Outside a sphere, treat the mass as concentrated at the centre.
Gravitational field decreases as $1/r^2$ .

Common Mistakes

Using $x$ instead of $x^2$ .
Forgetting that $x$ is from the centre of the Sun (the question states this explicitly).
Power-of-ten errors when squaring $10^{11}$ .

Things to Be Careful About

Keep units consistent (SI units).
Square both the number and the power of ten: $(10^{11})^2 = 10^{22}$ .
Quote $g$ in $\text{N kg}^{-1}$ (or $\text{m s}^{-2}$ ).

Techniques used

apply Newton's law of gravitation to a test massuse the inverse-square field strength formulasubstitute values carefully in standard form

(ii)

determine the gravitational potential energy $E_P$ of the probe.

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

DifficultyMedium-Easy

Worked solution

Working

Gravitational potential energy (zero at infinity):

E_P = -\frac{GMm}{x}

E_P = -\frac{(6.67\times 10^{-11})(1.99\times 10^{30})(2.63)}{1.47\times 10^{11}} = -2.38\times 10^{9}

Answer

$E_P = -2.4 \times 10^{9}\ \text{J}$

Final answer

-2.4 × 10^9 J

Detailed explanation

Background Concept

Taking gravitational potential energy to be zero at infinity, the gravitational potential energy of a mass $m$ at distance $r$ from a spherical mass $M$ is

E_P = -\frac{GMm}{r}

The negative sign arises because work must be done against the attractive gravitational force to move the mass from $r$ out to infinity.

Understanding the Question

Given:

$M = 1.99 \times 10^{30}\ \text{kg}$ (Sun)
Probe mass $m = 2.63\ \text{kg}$
Position $x = 1.47 \times 10^{11}\ \text{m}$ from the Sun’s centre

Find the probe’s gravitational potential energy due to the Sun (with the usual reference $E_P = 0$ at infinity).

Approach

Use $E_P = -GMm/x$ and substitute values in SI units.

Step-by-Step Reasoning

Start with

E_P = -\frac{GMm}{x}

Compute $GM$ (as in part (b)(i)):

GM \approx 1.33\times 10^{20}

Multiply by $m$ :

GMm \approx (1.33\times 10^{20})(2.63) \approx 3.49\times 10^{20}

Divide by $x$ :

\frac{GMm}{x} \approx \frac{3.49\times 10^{20}}{1.47\times 10^{11}} \approx 2.38\times 10^{9}

Apply the negative sign:

E_P \approx -2.4\times 10^{9}\ \text{J}

Key Takeaways

$E_P$ in a gravitational field is negative when zero is defined at infinity.
For a spherical mass, use the point-mass formula $-GMm/r$ outside the sphere.

Common Mistakes

Missing the negative sign (very common).
Using $r^2$ in the denominator (confusing with the $g$ formula).

Things to Be Careful About

State or imply the reference level (here it is the standard “zero at infinity”).
Check that the magnitude is sensible: large mass $M$ but huge distance $x$ gives a large-but-not-astronomical value for a small probe mass.

Techniques used

use the gravitational potential energy expression for a point masssubstitute values with correct sign conventionhandle standard form arithmetic

(c)

7M

(i)

Show that, for any particular value of $x$ , the numerical values of $g$ and $F$ are related by

g = \frac{4\pi GM}{L} F

where $M$ is the mass of the Sun, $L$ is the luminosity of the Sun and $G$ is the gravitational constant.

DifficultyMedium

Worked solution

Working

For distance $x$ from the Sun’s centre,

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

Radiant flux intensity from luminosity $L$ :

F = \frac{L}{4\pi x^2}

So

\frac{1}{x^2} = \frac{4\pi F}{L}

Substitute into $g$ :

g = GM\left(\frac{4\pi F}{L}\right) = \frac{4\pi GM}{L}F

Answer

g = \frac{4\pi GM}{L}F

Final answer

g = (4πGM/L) F

Detailed explanation

Background Concept

Two key inverse-square results for a spherical source:

Gravitational field strength due to a spherical mass $M$ at distance $x$ :

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

Radiant flux intensity (power per unit area) at distance $x$ from a star of luminosity $L$ :

F = \frac{L}{4\pi x^2}

The $4\pi x^2$ is the surface area of a sphere of radius $x$ over which the power spreads.

Understanding the Question

You must show that, at the same location of the probe (same $x$ ), the numerical values of $g$ and $F$ are related by a linear equation

g = \frac{4\pi GM}{L}F

So you need to eliminate $x$ between the expressions for $g$ and $F$ .

Approach

Write down both inverse-square laws.
Rearrange one (usually the radiation one) to express $1/x^2$ .
Substitute into the other.

Step-by-Step Reasoning

Start with the two equations:

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

F = \frac{L}{4\pi x^2}

Rearrange the flux equation for $1/x^2$ :

F = \frac{L}{4\pi x^2} \Rightarrow 4\pi x^2 F = L \Rightarrow \frac{1}{x^2} = \frac{4\pi F}{L}

Substitute into the expression for $g$ :

g = GM\left(\frac{1}{x^2}\right) = GM\left(\frac{4\pi F}{L}\right) = \frac{4\pi GM}{L}F

This shows $g$ is directly proportional to $F$ (straight line through the origin on a $g$ – $F$ graph).

Key Takeaways

Both gravity and radiation from a spherical source follow inverse-square behaviour.
Eliminating the common $x^2$ gives a linear relation between two measurable quantities.

Common Mistakes

Using $F = L/(2\pi x^2)$ instead of $4\pi x^2$ .
Algebra slip when rearranging for $1/x^2$ .
Forgetting that both formulas use the same distance measured from the centre.

Things to Be Careful About

Keep $x$ consistent: it is the distance from the centre of the Sun.
The relationship predicts the line passes through the origin because both $g$ and $F$ go to zero as $x \to \infty$ .

Techniques used

use inverse-square expressions for gravitational field and radiant fluxeliminate the distance variable by substitutionidentify the proportionality constant as a gradient

(ii)

Fig. 2.1 shows the variation of $g$ with $F$ .

Fig. 2.1

Determine a value for the luminosity $L$ of the Sun. Give a unit with your answer.

$L$ = ______ unit ______

DifficultyMedium

Worked solution

Working

From the graph, using $(F, g) = (2.0, 8)$ in the plotted scales:

\text{gradient} = \frac{8\times 10^{-3}}{2.0\times 10^{3}} = 4.0\times 10^{-6}

From (c)(i),

\text{gradient} = \frac{4\pi GM}{L} \Rightarrow L = \frac{4\pi GM}{\text{gradient}}

L = \frac{4\pi (6.67\times 10^{-11})(1.99\times 10^{30})}{4.0\times 10^{-6}} = 4.17\times 10^{26}

Answer

$L = 4.2 \times 10^{26}\ \text{W}$

Final answer

4.2 × 10^26 W

Detailed explanation

Background Concept

From part (c)(i), we have a linear relationship:

g = \left(\frac{4\pi GM}{L}\right)F

Comparing with $g = (\text{gradient})\,F$ , the gradient of the $g$ – $F$ graph is

\text{gradient} = \frac{4\pi GM}{L}

So if we can read the gradient, we can rearrange to find the luminosity $L$ .

Understanding the Question

The graph shows $g$ plotted against $F$ with scaled axes:

vertical axis labelled $g / 10^{-3}\ \text{N kg}^{-1}$
horizontal axis labelled $F / 10^{3}\ \text{W m}^{-2}$

The line passes through $(0,0)$ and approximately $(2.0, 8)$ in these plotted units.

Approach

Determine the numerical gradient using two points on the straight line.
Convert that gradient into SI units (undo the axis scaling).
Use $\text{gradient} = 4\pi GM/L$ and rearrange for $L$ .

Step-by-Step Reasoning

Gradient from the plotted coordinates:

Plotted change in $g$ is $8$ units of $(10^{-3}\ \text{N kg}^{-1})$ , so

\Delta g = 8\times 10^{-3}\ \text{N kg}^{-1}

Plotted change in $F$ is $2.0$ units of $(10^{3}\ \text{W m}^{-2})$ , so

\Delta F = 2.0\times 10^{3}\ \text{W m}^{-2}

Hence

\text{gradient} = \frac{\Delta g}{\Delta F} = \frac{8\times 10^{-3}}{2.0\times 10^{3}} = 4.0\times 10^{-6}

Use the theoretical gradient:

\text{gradient} = \frac{4\pi GM}{L}

Rearrange:

L = \frac{4\pi GM}{\text{gradient}}

Substitute $G = 6.67\times 10^{-11}$ and $M = 1.99\times 10^{30}$ :

L = \frac{4\pi (6.67\times 10^{-11})(1.99\times 10^{30})}{4.0\times 10^{-6}} \approx 4.2\times 10^{26}\ \text{W}

Key Takeaways

A straight-line graph through the origin suggests direct proportionality.
Always undo axis scaling before calculating a physical gradient.
The gradient links measured quantities to constants like $L$ .

Common Mistakes

Using the plotted numbers (e.g. $8/2$ ) as the gradient without including the $10^{-3}$ and $10^{3}$ scale factors.
Inverting the gradient (using $\Delta F/\Delta g$ by mistake).
Forgetting the factor $4\pi$ .

Things to Be Careful About

The unit of luminosity is the watt (W).
Choose points far apart on the line to reduce percentage uncertainty in the gradient.
Keep powers of ten consistent: here the gradient is very small because $g$ is tiny compared with $F$ in SI units.

Techniques used

find the gradient from a straight-line graphconvert from scaled axes to actual physical unitsrearrange the gradient relationship to solve for luminosity

(iii)

Use your answer in (c)(ii) to determine the radius $r$ of the Sun.

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

DifficultyMedium

Worked solution

Working

Stefan–Boltzmann law:

L = 4\pi r^2 \sigma T^4

r = \sqrt{\frac{L}{4\pi \sigma T^4}}

Using $L = 4.17\times 10^{26}\ \text{W}$ , $\sigma = 5.67\times 10^{-8}$ and $T = 5780\ \text{K}$ :

r = \sqrt{\frac{4.17\times 10^{26}}{4\pi (5.67\times 10^{-8})(5780)^4}} = 7.25\times 10^{8}

Answer

$r = 7.2 \times 10^{8}\ \text{m}$

Final answer

7.2 × 10^8 m

Detailed explanation

Background Concept

A star can be approximated as a black-body radiator. The Stefan–Boltzmann law gives the luminosity $L$ (total power emitted) in terms of its surface area and surface temperature:

L = 4\pi r^2 \sigma T^4

where

$r$ is the star’s radius,
$T$ is surface temperature,
$\sigma$ is the Stefan–Boltzmann constant ( $5.67\times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ ).

Understanding the Question

You are given the Sun’s surface temperature $T = 5780\ \text{K}$ and, from part (c)(ii), its luminosity $L$ . You must calculate the Sun’s radius $r$ .

Approach

Rearrange the Stefan–Boltzmann law for $r$ :

r = \sqrt{\frac{L}{4\pi \sigma T^4}}

Then substitute values in SI units.

Step-by-Step Reasoning

Start with

L = 4\pi r^2 \sigma T^4

Rearrange:

r^2 = \frac{L}{4\pi \sigma T^4}

r = \sqrt{\frac{L}{4\pi \sigma T^4}}

Substitute the numbers (taking $L \approx 4.2\times 10^{26}\ \text{W}$ from the graph):

First compute $T^4$ ; since $T$ is a few thousand kelvin, $T^4$ is a very large number.
Evaluate the denominator $4\pi \sigma T^4$ (this represents total power emitted per unit $r^2$ ).

After substitution you obtain

r \approx 7.2\times 10^{8}\ \text{m}

This is of the correct order of magnitude for the solar radius (about $7\times 10^8\ \text{m}$ ).

Key Takeaways

Luminosity depends strongly on temperature: $L \propto T^4$ .
If you know $L$ and $T$ , you can estimate radius using Stefan–Boltzmann.

Common Mistakes

Forgetting the square root when solving for $r$ .
Using $L = \pi r^2 \sigma T^4$ (missing the factor of $4$ for total surface area).
Not keeping $T$ in kelvin.

Things to Be Careful About

Use consistent SI units throughout.
Because $T^4$ is very large, calculator entry errors are common; bracket the entire denominator.
The final unit for radius must be metres (m).

Techniques used

apply the Stefan-Boltzmann lawrearrange to make the radius the subjectevaluate a power and a square root in standard form

Q3MediumTemperatureThermodynamics

(a)

Define specific latent heat.

DifficultyEasy

Worked solution

Answer

Specific latent heat is the thermal energy required per unit mass to change the state of a substance with no change in temperature.

Final answer

Specific latent heat is the thermal energy required per unit mass to change the state of a substance with no change in temperature.

Detailed explanation

Background Concept

When a substance changes state (e.g. melting or boiling), energy supplied does not raise its temperature; instead it is used to change the molecular arrangement and intermolecular potential energy. The specific latent heat $L$ quantifies this energy requirement per unit mass.

The relationship is

Q = mL

where $Q$ is the thermal energy transferred, $m$ is the mass, and $L$ is the specific latent heat (units $\text{J kg}^{-1}$ ).

Understanding the Question

You are asked to define specific latent heat. This means stating what it is, including the key condition that the temperature does not change during the state change, and that it is an energy per unit mass.

Approach

Give a concise definition that includes:

energy required (or released),
per unit mass,
to change state,
at constant temperature.

Step-by-Step Reasoning

“Specific” means per unit mass (i.e. per $1\,\text{kg}$ ).
“Latent heat” refers to energy associated with change of state.
During a phase change, temperature remains constant, so you must say “no change in temperature”.

A complete definition is: thermal energy required per unit mass to change state with no temperature change.

Key Takeaways

Use $Q = mL$ .
Temperature stays constant during the phase change.
Must include “per unit mass”.

Common Mistakes

Forgetting “per unit mass”.
Missing the condition “no temperature change”.
Saying “heat” without clarifying it is energy transferred.

Things to Be Careful About

Do not confuse specific latent heat with specific heat capacity (which involves temperature change).
Units should be $\text{J kg}^{-1}$ (or $\text{kJ kg}^{-1}$ if stated).

Techniques used

state the definition of a specific latent heatinclude condition of no temperature changeexpress definition per unit mass with correct units

(b)

A dish containing $7.2 \times 10^{-5}\text{ m}^3$ of a substance rests on a laboratory bench. The substance is initially a liquid of density $710\text{ kg m}^{-3}$ . Atmospheric pressure is $1.0 \times 10^5\text{ Pa}$ .

The liquid is heated at its boiling point so that it completely vaporises. The increase in the internal energy of the substance during this process is $17.6\text{ kJ}$ . The final volume of the vapour is $0.017\text{ m}^3$ .

6M

(i)

Show that the magnitude of the work done on the substance when it vaporises is $1.7\text{ kJ}$ .

DifficultyMedium-Easy

Worked solution

Working

Change in volume:

\Delta V = 0.017 - 7.2 \times 10^{-5} = 1.6928 \times 10^{-2}\ \text{m}^3

Work done (magnitude) against atmospheric pressure:

W = p\Delta V = (1.0 \times 10^5)(1.6928 \times 10^{-2}) = 1.69 \times 10^3\ \text{J}

W \approx 1.7\ \text{kJ}

Answer

$1.7\ \text{kJ}$

Final answer

1.7 kJ

Detailed explanation

Background Concept

When a substance expands at (approximately) constant external pressure $p$ , it does work on the surroundings. The work done by the gas/substance is

W_{\text{by}} = p\Delta V

If the system expands, $\Delta V>0$ so $W_{\text{by}}>0$ . The work done on the system is the negative of this:

W_{\text{on}} = -p\Delta V

Questions often ask for the magnitude, which is just $p\Delta V$ .

Understanding the Question

The liquid boils and becomes vapour, so its volume increases from $7.2 \times 10^{-5}\ \text{m}^3$ to $0.017\ \text{m}^3$ while pushing back the atmosphere at $1.0 \times 10^5\ \text{Pa}$ . You must show that the work involved has magnitude $1.7\ \text{kJ}$ .

Approach

Find $\Delta V = V_\text{final} - V_\text{initial}$ .
Use $W = p\Delta V$ .
Convert from joules to kilojoules.

Step-by-Step Reasoning

Compute the volume change:

\Delta V = 0.017 - 7.2\times 10^{-5}

\Delta V = 0.016928\ \text{m}^3 = 1.6928\times 10^{-2}\ \text{m}^3

Now multiply by atmospheric pressure:

W = p\Delta V = (1.0\times 10^5)(1.6928\times 10^{-2})

W = 1.6928\times 10^3\ \text{J}

Convert to kJ:

1.6928\times 10^3\ \text{J} = 1.6928\ \text{kJ} \approx 1.7\ \text{kJ}

This matches the required value.

Key Takeaways

Expansion at constant pressure: $W = p\Delta V$ .
For expansion, work is done by the substance on the atmosphere.
“Magnitude” avoids sign convention issues.

Common Mistakes

Using $V_\text{initial}-V_\text{final}$ and getting a negative magnitude.
Forgetting to convert $\text{J}$ to $\text{kJ}$ .
Using the vapour volume only (not the change in volume).

Things to Be Careful About

The pressure is in pascals ( $\text{Pa} = \text{N m}^{-2}$ ), so the volume must be in $\text{m}^3$ .
Keep enough significant figures during calculation; round at the end to $1.7\ \text{kJ}$ .

Techniques used

calculate the change in volumeapply work done against constant external pressureconvert joules to kilojoules and round appropriately

(ii)

Use the information in (b)(i) to calculate the thermal energy $Q$ , in kJ, supplied to the substance to cause it to vaporise.

$Q$ = ______ $\text{kJ}$

DifficultyMedium-Easy

Worked solution

Working

First law (with $W$ = work done on the substance):

\Delta U = Q + W

Expansion so $W = -1.7\ \text{kJ}$ .

17.6 = Q - 1.7

Q = 19.3\ \text{kJ}

Answer

$Q = 19.3\ \text{kJ}$

Final answer

19.3 kJ

Detailed explanation

Background Concept

The first law of thermodynamics links the change in internal energy to heating and work:

\Delta U = Q + W

where:

$\Delta U$ is the increase in internal energy of the system,
$Q$ is thermal energy transferred to the system,
$W$ is work done on the system.

If the system expands and pushes back the atmosphere, it does work on the surroundings, so the work done on the system is negative.

Understanding the Question

You are told the increase in internal energy during vaporisation is $17.6\ \text{kJ}$ . From part (i), the magnitude of work associated with expansion is $1.7\ \text{kJ}$ . You must find the thermal energy supplied $Q$ .

Approach

Use the first law with the paper’s stated convention ( $W$ = work done on the substance). Because the volume increases, set $W=-1.7\ \text{kJ}$ , then solve for $Q$ .

Step-by-Step Reasoning

Start with:

\Delta U = Q + W

Given $\Delta U = +17.6\ \text{kJ}$ .

During boiling the substance expands, so it does work on the atmosphere; therefore work done on the substance is negative:

W = -1.7\ \text{kJ}

Substitute:

17.6 = Q + (-1.7)

So:

Q = 17.6 + 1.7 = 19.3\ \text{kJ}

This means the heater supplies enough energy both to raise internal energy and to provide the mechanical work of expansion.

Key Takeaways

Use $\Delta U = Q + W$ with correct sign convention.
Expansion implies $W_{\text{on}}<0$ .
Heating must cover both internal energy increase and expansion work.

Common Mistakes

Taking $W=+1.7\ \text{kJ}$ and getting $Q=15.9\ \text{kJ}$ .
Mixing conventions ( $\Delta U = Q - p\Delta V$ vs $\Delta U = Q + W$ ) without consistency.

Things to Be Careful About

The question in (i) gave the magnitude of the work; you must decide the sign in (ii).
Keep units consistent (all in $\text{kJ}$ here).

Techniques used

apply the first law of thermodynamicsuse the correct sign convention for work during expansionrearrange to obtain the thermal energy transfer

(iii)

Use your answer in (b)(ii) to determine a value for the specific latent heat of vaporisation $L_V$ , in $\text{kJ kg}^{-1}$ , of the substance.

$L_V$ = ______ $\text{kJ kg}^{-1}$

DifficultyMedium

Worked solution

Working

Mass of liquid:

m = \rho V = 710 \times 7.2\times 10^{-5} = 5.11\times 10^{-2}\ \text{kg}

Specific latent heat of vaporisation:

L_V = \frac{Q}{m} = \frac{19.3\ \text{kJ}}{5.11\times 10^{-2}\ \text{kg}} = 3.78\times 10^2\ \text{kJ kg}^{-1}

Answer

$L_V \approx 3.78\times 10^2\ \text{kJ kg}^{-1}$

Final answer

3.78×10^2 kJ kg^-1

Detailed explanation

Background Concept

For a phase change at constant temperature, the thermal energy transferred is related to the mass and the specific latent heat:

Q = mL

So

L = \frac{Q}{m}

To find mass from density, use

\rho = \frac{m}{V} \quad \Rightarrow \quad m = \rho V

Understanding the Question

You have already found the thermal energy supplied to vaporise the liquid: $Q = 19.3\ \text{kJ}$ . You are given the liquid density $\rho = 710\ \text{kg m}^{-3}$ and the initial liquid volume $V = 7.2\times 10^{-5}\ \text{m}^3$ , which lets you find the mass that vaporised. Then compute $L_V$ .

Approach

Calculate $m = \rho V$ using the liquid volume (because that’s the known volume of the material before it changes state).
Use $L_V = Q/m$ .
Keep units in $\text{kJ kg}^{-1}$ since $Q$ is in kJ.

Step-by-Step Reasoning

Find the mass:

m = \rho V = (710)(7.2\times 10^{-5})\ \text{kg}

m = 5.112\times 10^{-2}\ \text{kg}

Now use $Q = mL_V$ :

L_V = \frac{Q}{m} = \frac{19.3\ \text{kJ}}{5.112\times 10^{-2}\ \text{kg}}

L_V = 3.78\times 10^2\ \text{kJ kg}^{-1}

This is the specific latent heat of vaporisation.

Key Takeaways

Use $m = \rho V$ to find how much substance changed state.
Then use $L = Q/m$ .
Be consistent with energy units (kJ gives $\text{kJ kg}^{-1}$ ).

Common Mistakes

Using the final vapour volume ( $0.017\ \text{m}^3$ ) with the liquid density to find mass (wrong: density changes enormously).
Using $\Delta U$ instead of $Q$ in $L = Q/m$ (latent heat is based on thermal energy supplied).
Converting $Q$ to joules but leaving the answer in $\text{kJ kg}^{-1}$ (unit mismatch).

Things to Be Careful About

Significant figures: input data are typically 2 s.f. (e.g. $7.2\times 10^{-5}$ ), so $L_V$ to 3 s.f. is reasonable ( $3.78\times 10^2$ ).
Ensure you use the same mass that actually vaporised (the whole initial liquid).

Techniques used

calculate mass from density and volumeapply the latent heat equation Q = mLconvert units to kJ kg^{-1}

(c)

The substance in (b) has a specific latent heat of fusion $L_F$ .

Suggest and explain whether $L_F$ is likely to be less than, the same as, or greater than the answer in (b)(iii).

DifficultyMedium

Worked solution

Answer

$L_F$ is likely to be less than $L_V$ .

On melting, intermolecular bonds are only partially overcome so molecules become mobile but remain close together, so the increase in intermolecular potential energy is smaller.

On vaporisation, molecules separate much more and the substance also expands greatly, so more energy is needed (including work against atmospheric pressure).

Final answer

LF is less than LV.

Detailed explanation

Background Concept

Latent heats measure the energy needed to change state at constant temperature:

Fusion (solid (\to) liquid): energy mainly goes into loosening the solid lattice so particles can move past each other.
Vaporisation (liquid (\to) gas): energy goes into separating particles much further, greatly increasing intermolecular potential energy, and typically doing significant expansion work against the surroundings.

Generally,

L_V > L_F

for the same substance.

Understanding the Question

You found a numerical value for the specific latent heat of vaporisation $L_V$ in (b)(iii). Now you must suggest and explain how the specific latent heat of fusion $L_F$ compares with that value.

Approach

Use qualitative particle/energy arguments:

Compare the change in molecular separation and intermolecular potential energy.
Mention that vaporisation involves much greater volume change and therefore work done against atmospheric pressure.

Step-by-Step Reasoning

In a solid, particles are in fixed positions with strong intermolecular attractions.
During melting, energy is used to break enough of these bonds to allow particles to move, but particles in the liquid remain close together. So the increase in intermolecular potential energy is relatively modest.
During boiling, particles must become essentially independent in the gas phase, so attractions are overcome much more completely; particle separation increases dramatically. This requires a much larger increase in intermolecular potential energy.
Additionally, boiling usually involves a large increase in volume, so the substance does work on the surroundings ( $p\Delta V$ ), increasing the required thermal energy. (This is consistent with part (b)(i), where a non-negligible work term existed.)

Therefore $L_F$ should be less than $L_V$ .

Key Takeaways

$L$ is linked to changes in intermolecular potential energy.
Vaporisation needs more energy than fusion because it separates molecules far more and often includes significant expansion work.

Common Mistakes

Saying they are the same because “both are changes of state” (ignores different energy changes).
Claiming $L_F > L_V$ without justification.
Only stating “less than” without explaining in terms of molecular separation/forces.

Things to Be Careful About

Make the comparison explicitly: “less than”, “same as”, or “greater than”.
Include an explanation tied to physics (intermolecular forces, potential energy, expansion work), not vague statements like “takes more heat”.

Techniques used

compare energy changes required for fusion and vaporisationrelate latent heat to intermolecular potential energy changesuse qualitative reasoning about molecular separation and work of expansion

Q4MediumIdeal Gases

(a)

State three of the basic assumptions of the kinetic theory of gases.

DifficultyEasy

Worked solution

Answer

Any three of:

Gas contains a large number of identical molecules (treated as point particles).
Molecular volume is negligible compared with the volume of the container.
Molecules move in constant random motion and obey Newton’s laws.
No intermolecular forces except during collisions.
Collisions between molecules and with the walls are perfectly elastic (collision time negligible).

Final answer

Any three valid kinetic theory assumptions (see solution).

Detailed explanation

Background Concept

The kinetic theory of gases is a model that explains macroscopic gas behaviour (pressure, temperature) by assuming a simple picture of microscopic particles (molecules/atoms). Because it is a model, it starts from a set of simplifying assumptions (e.g. point particles, no forces) that make it possible to derive results such as

pV = NkT

and

pV = \frac{1}{3}Nm\langle c^2 \rangle.

These derivations only work if the assumptions are stated clearly.

Understanding the Question

The question asks for three basic assumptions used in kinetic theory. It is not asking for derived equations, and it is not asking for conditions for an ideal gas law; it wants statements about what molecules are like and how they interact.

Approach

Recall the standard list of kinetic theory assumptions. Select any three distinct ones and write them as short, unambiguous statements.

Step-by-Step Reasoning

Start with particle description: many identical molecules, treated as point particles (so their size is negligible).
State the nature of motion: random motion, obeying Newton’s laws (straight lines at constant speed between collisions).
State interactions: no intermolecular forces except during collisions.
State collision behaviour: collisions are perfectly elastic (and collision time is negligible compared to time between collisions).

Any three of these earn the marks (provided they are clearly different points).

Key Takeaways

Kinetic theory assumptions describe: (i) particle size, (ii) particle motion, (iii) forces between particles, (iv) collision properties.
You only need three distinct assumptions for full credit.

Common Mistakes

Stating a gas law (e.g. “ $pV \propto T$ ”) instead of an assumption about molecules.
Saying “molecules move fast” (too vague) rather than “random motion” and “straight lines between collisions”.
Forgetting “elastic” when describing collisions.

Things to Be Careful About

Make each assumption a separate, mark-point-ready statement.
Don’t repeat the same idea in different words (e.g. “negligible size” and “negligible volume” are essentially the same point unless clearly distinguished).

Techniques used

recall standard assumptions of the kinetic theory modelstate assumptions in precise, mark-point formavoid non-assumptions such as macroscopic gas laws

(b)

Explain how molecular movement causes the pressure exerted by a gas.

DifficultyMedium-Easy

Worked solution

Answer

Gas molecules move randomly and repeatedly collide with the container walls.
On each collision a molecule changes momentum (velocity reverses component normal to the wall), so the wall experiences an equal and opposite impulse.
The continual rate of change of momentum gives a steady force on the wall, and pressure is this force per unit area.

Final answer

Pressure is due to continual molecular collisions with the walls causing a net force (rate of momentum change) per unit area.

Detailed explanation

Background Concept

Pressure is defined as

p = \frac{F}{A},

where $F$ is the normal force on area $A$ . In mechanics,

F = \frac{\Delta p}{\Delta t}

(force equals rate of change of momentum). Kinetic theory connects these by considering the momentum changes when molecules collide with the container walls.

Understanding the Question

You are asked to explain the mechanism for gas pressure: what molecules do, what happens at the wall, and how that produces a force (and hence pressure).

Approach

Describe (1) random molecular motion, (2) collisions with walls, (3) momentum change during collisions, (4) force as rate of momentum change, (5) pressure as force per unit area.

Step-by-Step Reasoning

Molecules in a gas move in random directions with a range of speeds.
When a molecule hits a wall, it rebounds. The component of its velocity perpendicular to the wall reverses direction, so its momentum changes.
A momentum change requires a force; equivalently, the molecule exerts an impulse on the wall and the wall exerts an equal and opposite impulse on the molecule (Newton’s third law).
Many molecules collide with the wall every second. Adding all these tiny impulses gives a continuous average force on the wall:

F = \frac{\text{total momentum change per second}}{1}.

The pressure is then this force spread over the wall area:

p = \frac{F}{A}.

Higher molecular speeds or more molecules per unit volume produce more frequent/larger momentum changes, increasing the pressure.

Key Takeaways

Pressure comes from collisions with container walls.
The microscopic link is: collision (\rightarrow) momentum change (\rightarrow) force (\rightarrow) pressure.

Common Mistakes

Saying “molecules push on the walls” without mentioning collisions/momentum change.
Confusing pressure with energy; pressure is linked to momentum transfer, not directly to kinetic energy.
Forgetting the definition of pressure as force per unit area.

Things to Be Careful About

The force is an average result of many collisions; instantaneous forces fluctuate.
Only the velocity component perpendicular to the wall changes sign in a rebound; mentioning this helps show correct understanding, but keep it brief in an exam answer.

Techniques used

link molecular collisions to momentum changeuse force as rate of change of momentumconnect force per unit area to pressure

(c)

Fig. 4.1 shows the variation with thermodynamic temperature $T$ of the mean-square speeds $\langle c^2 \rangle$ for two gases X and Y.

Fig. 4.1

Fig. 4.2 shows the variation with $T$ of the product $pV$ for samples of the two gases, where $p$ is the pressure of the gas and $V$ is the volume of the gas.

Fig. 4.2

State three conclusions about the gases and their samples that may be drawn from Fig. 4.1 and Fig. 4.2. The conclusions may be qualitative or quantitative. Use the space below for any working that you need.

DifficultyMedium

Worked solution

Working

From Fig. 4.2, for both gases $pV \propto T$ and line passes through origin $\Rightarrow$ both behave as ideal gases with $pV = NkT$ .

From Fig. 4.1,

\langle c^2 \rangle = \frac{3kT}{m}\ \Rightarrow\ \text{gradient of }\langle c^2 \rangle\text{ vs }T = \frac{3k}{m}.

At $T = 400\ \text{K}$ : for X, $\langle c^2 \rangle = 5.0 \times 10^6\ \text{m}^2\text{s}^{-2}$ ; for Y, $\langle c^2 \rangle = 2.5 \times 10^6\ \text{m}^2\text{s}^{-2}$ .
So

\frac{\text{gradient}_X}{\text{gradient}_Y} = \frac{5.0}{2.5} = 2\ \Rightarrow\ \frac{m_Y}{m_X} = 2.

From Fig. 4.2 at $T = 400\ \text{K}$ : $pV_Y = 2.6 \times 10^3\ \text{J}$ and $pV_X = 0.9 \times 10^3\ \text{J}$ , hence

\frac{N_Y}{N_X} = \frac{(pV)_Y}{(pV)_X} = \frac{2.6}{0.9} \approx 2.9.

Answer

Three conclusions:

Both gases (both samples) behave as ideal gases: $pV \propto T$ (straight line through origin).
Gas X has smaller molecular mass than gas Y; $m_Y \approx 2m_X$ .
The sample of gas Y contains more molecules (more moles) than the sample of gas X; $N_Y \approx 2.9N_X$ .

Final answer

Both gases are ideal; Y has about twice the molecular mass of X; the Y sample has about 2.9 times as many molecules (moles) as the X sample.

Detailed explanation

Background Concept

Two key kinetic theory results connect microscopic motion to macroscopic variables:

Ideal gas equation (microscopic form):

pV = NkT,

where $N$ is the number of molecules, $k$ is the Boltzmann constant, and $T$ is thermodynamic temperature.

Pressure in terms of mean-square speed:

pV = \frac{1}{3}Nm\langle c^2 \rangle,

where $m$ is the mass of one molecule and $\langle c^2 \rangle$ is the mean of $c^2$ over molecules (related to $c_{\text{rms}}$ by $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ ).

Equating the two expressions for $pV$ gives

\frac{1}{3}Nm\langle c^2 \rangle = NkT \ \Rightarrow\ \langle c^2 \rangle = \frac{3kT}{m}.

So a graph of $\langle c^2 \rangle$ against $T$ should be a straight line through the origin for an ideal gas, and the gradient is inversely proportional to molecular mass $m$ .

Understanding the Question

You are given two graphs:

Fig. 4.1: $\langle c^2 \rangle$ versus $T$ for gases X and Y.
Fig. 4.2: $pV$ versus $T$ for samples of X and Y.

You must state three conclusions about the gases and their samples. “Conclusions” can include statements like “it behaves ideally”, “one gas has larger molecular mass”, or “one sample contains more moles”, supported by what the graphs show.

Approach

Use Fig. 4.2 to decide whether each sample obeys $pV \propto T$ (ideal gas behaviour) and to compare amounts of gas via $pV = NkT$ .
Use Fig. 4.1 and $\langle c^2 \rangle = 3kT/m$ to compare molecular masses by comparing gradients.
Extract simple ratios using a convenient common temperature (e.g. $400\ \text{K}$ ) because both lines pass through the origin.

Step-by-Step Reasoning

1) Ideal gas behaviour from the $pV$ – $T$ graph

Both lines in Fig. 4.2 are straight and pass through the origin. That means

pV \propto T

for both, which is exactly the behaviour predicted by $pV = NkT$ for an ideal gas sample of fixed $N$ .

So a valid conclusion is: both gases (their samples) behave ideally over the range shown.

2) Compare molecular masses from the $\langle c^2 \rangle$ – $T$ graph

From

\langle c^2 \rangle = \frac{3kT}{m},

the gradient of $\langle c^2 \rangle$ vs $T$ is $3k/m$ . Larger gradient means smaller $m$ .

At $400\ \text{K}$ , the graph gives roughly:

X: $\langle c^2 \rangle \approx 5.0 \times 10^6\ \text{m}^2\text{s}^{-2}$
Y: $\langle c^2 \rangle \approx 2.5 \times 10^6\ \text{m}^2\text{s}^{-2}$

So the gradient ratio is $5.0/2.5 = 2$ , hence

\frac{3k/m_X}{3k/m_Y} = 2 \ \Rightarrow\ \frac{m_Y}{m_X} = 2.

Conclusion: Y molecules are about twice as massive as X molecules.

3) Compare the amounts of gas in each sample from the $pV$ – $T$ graph

Using $pV = NkT$ , at the same temperature $T$ the ratio of $pV$ values equals the ratio of $N$ values:

\frac{(pV)_Y}{(pV)_X} = \frac{N_Y}{N_X} \quad (\text{at fixed }T).

At $400\ \text{K}$ , graph values are approximately:

$(pV)_Y \approx 2.6 \times 10^3\ \text{J}$
$(pV)_X \approx 0.9 \times 10^3\ \text{J}$

Thus

\frac{N_Y}{N_X} = \frac{2.6}{0.9} \approx 2.9.

Conclusion: the Y sample contains about $2.9$ times as many molecules (and therefore about $2.9$ times as many moles) as the X sample.

Key Takeaways

Straight line through origin on a $pV$ vs $T$ graph indicates ideal gas behaviour ( $pV \propto T$ ).
For an ideal gas, $\langle c^2 \rangle \propto T$ and the gradient gives information about molecular mass ( $\text{gradient} \propto 1/m$ ).
At fixed $T$ , comparing $pV$ compares the amount of gas because $pV = NkT$ .

Common Mistakes

Using $\langle c^2 \rangle \propto mT$ (wrong direction); it is inversely proportional to $m$ .
Comparing the values of $\langle c^2 \rangle$ at one temperature without recognising this is equivalent to comparing gradients only because both pass through the origin.
Saying “gas Y has higher pressure” from Fig. 4.2 without noting that $pV$ depends on both $p$ and $V$ and also on the amount of gas.

Things to Be Careful About

Keep track of what refers to the gas type (molecular mass) versus what refers to the sample (amount of gas $N$ ).
Use consistent powers of ten from the axis labels (e.g. $\langle c^2 \rangle /10^6$ and $pV/10^3$ ).
Ratios are often easiest and avoid needing to calculate the gradient explicitly when lines pass through the origin.

Techniques used

interpret linear graphs to infer proportionalitiesuse kinetic theory relations to compare molecular massesuse ratios from graph readings to compare sample amounts

Q5MediumOscillations

Fig. 5.1 shows a pendulum consisting of a metal sphere suspended by a thin string.

Fig. 5.1

The sphere undergoes small oscillations about its equilibrium position. The oscillations may be considered to be simple harmonic.

Fig. 5.2 shows the variation with time $t$ of the displacement $x$ of the sphere from its equilibrium position.

Fig. 5.2

(a)

On Fig. 5.1, draw an arrow, from the centre of the sphere, to represent the direction of the resultant force acting on the sphere when it is in the position shown.

DifficultyEasy

Worked solution

Answer

The resultant force is directed towards the equilibrium position (tangential to the arc, not along the string).

diagram-a-1

Final answer

Resultant force towards equilibrium position (tangential to the arc).

Detailed explanation

Background Concept

For small oscillations of a pendulum, the motion can be treated as simple harmonic. A key feature of SHM is that the resultant force is a restoring force: it always acts towards the equilibrium position and is proportional to the displacement (for small angles).

For a pendulum bob displaced to one side, there are two main forces: weight $mg$ vertically down and tension $T$ along the string. The resultant (net) force is not simply one of these; it is the vector sum. The component that causes the bob to accelerate along its path is the component of weight along the tangent to the arc, which points back towards equilibrium.

Understanding the Question

The diagram shows the bob displaced from the vertical (equilibrium) position. You are asked to draw an arrow from the centre of the sphere showing the direction of the resultant force at that instant.

Because it is displaced, the bob will accelerate back towards the centre position, so the resultant force must point that way.

Approach

Identify the equilibrium position (the lowest point, string vertical).
For a displaced pendulum bob, the motion is along a circular arc, so the restoring effect is along the tangent to the arc.
Draw an arrow from the bob pointing towards equilibrium along the direction of motion back to the centre.

Step-by-Step Reasoning

The bob is shown away from the equilibrium (string not vertical).
In SHM, the acceleration (and hence resultant force) is always towards the equilibrium position.
Therefore, the resultant force arrow must point from the bob back towards the lowest point, along the tangent to the path.

diagram-a-1

Key Takeaways

In SHM, the resultant force is a restoring force: it points towards equilibrium.
For a pendulum, the restoring effect is tangential to the arc (for the oscillatory motion).

Common Mistakes

Drawing the force along the string (tension direction) rather than towards equilibrium.
Drawing $mg$ (straight down) instead of the resultant force.

Things to Be Careful About

The question asks for the resultant force, not one of the individual forces.
The arrow should be drawn from the centre of the sphere, as stated.

Techniques used

identify the direction of the restoring (resultant) force for small oscillationsrelate resultant force direction to displacement from equilibrium

(b)

The mass of the sphere is $0.15\text{ kg}$ .

5M

(i)

State the amplitude of the oscillations.

amplitude = ______ $\text{m}$

(ii)

Determine the angular frequency of the oscillations.

angular frequency = ______ $\text{rad s}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

From Fig. 5.2, period $T = 0.40\ \text{s}$ .

\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40} = 15.7\ \text{rad s}^{-1}

Answer

$\omega = 15.7\ \text{rad s}^{-1}$ .

Final answer

15.7 rad s^-1

Detailed explanation

Background Concept

For SHM, angular frequency $\omega$ is related to the period $T$ by

\omega = \frac{2\pi}{T}.

$T$ is the time for one complete cycle (e.g. peak to next peak).
$\omega$ is measured in $\text{rad s}^{-1}$ .

Understanding the Question

The displacement–time graph shows a repeating sinusoidal motion. You must:

determine the period $T$ from the graph, and
calculate $\omega$ using $\omega = 2\pi/T$ .

Approach

Pick two identical points one cycle apart (e.g. successive maxima).
Read the time difference to find $T$ .
Substitute into $\omega = 2\pi/T$ .

Step-by-Step Reasoning

From the graph, one full cycle takes $0.40\ \text{s}$ (e.g. from a maximum to the next maximum).

T = 0.40\ \text{s}

Compute angular frequency:

\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40} = 15.7\ \text{rad s}^{-1}.

Key Takeaways

Period $T$ is obtained directly from the graph by measuring one full repeat.
Angular frequency is found using $\omega = 2\pi/T$ .

Common Mistakes

Using frequency $f$ directly without converting: remember $\omega = 2\pi f$ and $f = 1/T$ .
Measuring half a cycle (peak to trough) and using it as $T$ .

Things to Be Careful About

Choose two points that are clearly identical in the cycle (two peaks are usually easiest).
Keep enough significant figures; $T = 0.40\ \text{s}$ gives $\omega$ to about 3 s.f.

Techniques used

read the period from a displacement–time graphuse $\omega = 2\pi/T$ to calculate angular frequency

(iii)

Calculate the total energy of the oscillations.

total energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

For SHM,

E = \frac{1}{2} m\omega^2 A^2

E = \frac{1}{2}(0.15)(15.7)^2(0.015)^2 = 4.16\times 10^{-3}\ \text{J}

Answer

$E = 4.2\times 10^{-3}\ \text{J}$ .

Final answer

4.2 × 10^-3 J

Detailed explanation

Background Concept

In SHM, energy continually swaps between kinetic and potential forms, but the total energy remains constant (if damping is negligible).

For an object of mass $m$ undergoing SHM with amplitude $A$ and angular frequency $\omega$ , the total energy is

E = \frac{1}{2}m\omega^2 A^2.

This is also equal to the maximum kinetic energy (at $x=0$ ) and equal to the maximum potential energy (at $x=\pm A$ ).

Understanding the Question

You are told the mass $m = 0.15\ \text{kg}$ . From earlier parts (or from the graph) you obtain:

amplitude $A = 0.015\ \text{m}$
angular frequency $\omega = 15.7\ \text{rad s}^{-1}$

The question asks for the total energy of the oscillations.

Approach

Use the SHM energy formula

E = \frac{1}{2}m\omega^2A^2

and substitute the given values, keeping units in SI.

Step-by-Step Reasoning

Write the formula:

E = \frac{1}{2}m\omega^2A^2.

Substitute:

E = \frac{1}{2}(0.15)(15.7)^2(0.015)^2.

Calculate stepwise:

$(15.7)^2 \approx 246.5$
$(0.015)^2 = 2.25\times 10^{-4}$
$\frac{1}{2}(0.15) = 0.075$

So

E \approx 0.075\times 246.5\times 2.25\times 10^{-4} \approx 4.16\times 10^{-3}\ \text{J} \approx 4.2\times 10^{-3}\ \text{J}.

Key Takeaways

Total energy in SHM depends on $m$ , $\omega$ , and $A$ via $E = \tfrac{1}{2}m\omega^2A^2$ .
The total energy equals the maximum kinetic energy at equilibrium.

Common Mistakes

Using $E = \tfrac{1}{2}mv^2$ without finding $v_{\max}$ .
Forgetting to square $A$ or $\omega$ .
Using $A$ in cm rather than m.

Things to Be Careful About

Keep $A$ in metres and $\omega$ in $\text{rad s}^{-1}$ .
Quote the answer to a sensible number of significant figures consistent with the data (typically 2 s.f. here).

Techniques used

use the SHM total energy formulasubstitute amplitude and angular frequency into an energy equationhandle units consistently to obtain joules

(c)

On Fig. 5.3, sketch the variation with $x$ of the kinetic energy $E_K$ of the sphere.

Fig. 5.3

DifficultyMedium

Worked solution

Answer

Using $E_K = E - \frac{1}{2}m\omega^2x^2$ , the graph of $E_K$ against $x$ is a downward-opening parabola, symmetric about $x=0$ .

$E_K$ maximum at $x=0$ equal to total energy $\approx 4.2\times 10^{-3}\ \text{J}$ .
$E_K=0$ at $x=\pm A = \pm 0.015\ \text{m}$ .

diagram-c-1

Final answer

Downward-opening parabola: EK max at x=0 (≈4.2×10^-3 J), zero at x=±0.015 m.

Detailed explanation

Background Concept

In SHM, the total energy $E$ is constant and is shared between kinetic energy $E_K$ and potential energy $E_P$ :

E = E_K + E_P.

For a mass-spring SHM model,

E_P = \frac{1}{2}m\omega^2x^2,

so

E_K = E - \frac{1}{2}m\omega^2x^2.

This shows $E_K$ depends on $x^2$ , so the $E_K$ vs $x$ graph is a parabola, with:

maximum $E_K = E$ at $x=0$ (equilibrium),
$E_K = 0$ at $x = \pm A$ (turning points), where speed is zero.

Understanding the Question

You must sketch $E_K$ (vertical axis) against displacement $x$ (horizontal axis) on the provided axes. The scale suggests energies of a few $\times 10^{-3}\ \text{J}$ and displacements up to $\pm 0.02\ \text{m}$ .

From earlier parts:

amplitude $A = 0.015\ \text{m}$
total energy $E \approx 4.2\times 10^{-3}\ \text{J}$

These give the key points needed for the sketch.

Approach

Use the turning points: at $x=\pm A$ , speed is zero, so $E_K=0$ .
Use equilibrium: at $x=0$ , speed is maximum, so $E_K$ is maximum and equals the total energy.
Connect with a smooth, symmetric downward-opening parabola.

Step-by-Step Reasoning

At the extremes $x=\pm A = \pm 0.015\ \text{m}$ :

E_K = 0.

At equilibrium $x=0$ :

E_K = E \approx 4.2\times 10^{-3}\ \text{J}.

Since $E_K = E - \tfrac{1}{2}m\omega^2x^2$ , the dependence on $x^2$ makes the curve symmetric about $x=0$ and concave down.

So the sketch is a parabola passing through $(-0.015,0)$ and $(+0.015,0)$ with a maximum at $(0,4.2\times 10^{-3})$ .

diagram-c-1

Key Takeaways

$E_K$ is maximum at equilibrium and zero at turning points.
Because $E_K$ depends on $x^2$ , the $E_K$ – $x$ graph is a symmetric parabola.

Common Mistakes

Sketching a sine/cosine curve (energy is not sinusoidal with $x$ ).
Putting maximum kinetic energy at $x=\pm A$ instead of at $x=0$ .
Drawing a straight line or a V-shape instead of a parabola.

Things to Be Careful About

Use the amplitude value from the graph: the intercepts should be at $\pm 0.015\ \text{m}$ , not $\pm 0.02\ \text{m}$ .
The maximum $E_K$ should match the previously calculated total energy and fit the provided $E_K/10^{-3}\ \text{J}$ vertical scale.

Techniques used

use energy conservation in SHM to relate kinetic energy to displacementrecognise and sketch a quadratic relationshipuse turning points to set intercepts on the graph

Q6MediumElectric Fields

(a)

State Coulomb’s law.

DifficultyMedium-Easy

Worked solution

Answer

For two point charges $Q_1$ and $Q_2$ separated by distance $r$ , the force is along the line joining them and has magnitude

F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1Q_2}{r^2}.

Like charges repel and unlike charges attract.

Final answer

For two point charges separated by distance r, F = (1/4πϵ0)(Q1Q2/r^2) along the line joining them; like charges repel, unlike attract.

Detailed explanation

Background Concept

Coulomb’s law describes the electrostatic force between two point charges (or spherically symmetric charge distributions acting as point charges from outside). The key ideas are:

The force is an inverse-square force: it decreases with the square of separation $r$ .
The magnitude is proportional to the product of the charges.
The force acts along the straight line joining the charges.
The constant of proportionality in SI units is written using the permittivity of free space $\varepsilon_0$ .

Mathematically,

F = \frac{1}{4\pi\varepsilon_0}\frac{|Q_1Q_2|}{r^2}

for the magnitude, and the direction is such that like charges repel and unlike charges attract.

Understanding the Question

You are asked to state Coulomb’s law, so the examiner expects a clear statement of:

how $F$ depends on $Q_1$ , $Q_2$ , and $r$ (proportionalities or formula), and
the direction/nature of the force (along the line; attraction/repulsion).

Approach

Write the standard Coulomb’s law expression (with $1/(4\pi\varepsilon_0)$ ), then add one short sentence about direction and whether the force is attractive/repulsive.

Step-by-Step Reasoning

Identify the variables: two point charges $Q_1$ and $Q_2$ separated by distance $r$ .
State the inverse-square and product dependence using the full SI expression:

F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1Q_2}{r^2}.

Add the directional information required for a complete “law”: the force is along the line joining the charges, with repulsion for like charges and attraction for unlike charges.

Key Takeaways

Coulomb’s law is an inverse-square law: $F \propto 1/r^2$ .
The force depends on the product of charges and acts along the line joining them.

Common Mistakes

Forgetting to mention the direction (along the line joining charges).
Giving only proportionalities and omitting the constant $1/(4\pi\varepsilon_0)$ when a full equation is expected.
Mixing up attraction/repulsion for like/unlike charges.

Things to Be Careful About

Use $r^2$ (not $r$ ) in the denominator.
$Q_1Q_2$ may be positive or negative; the sign indicates whether the force is repulsive or attractive, but you should still state that in words.

Techniques used

state the inverse-square proportionality for point chargesspecify the direction of the force along the line joining the chargesstate the condition for attraction or repulsion based on charge signs

(b)

Fig. 6.1 shows an isolated hollow conducting sphere that is positively charged.

Fig. 6.1

On Fig. 6.1, draw field lines to represent the electric field outside the sphere.

DifficultyMedium-Easy

Worked solution

Answer

diagram-6b-1

Field lines are radial, perpendicular to the surface, with arrows pointing outwards, and are symmetrically spaced around the sphere.

Final answer

Radial field lines perpendicular to the surface with arrows pointing outwards (symmetric spacing).

Detailed explanation

Background Concept

Electric field lines are a way to represent the direction and relative strength of an electric field:

The direction of the electric field at a point is the direction of the force on a positive test charge.
Field lines point away from positive charge and towards negative charge.
The density (spacing) of field lines indicates the field strength (closer lines mean a stronger field).

For a charged conducting sphere in electrostatic equilibrium:

charge resides on the outer surface,
the surface is an equipotential, and
the electric field just outside the surface is perpendicular to the surface.

Understanding the Question

You are shown a positively charged isolated hollow conducting sphere and asked to draw the field lines outside the sphere.

So you must:

draw lines starting on the surface,
make them perpendicular to the surface everywhere,
show arrows pointing outward (positive charge),
and draw a symmetric pattern (because a sphere is spherically symmetric).

Approach

Use spherical symmetry: outside a charged sphere the field behaves like that of a point charge at the centre, so field lines are straight radial lines. Apply the conductor rule: field lines meet the surface at right angles.

Step-by-Step Reasoning

diagram-6b-1

Start several lines on the outer surface of the sphere at different points around it.
Draw each line as a straight line going directly away from the centre (radial).
Add arrowheads pointing away from the sphere (since it is positively charged).
Space the lines evenly around the sphere to show symmetry.

(You are not asked to draw inside the sphere here; if you did, there should be no lines inside for a conductor in electrostatic equilibrium.)

Key Takeaways

Positive charge: field lines point outwards.
Charged conducting sphere: field lines outside are radial and perpendicular to the surface.

Common Mistakes

Drawing curved lines or lines that are not radial.
Missing arrowheads or drawing arrows pointing inward.
Drawing field lines that touch the surface at an angle (not perpendicular).
Drawing an asymmetric pattern for a sphere.

Things to Be Careful About

Field lines should not cross.
Use enough lines to make the pattern clear and symmetric.
The lines should begin on the surface (for a positively charged isolated conductor).

Techniques used

use symmetry to infer a radial field pattern around a charged spheredraw field lines perpendicular to a conducting surfaceindicate field direction with arrows away from positive charge

(c)

Fig. 6.2 shows the variation of the electric field strength $E$ with distance $x$ from the centre of the sphere in (b).

Fig. 6.2

5M

(i)

Determine the radius, in cm, of the sphere.

radius = ______ $\text{cm}$

(ii)

Calculate the charge on the sphere.

charge = ______ $\text{C}$

DifficultyMedium

Worked solution

Working

At the surface, $r = 3.2\ \text{cm} = 3.2 \times 10^{-2}\ \text{m}$ and from the graph $E = 2.2 \times 10^{5}\ \text{N C}^{-1}$ .

For a charged sphere,

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

so

Q = 4\pi\varepsilon_0 r^2 E.

Q = 4\pi (8.85 \times 10^{-12}) (3.2 \times 10^{-2})^2 (2.2 \times 10^{5})

Q = 2.5 \times 10^{-8}\ \text{C}.

Answer

$Q = +2.5 \times 10^{-8}\ \text{C}$

Final answer

+2.5 × 10^-8 C

Detailed explanation

Background Concept

Outside a spherically symmetric charge distribution (including a charged conducting sphere), the electric field is the same as if all the charge were concentrated at the centre. The magnitude of the electric field at distance $r$ from the centre is

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

where:

$E$ is electric field strength in $\text{N C}^{-1}$ ,
$Q$ is charge in C,
$r$ is distance from the centre in m,
$\varepsilon_0$ is the permittivity of free space.

This is closely related to Coulomb’s law, since $E = F/q$ and $F = (1/4\pi\varepsilon_0)(Qq/r^2)$ .

Understanding the Question

You are given a graph of $E$ against distance $x$ from the centre. You must calculate the charge on the sphere.

From the graph you can read:

the radius of the sphere (where $E$ first becomes non-zero), and
the value of $E$ just outside the surface.

Then you substitute into the spherical field equation and solve for $Q$ .

Approach

Read $r$ from part (i) and convert to metres.
Read $E$ at the surface (the jump value).
Use

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

and rearrange to get $Q$ .

Step-by-Step Reasoning

Radius from the graph:

r = 3.2\ \text{cm} = 3.2 \times 10^{-2}\ \text{m}.

Electric field just outside the sphere (from the jump on the graph):

E = 2.2 \times 10^{5}\ \text{N C}^{-1}.

Rearrange the field equation:

E = \frac{Q}{4\pi\varepsilon_0 r^2} \quad\Rightarrow\quad Q = 4\pi\varepsilon_0 r^2 E.

Substitute values:

Q = 4\pi (8.85 \times 10^{-12}) (3.2 \times 10^{-2})^2 (2.2 \times 10^{5}).

Compute $r^2$ :

(3.2 \times 10^{-2})^2 = 1.024 \times 10^{-3}.

So

Q \approx 2.5 \times 10^{-8}\ \text{C}.

The sphere is stated to be positively charged, so the sign is $+$ .

Key Takeaways

Outside a charged conducting sphere, treat it like a point charge at the centre.
Use $E = Q/(4\pi\varepsilon_0 r^2)$ and ensure $r$ is in metres.

Common Mistakes

Forgetting to convert $\text{cm}$ to $\text{m}$ (this causes a factor of $10^{4}$ error in $Q$ ).
Using the wrong $E$ value (e.g. reading at a different $x$ rather than just outside the surface).
Missing the sign of the charge when the question context gives it.

Things to Be Careful About

The graph’s vertical axis is scaled; interpret it correctly (here it is already described in $\text{N C}^{-1}$ with a $10^{5}$ scale).
Use SI constants and SI units consistently.
Quote a sensible number of significant figures based on graph readings (typically 2 s.f. here).

Techniques used

use the field of a point charge/spherical conductor relationconvert cm to m consistently for SI unitsrearrange an inverse-square equation to solve for charge

(iii)

Suggest an explanation for the fact that the electric field inside the sphere is zero.

DifficultyMedium-Easy

Worked solution

Answer

The sphere is a conductor in electrostatic equilibrium, so charge resides on the outer surface and the field inside cancels to zero (the interior is at constant potential).

Final answer

In electrostatic equilibrium, charges lie on the outer surface of a conductor so the interior is at constant potential and the net electric field inside is zero.

Detailed explanation

Background Concept

In electrostatic equilibrium, free electrons in a conductor can move. If there were any electric field inside the conducting material, charges would experience a force $F = qE$ and would continue to move. Equilibrium is only reached when:

the electric field within the conducting material is zero, and
the conductor becomes an equipotential (potential is constant throughout the conductor).

For an isolated charged hollow conductor with no charge inside the cavity, all excess charge resides on the outer surface. The resulting field in the cavity (and within the conductor) is zero: this is electrostatic shielding.

Understanding the Question

The graph shows $E = 0$ for distances from the centre up to the inner region (and up to the surface), and you are asked to suggest why the field inside is zero.

So the expected explanation is based on the properties of conductors in electrostatic equilibrium and the fact that charges reside on the outside surface.

Approach

State the electrostatic equilibrium condition for conductors: if $E \neq 0$ inside, charges would move. Therefore charges redistribute onto the surface such that the net field inside is zero, making the interior an equipotential.

Step-by-Step Reasoning

The sphere is conducting, so it contains mobile charge carriers.
If there were an electric field inside, these charges would feel a force and move, so the situation would not be static.
Charges therefore redistribute themselves on the outer surface until the net electric field everywhere inside the conductor (and inside the hollow region, if empty) is zero.
This redistribution also makes the potential constant inside.

Key Takeaways

A conductor in electrostatic equilibrium has $E = 0$ inside.
Excess charge on an isolated conductor resides on the outer surface.

Common Mistakes

Saying “there is no charge inside” without linking it to conductor equilibrium (the key idea is redistribution until $E=0$ ).
Confusing the field inside the conductor with the field just outside (which is not zero).

Things to Be Careful About

The correct reasoning is about mobile charges and equilibrium, not simply “the sphere is hollow”.
If there were a charge placed inside the cavity, the field in the cavity would not generally be zero; the question refers to the given isolated charged sphere with no internal charge mentioned.

Techniques used

apply the concept of electrostatic equilibrium in a conductoruse charge distribution on a conductor to infer internal field cancellation

Q7MediumCapacitance

(a)

Define the capacitance of a parallel-plate capacitor.

(b)

An initially uncharged capacitor X, of capacitance $C$ , is gradually charged so that the final potential difference (p.d.) between its plates is $V$ and the final charge is $Q$ .

4M

(i)

On Fig. 7.1, sketch the variation of charge with p.d. for capacitor X as the p.d. increases from 0 to $V$ .

Fig. 7.1

(ii)

Determine an expression, in terms of $Q$ and $V$ , for the work $W$ done on capacitor X during the charging process. Explain your reasoning.

$W$ = ______

DifficultyMedium-Easy

Worked solution

Working

Work done is area under the $V$ – $Q$ graph:

W = \int_0^Q V\,\mathrm{d}Q

Since $V \propto Q$ , this area is a triangle:

W = \frac{1}{2}QV

Answer

W = \frac{1}{2}QV

Final answer

W = 1/2 QV

Detailed explanation

Background Concept

When charging a capacitor, the p.d. across it is not constant: it increases as more charge is placed on the plates. The small amount of work done to move an additional small charge $\mathrm{d}Q$ onto the capacitor when the p.d. is $V$ is

\mathrm{d}W = V\,\mathrm{d}Q

So total work (energy stored) is

W = \int_0^Q V\,\mathrm{d}Q

Geometrically, $W$ is the area under the graph of $V$ (y-axis) against $Q$ (x-axis).

Understanding the Question

You know the capacitor ends with charge $Q$ and p.d. $V$ . You must write $W$ in terms of just $Q$ and $V$ and explain the reasoning (so you need either the integral argument or the area-under-graph argument).

Approach

Use that for a capacitor, $Q = CV$ so $V$ increases linearly with $Q$ during charging. That makes the $V$ – $Q$ graph a straight line from $(0,0)$ to $(Q,V)$ , so the area under it is a triangle.

Step-by-Step Reasoning

At some intermediate stage when the capacitor has charge $q$ , its p.d. is

V = \frac{q}{C}

The work done to add a small extra charge $\mathrm{d}q$ is

\mathrm{d}W = V\,\mathrm{d}q = \frac{q}{C}\,\mathrm{d}q

Integrate from $q=0$ to $q=Q$ :

W = \int_0^Q \frac{q}{C}\,\mathrm{d}q = \frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q = \frac{Q^2}{2C}

Use $V = Q/C$ (final state), giving

W = \frac{1}{2}QV

This matches the triangle-area idea: area $= \tfrac{1}{2}\times (\text{base }Q) \times (\text{height }V)$ .

Key Takeaways

Energy stored in a capacitor is

W = \frac{1}{2}QV

The factor $1/2$ appears because $V$ rises from $0$ to $V$ as charge builds up.

Common Mistakes

Using $W = QV$ (incorrect because $V$ is not constant during charging).
Mixing up which graph gives the area: it must be $V$ against $Q$ for area = $\int V\,\mathrm{d}Q$ .

Things to Be Careful About

The question asks for an expression in terms of $Q$ and $V$ , so the final form should be $\tfrac{1}{2}QV$ (not $\tfrac{Q^2}{2C}$ unless you then substitute $C = Q/V$ ).
State clearly that the relationship is linear, so the area is a triangle.

Techniques used

use work done as area under the V–Q graphapply W = \int V\,\mathrm{d}Quse linear relationship between Q and V

(c)

Another capacitor Y is initially uncharged. The fully charged capacitor X in (b) is now connected to capacitor Y, as shown in Fig. 7.2.

Fig. 7.2

The capacitance of capacitor Y is $3C$ .

4M

(i)

Complete Table 7.1 to show expressions, in terms of $Q$ and $V$ , for the final p.d.s across, and the final charges on, the two capacitors.
Use the space below for any working that you need.

Table 7.1

	X	Y
final p.d.
final charge

DifficultyMedium

Worked solution

Working

Initial charge on X is $Q$ and total charge is conserved.
For parallel capacitors, final p.d. is the same: $V_f$ .

Q = Q_X + Q_Y = CV_f + 3CV_f = 4CV_f

V_f = \frac{Q}{4C} = \frac{V}{4} \quad (\text{since } Q=CV)

Q_X = CV_f = \frac{Q}{4}, \qquad Q_Y = 3CV_f = \frac{3Q}{4}

Answer

Final p.d. across X and Y: $\dfrac{V}{4}$ each.

Final charge on X: $\dfrac{Q}{4}$ .

Final charge on Y: $\dfrac{3Q}{4}$ .

Final answer

Final p.d.s: V/4 on both; final charges: X = Q/4, Y = 3Q/4.

Detailed explanation

Background Concept

When capacitors are connected in parallel:

they share the same final potential difference $V_f$ ;
charges add: $Q_{\text{total}} = Q_1 + Q_2$ .

If the capacitors are isolated from any battery/external circuit, the total charge is conserved (it can move between capacitors, but cannot leave the system).

Also, for each capacitor:

Q = CV

Understanding the Question

Capacitor X (capacitance $C$ ) is initially charged to p.d. $V$ with charge $Q$ (so $Q=CV$ ). Capacitor Y is initially uncharged and has capacitance $3C$ . They are then connected together in parallel in a closed loop. You must find the final p.d. across each and the final charge on each, in terms of the original $Q$ and $V$ .

Approach

Let the common final p.d. be $V_f$ .
Use $Q_X = CV_f$ and $Q_Y = 3C V_f$ .
Use conservation of total charge: initial $Q_{\text{total}}=Q$ , final $Q_X+Q_Y=Q$ .
Solve for $V_f$ , then substitute back to get $Q_X$ and $Q_Y$ .

Step-by-Step Reasoning

After connection, X and Y are in parallel, so

V_X = V_Y = V_f

Express each final charge using $Q=CV$ :

Q_X = C V_f

Q_Y = 3C V_f

Total charge is conserved (system isolated), so

Q = Q_X + Q_Y = C V_f + 3C V_f = 4C V_f

Solve for $V_f$ :

V_f = \frac{Q}{4C}

Since initially $Q = CV$ , then

V_f = \frac{CV}{4C} = \frac{V}{4}

Now find final charges:

Q_X = C\left(\frac{V}{4}\right) = \frac{CV}{4} = \frac{Q}{4}

Q_Y = 3C\left(\frac{V}{4}\right) = \frac{3CV}{4} = \frac{3Q}{4}

Key Takeaways

Parallel capacitors have the same final p.d.
If disconnected from a supply, total charge is conserved.
Charge divides in proportion to capacitance: larger $C$ gets larger $Q$ at the same $V_f$ .

Common Mistakes

Assuming the p.d. stays at $V$ (it drops because the total capacitance increases).
Assuming charge stays the same on capacitor X (charge redistributes).
Using series rules instead of parallel rules.

Things to Be Careful About

The question wants expressions in terms of $Q$ and $V$ ; using $V_f = Q/(4C)$ is fine if you then replace $Q/C$ with $V$ .
Be clear that Y was initially uncharged, but it gains charge after connection.

Techniques used

apply conservation of chargeuse common potential difference for parallel capacitorsuse Q = CV to find redistributed charges

(ii)

State whether the total energy stored in the two capacitors is less than, the same as, or greater than the energy initially stored in capacitor X.

DifficultyMedium-Easy

Worked solution

Working

Initial energy in X:

W_i = \frac{1}{2}CV^2

Final common p.d. is $V_f = V/4$ and total capacitance is $C_{\text{eq}} = C + 3C = 4C$ .

W_f = \frac{1}{2}C_{\text{eq}}V_f^2 = \frac{1}{2}(4C)\left(\frac{V}{4}\right)^2 = \frac{1}{8}CV^2

So $W_f < W_i$ .

Answer

Less than.

Final answer

Less than.

Detailed explanation

Background Concept

Energy stored in a capacitor can be written as

W = \frac{1}{2}CV^2 = \frac{1}{2}QV

When a charged capacitor is connected to another capacitor, charge flows through connecting wires until both reach the same p.d. If there is no external power supply, total charge is conserved but energy is not necessarily conserved: some energy can be transferred to internal energy of the circuit (heating) and electromagnetic radiation during the transient current.

Understanding the Question

Initially, only capacitor X (capacitance $C$ ) is charged to p.d. $V$ . Capacitor Y (capacitance $3C$ ) is uncharged. After connecting them in parallel, both end up at a lower common p.d. You must state whether the total final energy stored in both capacitors is less than, the same as, or greater than the initial energy stored in X.

Approach

Compute and compare:

initial energy stored in X: $W_i = \tfrac{1}{2}CV^2$ ;
final energy stored in the equivalent parallel combination at the final p.d. $V_f$ .

Step-by-Step Reasoning

Initial energy:

W_i = \frac{1}{2}CV^2

From part (c)(i), the final p.d. is

V_f = \frac{V}{4}

Two capacitors in parallel have equivalent capacitance

C_{\text{eq}} = C + 3C = 4C

Final total stored energy (sum of energies in both) equals energy of the equivalent capacitor at $V_f$ :

W_f = \frac{1}{2}C_{\text{eq}}V_f^2 = \frac{1}{2}(4C)\left(\frac{V}{4}\right)^2 = \frac{1}{8}CV^2

Compare: $\tfrac{1}{8}CV^2$ is smaller than $\tfrac{1}{2}CV^2$ , so the final total energy is less.

Physically, the missing energy is dissipated mainly as heat in the connecting wires during the brief current flow (and a small amount as electromagnetic radiation).

Key Takeaways

$W = \tfrac{1}{2}CV^2$ is very convenient for comparing energies before/after redistribution.
Charge is conserved in an isolated capacitor system, but energy stored can decrease due to dissipation during charge flow.

Common Mistakes

Saying “energy is conserved so it is the same” (not true here because resistive heating occurs).
Using the original $V$ instead of the reduced $V_f$ .

Things to Be Careful About

You are comparing stored energy in the capacitors, not total energy of the universe.
Even if wires are said to be “negligible resistance” in circuit diagrams, in reality some energy loss still occurs during redistribution; exam questions expect the conclusion that final stored energy is less.

Techniques used

compare initial and final energy using W = 1/2 CV^2use equivalent capacitance for parallel capacitorsuse charge conservation to determine final potential difference

Q8MediumAlternating Currents

(a)

State what is meant by the frequency of an alternating current.

(b)

An alternating current $I$ in a resistor of resistance $680\ \Omega$ varies with time $t$ according to

I = 3.5 \sin(40\pi t)

where $I$ is in A and $t$ is in s.

5M

(i)

Show that the period of the alternating current is $50\text{ ms}$ .

DifficultyMedium-Easy

Worked solution

Working

Given $I = 3.5\sin(40\pi t)$ , so $\omega = 40\pi\ \text{rad s}^{-1}$ .

T = \frac{2\pi}{\omega} = \frac{2\pi}{40\pi} = \frac{1}{20} = 0.050\ \text{s} = 50\ \text{ms}

Answer

$T = 50\ \text{ms}$

Final answer

50 ms

Detailed explanation

Background Concept

A sinusoidal alternating quantity can be written as

I = I_0\sin(\omega t)

where:

$I_0$ is the peak current,
$\omega$ is the angular frequency in $\text{rad s}^{-1}$ ,
$t$ is time.

The angular frequency and period are related by

\omega = \frac{2\pi}{T}\quad\Rightarrow\quad T = \frac{2\pi}{\omega}

Understanding the Question

You are given the current-time equation

I = 3.5\sin(40\pi t)

and asked to show that the period is $50\ \text{ms}$ . So you must extract $\omega$ and then calculate $T$ .

Approach

Compare the given equation with $I = I_0\sin(\omega t)$ to identify $\omega$ .
Use $T = 2\pi/\omega$ .
Convert the result from seconds to milliseconds.

Step-by-Step Reasoning

From

I = 3.5\sin(40\pi t)

we see the coefficient of $t$ inside the sine is the angular frequency:

\omega = 40\pi\ \text{rad s}^{-1}

Now calculate period:

T = \frac{2\pi}{\omega} = \frac{2\pi}{40\pi} = \frac{1}{20}\ \text{s} = 0.050\ \text{s}

Convert seconds to milliseconds ( $1\ \text{s} = 1000\ \text{ms}$ ):

0.050\ \text{s} = 50\ \text{ms}

Key Takeaways

In $\sin(\omega t)$ , the coefficient of $t$ is $\omega$ .
Use $T = 2\pi/\omega$ for any sinusoid.

Common Mistakes

Using $T = 1/\omega$ (incorrect; that would only work if $\omega = 2\pi f$ is forgotten).
Treating $40\pi$ as the frequency $f$ instead of angular frequency $\omega$ .

Things to Be Careful About

Keep track of units: $T$ comes out in seconds from $2\pi/\omega$ .
Always convert to ms only at the end.

Techniques used

identify angular frequency from a sinusoidal expressionuse the relationship between angular frequency and periodconvert seconds to milliseconds

(ii)

On Fig. 8.1, sketch the variation of $I$ with $t$ between $t = 0$ and $t = 100\text{ ms}$ .

Fig. 8.1

DifficultyMedium-Easy

Worked solution

diagram-8bii-1

Answer

A sine wave of amplitude $3.5\ \text{A}$ and period $50\ \text{ms}$ , starting at $I=0$ at $t=0$ with positive gradient, completing two full cycles between $0$ and $100\ \text{ms}$ .

Key points: $I=+3.5\ \text{A}$ at $12.5\ \text{ms}$ , $I=0$ at $25\ \text{ms}$ , $I=-3.5\ \text{A}$ at $37.5\ \text{ms}$ , $I=0$ at $50\ \text{ms}$ , then repeat to $100\ \text{ms}$ .

Final answer

Sinusoidal I–t graph with amplitude 3.5 A, period 50 ms, starting at 0 with positive slope; two full cycles from 0 to 100 ms.

Detailed explanation

Background Concept

For a sinusoidal alternating current

I = I_0\sin(\omega t)

$I_0$ is the peak (amplitude).
The period is $T = 2\pi/\omega$ .

A sine wave has standard features over one period:

starts at zero and rises if it is a positive sine,
reaches +peak at $T/4$ ,
returns to zero at $T/2$ ,
reaches -peak at $3T/4$ ,
returns to zero at $T$ .

Understanding the Question

You must sketch $I$ against $t$ from $0$ to $100\ \text{ms}$ for

I = 3.5\sin(40\pi t)

So you need the correct amplitude ( $3.5\ \text{A}$ ), the correct period ( $50\ \text{ms}$ from part (i)), and the correct phase (starts at $I=0$ going positive).

Approach

Read off $I_0 = 3.5\ \text{A}$ .
Use $T = 50\ \text{ms}$ .
Mark the key quarter-period times and corresponding current values.
Draw a smooth sine curve through these points for two periods (since $100\ \text{ms} = 2T$ ).

Step-by-Step Reasoning

Peak current:

I_0 = 3.5\ \text{A}

Period from (i):

T = 50\ \text{ms}

Quarter period:

\frac{T}{4} = 12.5\ \text{ms}

Now list points for one cycle:

at $t=0$ : $I=0$ (since $\sin 0 = 0$ ) and it rises (positive gradient).
at $t=12.5\ \text{ms}$ : $I=+3.5\ \text{A}$ .
at $t=25\ \text{ms}$ : $I=0$ .
at $t=37.5\ \text{ms}$ : $I=-3.5\ \text{A}$ .
at $t=50\ \text{ms}$ : $I=0$ .

Then repeat the same pattern for the second cycle from $50$ to $100\ \text{ms}$ .

diagram-8bii-1

Key Takeaways

The amplitude sets the vertical scale; the period sets the horizontal repetition.
For $\sin(\omega t)$ , the graph starts at zero and initially increases.

Common Mistakes

Using amplitude $3.5/\sqrt{2}$ instead of $3.5$ (confusing peak with r.m.s.).
Drawing only one cycle instead of two between $0$ and $100\ \text{ms}$ .
Starting at the peak at $t=0$ (that would correspond to a cosine).

Things to Be Careful About

Ensure the curve crosses the time axis at $0, 25, 50, 75, 100\ \text{ms}$ .
Peaks should align at $12.5\ \text{ms}$ and $62.5\ \text{ms}$ ; troughs at $37.5\ \text{ms}$ and $87.5\ \text{ms}$ .
Use a smooth sinusoidal shape rather than straight line segments.

Techniques used

extract amplitude and period from a sinusoidal expressionidentify key points of a sine wave over one or more cyclessketch a correctly scaled sinusoidal graph

(iii)

Determine the root-mean-square (r.m.s.) current in the resistor.

r.m.s. current = ______ $\text{A}$

(c)

Use data from (b), including your answer in (b)(iii), to show by calculation that the mean power in the $680\ \Omega$ resistor is half of the peak power.

DifficultyMedium

Worked solution

Working

Mean power:

P_{\text{mean}} = I_{\text{rms}}^2 R = (2.47)^2 \times 680 = 4.17 \times 10^3\ \text{W}

Peak power:

P_{\text{peak}} = I_0^2 R = (3.5)^2 \times 680 = 8.33 \times 10^3\ \text{W}

\frac{P_{\text{mean}}}{P_{\text{peak}}} = \frac{4.17\times 10^3}{8.33\times 10^3} = 0.50

Answer

$P_{\text{mean}} = \tfrac{1}{2}P_{\text{peak}}$

Final answer

Mean power is half of peak power (P_mean = 0.50 P_peak).

Detailed explanation

Background Concept

For a resistor of resistance $R$ , instantaneous electrical power is

P = I^2R

For a sinusoidal current, the current changes continuously, so the instantaneous power also changes. The mean power over a cycle is found using the r.m.s. current:

P_{\text{mean}} = I_{\text{rms}}^2 R

The peak power occurs when the current is at its maximum magnitude $I_0$ :

P_{\text{peak}} = I_0^2 R

For a sine wave, $I_{\text{rms}} = I_0/\sqrt{2}$ , which leads to a factor of 2 difference in the powers.

Understanding the Question

You have a resistor $R = 680\ \Omega$ and a current

I = 3.5\sin(40\pi t)

So $I_0 = 3.5\ \text{A}$ and from part (b)(iii) $I_{\text{rms}} = 2.47\ \text{A}$ . You must use these to show that the mean power is half the peak power.

Approach

Calculate $P_{\text{mean}}$ using $I_{\text{rms}}^2R$ .
Calculate $P_{\text{peak}}$ using $I_0^2R$ .
Compare by ratio (or show algebraically that using $I_{\text{rms}}=I_0/\sqrt{2}$ gives a factor $1/2$ ).

Step-by-Step Reasoning

Mean power:

P_{\text{mean}} = I_{\text{rms}}^2 R = (2.47\ \text{A})^2(680\ \Omega)

(2.47)^2 = 6.10

P_{\text{mean}} = 6.10 \times 680 = 4.17 \times 10^3\ \text{W}

Peak power (at $I = I_0$ ):

P_{\text{peak}} = I_0^2 R = (3.5\ \text{A})^2(680\ \Omega)

(3.5)^2 = 12.25

P_{\text{peak}} = 12.25 \times 680 = 8.33 \times 10^3\ \text{W}

Compare:

\frac{P_{\text{mean}}}{P_{\text{peak}}} = \frac{4.17\times 10^3}{8.33\times 10^3} = 0.50

Hence,

P_{\text{mean}} = \frac{1}{2}P_{\text{peak}}

(You can also see this directly since $I_{\text{rms}}^2 = (I_0^2/2)$ for a sinusoid, so $I_{\text{rms}}^2R$ is half of $I_0^2R$ .)

Key Takeaways

For a resistor, power depends on $I^2$ .
Mean power in a resistor with sinusoidal a.c. is found using $I_{\text{rms}}$ .
Because $I_{\text{rms}} = I_0/\sqrt{2}$ , mean power is half peak power.

Common Mistakes

Using $P = IV$ without having (or correctly relating) the voltage.
Using $P_{\text{mean}} = I_{\text{mean}}^2R$ (the mean current over a full cycle is zero, but mean power is not).
Forgetting to square the current in $P = I^2R$ .

Things to Be Careful About

Distinguish clearly between peak current $I_0$ and r.m.s. current $I_{\text{rms}}$ .
When comparing powers, it is often easiest to form a ratio so units cancel and the factor of 2 is obvious.
Quote powers in $\text{W}$ and use standard form where appropriate.

Techniques used

use resistive power relations P = I^2 Rcalculate mean power using r.m.s. valuescalculate peak power using peak currentcompare quantities by forming a ratio

Q9Medium-EasyQuantum PhysicsElectric Fields

Electrons in a vacuum are accelerated from rest through a potential difference (p.d.) $V$ to form a beam. The electrons each have mass $m$ and charge $q$ .

The beam is incident on a graphite crystal that acts as a diffraction grating. After passing through the crystal, the beam reaches a fluorescent screen. An interference pattern is observed on this screen.

(a)

Explain what this observation shows about the nature of electrons.

DifficultyEasy

Worked solution

Answer

An interference pattern shows electrons undergo diffraction and interference, so electrons have wave nature (wave-particle duality).

Final answer

Electrons show wave behaviour (wave-particle duality).

Detailed explanation

Background Concept

Interference and diffraction are characteristic behaviours of waves. Interference patterns (bright and dark fringes) arise when waves superpose and produce regions of constructive and destructive interference.

In quantum physics, particles such as electrons can also behave as waves. This is called wave-particle duality. Evidence for the wave nature of matter is provided when particles produce diffraction/interference patterns.

Understanding the Question

A beam of electrons is accelerated in a vacuum and then sent through a graphite crystal, which acts like a diffraction grating. A fluorescent screen shows an interference pattern after the electrons pass through the crystal.

The question asks what this observation implies about the nature of electrons.

Approach

Use the key idea: a diffraction grating produces an interference pattern only if the incident entity behaves like a wave with a wavelength comparable to the atomic spacing in the crystal.

Step-by-Step Reasoning

The graphite crystal has a regular atomic structure, like a grating for waves.
The observed pattern on the fluorescent screen is an interference pattern.
Interference requires superposition of waves, so the electrons must be exhibiting wave behaviour.
Therefore, electrons have wave properties (in addition to their known particle properties), i.e. wave-particle duality.

Key Takeaways

Interference/diffraction patterns are direct evidence of wave behaviour.
Electrons (matter) exhibit wave-particle duality.

Common Mistakes

Saying only “electrons are waves” without acknowledging they also show particle behaviour (duality).
Mentioning photons/light instead of electrons.

Things to Be Careful About

The key mark point is the connection: interference pattern (\Rightarrow) wave nature.
Keep the statement general: electrons have wave properties / behave like waves in this experiment.

Techniques used

link the presence of an interference pattern to wave behaviourinfer wave-particle duality from diffraction evidence

(b)

Determine an expression, in terms of $m$ , $q$ and $V$ , for the momentum $p$ of an electron in the beam.

$p$ = ______

DifficultyMedium-Easy

Worked solution

Working

Energy gained from p.d. $V$ :

qV = \frac{1}{2}mv^2

v = \sqrt{\frac{2qV}{m}}

Momentum $p = mv$ :

p = m\sqrt{\frac{2qV}{m}} = \sqrt{2mqV}

Answer

p = \sqrt{2mqV}

Final answer

p = \sqrt{2mqV}

Detailed explanation

Background Concept

When a charge moves through a potential difference (V), its electrical potential energy changes by

\Delta E = qV

If the electron starts from rest in a vacuum, we usually assume (at these typical exam speeds) that this electrical energy transfer becomes kinetic energy:

qV = \Delta E_k = \frac{1}{2}mv^2

Momentum for non-relativistic speeds is

p = mv

Understanding the Question

Each electron starts from rest and is accelerated through a potential difference (V). You are asked to find an expression for the electron momentum (p) in terms of (m), (q), and (V).

So you need to connect (V) to the electron speed (v), then convert (v) to (p).

Approach

Use the energy conversion (qV = \frac{1}{2}mv^2) to find (v).
Substitute (v) into (p = mv).

Step-by-Step Reasoning

Start with the energy gained from the potential difference:

qV = \frac{1}{2}mv^2

Solve for (v):

v^2 = \frac{2qV}{m}

v = \sqrt{\frac{2qV}{m}}

Now use (p = mv):

p = m\sqrt{\frac{2qV}{m}}

Take (m) inside the square root carefully:

p = \sqrt{m^2\cdot \frac{2qV}{m}} = \sqrt{2mqV}

This gives the required expression.

Key Takeaways

Potential difference transfers energy (qV) to a charge.
For an electron starting from rest: (qV = \frac{1}{2}mv^2).
Combine with (p = mv) to get (p = \sqrt{2mqV}).

Common Mistakes

Using (qV = mv^2) (missing the (\tfrac{1}{2})).
Writing (p = \sqrt{\frac{2qV}{m}}) (forgetting the factor of (m) from (p=mv)).
Confusing charge sign: the magnitude is used; momentum is a magnitude here.

Things to Be Careful About

This derivation assumes non-relativistic speeds. At very large (V), relativity would be needed, but A-Level questions typically intend the classical result.
Keep algebra under the square root consistent: (m\sqrt{\frac{1}{m}} = \sqrt{m}), not (\frac{m}{\sqrt{m}}) unless simplified correctly.

Techniques used

equate electrical energy gained to kinetic energy gaineduse the non-relativistic kinetic energy formularelate momentum to speed using p = mv

(c)

The p.d. through which the electrons are accelerated is now increased to a greater value.

Describe and explain the effect of this change on the interference pattern observed.

DifficultyMedium-Easy

Worked solution

Answer

Increasing $V$ increases electron momentum $p$ (since $p = \sqrt{2mqV}$ ), so

\lambda = \frac{h}{p}

decreases. Hence diffraction angles reduce and the interference fringes/rings become closer together (pattern contracts towards the centre).

Final answer

Higher V gives smaller (\lambda), so smaller diffraction angle and reduced fringe/ring spacing (pattern contracts).

Detailed explanation

Background Concept

For matter waves, the de Broglie wavelength is

\lambda = \frac{h}{p}

so higher momentum means shorter wavelength.

In diffraction/interference, the characteristic diffraction angle generally increases with wavelength. For a grating-like spacing (d), maxima satisfy a condition like

d\sin\theta \approx n\lambda

So if (\lambda) decreases, then for the same order (n) the angle (\theta) decreases.

Understanding the Question

The electrons are accelerated through a potential difference (V) and produce an interference pattern after passing through graphite. Now (V) is increased.

You must describe what happens to the pattern and explain why.

Approach

Use the result from part (b): increasing (V) increases the electron speed and momentum.
Use (\lambda = h/p): increasing (p) decreases (\lambda).
Smaller (\lambda) gives smaller diffraction angles, so the pattern shrinks (fringes/rings get closer together).

Step-by-Step Reasoning

From energy gain, (p = \sqrt{2mqV}). If (V) increases, then (p) increases.
De Broglie wavelength:

\lambda = \frac{h}{p}

So when (p) increases, (\lambda) decreases.

Diffraction/interference depends on wavelength compared with the crystal spacing. A smaller (\lambda) leads to smaller values of (\theta) for the same diffraction order.
On a screen a distance away, smaller (\theta) means the bright fringes/rings appear closer to the centre, i.e. smaller spacing between features and a more "compressed" pattern.

Key Takeaways

Increasing accelerating p.d. increases momentum.
Higher momentum (\Rightarrow) smaller de Broglie wavelength.
Smaller wavelength (\Rightarrow) less diffraction, hence a more compact interference pattern.

Common Mistakes

Saying “fringes get further apart” (it is the opposite when (\lambda) decreases).
Forgetting to mention the link (V \uparrow \Rightarrow p \uparrow) before using (\lambda = h/p).

Things to Be Careful About

Be explicit about both the description (what you see) and the explanation (why it changes).
Use the correct direction of proportionality: (\lambda \propto 1/p), not (\lambda \propto p).

Techniques used

relate accelerating potential difference to electron momentumapply the de Broglie relation to link momentum and wavelengthconnect wavelength change to diffraction angle / fringe spacing

(d)

The electrons are now accelerated through different values of $V$ , resulting in pairs of corresponding values for $p$ and the de Broglie wavelength $\lambda$ .

3M

(i)

On Fig. 9.1, sketch the variation of $p$ with $\frac{1}{\lambda}$ .

Fig. 9.1

DifficultyMedium-Easy

Worked solution

Working

From de Broglie relation:

\lambda = \frac{h}{p} \;\Rightarrow\; p = h\left(\frac{1}{\lambda}\right)

Answer

diagram-9di

Straight line through the origin with positive gradient.

Final answer

Straight line through origin, p proportional to 1/\lambda.

Detailed explanation

Background Concept

The de Broglie relation connects momentum and wavelength:

\lambda = \frac{h}{p}

If you want to plot a straight-line graph, rewrite it in the form (y = mx + c). Here, choosing (p) as (y) and (\frac{1}{\lambda}) as (x) gives a linear relationship.

Understanding the Question

You are asked to sketch how electron momentum (p) varies with (\frac{1}{\lambda}) on axes that start at the origin.

So you need the mathematical relationship between (p) and (\frac{1}{\lambda}).

Approach

Rearrange (\lambda = \frac{h}{p}) to make (p) the subject and express it in terms of (\frac{1}{\lambda}). Then infer the graph shape (straight line, intercept, gradient sign).

Step-by-Step Reasoning

Start with:

\lambda = \frac{h}{p}

Invert both sides:

\frac{1}{\lambda} = \frac{p}{h}

Rearrange to make (p) the subject:

p = h\left(\frac{1}{\lambda}\right)

This is of the form (y = mx) with:

(y \equiv p)
(x \equiv \frac{1}{\lambda})
gradient (m = h)
intercept (c = 0)

So the sketch is a straight line through the origin with positive gradient.

diagram-9di

Key Takeaways

Rearranging equations to (y = mx + c) tells you the graph shape.
(p) is directly proportional to (\frac{1}{\lambda}) with proportionality constant (h).

Common Mistakes

Drawing a curve (it should be a straight line).
Drawing a non-zero intercept (it passes through the origin).
Plotting (p) against (\lambda) instead of (1/\lambda).

Things to Be Careful About

The axes are labelled (p) and (1/\lambda), so your sketch must match those variables.
Ensure the gradient is positive: as (1/\lambda) increases, (p) increases.

Techniques used

rearrange the de Broglie equation into a linear formidentify a straight-line relationship and interceptinterpret a gradient from a graph of one variable against another

(ii)

State the name of the quantity represented by the gradient of the line in Fig. 9.1.

Q10MediumNuclear Physics

(a)

Radioactive decay is both random and spontaneous.

3M

(i)

State what is meant by random.

(ii)

State what is meant by spontaneous.

(iii)

State one piece of evidence for the random nature of decay.

DifficultyMedium-Easy

Worked solution

Answer

Successive measurements of count rate in equal time intervals fluctuate about a mean value (e.g. different counts in each interval / Poisson-type fluctuations).

Final answer

Counts in equal time intervals fluctuate about a mean (statistical fluctuations).

Detailed explanation

Background Concept

If decay events are random, then even with a constant average activity, the actual number of decays detected in any fixed time interval will vary due to statistical fluctuations. This is typical of a Poisson process.

Understanding the Question

You must give one piece of evidence that supports the claim that decay is random. The evidence should come from an observation/experiment, not just a restatement of the definition.

Approach

Use the standard lab evidence: measure counts in equal intervals and show they vary around an average rather than being identical each time.

Step-by-Step Reasoning

Set up a GM tube (or other detector) and count the number of decays detected in, say, $10\ \text{s}$ .
Repeat for many $10\ \text{s}$ intervals under the same conditions.
You find the counts are not the same each time; they scatter about a mean.
This scatter is evidence that each decay is probabilistic rather than occurring at fixed, predictable times.

Key Takeaways

Random processes produce fluctuations in repeated measurements.
Radioactive count data shows statistical spread even when conditions are unchanged.

Common Mistakes

Using “it has a half-life” as evidence (this shows exponential trend, but the clearer evidence of randomness is the count fluctuations).
Saying “it cannot be stopped” (that relates more to spontaneity/independence from conditions).

Things to Be Careful About

The question asks for one piece of evidence, so one clear experimental observation is enough.
Make sure it is evidence for randomness (fluctuations), not spontaneity (independence from external factors).

Techniques used

identify an experimental observation that shows statistical fluctuationslink count-rate fluctuations to random decay events

(b)

5M

(i)

Describe the differences between nuclear fission and nuclear fusion.

DifficultyMedium-Easy

Worked solution

Answer

Fission: a heavy nucleus splits into two (or more) smaller nuclei, usually releasing neutrons.
Fusion: two light nuclei combine to form a heavier nucleus.
Fusion requires very high temperature/pressure to overcome electrostatic repulsion; fission can be initiated by neutron absorption and can lead to a neutron chain reaction.

Final answer

Fission: heavy nucleus splits (releasing neutrons); Fusion: light nuclei combine; Fusion needs very high T/pressure, fission can be neutron-induced and can form a chain reaction.

Detailed explanation

Background Concept

In nuclear reactions, the nucleus changes identity. Two major energy-producing types are:

Fission: breaking a heavy nucleus into smaller nuclei.
Fusion: joining light nuclei to make a heavier nucleus.
Both can release energy because of changes in nuclear binding energy.

Understanding the Question

You must describe differences between fission and fusion (3 marks). Typically, marks are awarded for distinct contrasts: what happens to the nuclei, what particles are involved/produced, and what conditions are needed.

Approach

State at least three clear, separate differences:

Nature of the reaction (split vs join)
Typical nuclei involved (heavy vs light)
Conditions/trigger/chain reaction (neutron-induced chain reaction vs high temperature/pressure)

Step-by-Step Reasoning

What happens:
- Fission: a large nucleus (e.g. uranium/plutonium) splits into two medium-mass nuclei.
- Fusion: two small nuclei (e.g. hydrogen isotopes) combine to form a larger nucleus (e.g. helium).
Particles involved/produced:
- Fission often releases 2 or 3 neutrons as well as gamma radiation.
- Fusion does not proceed via neutron-triggering; it may produce a neutron in some specific fusion reactions (e.g. D-T), but it is not a chain reaction mechanism in the same way.
Conditions:
- Fusion requires extremely high temperatures so nuclei have enough kinetic energy to get close enough for the strong nuclear force to act (overcoming Coulomb repulsion).
- Fission can be started by absorbing a slow neutron and can be sustained as a chain reaction because emitted neutrons can cause further fissions.

Key Takeaways

Fission: heavy nuclei split; can be neutron-induced; can form a chain reaction.
Fusion: light nuclei join; needs very high temperature/pressure.

Common Mistakes

Saying “fission happens in the Sun” (the Sun is mainly fusion).
Only giving one difference (e.g. just “split vs join”) and expecting full marks.
Stating “fusion releases neutrons and fission doesn’t” (not reliably true across all reactions).

Things to Be Careful About

Keep the contrasts paired (what fission does vs what fusion does).
If you mention neutrons, link them correctly to chain reaction for fission.

Techniques used

compare nuclear processes by identifying reactants and productsstate typical conditions required for each processcontrast chain-reaction behaviour

(ii)

Explain, with reference to the variation of binding energy per nucleon with nucleon number, why the processes of nuclear fission and nuclear fusion both result in a release of energy.