Physics 9702 / 42 — May/June 2025

(b)

The rear wheel and the pedals of a bicycle are connected by a chain that passes around two cogs (toothed wheels), as shown in Fig. 1.1.

Fig. 1.1

The small cog has a radius of $0.038\ \text{m}$ and is fixed to the rear wheel so that it rotates with it.
The large cog has a radius of $0.15\ \text{m}$ and is fixed to the pedals so that it rotates with them.
The rear wheel has a radius of $0.46\ \text{m}$ .

The bicycle is being pedalled so that it moves in a straight line at a constant speed of $17\ \text{m s}^{-1}$ .

7M

(i)

Calculate the angular speed of the rear wheel.

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

DifficultyMedium-Easy

Worked solution

Working

For the rear wheel,

v = r\omega

\omega = \frac{v}{r} = \frac{17}{0.46} = 3.70 \times 10^{1}\ \text{rad s}^{-1}

Answer

$\omega \approx 37\ \text{rad s}^{-1}$

Final answer

37 rad s^-1

Detailed explanation

Background Concept

For a point on the rim of an object rotating with angular speed $\omega$ , the tangential (linear) speed $v$ is related by

v = r\omega

where $r$ is the radius from the axis of rotation.

This applies to rolling motion too: if a wheel rolls without slipping, the speed of the bicycle equals the tangential speed of the rim relative to the wheel’s centre.

Understanding the Question

The bicycle moves at constant speed $17\ \text{m s}^{-1}$ . The rear wheel radius is $0.46\ \text{m}$ . You are asked for the angular speed of the rear wheel, so you must connect the linear speed of the bicycle to the wheel’s rotation.

Approach

Use $v = r\omega$ and rearrange for $\omega$ . Substitute the given speed and wheel radius.

Step-by-Step Reasoning

Start with

v = r\omega

Rearrange:

\omega = \frac{v}{r}

Substitute $v = 17\ \text{m s}^{-1}$ and $r = 0.46\ \text{m}$ :

\omega = \frac{17}{0.46} = 36.96\ \text{rad s}^{-1}

To suitable significant figures:

\omega \approx 37\ \text{rad s}^{-1}

Key Takeaways

Use $v = r\omega$ to convert between linear speed and angular speed.
Radians are dimensionless, so $\omega$ has unit $\text{rad s}^{-1}$ .

Common Mistakes

Using the radius of the small cog instead of the wheel radius.
Writing $\omega = r/v$ (wrong rearrangement).
Giving unit $\text{s}^{-1}$ only; the expected unit is $\text{rad s}^{-1}$ .

Things to Be Careful About

Ensure $r$ is in metres (it is already).
Use the wheel radius given ( $0.46\ \text{m}$ ), not diameter.

Techniques used

use the relation between tangential speed and angular speedsubstitute numerical values with consistent units

(ii)

Calculate the period of rotation of the small cog.

period = ______ $\text{s}$

DifficultyMedium-Easy

Worked solution

Working

Small cog rotates with the rear wheel, so $\omega = 37\ \text{rad s}^{-1}$ .

\omega = \frac{2\pi}{T} \Rightarrow T = \frac{2\pi}{\omega} = \frac{2\pi}{36.96} = 0.170\ \text{s}

Answer

$T \approx 0.17\ \text{s}$

Final answer

0.17 s

Detailed explanation

Background Concept

For uniform circular motion, the angular speed $\omega$ and the period $T$ (time for one full revolution) are related by

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

This comes from the fact that one full revolution corresponds to an angular displacement of $2\pi$ radians.

Understanding the Question

The small cog is fixed to the rear wheel, so it rotates with the rear wheel (same angular speed, same period). You are asked for the period of the small cog.

Approach

Use the angular speed from part (b)(i).
Convert angular speed to period using $T = 2\pi/\omega$ .

Step-by-Step Reasoning

Because the small cog is rigidly attached to the rear wheel:

\omega_{\text{cog}} = \omega_{\text{wheel}}

Using $\omega \approx 36.96\ \text{rad s}^{-1}$ ,

T = \frac{2\pi}{\omega} = \frac{2\pi}{36.96} = 0.1699\ \text{s}

So

T \approx 0.17\ \text{s}

Key Takeaways

One revolution is $2\pi$ radians.
Fixed (rigidly connected) components share the same angular speed and period.

Common Mistakes

Using $T = \omega/2\pi$ instead of $T = 2\pi/\omega$ .
Assuming the chain makes the small cog rotate differently from the wheel, despite them being fixed together.

Things to Be Careful About

Carry enough significant figures from (b)(i) until the end; rounding early can shift the last digit slightly.
Period is in seconds; do not attach $\text{rad}$ to $T$ .

Techniques used

use the relation between period and angular speedsubstitute and evaluate using calculator

(iii)

Show that the distance moved by point X on the chain during one full rotation of the small cog is $0.24\ \text{m}$ .

DifficultyEasy

Worked solution

Working

Distance moved by the chain in one rotation of the small cog is its circumference:

s = 2\pi r = 2\pi(0.038) = 0.239\ \text{m} \approx 0.24\ \text{m}

Answer

$0.24\ \text{m}$

Final answer

0.24 m

Detailed explanation

Background Concept

When a wheel or cog rotates, a point on its rim moves through an arc length. For a full rotation, the arc length is the circumference:

\text{circumference} = 2\pi r

For a chain wrapped around a cog without slipping, the linear distance the chain moves equals the arc length that passes the contact point.

Understanding the Question

Point $X$ is on the chain. During one full rotation of the small cog, the length of chain that passes over the cog is one full circumference of that cog. The question asks you to show this distance is $0.24\ \text{m}$ using the given small-cog radius $0.038\ \text{m}$ .

Approach

Calculate the circumference of the small cog using $2\pi r$ , then round appropriately.

Step-by-Step Reasoning

For one complete revolution of the small cog, the chain advances by one circumference:

s = 2\pi r

Substitute $r = 0.038\ \text{m}$ :

s = 2\pi(0.038) = 0.23876\ \text{m}

Rounding to 2 significant figures (consistent with the given radius),

s \approx 0.24\ \text{m}

Key Takeaways

One full rotation corresponds to one circumference of travel at the rim.
For a chain drive (no slipping), chain distance moved equals rim arc length.

Common Mistakes

Using diameter instead of radius (giving $\pi d$ is fine only if $d$ is used correctly).
Using the rear wheel radius instead of the small cog radius.
Rounding to $0.23\ \text{m}$ (too much rounding down for typical sig figs).

Things to Be Careful About

Keep units in metres throughout.
The value $0.24\ \text{m}$ is a rounded version of $2\pi(0.038)$ ; stating both shows “show that” clearly.

Techniques used

use circumference to find distance moved in one rotationapply arc length of a circle s = 2 pi r for one revolution

(iv)

Use the information in (b)(iii) to determine the angle through which the large cog rotates during one full rotation of the small cog.

angle = ______ $\text{rad}$

DifficultyMedium-Easy

Worked solution

Working

Chain distance for one rotation of small cog: $s = 0.24\ \text{m}$ .

For the large cog,

s = r\theta

\theta = \frac{s}{r} = \frac{0.24}{0.15} = 1.6\ \text{rad}

Answer

$1.6\ \text{rad}$

Final answer

1.6 rad

Detailed explanation

Background Concept

Radian measure links angular displacement $\theta$ to arc length $s$ for a circle of radius $r$ :

\theta = \frac{s}{r} \quad \text{or} \quad s = r\theta

In a chain drive with no slipping, the distance the chain moves is the same everywhere along the chain. That same distance corresponds to an arc length that has passed around each cog.

Understanding the Question

From (b)(iii), during one full rotation of the small cog the chain moves a distance $s = 0.24\ \text{m}$ . The large cog has radius $0.15\ \text{m}$ . The question asks for the angle through which the large cog turns when that length of chain passes around it.

Approach

Treat the chain movement $0.24\ \text{m}$ as an arc length on the circumference of the large cog. Use $s = r\theta$ and rearrange to $\theta = s/r$ .

Step-by-Step Reasoning

Chain displacement for one full small-cog turn: $s = 0.24\ \text{m}$ .
On the large cog, that means an arc length $s$ has moved past the chain contact.
Use

s = r\theta

with $r = 0.15\ \text{m}$ :

\theta = \frac{s}{r} = \frac{0.24}{0.15} = 1.6

So the angular displacement of the large cog is

\theta = 1.6\ \text{rad}

This is less than $2\pi$ rad, which makes sense: the large cog has a larger radius, so the same chain length corresponds to a smaller fraction of a full revolution.

Key Takeaways

Use $s=r\theta$ to convert between chain travel (a linear distance) and angular rotation.
Bigger radius means smaller angle for the same arc length.

Common Mistakes

Using $\theta = r/s$ instead of $s/r$ .
Using the radius of the small cog ( $0.038\ \text{m}$ ) instead of the large cog radius ( $0.15\ \text{m}$ ).
Assuming the large cog rotates through $2\pi$ just because the small cog did.

Things to Be Careful About

Use the distance from (b)(iii) (rounded consistently) to match the expected answer.
Ensure the angle is in radians, not revolutions or degrees.

Techniques used

equate chain displacement to arc length on the large coguse the radian relation s = r thetarearrange to find angular displacement

(c)

The chain of the bicycle in (b) is moved onto a smaller cog fixed to the rear wheel. The speed of the bicycle does not change.

Explain, without calculation, the effect of this change on the angular speed of the pedals.

DifficultyMedium-Easy

Worked solution

Answer

With the bicycle speed unchanged, the rear wheel angular speed is unchanged.
Moving the chain to a smaller rear cog means the chain (tangential) speed is smaller ( $v = r\omega$ with smaller $r$ and same $\omega$ ).
Since $v = r\omega$ for the front cog and its radius is unchanged, the pedals must have a smaller angular speed.

Final answer

Angular speed of the pedals decreases.

Detailed explanation

Background Concept

For any rotating object,

v = r\omega

At the rim of a cog, $v$ is the tangential speed of the teeth (and of the chain if there is no slipping). In a bicycle chain system:

The chain speed is the same everywhere along the chain.
That chain speed must equal the tangential speed at the edge of each cog.

So, for the rear cog:

v_{\text{chain}} = r_{\text{rear cog}}\,\omega_{\text{rear wheel}}

and for the front (pedal) cog:

v_{\text{chain}} = r_{\text{front cog}}\,\omega_{\text{pedals}}

Understanding the Question

The chain is moved onto a smaller cog at the rear wheel, while the bicycle’s forward speed stays the same. The question asks, without calculation, how the angular speed of the pedals changes.

Key points hidden in the wording:

“Speed of the bicycle does not change” means the rear wheel still rotates at the same angular speed as before.
Only the rear cog radius has changed (it is now smaller).

Approach

Keep in mind rear wheel angular speed stays the same because bicycle speed is fixed.
Use $v=r\omega$ qualitatively at the rear cog: if $\omega$ is fixed but $r$ decreases, the tangential speed (chain speed) must decrease.
Use $v=r\omega$ at the front cog: if $v$ decreases but $r$ is unchanged, then $\omega$ of the pedals must decrease.

Step-by-Step Reasoning

Bicycle speed unchanged $\Rightarrow$ rear wheel angular speed unchanged (same wheel radius).
Rear cog is fixed to the wheel, so it has the same angular speed as the wheel.
Chain speed equals tangential speed at the rear cog:
- With the same $\omega$ but a smaller $r$ , the tangential speed $v = r\omega$ is smaller.
- Therefore the chain moves more slowly.
At the front cog (radius unchanged), the chain speed is again $v = r\omega$ $v = r ω$ :
- If $v$ is smaller and $r$ is the same, then $\omega$ must be smaller.
So the pedals rotate more slowly (smaller angular speed).

Key Takeaways

A smaller driving/driven radius changes angular speed ratios.
If the same angular speed is applied to a smaller radius, the linear (chain) speed is reduced.
With fixed front-cog radius, reduced chain speed means reduced pedal angular speed.

Common Mistakes

Saying the pedals must rotate faster because the rear cog is smaller (that would be true for changing the front cog or for keeping chain speed fixed, but here wheel speed is fixed by bicycle speed).
Assuming chain speed is fixed by bicycle speed directly (it is fixed by the tangential speed at the rear cog, which depends on the cog radius).

Things to Be Careful About

Distinguish clearly between what is unchanged (bicycle speed, rear wheel angular speed) and what changes (rear cog radius, chain speed, pedal angular speed).
Keep the direction of proportionality straight: at fixed $\omega$ , smaller $r$ gives smaller $v$ ; at fixed $r$ , smaller $v$ gives smaller $\omega$ .

Techniques used

reason qualitatively using v = r omegause the fact that chain speed matches the tangential speed of each cogcompare the effect of changing a radius while keeping another speed fixed

Q2MediumGravitational FieldsMagnetic Fields

(a)

4M

(i)

State what is represented by a gravitational field line.

DifficultyEasy

Worked solution

Answer

A gravitational field line shows the direction of the gravitational force on a small test mass (i.e. the direction of $\vec{g}$ ; tangent to the line).

Closer spacing (greater density) of field lines represents a stronger gravitational field.

Final answer

A gravitational field line gives the direction of gravitational force on a test mass (direction of g); line density indicates field strength.

Detailed explanation

Background Concept

The gravitational field at a point is described by the gravitational field strength $\vec{g}$ .

It is defined as force per unit mass:

\vec{g} = \frac{\vec{F}}{m}

So $\vec{g}$ is a vector: it has both magnitude and direction. Field lines are a diagrammatic way to represent a vector field.

Understanding the Question

You are asked what a gravitational field line represents. This is not asking for an equation, but for the meaning of the arrows/lines on a field diagram.

Approach

State (1) what direction the line indicates and (2) what the spacing of lines tells you about strength, because these are the two standard pieces of information encoded in field-line diagrams.

Step-by-Step Reasoning

Because $\vec{g}$ is in the same direction as the force on a small test mass, the direction of the gravitational field at any point is the direction a small mass would accelerate.
On a field-line diagram, the direction of the field at a point is given by the tangent to the field line (with arrows showing direction).
Where the field is stronger, we represent that by drawing field lines closer together (greater line density), indicating larger magnitude of $g$ .

Key Takeaways

A field line’s tangent gives the direction of the field vector.
Line density represents field strength.

Common Mistakes

Saying a field line shows the path a mass will take (it does not necessarily; motion depends on initial velocity).
Forgetting that field strength is force per unit mass.

Things to Be Careful About

Always mention direction (tangent/arrows) for one mark and spacing/density for the other mark when 2 marks are available.

Techniques used

state the definition of gravitational field strengthinterpret the meaning of tangents to field linesrelate line density to field strength

(ii)

The Earth may be considered as a uniform sphere, as shown in Fig. 2.1.

Fig. 2.1

On Fig. 2.1, draw field lines to represent the Earth’s gravitational field outside the Earth.

DifficultyMedium-Easy

Worked solution

grav-field-outside-earth

Final answer

Radial field lines directed towards the Earth’s centre (inward), symmetric all around.

Detailed explanation

Background Concept

For a spherically symmetric mass like a uniform sphere, the gravitational field outside behaves as if all mass were concentrated at the centre. The field direction is always towards the centre, and its magnitude depends only on distance $r$ from the centre:

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

Field lines for gravity therefore:

point towards the mass (gravity is attractive),
are radial for a spherical mass,
are denser nearer the mass (because $g$ is larger for smaller $r$ ).

Understanding the Question

You are given a circle representing Earth and asked to draw the gravitational field lines outside it. So you should not draw anything inside the Earth; only the external pattern is required.

Approach

Use spherical symmetry:

Draw several straight lines radiating symmetrically around the Earth.
Put arrowheads pointing inwards (towards the centre).
(Optionally) show lines slightly closer together nearer Earth to indicate stronger field.

Step-by-Step Reasoning

Because the field must point towards the Earth’s centre, each field line is along a radius.
At every point on the surface the field is perpendicular to the surface (since the surface of a sphere is an equipotential), so the lines meet the surface at right angles.
Add arrowheads pointing towards the Earth to show the attractive direction.

grav-field-outside-earth

Key Takeaways

Outside a uniform sphere, gravitational field lines are radial and inward.
Symmetry means the pattern looks the same all around the sphere.

Common Mistakes

Drawing circular lines around the Earth (that would be wrong for gravity).
Drawing arrows pointing outward.
Drawing only two lines (too few to represent a field pattern).

Things to Be Careful About

Ensure lines are straight and meet the surface normally.
Ensure arrows point towards the Earth (gravity attracts).

Techniques used

use spherical symmetry to infer a radial field patterndraw field lines perpendicular to an equipotential surfaceindicate direction of a force field with arrows

(b)

The Earth’s magnetic field may be considered as being due to the Earth acting as a long solenoid, as shown in Fig. 2.2.

Fig. 2.2

The magnetic poles do not align with the geographic poles, which are on the axis of rotation.

Fig. 2.3 is a copy of Fig. 2.2 without the labels but with two magnetic field lines shown.

Fig. 2.3

3M

(i)

On Fig. 2.3, label the magnetic poles with the letters N and S to indicate which one is the magnetic N pole and which one is the magnetic S pole.

DifficultyMedium-Easy

Worked solution

Answer

Outside the Earth, magnetic field lines go from $N$ to $S$ .

So the pole where the shown field line leaves the Earth is $N$ , and the pole where the shown field line enters the Earth is $S$ .

Final answer

Label the pole where field lines leave as N, and where they enter as S.

Detailed explanation

Background Concept

Magnetic field lines are drawn using a standard convention:

Outside a magnet, field lines go from the magnetic north pole ( $N$ ) to the magnetic south pole ( $S$ ).
Inside the magnet, they return from $S$ to $N$ to form continuous loops.

So if you can see the direction of the arrows on field lines outside the object, you can identify which end is $N$ (lines emerge) and which is $S$ (lines enter).

Understanding the Question

In Fig. 2.3 you are shown two magnetic field lines outside Earth: one is entering the Earth at one magnetic pole and one is leaving at the other. You must label which pole is magnetic $N$ and which is magnetic $S$ .

Approach

Use the field-line direction convention:

emerging/outgoing lines indicate $N$ ,
incoming/entering lines indicate $S$ .

Step-by-Step Reasoning

Inspect the given field lines: at one end, the external field line is directed into the Earth surface (it terminates/enters there). That end must be the magnetic south pole $S$ .
At the other end, the external field line is directed away from the Earth (it emerges/leaves there). That end must be the magnetic north pole $N$ .

Key Takeaways

Outside a magnet, field lines go $N \to S$ .
Use “emerges from $N$ ” and “enters $S$ ” to label poles.

Common Mistakes

Reversing the convention (labelling the entry point as $N$ ).
Forgetting that the convention refers to the field direction outside the magnet.

Things to Be Careful About

Make sure you are using the arrows/direction of the drawn field lines, not the geographic axis labels (which are different in this question).

Techniques used

use the convention for magnetic field-line directioninfer pole labels from field lines entering or leaving

(ii)

On Fig. 2.3, draw field lines to represent the Earth’s magnetic field outside the Earth.

DifficultyMedium-Easy

Worked solution

earth-magnetic-field-lines

Final answer

Dipole-like field lines outside Earth, leaving N and entering S, symmetric about the magnetic axis.

Detailed explanation

Background Concept

A long solenoid (and similarly a bar magnet) produces a dipole-like magnetic field:

Field lines form closed loops.
Outside the solenoid/magnet, field lines go from $N$ to $S$ .
The field is strongest near the poles, represented by closer field lines.

Earth’s magnetic field is often approximated by a tilted dipole, so the field pattern around Earth resembles that of a bar magnet whose axis is not the same as the rotation axis.

Understanding the Question

You are given Fig. 2.3 with two magnetic field lines already shown outside the Earth. You must add more lines to represent the external magnetic field pattern.

Approach

Keep the pattern consistent with a dipole: curved loops connecting the two poles.
Ensure the direction is from $N$ to $S$ outside Earth.
Draw a symmetric set of lines about the magnetic axis (the solenoid axis), not necessarily the geographic axis.

Step-by-Step Reasoning

Identify $N$ and $S$ from part (b)(i) (lines emerge from $N$ and enter $S$ ).
Add several more curved lines outside Earth: they should start at the $N$ pole region, bow outward around the Earth, and enter at the $S$ pole region.
Make lines closer together near the poles to indicate stronger field there.

earth-magnetic-field-lines

Key Takeaways

Earth’s external magnetic field is drawn like a tilted dipole.
External field direction is $N \to S$ .

Common Mistakes

Drawing magnetic field lines radial like gravity.
Drawing arrows going from $S$ to $N$ outside the Earth.
Drawing lines symmetric about the rotation axis instead of the magnetic axis in the diagram.

Things to Be Careful About

Lines must not start or end in empty space; they should connect from $N$ to $S$ externally.
Ensure your added lines match the curvature of the two given lines (consistency).

Techniques used

represent a dipole magnetic field with continuous loopsapply the N-to-S field-line direction conventionshow relative field strength via line density

(c)

An observer moves around the surface of the Earth.

5M

(i)

Use your answer in (a)(ii) to explain why the observed gravitational field of the Earth does not vary around the surface.

DifficultyMedium-Easy

Worked solution

Answer

In (a)(ii) the field lines are radial and symmetrically/equally spaced around the Earth.

An observer on the surface is always at the same distance $R$ from the centre, so $g = GM/R^2$ is the same everywhere (direction always towards the centre).

Final answer

Because the field lines are symmetric and the radius is constant on the surface, g has the same magnitude everywhere (towards the centre).

Detailed explanation

Background Concept

For a spherically symmetric mass, the gravitational field outside depends only on distance from the centre:

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

Field lines provide a visual version of this: if the pattern is the same in every direction and the line density at a given radius is the same, then the field strength at that radius is the same.

Understanding the Question

You are told an observer moves around Earth’s surface. The question asks you to use your diagram from (a)(ii) to explain why the observed gravitational field does not vary around the surface.

So you must connect:

field-line symmetry (from your drawing),
constant distance from the centre on the surface,
to constant gravitational field strength.

Approach

Use two linked arguments:

On a sphere, moving around the surface keeps $r$ constant.
Your field lines show a symmetric, radial field, so the field magnitude at that fixed radius is the same everywhere.

Step-by-Step Reasoning

On Earth’s surface, the observer is always at $r = R$ from Earth’s centre.
Since $g$ depends only on $r$ (not on direction around the sphere), $g = GM/R^2$ has one value everywhere on the surface.
The field-line diagram reflects this: at the same distance from the centre, the spacing of field lines is uniform all around, indicating the same field strength.
The direction of $\vec{g}$ always points towards the centre, so as you move, the direction changes with your position but the observed local direction is always “downwards towards the centre.”

Key Takeaways

For a spherical mass, $g$ depends only on radius $r$ .
Constant radius around the surface means constant field magnitude.

Common Mistakes

Saying “it doesn’t vary because Earth is uniform” without linking to constant distance from the centre.
Confusing direction and magnitude: the vector direction in space changes around the Earth, but locally it is always towards the centre.

Things to Be Careful About

The question wants you to use the field-line diagram idea: mention symmetry/equal spacing (line density) and radial direction.
Ignore small real-world effects (rotation, non-uniform density) since the Earth is stated to be a uniform sphere here.

Techniques used

use symmetry of field-line diagrams to infer constancy around a surfacerelate equal distance from centre to constant field magnitudelink field-line density to field strength

(ii)

With reference to your answer in (b)(ii), describe how the observed magnetic field of the Earth varies around the surface.

DifficultyMedium

Worked solution

Answer

The magnetic field is not the same all around the surface because the dipole axis is tilted.

Field direction changes with position: near the magnetic poles the field is more nearly radial/vertical, but near the equatorial region it is more nearly horizontal.

Field strength also varies: it is stronger near the magnetic poles (field lines closer together) and weaker further from the poles (field lines more widely spaced).

Final answer

Magnetic field varies in both direction and strength around the surface: strongest and more vertical near magnetic poles; weaker and more horizontal near equator, due to tilted dipole.

Detailed explanation

Background Concept

A dipole magnetic field (like a bar magnet/solenoid) has:

field lines emerging from $N$ and entering $S$ outside,
strongest field near the poles (high line density),
a direction that changes from place to place: near poles it is more vertical; around the mid-region it is more horizontal.

Unlike gravity for a uniform sphere, a dipole field is not the same at every point on a spherical surface because the field is not spherically symmetric.

Understanding the Question

You must describe how the observed magnetic field varies around Earth’s surface, referring to your external field-line diagram from (b)(ii). That means you should interpret what the line pattern implies about:

how the direction of $\vec{B}$ changes,
how the magnitude of $B$ changes,
and link this to the fact the magnetic axis is tilted relative to the rotation axis.

Approach

From a field-line diagram:

Direction of $\vec{B}$ is along the tangent to the field line.
Strength is indicated by how close the field lines are.
So, compare what the field lines look like near poles vs near the equator, and note that because the dipole is tilted, these effects occur at different geographic locations.

Step-by-Step Reasoning

Direction variation: As you move around the surface, the tangent to the field lines changes. Near the magnetic poles, the lines go into/out of Earth almost perpendicular to the surface, so the field is more vertical. Near the equatorial region of the dipole, the lines run more parallel to the surface, so the field is more horizontal.
Strength variation: Where field lines are closer together (near the magnetic poles) the field is stronger. Where the lines spread out (around the sides/equator) the field is weaker.
Not constant around the Earth: Because the magnetic dipole axis is tilted relative to the rotation axis, an observer moving around the surface (especially along a line of latitude relative to the geographic poles) will find that both the direction (inclination) and magnitude of $B$ change with position.

Key Takeaways

For magnetic fields, the field-line pattern shows both changing direction (tangent) and changing strength (line density).
A dipole field is not spherically symmetric, so it cannot be the same everywhere on a sphere.

Common Mistakes

Stating only “it varies” without saying how (direction and/or strength).
Mixing up gravity and magnetism and claiming magnetic field is constant because Earth is spherical.
Saying field lines are strongest where they are furthest apart.

Things to Be Careful About

The question asks for variation “around the surface”: make sure you mention both poles and equatorial regions (relative to the magnetic axis).
Be explicit that the tilt causes the pattern not to align with geographic poles, so different locations experience different field direction/strength.

Techniques used

use a dipole field-line pattern to infer directional variationrelate field-line density to changes in field strengthcompare field behaviour at poles versus equator

Q3MediumTemperatureThermodynamics

(a)

Define specific heat capacity.

DifficultyEasy

Worked solution

Answer

Specific heat capacity is the thermal energy required to raise the temperature of $1.0\ \text{kg}$ of a substance by $1.0\ \text{K}$ (or $1.0\ ^\circ\text{C}$ ), without change of state.

Final answer

Thermal energy required to raise the temperature of 1 kg of a substance by 1 K (no change of state).

Detailed explanation

Background Concept

Specific heat capacity $c$ tells you how much energy is needed to warm up a substance.

It is defined through the relationship

E = mc\Delta T

where:

$E$ is the thermal energy transferred to the substance (or energy required),
$m$ is the mass,
$\Delta T$ is the temperature rise,
$c$ is the specific heat capacity.

The phrase “without change of state” matters because during melting/boiling, energy goes into latent heat rather than increasing temperature.

Understanding the Question

You are asked to define specific heat capacity. A full definition must include:

energy per unit mass,
per unit temperature rise,
and the condition of no change of state.

Approach

Give the standard Cambridge definition: energy required to raise the temperature of $1\ \text{kg}$ by $1\ \text{K}$ (or $1\ ^\circ\text{C}$ ), with no state change.

Step-by-Step Reasoning

“Specific” means per unit mass ( $1\ \text{kg}$ ).
“Heat capacity” refers to energy needed for a temperature increase, so specify per unit temperature rise ( $1\ \text{K}$ ).
Add “without change of state” because otherwise the definition becomes ambiguous (energy could be used for melting/boiling).

Key Takeaways

Specific heat capacity is energy per kg per kelvin.
It applies when temperature changes and the substance stays in the same phase.

Common Mistakes

Forgetting to mention 1 kg (per unit mass).
Forgetting 1 K (per unit temperature rise).
Omitting “no change of state”.

Things to Be Careful About

$1\ \text{K}$ temperature rise is the same size as $1\ ^\circ\text{C}$ rise; the difference is what matters.
Don’t confuse specific heat capacity with specific latent heat (which is energy to change state at constant temperature).

Techniques used

state a definition with per-unit-mass and per-unit-temperature-risespecify the condition of no change of state

(b)

A block of aluminium has a volume of $3.612 \times 10^{-3}\ \text{m}^3$ at a temperature of $0\ ^{\circ}\text{C}$ .

Aluminium has a density of $2.700 \times 10^3\ \text{kg m}^{-3}$ at $0\ ^{\circ}\text{C}$ .
It has a density of $2.620 \times 10^3\ \text{kg m}^{-3}$ at $500\ ^{\circ}\text{C}$ .

The block is heated so that its temperature increases from $0\ ^{\circ}\text{C}$ to $500\ ^{\circ}\text{C}$ at an atmospheric pressure of $1.01 \times 10^5\ \text{Pa}$ .

The increase in internal energy of the block is $4.38\ \text{MJ}$ .

10M

(i)

Calculate the mass of the block.

mass = ______ $\text{kg}$

DifficultyMedium-Easy

Worked solution

Working

m = \rho V

m = (2.700 \times 10^{3})(3.612 \times 10^{-3})

m = 9.7524\ \text{kg}

Answer

m = 9.75\ \text{kg}

Final answer

9.75 kg

Detailed explanation

Background Concept

Density $\rho$ is defined as mass per unit volume:

\rho = \frac{m}{V}

Rearranging gives

m = \rho V

This works as long as the density given corresponds to the temperature at which the volume is stated.

Understanding the Question

At $0\ ^\circ\text{C}$ you are given:

volume $V = 3.612 \times 10^{-3}\ \text{m}^3$
density $\rho = 2.700 \times 10^3\ \text{kg m}^{-3}$

You must find the mass of the aluminium block.

Approach

Use $m = \rho V$ directly, since both $\rho$ and $V$ are given at the same temperature.

Step-by-Step Reasoning

Start with

m = \rho V

Substitute the given values:

m = (2.700 \times 10^3)(3.612 \times 10^{-3})

Combine powers of ten: $10^3 \times 10^{-3} = 10^0 = 1$ .
Multiply the numbers: $2.700 \times 3.612 = 9.7524$ .

So

m = 9.7524\ \text{kg}

Quote to 3 s.f. (matching the data):

m = 9.75\ \text{kg}

Key Takeaways

Use $m = \rho V$ when density and volume are known.
Make sure density corresponds to the same conditions as the volume (here, $0\ ^\circ\text{C}$ ).

Common Mistakes

Using the density at $500\ ^\circ\text{C}$ instead of at $0\ ^\circ\text{C}$ .
Incorrectly handling powers of ten.

Things to Be Careful About

Both quantities are already in SI units ( $\text{m}^3$ and $\text{kg m}^{-3}$ ), so no conversion is needed.
Round at the end to avoid rounding errors.

Techniques used

use the definition of densitysubstitute values in SI unitsapply significant figures appropriately

(ii)

Show that the volume of the block at a temperature of $500\ ^{\circ}\text{C}$ is $3.722 \times 10^{-3}\ \text{m}^3$ .

DifficultyMedium-Easy

Worked solution

Working

Mass is constant, so

V_{500} = \frac{m}{\rho_{500}}

V_{500} = \frac{9.7524}{2.620 \times 10^{3}} = 3.722\times 10^{-3}\ \text{m}^3

Answer

V_{500} = 3.722 \times 10^{-3}\ \text{m}^3

Final answer

3.722 × 10^-3 m^3

Detailed explanation

Background Concept

For a solid block, the mass stays the same when it is heated (no material is added/removed). Density changes because the volume changes.

Using

\rho = \frac{m}{V}

we can write

V = \frac{m}{\rho}

So if $\rho$ decreases on heating, $V$ must increase.

Understanding the Question

You have already found the mass $m$ in (b)(i). At $500\ ^\circ\text{C}$ the density is given as

\rho_{500} = 2.620 \times 10^3\ \text{kg m}^{-3}

You must show that the new volume is $3.722 \times 10^{-3}\ \text{m}^3$ .

Approach

Use $V_{500} = m/\rho_{500}$ with the same mass from part (i).

Step-by-Step Reasoning

Mass conservation: $m$ is unchanged from $0\ ^\circ\text{C}$ to $500\ ^\circ\text{C}$ .
Apply the density relation at $500\ ^\circ\text{C}$ :

V_{500} = \frac{m}{\rho_{500}}

Substitute:

V_{500} = \frac{9.7524}{2.620 \times 10^3}

Evaluate:

V_{500} \approx 3.722 \times 10^{-3}\ \text{m}^3

This matches the value to be shown.

Key Takeaways

Heating changes density mainly by changing volume; mass remains constant.
Use $V = m/\rho$ when density at a new temperature is provided.

Common Mistakes

Recalculating mass using the $500\ ^\circ\text{C}$ density (mass should not change).
Forgetting the $\times 10^3$ in the density.

Things to Be Careful About

Keep enough significant figures from part (i) to reproduce the stated volume.
Don’t round the mass too aggressively before substituting.

Techniques used

assume mass is conserveduse density-volume relationship to find a new volumesubstitute numerical values and round appropriately

(iii)

Use the information in (b)(ii) to determine the magnitude of the work done on the block when its temperature is raised from $0\ ^{\circ}\text{C}$ to $500\ ^{\circ}\text{C}$ .

work done = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

\Delta V = 3.722\times 10^{-3} - 3.612\times 10^{-3} = 1.10\times 10^{-4}\ \text{m}^3

Magnitude of work done:

|W| = p\Delta V = (1.01\times 10^5)(1.10\times 10^{-4}) = 11.1\ \text{J}

Answer

\text{work done} = 11.1\ \text{J}

Final answer

11.1 J

Detailed explanation

Background Concept

When an object expands against an external pressure $p$ , energy is transferred as mechanical work.

At constant external pressure, the magnitude of work associated with a volume change is

|W| = p\Delta V

The sign depends on whether work is done on the system or by the system; this is dealt with explicitly in part (iv).

Understanding the Question

You are told the block is heated at atmospheric pressure

p = 1.01 \times 10^5\ \text{Pa}

From (b)(ii), the volume increases from $3.612\times 10^{-3}$ to $3.722\times 10^{-3}\ \text{m}^3$ .

You must find the magnitude of the work done on the block during this heating.

Approach

Find the volume increase $\Delta V$ .
Use $|W| = p\Delta V$ .

Step-by-Step Reasoning

Compute the volume change:

\Delta V = V_{500} - V_0 = 3.722\times 10^{-3} - 3.612\times 10^{-3}

\Delta V = 0.110\times 10^{-3} = 1.10\times 10^{-4}\ \text{m}^3

Use constant pressure work magnitude:

|W| = p\Delta V = (1.01\times 10^5)(1.10\times 10^{-4})

Multiply: $10^5\times 10^{-4} = 10^1$ , so the result is of order $10$ joules.

|W| = 11.1\ \text{J}

Key Takeaways

Expansion at constant external pressure transfers energy as $p\Delta V$ work.
Even a small fractional expansion of a solid can be found from density/volume data.

Common Mistakes

Using $V$ instead of $\Delta V$ .
Forgetting that $1\ \text{Pa} = 1\ \text{N m}^{-2}$ and that $\text{Pa}\cdot\text{m}^3 = \text{J}$ .
Giving the signed value when the question asks for magnitude.

Things to Be Careful About

Keep powers of ten consistent when subtracting volumes.
This work is very small compared with $\Delta U$ in part (v), but it still affects the sign logic.

Techniques used

compute a change in volume from two stated volumesuse work done at constant external pressuretake magnitude as requested

(iv)

Explain whether the work done on the block is positive or negative.

DifficultyMedium-Easy

Worked solution

Answer

The block expands as it is heated, so it pushes back the atmosphere and does work on the surroundings.

Therefore the work done on the block is negative ( $W < 0$ ).

Final answer

Negative (the block expands and does work on the surroundings).

Detailed explanation

Background Concept

In the Cambridge convention for the first law,

\Delta U = q + W

where:

$q$ is thermal energy transferred to the system,
$W$ is work done on the system.

So:

Compression (surroundings do work on the system) gives $W>0$ .
Expansion (system does work on surroundings) gives $W<0$ .

Understanding the Question

The block is heated from $0\ ^\circ\text{C}$ to $500\ ^\circ\text{C}$ at atmospheric pressure. From part (ii) its volume increases, meaning it expands.

You must decide whether the work done on the block is positive or negative.

Approach

Decide whether the surroundings are doing work on the block or the block is doing work on the surroundings.

If the block expands against external pressure, the block must push air away: that is work done by the block.
Hence work done on the block is negative.

Step-by-Step Reasoning

Heating causes aluminium to expand: $V$ increases.
External pressure resists expansion, so the block must exert a force on the surrounding air.
The block therefore transfers energy mechanically to the surroundings.
With $W$ defined as work done on the block, energy leaving the block as work corresponds to

W < 0

Key Takeaways

Always check the sign convention used in the syllabus: here $W$ is work done on the system.
Expansion implies $W$ (on the system) is negative.

Common Mistakes

Saying “work done is positive because the volume increases” (sign depends on definition of $W$ ).
Mixing up “work done by the block” with “work done on the block”.

Things to Be Careful About

Part (iii) asked for the magnitude, so it was positive there; in part (iv) you must now state the sign.
Be explicit about which direction the energy transfer occurs (to surroundings during expansion).

Techniques used

interpret the sign of work from expansion or compressionapply the convention that W is work done on the system

(v)

Use the first law of thermodynamics to determine, to three significant figures, a value for the specific heat capacity of aluminium. Explain your reasoning. Give a unit with your answer.

specific heat capacity = ______ unit ______

DifficultyMedium

Worked solution

Working

From (iii) and (iv), for expansion

W = -p\Delta V = -11.1\ \text{J}

First law:

\Delta U = q + W \Rightarrow q = \Delta U - W

q = 4.38\times 10^{6} - (-11.1) = 4.380011\times 10^{6}\ \text{J}

Temperature rise: $\Delta T = 500\ \text{K}$ .

q = mc\Delta T \Rightarrow c = \frac{q}{m\Delta T}

c = \frac{4.380011\times 10^{6}}{(9.75)(500)} = 8.98\times 10^{2}\ \text{J kg}^{-1}\ \text{K}^{-1}

Answer

c = 898\ \text{J kg}^{-1}\ \text{K}^{-1}

Final answer

898 J kg^-1 K^-1

Detailed explanation

Background Concept

For heating a solid with no change of state, the thermal energy transferred to it is

q = mc\Delta T

However, if the solid expands against external pressure, some energy transfer also occurs as work. The first law (Cambridge convention) is

\Delta U = q + W

where $W$ is work done on the system. For expansion, $W$ is negative.

Understanding the Question

Given:

$\Delta U = 4.38\ \text{MJ} = 4.38\times 10^6\ \text{J}$
$p = 1.01\times 10^5\ \text{Pa}$
$V$ increases from $3.612\times 10^{-3}$ to $3.722\times 10^{-3}\ \text{m}^3$
$m$ from (i) is $9.75\ \text{kg}$
temperature rises from $0\ ^\circ\text{C}$ to $500\ ^\circ\text{C}$ , so

\Delta T = 500\ \text{K}

You must use the first law to find $c$ and explain the reasoning (i.e. show how you obtained $q$ from $\Delta U$ and $W$ ).

Approach

Find the signed work $W$ using the result from (iii) and the sign from (iv).
Rearrange the first law to find $q$ .
Use $q = mc\Delta T$ to solve for $c$ .

Step-by-Step Reasoning

From (iii), the magnitude of the work associated with expansion is $11.1\ \text{J}$ .
From (iv), because the block expands it does work on the surroundings, so work done on the block is negative:

W = -11.1\ \text{J}

Apply the first law:

\Delta U = q + W

Rearrange for $q$ :

q = \Delta U - W

Substitute values:

q = 4.38\times 10^6 - (-11.1) = 4.380011\times 10^6\ \text{J}

(Notice this is essentially $4.38\times 10^6\ \text{J}$ because $11\ \text{J}$ is tiny compared with $4.38\times 10^6\ \text{J}$ , but including it shows correct use of the law.)

Now use the definition of specific heat capacity:

q = mc\Delta T

So

c = \frac{q}{m\Delta T}

Substitute $m = 9.75\ \text{kg}$ and $\Delta T = 500\ \text{K}$ :

c = \frac{4.380011\times 10^6}{(9.75)(500)}

Compute denominator: $(9.75)(500)=4875$ .

c = 8.98\times 10^2\ \text{J kg}^{-1}\ \text{K}^{-1} = 898\ \text{J kg}^{-1}\ \text{K}^{-1}

Key Takeaways

First law links internal energy change to heat transfer and work: $\Delta U = q + W$ .
For expansion, work done on the system is negative.
Once $q$ is found, use $q = mc\Delta T$ to obtain $c$ .

Common Mistakes

Using $\Delta U = q - W$ (wrong sign convention for this syllabus).
Taking $W$ as $+p\Delta V$ for expansion when $W$ is defined as work done on the system.
Using $\Delta T = 500\ ^\circ\text{C}$ as if it were different from $500\ \text{K}$ (temperature difference is the same, but you must write it consistently).
Forgetting to convert $4.38\ \text{MJ}$ to joules.

Things to Be Careful About

Quote $c$ to three significant figures as requested, and include the unit $\text{J kg}^{-1}\ \text{K}^{-1}$ .
Keep track of whether a quantity is a magnitude (part iii) or a signed value (part iv/v).
The work term here is extremely small compared with $\Delta U$ ; a small arithmetic slip in $W$ will not change the final $c$ much, but the reasoning/sign is what the marks test.

Techniques used

apply the first law of thermodynamics with a consistent sign conventionrelate heat transfer to temperature rise using mcΔTcombine results from earlier parts to calculate a material constant

(c)

Without further calculation, suggest with a reason how doubling the pressure in (b) is likely to affect the answer in (b)(v).

DifficultyMedium-Easy

Worked solution

Answer

Doubling $p$ doubles the magnitude of $p\Delta V$ , so the (negative) $W$ becomes about twice as large in magnitude.

From $\Delta U = q + W$ , $q = \Delta U - W$ therefore increases slightly, so the calculated $c$ would increase very slightly (negligible change because $|W| \ll \Delta U$ ).

Final answer

Calculated specific heat capacity would be very slightly larger (effect negligible).

Detailed explanation

Background Concept

At constant external pressure, the work term scales as

W \propto p\Delta V

With the first law (Cambridge convention)

\Delta U = q + W

rearranging gives

q = \Delta U - W

So changing $W$ changes the required heat transfer $q$ for the same $\Delta U$ .

Understanding the Question

You are asked to predict, without calculation, what happens to the answer for $c$ in (b)(v) if the external pressure is doubled.

The key idea is how $W$ depends on pressure and then how $W$ affects $q$ and hence $c$ .

Approach

Note that $\Delta V$ for the expansion is essentially a material property for a given temperature change (so assume it is roughly unchanged).
Doubling $p$ doubles the magnitude of $p\Delta V$ .
Use the sign of $W$ (negative for expansion) in $q = \Delta U - W$ .
Since $c \propto q$ , infer how $c$ changes.

Step-by-Step Reasoning

The block expands when heated, so $W$ (work done on the block) is negative.
Magnitude of the work is $|W| = p\Delta V$ . If $p$ is doubled and $\Delta V$ is approximately the same, then $|W|$ doubles.
Because $W$ is negative, making its magnitude larger means $W$ becomes more negative.
From

q = \Delta U - W

if $W$ becomes more negative, $-W$ becomes more positive, so $q$ increases slightly.
5. Since

c = \frac{q}{m\Delta T}

an increase in $q$ gives a (very slightly) larger value of $c$ .
6. The change is negligible because, in this question, $|W|$ is only of order $10\ \text{J}$ whereas $\Delta U$ is of order $10^6\ \text{J}$ .

Key Takeaways

$p\Delta V$ work scales linearly with pressure.
For expansion, work done on the system is negative.
If $\Delta U$ is fixed, changing $W$ changes the required $q$ , and hence the inferred $c$ .

Common Mistakes

Saying $c$ decreases because “more work is done” without considering the sign.
Assuming doubling pressure doubles $\Delta U$ (it doesn’t; $\Delta U$ is given as fixed in the question stem).

Things to Be Careful About

The question asks “likely to affect” and “without further calculation”: a qualitative statement with a reason is sufficient.
You must mention that the effect is very small in this case because $|W|$ is tiny compared with $\Delta U$ .

Techniques used

predict how pΔV work scales with pressureinfer the effect on calculated heat transfer using the first lawjudge whether the effect is significant by comparing magnitudes

Q4Medium-EasyIdeal Gases

(a)

The equation of state for an ideal gas may be written as

pVA = NBT

where $p$ is the pressure of the gas, $V$ is the volume of the gas, $A$ is the Avogadro constant, $B$ is another constant and $N$ is the number of molecules of the gas.

2M

(i)

State the meaning, in the equation, of the symbol $T$ .

(ii)

Identify the constant $B$ .

(b)

The product $pV$ for an ideal gas is also given by

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

4M

(i)

State the meanings, in this equation, of the symbols $m$ and $\langle c^2 \rangle$ .

$m$ : ______
$\langle c^2 \rangle$ : ______

DifficultyMedium-Easy

Worked solution

Answer

$m$ : mass of one molecule.

$\langle c^2 \rangle$ : mean value of $c^2$ (average of the square of the molecular speed).

Final answer

m: mass of one molecule; <c^2>: mean square speed (average of c^2).

Detailed explanation

Background Concept

In kinetic theory, molecules in a gas have a distribution of speeds. We often work with averages:

$c$ is the speed of one molecule.
$\langle c^2 \rangle$ means “the average of $c^2$ over all molecules”.

The equation

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

relates macroscopic quantities ( $p$ , $V$ ) to microscopic motion.

Understanding the Question

The question gives the kinetic theory expression for $pV$ and asks for the meanings of:

$m$
$\langle c^2 \rangle$

So you must state what physical quantity each symbol represents.

Approach

Use the standard interpretation of symbols in the kinetic theory equation:

$m$ is particle mass.
$\langle c^2 \rangle$ is an average of the square of the speed.

Step-by-Step Reasoning

$N$ counts molecules, so $m$ must be the mass per molecule (not per mole).
Angle brackets $\langle \ \rangle$ denote an average.
Therefore $\langle c^2 \rangle$ is the mean of $c^2$ across all molecules: the mean square speed.

Key Takeaways

$\langle c^2 \rangle$ is a mean square quantity, not the square of the mean speed.
RMS speed is related by $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ .

Common Mistakes

Writing “ $\langle c^2 \rangle$ is the r.m.s. speed”: the r.m.s. speed is the square root of $\langle c^2 \rangle$ .
Taking $m$ to be molar mass instead of mass of one molecule.

Things to Be Careful About

Be precise with wording: “mean square speed” (for $\langle c^2 \rangle$ ) vs “root-mean-square speed” (for $\sqrt{\langle c^2 \rangle}$ ).

Techniques used

interpret symbols in the kinetic theory equationdistinguish mean square quantity from root-mean-square quantity

(ii)

Use the equations in (a) and (b) to derive an expression, in terms of $A$ , $B$ and $T$ , for the mean kinetic energy $E_K$ of a molecule of the gas.

$E_K$ = ______

DifficultyMedium-Easy

Worked solution

Working

From (a):

pVA = NBT \Rightarrow pV = \frac{NBT}{A}

From (b):

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

Equate:

\frac{NBT}{A} = \frac{1}{3}Nm\langle c^2 \rangle

m\langle c^2 \rangle = \frac{3BT}{A}

Mean kinetic energy per molecule:

E_K = \frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}\frac{B}{A}T

Answer

E_K = \frac{3}{2}\frac{B}{A}T

Final answer

EK = (3/2)(B/A)T

Detailed explanation

Background Concept

There are two key results being connected:

Ideal gas equation (in a molecular form): $pV$ is proportional to $NT$ .
Kinetic theory result:

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

The mean translational kinetic energy of one molecule is

E_K = \frac{1}{2}m\langle c^2 \rangle

so if you can express $m\langle c^2 \rangle$ in terms of $T$ , you immediately get $E_K$ in terms of $T$ .

Understanding the Question

You are told two different expressions for $pV$ :

From (a): $pVA = NBT$ .
From (b): $pV = \frac{1}{3}Nm\langle c^2 \rangle$ .

You must combine them to find $E_K$ (mean kinetic energy of a molecule) in terms of $A$ , $B$ , and $T$ only.

Approach

Rearrange the equation in (a) to make $pV$ the subject.
Set that equal to the kinetic theory expression for $pV$ .
Rearrange to find $m\langle c^2 \rangle$ .
Use $E_K = \frac{1}{2}m\langle c^2 \rangle$ .

Step-by-Step Reasoning

From (a):

pVA = NBT

Divide by $A$ :

pV = \frac{NBT}{A}

From (b):

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

Since both equal the same quantity $pV$ , equate them:

\frac{NBT}{A} = \frac{1}{3}Nm\langle c^2 \rangle

Cancel $N$ from both sides:

\frac{BT}{A} = \frac{1}{3}m\langle c^2 \rangle

Multiply by $3$ :

m\langle c^2 \rangle = \frac{3BT}{A}

Now convert mean square speed into mean kinetic energy per molecule:

E_K = \frac{1}{2}m\langle c^2 \rangle = \frac{1}{2}\left(\frac{3BT}{A}\right) = \frac{3}{2}\frac{B}{A}T

This matches the well-known result $E_K = \frac{3}{2}kT$ because $k = \frac{B}{A}$ .

Key Takeaways

If two expressions equal the same physical quantity (here $pV$ ), you can set them equal to eliminate unwanted variables.
The kinetic theory link between microscopic motion and temperature leads to $E_K \propto T$ .

Common Mistakes

Using $E_K = \frac{1}{2}m c_{\text{rms}}^2$ but then forgetting that $c_{\text{rms}}^2 = \langle c^2 \rangle$ .
Forgetting to first rearrange (a) to get $pV$ alone.
Losing the factor of $\frac{1}{3}$ when equating the two $pV$ expressions.

Things to Be Careful About

Be clear that $E_K$ is per molecule (not total kinetic energy of all molecules).
Algebraic cancellation: cancelling $N$ is valid because it appears as a factor on both sides.
Keeping symbols consistent: $A$ is $N_A$ and $B$ is $R$ , so $\frac{B}{A} = k$ .

Techniques used

equate two expressions for the same physical quantityrearrange algebra to isolate a required variablerelate mean kinetic energy to mean square speed

(c)

On Fig. 4.1, sketch the variation with $T$ of the root-mean-square (r.m.s.) speed of the molecules of an ideal gas.

Fig. 4.1

DifficultyMedium-Easy

Worked solution

diagram-c1

Answer

$c_{\text{rms}} \propto \sqrt{T}$ : curve passes through the origin and increases with decreasing gradient (concave down).

Final answer

c_rms ∝ √T: through origin, increasing concave-down curve.

Detailed explanation

Background Concept

For an ideal gas, kinetic theory gives

\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT

so

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

The root-mean-square speed is defined by

c_{\text{rms}} = \sqrt{\langle c^2 \rangle}

Hence

c_{\text{rms}} = \sqrt{\frac{3k}{m}}\,\sqrt{T}

So $c_{\text{rms}}$ is proportional to $\sqrt{T}$ .

Understanding the Question

You are asked to sketch how r.m.s. speed varies with $T$ on axes that both start at $0$ . This means your sketch should show:

whether it passes through the origin,
whether it is a straight line or a curve,
whether the gradient increases or decreases.

Approach

Use the relationship $c_{\text{rms}} \propto \sqrt{T}$ . A square-root graph starts at the origin, rises quickly at first, and then rises more slowly (decreasing gradient).

Step-by-Step Reasoning

From kinetic theory, $\langle c^2 \rangle \propto T$ .
Taking the square root gives $c_{\text{rms}} \propto \sqrt{T}$ .
At $T = 0$ , $\sqrt{T} = 0$ , so the curve passes through $(0,0)$ .
As $T$ increases, $\sqrt{T}$ increases but with a decreasing slope, so the curve is concave down.

diagram-c1

Key Takeaways

For an ideal gas, molecular speeds scale as the square root of absolute temperature.
Knowing a functional dependence ( $\sqrt{T}$ ) lets you sketch the correct qualitative graph.

Common Mistakes

Drawing a straight line ( $c_{\text{rms}} \propto T$ ): that would correspond to $\langle c^2 \rangle \propto T^2$ , which is incorrect.
Drawing a curve that does not pass through the origin when axes begin at $0$ .
Sketching a curve that is concave up (increasing gradient), which is not the shape of $\sqrt{T}$ .

Things to Be Careful About

$T$ must be in kelvin; the square-root relationship is with absolute temperature.
The question asks for a sketch (shape), not numerical values or a detailed scale.

Techniques used

use a proportionality to determine graph shaperelate root-mean-square speed to temperaturesketch a curve with correct intercept and changing gradient

Q5MediumOscillations

(a)

State what is meant by simple harmonic motion.

DifficultyEasy

Worked solution

Answer

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

a \propto -x \quad \text{or} \quad a = -\omega^2 x

Final answer

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

Detailed explanation

Background Concept

In oscillations, simple harmonic motion (SHM) is a special type of periodic motion where the restoring effect gets stronger the further you are from equilibrium.

The defining mathematical condition is:

a = -\omega^2 x

where:

$x$ is the displacement from equilibrium,
$a$ is the acceleration,
$\omega$ is the angular frequency,
the negative sign means the acceleration is always towards equilibrium (opposite to the displacement).

Understanding the Question

You are asked to state what is meant by SHM. That means you must give the definition clearly (not derive anything), ideally including the idea of proportionality and direction.

Approach

Give the core statement: acceleration is proportional to displacement and directed towards equilibrium. Writing the standard equation $a=-\omega^2 x$ makes this precise.

Step-by-Step Reasoning

In SHM, when the object is displaced by $x$ , a restoring acceleration acts.
The size of this acceleration increases in direct proportion to $|x|$ .
The acceleration always points back towards equilibrium, so it has the opposite sign to $x$ .
Hence the defining relation is $a=-\omega^2 x$ .

Key Takeaways

SHM is defined by an acceleration–displacement relationship.
“Towards equilibrium” is captured by the negative sign in $a=-\omega^2 x$ .

Common Mistakes

Saying only “motion is sinusoidal” without linking acceleration to displacement.
Missing the idea that the acceleration is towards equilibrium.
Writing $a \propto x$ (wrong sign) without the negative.

Things to Be Careful About

The definition uses acceleration, not velocity.
“Proportional to displacement” and “directed towards equilibrium” are both required for full credit.

Techniques used

state the defining condition for simple harmonic motionexpress the condition using the acceleration–displacement equation

(b)

A block is suspended by a spring. The block oscillates vertically with simple harmonic motion.

The velocity $v$ of the block varies with time $t$ according to

v = 0.56 \cos 16t

where $v$ is in $\text{m s}^{-1}$ and $t$ is in $\text{s}$ .

7M

(i)

Calculate the period of the oscillation.

period = ______ $\text{s}$

DifficultyEasy

Worked solution

Working

From $v = 0.56\cos 16t$ , the angular frequency is $\omega = 16\ \text{rad s}^{-1}$ .

T = \frac{2\pi}{\omega} = \frac{2\pi}{16} = 0.393\ \text{s}

Answer

$0.39\ \text{s}$

Final answer

0.39 s

Detailed explanation

Background Concept

For SHM, displacement, velocity, and acceleration are sinusoidal functions with the same angular frequency $\omega$ .

If a quantity is written in the form $\cos(\omega t)$ or $\sin(\omega t)$ , then the coefficient of $t$ is the angular frequency $\omega$ (in $\text{rad s}^{-1}$ ).

Period and angular frequency are related by:

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

Understanding the Question

You are given the velocity-time equation:

v = 0.56\cos 16t

and asked to calculate the period $T$ of the oscillation. The number $16$ multiplying $t$ is the key.

Approach

Read off $\omega$ from the cosine.
Substitute into $T = 2\pi/\omega$ .

Step-by-Step Reasoning

Compare $v = 0.56\cos 16t$ with $v = v_{\max}\cos(\omega t)$ , so $\omega = 16\ \text{rad s}^{-1}$ .
Compute the period:

T = \frac{2\pi}{16} = 0.392699...\ \text{s}

So $T \approx 0.39\ \text{s}$ .

Key Takeaways

In $\cos(\omega t)$ or $\sin(\omega t)$ , the coefficient of $t$ is $\omega$ .
Always use $T = 2\pi/\omega$ when $\omega$ is given.

Common Mistakes

Treating $16$ as the frequency $f$ (in Hz). It is not $f$ ; it is $\omega$ .
Using $T = 1/16$ which would only be correct if $16$ were $f$ .

Things to Be Careful About

Units: $\omega$ is in $\text{rad s}^{-1}$ , $T$ in $\text{s}$ .
Rounding: giving $0.393\ \text{s}$ or $0.39\ \text{s}$ is appropriate.

Techniques used

identify angular frequency from a sinusoidal functionuse the relationship between period and angular frequency

(ii)

Determine the amplitude $x_0$ of the oscillation.

$x_0$ = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

From $v = 0.56\cos 16t$ :

v_{\max} = 0.56\ \text{m s}^{-1}, \quad \omega = 16\ \text{rad s}^{-1}

For SHM, $v_{\max} = \omega x_0$ .

x_0 = \frac{v_{\max}}{\omega} = \frac{0.56}{16} = 3.5 \times 10^{-2}\ \text{m}

Answer

$3.5 \times 10^{-2}\ \text{m}$

Final answer

3.5 × 10^-2 m

Detailed explanation

Background Concept

In SHM, displacement can be written as:

x = x_0 \sin(\omega t)

Differentiating gives velocity:

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

So the maximum speed is:

v_{\max} = \omega x_0

This is a standard SHM result: the larger the amplitude $x_0$ and the higher the angular frequency $\omega$ , the larger the peak speed.

Understanding the Question

You are given:

v = 0.56\cos 16t

This matches $v = \omega x_0\cos(\omega t)$ . So:

the amplitude of the cosine, $0.56$ , is the maximum speed $v_{\max}$ ,
the coefficient of $t$ , $16$ , is $\omega$ .

You must find $x_0$ .

Approach

Use $v_{\max} = \omega x_0$ and rearrange to $x_0 = v_{\max}/\omega$ .

Step-by-Step Reasoning

Identify the two key parameters:

v_{\max} = 0.56\ \text{m s}^{-1}

\omega = 16\ \text{rad s}^{-1}

Substitute into the SHM relation:

x_0 = \frac{0.56}{16} = 0.035\ \text{m}

Express neatly:

x_0 = 3.5 \times 10^{-2}\ \text{m}

This is $3.5\ \text{cm}$ .

Key Takeaways

For SHM: $v_{\max} = \omega x_0$ .
The amplitude of the velocity-time cosine/sine is the maximum speed.

Common Mistakes

Using $x_0 = 0.56$ (confusing displacement amplitude with velocity amplitude).
Using $\omega = 2\pi f$ incorrectly because $\omega$ is already given.

Things to Be Careful About

Keep units consistent: $x_0$ in $\text{m}$ .
Recognise that $0.035\ \text{m} = 3.5\ \text{cm}$ , which helps for the next graph part where the $x$ -axis is in cm.

Techniques used

identify maximum speed from a sinusoidal velocity functionuse the SHM relation v_max = ω x0rearrange to determine amplitude

(iii)

Use your answer in (b)(ii) to determine the equation for $v$ in terms of the displacement $x$ of the block, where $v$ is in $\text{m s}^{-1}$ and $x$ is in $\text{m}$ .

$v$ = ______

DifficultyMedium-Easy

Worked solution

Working

For SHM,

v^2 = \omega^2\left(x_0^2 - x^2\right)

With $\omega = 16\ \text{rad s}^{-1}$ and $x_0 = 0.035\ \text{m}$ ,

v = \pm 16\sqrt{(0.035)^2 - x^2}

Answer

v = \pm 16\sqrt{(0.035)^2 - x^2}

Final answer

v = ±16√((0.035)^2 − x^2)

Detailed explanation

Background Concept

In SHM there is a standard link between speed and displacement.

Starting with:

x = x_0 \sin(\omega t)

we can write the velocity as:

v = \omega x_0 \cos(\omega t)

Eliminate time using $\sin^2(\omega t) + \cos^2(\omega t) = 1$ :

\left(\frac{x}{x_0}\right)^2 + \left(\frac{v}{\omega x_0}\right)^2 = 1

Rearranging gives the commonly used form:

v^2 = \omega^2(x_0^2 - x^2)

The speed can be positive or negative depending on direction, so:

v = \pm \omega\sqrt{x_0^2 - x^2}

Understanding the Question

You have already found $x_0$ from part (ii). Now you must write an equation for $v$ in terms of $x$ (not $t$ ).

That means you must use a relationship that connects instantaneous speed to instantaneous displacement.

Approach

Use:

v = \pm \omega\sqrt{x_0^2 - x^2}

and substitute the numerical values of $\omega$ and $x_0$ .

Step-by-Step Reasoning

Take the SHM speed–displacement equation:

v^2 = \omega^2(x_0^2 - x^2)

Substitute $\omega=16$ and $x_0 = 0.035\ \text{m}$ :

v^2 = 16^2\left((0.035)^2 - x^2\right)

Square root both sides:

v = \pm 16\sqrt{(0.035)^2 - x^2}

The $\pm$ is important: for the same $x$ value, the block passes twice each cycle with opposite velocity directions.

Key Takeaways

You can eliminate time in SHM and relate $v$ directly to $x$ .
The $v$ – $x$ relationship forms an ellipse when plotted.

Common Mistakes

Forgetting the $\pm$ sign.
Writing $v = 16\sqrt{x^2 - x_0^2}$ (wrong sign inside the root).
Mixing units (e.g. using $x$ in cm while $x_0$ is in m) in the equation.

Things to Be Careful About

The expression under the square root must be non-negative, so it only makes sense for $|x| \le x_0$ .
Ensure $x$ is in $\text{m}$ because the question states $x$ is in $\text{m}$ for this part.

Techniques used

use the SHM speed–displacement relation v^2 = ω^2(x0^2 - x^2)substitute numerical values for ω and x0take the square root and include the ± sign

(iv)

On Fig. 5.1, sketch the variation of $v$ with $x$ .

Fig. 5.1

DifficultyMedium

Worked solution

Answer

diagram-1

Final answer

Ellipse on v–x axes with intercepts v = ±0.56 m s^-1 at x = 0 and x = ±3.5 cm at v = 0.

Detailed explanation

Background Concept

For SHM, velocity and displacement are related by:

\left(\frac{x}{x_0}\right)^2 + \left(\frac{v}{v_{\max}}\right)^2 = 1

This is the equation of an ellipse centered at the origin in an $x$ – $v$ plot.

Key features:

When $x = 0$ (passing equilibrium), $|v|$ is maximum: $v = \pm v_{\max}$ .
When $x = \pm x_0$ (turning points), $v = 0$ .
The graph is symmetric about both axes.

Understanding the Question

You must sketch $v$ against $x$ on axes where:

vertical axis is $v / \text{m s}^{-1}$ from about $-0.8$ to $+0.8$ ,
horizontal axis is $x / \text{cm}$ from $-6$ to $+6$ .

From earlier parts:

$v_{\max} = 0.56\ \text{m s}^{-1}$ ,
$x_0 = 0.035\ \text{m} = 3.5\ \text{cm}$ .

Approach

Mark the four key points first:

$(x, v) = (0, +0.56)$ and $(0, -0.56)$ ,
$(+3.5\ \text{cm}, 0)$ and $(-3.5\ \text{cm}, 0)$ .

Then draw a smooth ellipse through them, symmetric about both axes.

Step-by-Step Reasoning

Convert amplitude to cm because the provided axis is in cm:

0.035\ \text{m} = 3.5\ \text{cm}

Identify intercepts:

At turning points, $x=\pm x_0$ , so $v=0$ giving x-intercepts at $x=\pm 3.5\ \text{cm}$ .
At equilibrium, $x=0$ , so $v=\pm v_{\max} = \pm 0.56\ \text{m s}^{-1}$ giving y-intercepts at $v=\pm 0.56\ \text{m s}^{-1}$ .

Sketch:

The curve must be closed (because the motion repeats).
It must be smooth and symmetric about both axes.
It should look like an ellipse that just touches those intercepts.

diagram-1

Key Takeaways

A $v$ – $x$ graph for SHM is an ellipse.
The intercepts correspond to turning points ( $v=0$ ) and equilibrium ( $|v|$ maximum).

Common Mistakes

Drawing a sine curve (that would be $x$ vs $t$ or $v$ vs $t$ , not $v$ vs $x$ ).
Using $x_0 = 0.035$ directly on an axis labelled in cm (placing intercept at $0.035\ \text{cm}$ instead of $3.5\ \text{cm}$ ).
Drawing two separate curves instead of one closed loop.

Things to Be Careful About

Units on the axis: $x$ is in cm on the figure, so the amplitude must be plotted at $\pm 3.5\ \text{cm}$ .
Ensure the maximum velocities are at $x=0$ , not at the ends.
The sketch should be centered on the origin and symmetric (important for full sketch marks).

Techniques used

identify key intercepts and maxima for a v–x plot in SHMuse the speed–displacement relation to determine the curve shapesketch a symmetric ellipse with correct axis crossings

Q6MediumElectric FieldsQuantum PhysicsMedical Physics

Two parallel metal plates X and Y are separated by a distance of $0.041\ \text{m}$ , as shown in Fig. 6.1.

Fig. 6.1

There is a vacuum between the plates. An electron is at rest at the centre of plate X.

A potential difference (p.d.) of $58\ \text{kV}$ is applied across the plates. This causes the electron to accelerate towards plate Y.

(a)

On Fig. 6.1, use the symbols + and – to indicate which of plates X and Y is the positive plate and which is the negative plate.

DifficultyEasy

Worked solution

The electron accelerates towards plate Y, so the force on the electron is towards Y.

Since $F = qE$ and $q$ is negative for an electron, the electric field $E$ is opposite to the force.

Hence $E$ is towards X, so Y must be the positive plate and X the negative plate.

Answer

Plate Y is $+$ and plate X is $-$ .

Final answer

Plate Y is + and plate X is −.

Detailed explanation

Background Concept

The electric field direction is defined as the direction of the force on a positive test charge. In a uniform field between parallel plates, the field lines go from the positive plate to the negative plate.

For any charge in an electric field,

F = qE

If $q$ is positive, the force is in the same direction as $E$ .
If $q$ is negative (an electron), the force is opposite to $E$ .

Understanding the Question

An electron starts from rest at plate X and accelerates towards plate Y. We must label which plate is positive and which is negative.

Approach

Use the given acceleration direction to identify the force direction on the electron.
Use $F = qE$ with $q<0$ to relate force direction to field direction.
Use the fact that $E$ points from $+$ to $-$ to decide the plate signs.

Step-by-Step Reasoning

The electron accelerates towards Y, so its net force is towards Y (Newton's second law: acceleration is in direction of net force).
For an electron, $q=-e$ , so in $F=qE$ the force is opposite to the electric field direction.
Therefore the electric field must point towards X.
Since electric fields point from the positive plate to the negative plate, the plate the field points towards is negative.
So X is negative and Y is positive.

Key Takeaways

Electric field direction is defined using a positive charge.
Electrons move opposite to the electric field direction.

Common Mistakes

Assuming the electron accelerates in the same direction as the electric field (forgetting the electron is negative).
Reversing the rule that field lines go from $+$ to $-$ .

Things to Be Careful About

Always separate these two ideas: (i) acceleration direction equals force direction, (ii) field direction is force direction only for a positive charge.
Do not confuse the electron's motion direction with the field direction.

Techniques used

use the direction of acceleration to infer the direction of force on a chargerelate force direction to electric field direction for a negative chargeuse the convention that electric field lines run from positive to negative

(b)

4M

(i)

Calculate the electric field strength $E$ between the plates. Give a unit with your answer.

$E$ = ______ unit ______

DifficultyMedium-Easy

Worked solution

Working

E = \frac{V}{d} = \frac{58\times10^3}{0.041} = 1.41\times10^6

Answer

E = 1.4\times10^6\ \text{V m}^{-1}

Final answer

1.4 × 10^6 V m⁻1

Detailed explanation

Background Concept

Between large parallel plates with a potential difference $V$ and separation $d$ , the electric field is approximately uniform and has magnitude

E = \frac{V}{d}

Units: $V$ in volts, $d$ in metres, so $E$ is in $\text{V m}^{-1}$ (equivalently $\text{N C}^{-1}$ ).

Understanding the Question

We are given:

separation $d = 0.041\ \text{m}$
potential difference $V = 58\ \text{kV} = 58\times 10^3\ \text{V}$

We need the electric field strength between the plates.

Approach

Use the uniform-field relation $E = V/d$ , ensuring $V$ is in volts and $d$ in metres, then quote the unit.

Step-by-Step Reasoning

Convert the p.d.:

V = 58\ \text{kV} = 58\times10^3\ \text{V}

Apply the formula:

E = \frac{V}{d} = \frac{58\times10^3}{0.041}

Calculate:

E = 1.4146\times10^6\ \text{V m}^{-1} \approx 1.4\times10^6\ \text{V m}^{-1}

Key Takeaways

For uniform fields between parallel plates, $E$ is found from $V/d$ .
Always convert kV to V and keep $d$ in metres.

Common Mistakes

Using $58$ instead of $58\times10^3$ (forgetting kilo).
Using $d$ in cm or mm without converting to metres.
Omitting the unit.

Things to Be Careful About

Quote $E$ to a sensible number of significant figures (typically 2 s.f.).
Use $\text{V m}^{-1}$ or $\text{N C}^{-1}$ , not both together.

Techniques used

use the uniform-field relation between potential difference and plate separationconvert kilovolts to voltscalculate a derived quantity with correct SI units

(ii)

Determine the acceleration of the electron.

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

DifficultyMedium-Easy

Worked solution

Working

F = qE = eE

a = \frac{F}{m} = \frac{eE}{m}

a = \frac{(1.60\times10^{-19})(1.41\times10^{6})}{9.11\times10^{-31}} = 2.5\times10^{17}

Answer

a = 2.5\times10^{17}\ \text{m s}^{-2}

Final answer

2.5 × 10^17 m s⁻2

Detailed explanation

Background Concept

A charge $q$ in an electric field $E$ experiences a force

F = qE

The resulting acceleration is found from Newton's second law:

F = ma \quad \Rightarrow \quad a = \frac{F}{m}

For an electron, the magnitude of charge is $e = 1.60\times10^{-19}\ \text{C}$ and its mass is $m_e = 9.11\times10^{-31}\ \text{kg}$ .

Understanding the Question

We have already found the uniform field between the plates. The question asks for the acceleration of the electron due to the electric force.

Approach

Find the magnitude of the electric force using $F = qE$ (use magnitude $e$ ).
Convert that force into acceleration using $a = F/m_e$ .

Step-by-Step Reasoning

Using the field from part (i), $E \approx 1.41\times10^6\ \text{V m}^{-1}$ .

Force magnitude on the electron:

F = eE = (1.60\times10^{-19})(1.41\times10^6) = 2.26\times10^{-13}\ \text{N}

Acceleration magnitude:

a = \frac{F}{m_e} = \frac{2.26\times10^{-13}}{9.11\times10^{-31}} = 2.48\times10^{17}\ \text{m s}^{-2}

So to 2 s.f.:

a \approx 2.5\times10^{17}\ \text{m s}^{-2}

(Direction would be towards plate Y because the force on the electron is towards Y.)

Key Takeaways

Combine $F=qE$ with $F=ma$ to get $a = qE/m$ .
For electrons, direction is opposite to the field, but the magnitude uses $e$ .

Common Mistakes

Using $q = +e$ for an electron and then giving the wrong direction (or mixing sign errors into the magnitude).
Using the proton mass instead of the electron mass.
Carrying forward an incorrect $E$ from part (i) due to kV conversion errors.

Things to Be Careful About

Keep track of powers of ten: $10^{-19}\times10^6 = 10^{-13}$ .
The question asks for acceleration in $\text{m s}^{-2}$ ; ensure the final unit is correct.

Techniques used

apply the electric force equation F = qEapply Newton's second law a = F/msubstitute standard values for electron charge and mass

(c)

Many electrons are now accelerated from rest from plate X to plate Y in Fig. 6.1. When the electrons hit plate Y, the absorption of their kinetic energies results in the emission of electromagnetic waves.

6M

(i)

Show that the minimum wavelength of these electromagnetic waves is $21\ \text{pm}$ .

DifficultyMedium

Worked solution

Working

Maximum photon energy equals the electron energy gain:

E_{\gamma,\max} = eV = (1.60\times10^{-19})(58\times10^{3}) = 9.28\times10^{-15}\ \text{J}

E_{\gamma} = \frac{hc}{\lambda} \Rightarrow \lambda_{\min} = \frac{hc}{E_{\gamma,\max}}

\lambda_{\min} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{9.28\times10^{-15}} = 2.14\times10^{-11}\ \text{m}

\lambda_{\min} = 2.14\times10^{-11}\ \text{m} = 21\ \text{pm}

Answer

$\lambda_{\min} = 21\ \text{pm}$

Final answer

21 pm

Detailed explanation

Background Concept

Electrons accelerated through a potential difference $V$ gain kinetic energy equal to the electrical work done:

\Delta E = qV

For an electron, the magnitude of this energy gain is $eV$ .

When fast electrons strike a metal target, electromagnetic radiation can be produced. The maximum possible photon energy occurs if one electron transfers all its kinetic energy to a single photon:

E_{\gamma,\max} = eV

Photon energy is related to wavelength by

E_{\gamma} = hf = \frac{hc}{\lambda}

So the maximum photon energy corresponds to the minimum wavelength:

\lambda_{\min} = \frac{hc}{E_{\gamma,\max}}

Understanding the Question

Electrons start from rest at plate X and accelerate to plate Y through $58\ \text{kV}$ . When they hit plate Y, electromagnetic waves are emitted. We must show that the minimum wavelength is $21\ \text{pm}$ .

Approach

Calculate the maximum energy available from one electron: $eV$ .
Assume a photon can take up to this energy, so set $E_{\gamma,\max} = eV$ .
Use $E_{\gamma}=hc/\lambda$ to find $\lambda_{\min}$ .

Step-by-Step Reasoning

Energy gained by one electron:

E_{\text{electron}} = eV = (1.60\times10^{-19})(58\times10^{3})

E_{\text{electron}} = 9.28\times10^{-15}\ \text{J}

Minimum wavelength occurs at maximum photon energy:

\lambda_{\min} = \frac{hc}{E_{\gamma,\max}} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{9.28\times10^{-15}}

\lambda_{\min} = 2.14\times10^{-11}\ \text{m}

Convert metres to picometres ( $1\ \text{pm} = 10^{-12}\ \text{m}$ ):

2.14\times10^{-11}\ \text{m} = 21.4\times10^{-12}\ \text{m} = 21\ \text{pm}

Key Takeaways

Accelerating a charge through a p.d. gives energy $qV$ .
Minimum wavelength corresponds to maximum photon energy.
Use $E=hc/\lambda$ to link energy and wavelength.

Common Mistakes

Using $V=58\ \text{V}$ instead of $58\ \text{kV}$ .
Using $E = \frac{1}{2}mv^2$ unnecessarily (you are not asked to find $v$ ).
Failing to convert metres to picometres correctly.

Things to Be Careful About

Keep $h$ and $c$ in SI units so energy comes out in joules.
Check powers of ten carefully; wavelength should be of order $10^{-11}\ \text{m}$ for X-rays.

Techniques used

equate electrical energy gained by a charge to kinetic energyuse the photon energy relation E = hc/lambdarearrange to determine a minimum wavelength

(ii)

State the region of the electromagnetic spectrum that contains these waves.

(iii)

Explain how these electromagnetic waves may be used to form images of internal body structures.

DifficultyMedium-Easy

Worked solution

Answer

X-rays pass through the body and are absorbed (attenuated) by different amounts in different tissues.
Bone absorbs more than soft tissue, so the intensity reaching a film/detector varies across the beam, producing a contrast image of internal structures.

Final answer

Different tissues attenuate X-rays by different amounts; a detector records the transmitted intensity pattern to give a contrast image.

Detailed explanation

Background Concept

X-ray imaging relies on attenuation: as X-rays pass through matter, their intensity decreases because photons are absorbed or scattered.

A simple model is:

I = I_0 e^{-\mu x}

where $I_0$ is incident intensity, $I$ is transmitted intensity, $x$ is thickness, and $\mu$ depends on the material (denser / higher atomic number materials generally attenuate more strongly).

An image is formed because different regions in the body attenuate the X-rays differently, leading to different transmitted intensities at the detector.

Understanding the Question

The waves produced when electrons hit plate Y are X-rays. The question asks how these waves are used to form images of internal body structures (i.e. the basic principle behind radiographs).

Approach

Explain:

X-rays are partially transmitted through the body.
Different tissues absorb different fractions of the X-rays (contrast).
A detector (film or digital sensor) measures the transmitted intensity pattern to create the image.

Step-by-Step Reasoning

diagram-c-iii-1

A beam of X-rays is directed through the body towards a detector.
As the beam passes through, some photons are removed from the beam by absorption/scattering.
Bone (higher density and effective atomic number) attenuates X-rays more strongly than soft tissue, so fewer X-rays reach the detector behind bones.
The detector records a spatial pattern of intensity: regions with low transmitted intensity correspond to more absorption (appear lighter on a negative film / different shade on a digital image), and regions with high transmitted intensity correspond to less absorption.
This contrast reveals internal structures such as bones and some organs.

Key Takeaways

Imaging needs two ingredients: transmission through the body and different attenuation in different tissues.
The detector converts the transmitted intensity pattern into a visible image.

Common Mistakes

Saying only “X-rays go through the body” without mentioning different absorption/attenuation (no mechanism for contrast).
Claiming X-rays are “reflected” to form images (standard radiographs use transmission, not reflection).

Things to Be Careful About

Use the term attenuation/absorption and link it to tissue type (bone vs soft tissue).
Mention a detector/film; without it, there is no recorded image.

Techniques used

describe how differential absorption produces image contrastexplain the role of a detector in recording transmitted intensitylink material density/thickness to X-ray attenuation

Q7MediumCapacitance

Fig. 7.1 shows a circuit containing a capacitor of capacitance $C$ and a resistor of resistance $R$ .

Fig. 7.1

Initially, the switch is open and the potential difference (p.d.) across the capacitor is $12\ \text{V}$ .

The switch is closed at time $t = 0$ and the capacitor discharges through the resistor.

Fig. 7.2 shows the variation of the charge $Q$ on the capacitor with the p.d. $V_C$ across the capacitor as the capacitor discharges. Fig. 7.3 shows the variation of the current $I$ in the resistor with the p.d. $V_R$ across the resistor as the capacitor discharges.

Fig. 7.2

Fig. 7.3

(a)

State the relationship between $V_C$ and $V_R$ .

(b)

Determine:

6M

(i)

the capacitance $C$ , in $\mu\text{F}$

$C$ = ______ $\mu\text{F}$

DifficultyMedium-Easy

Worked solution

Working

From Fig. 7.2, using point $(V_C, Q) = (12\ \text{V}, 6\ \text{mC})$ :

C = \frac{Q}{V_C} = \frac{6.0 \times 10^{-3}}{12} = 5.0 \times 10^{-4}\ \text{F}

5.0 \times 10^{-4}\ \text{F} = 5.0 \times 10^{2}\ \mu\text{F}

Answer

C = 500\ \mu\text{F}

Final answer

$500\ \mu\text{F}$

Detailed explanation

Background Concept

Capacitance is defined by

C = \frac{Q}{V}

so a graph of $Q$ against $V$ for a capacitor is a straight line through the origin with gradient equal to $C$ .

Understanding the Question

Fig. 7.2 gives a straight-line graph of charge $Q$ (in mC) against capacitor p.d. $V_C$ (in V) as it discharges. You are asked to determine $C$ in $\mu\text{F}$ .

Approach

Pick a clear point on the straight line (the mark scheme typically expects you to use the labelled point), calculate $C = Q/V_C$ , then convert from farads to microfarads.

Step-by-Step Reasoning

Using the point shown: at $V_C = 12\ \text{V}$ , $Q = 6\ \text{mC}$ .

Convert charge to coulombs:

6\ \text{mC} = 6 \times 10^{-3}\ \text{C}

Now apply the definition:

C = \frac{Q}{V_C} = \frac{6 \times 10^{-3}}{12} = 5 \times 10^{-4}\ \text{F}

Convert to microfarads using $1\ \mu\text{F} = 10^{-6}\ \text{F}$ :

5 \times 10^{-4}\ \text{F} = 5 \times 10^{2}\ \mu\text{F} = 500\ \mu\text{F}

Key Takeaways

The gradient of a $Q$ – $V$ graph is the capacitance.
Always convert mC to C before using $C = Q/V$ .

Common Mistakes

Treating $6\ \text{mC}$ as $6\ \text{C}$ .
Inverting the ratio and calculating $V/Q$ .
Giving the answer in F without converting to $\mu\text{F}$ when asked.

Things to Be Careful About

Use a point actually on the straight line (here the labelled point makes this unambiguous).
Keep units consistent (C and V give F).

Techniques used

use the gradient of a straight-line graph to obtain a constant of proportionalityapply $C = Q/V$convert between milli-units and SI units

(ii)

the resistance $R$ , in $\text{k}\Omega$

$R$ = ______ $\text{k}\Omega$

(iii)

the time constant $\tau$ of the circuit.

$\tau$ = ______ $\text{s}$

(c)

Use Fig. 7.2, Fig. 7.3 and your answer in (a) to explain why the variation of $Q$ with $t$ is exponential in nature.

DifficultyMedium

Worked solution

Answer

From Fig. 7.2, $Q \propto V_C$ (straight line through origin).

From Fig. 7.3, $I \propto V_R$ (straight line through origin).

From (a), $V_C = V_R \Rightarrow I \propto V_C \Rightarrow I \propto Q$ .

Since

I = -\frac{\mathrm{d}Q}{\mathrm{d}t}

then

\frac{\mathrm{d}Q}{\mathrm{d}t} \propto -Q

so $Q$ decays exponentially with $t$ (i.e. $Q = Q_0 e^{-t/RC}$ ).

Final answer

Because $\mathrm{d}Q/\mathrm{d}t \propto -Q$ , $Q$ varies exponentially with $t$ .

Detailed explanation

Background Concept

In an RC discharge, the key ideas are:

Capacitor relation:

Q = CV_C

Resistor relation:

V_R = IR

Current is the rate of change of charge:

I = -\frac{\mathrm{d}Q}{\mathrm{d}t}

The minus sign shows that during discharge the charge on the capacitor is decreasing.

If you can show that the rate of decrease of $Q$ is proportional to $Q$ itself,

\frac{\mathrm{d}Q}{\mathrm{d}t} = -kQ

then the solution is an exponential decay:

Q = Q_0 e^{-kt}

For an RC circuit, $k = 1/RC$ .

Understanding the Question

You are given two straight-line graphs:

Fig. 7.2: $Q$ against $V_C$ .
Fig. 7.3: $I$ against $V_R$ .

You must use those graphs plus the relationship from (a) between $V_C$ and $V_R$ to explain why $Q$ vs $t$ is exponential.

Approach

Use each straight-line graph to extract a proportionality:

Straight line through the origin means “directly proportional”.
Then connect the proportionalities using $V_C = V_R$ to show $I \propto Q$ .
Finally use $I = -\mathrm{d}Q/\mathrm{d}t$ to obtain the differential equation whose solution is exponential.

Step-by-Step Reasoning

From Fig. 7.2, the graph of $Q$ against $V_C$ is a straight line through the origin. Therefore:

Q \propto V_C

(Equivalently, $Q = CV_C$ where $C$ is the gradient.)

From Fig. 7.3, the graph of $I$ against $V_R$ is also a straight line through the origin, so:

I \propto V_R

(Equivalently, $I = \frac{1}{R}V_R$ where $1/R$ is the gradient.)

From part (a) for this single-loop discharge circuit, the magnitudes of the p.d.s are equal at any instant:

V_C = V_R

So we can replace $V_R$ with $V_C$ in the proportionality from step 2:

I \propto V_R \Rightarrow I \propto V_C

Combine with step 1 ( $Q \propto V_C$ ): if both $I$ and $Q$ are proportional to $V_C$ , then $I$ is proportional to $Q$ :

I \propto Q

More explicitly using the circuit equations:

V_C = V_R \Rightarrow \frac{Q}{C} = IR \Rightarrow I = \frac{1}{RC}Q

Now connect current to charge change:

I = -\frac{\mathrm{d}Q}{\mathrm{d}t}

Substitute $I = \frac{1}{RC}Q$ :

-\frac{\mathrm{d}Q}{\mathrm{d}t} = \frac{1}{RC}Q

or

\frac{\mathrm{d}Q}{\mathrm{d}t} = -\frac{1}{RC}Q

This has the form $\mathrm{d}Q/\mathrm{d}t = -kQ$ , whose solution is an exponential decay:

Q = Q_0 e^{-t/RC}

Hence $Q$ varies exponentially with time.

Key Takeaways

Straight line through origin on a graph implies direct proportionality.
Using $V_C = V_R$ , the resistor’s $I$ – $V$ relation links directly to the capacitor’s $Q$ – $V$ relation.
Exponential decay comes from the differential equation $\mathrm{d}Q/\mathrm{d}t \propto -Q$ .

Common Mistakes

Claiming “it’s exponential because it’s an RC circuit” without showing the proportionality steps.
Missing the key link $I = -\mathrm{d}Q/\mathrm{d}t$ .
Forgetting to use the result from (a) to connect $V_C$ and $V_R$ .

Things to Be Careful About

The negative sign matters: $Q$ must decrease during discharge.
Be clear that the graphs are straight lines through the origin; that is what justifies the proportionalities.
If you write $I = Q/RC$ , remember this is the magnitude relation; the time-rate equation carries the negative sign via $I = -\mathrm{d}Q/\mathrm{d}t$ .

Techniques used

use proportionalities implied by straight-line graphscombine $V_C = V_R$ with $Q = CV$ and $V = IR$use $I = -\mathrm{d}Q/\mathrm{d}t$ to form a differential equationdeduce exponential behaviour from $\mathrm{d}Q/\mathrm{d}t \propto -Q$

Q8MediumAlternating Currents

Fig. 8.1 shows a circuit that produces rectification of an alternating input voltage.

Fig. 8.1

The input voltage $V_{IN}$ is sinusoidal. The rectified output voltage $V_{OUT}$ is applied across resistor $R$ .

The variation of $V_{IN}$ with time $t$ has amplitude $V_0$ and period $T$ , as shown in Fig. 8.2.

Fig. 8.2

The root-mean-square (r.m.s.) value of $V_{IN}$ is $6.0\ \text{V}$ .

(a)

2M

(i)

State the type of rectification produced by the circuit of Fig. 8.1.

DifficultyEasy

Worked solution

Answer

Half-wave rectification.

Final answer

Half-wave rectification.

Detailed explanation

Background Concept

Rectification is the process of converting an alternating voltage (a.c.), which changes polarity, into a voltage that has only one polarity (d.c. pulses).

A diode is a component that conducts current in only one direction (forward bias) and blocks current in the reverse direction (reverse bias). In an a.c. circuit:

During one half-cycle, the diode may be forward-biased and conduct.
During the opposite half-cycle, the diode is reverse-biased and does not conduct.

If only one half of each cycle reaches the load resistor, the circuit produces half-wave rectification.

Understanding the Question

The circuit in Fig. 8.1 has a single diode in series with a resistor $R$ , and the output $V_{OUT}$ is taken across $R$ . The question asks what type of rectification results from this arrangement.

Approach

Decide whether the diode arrangement allows current through $R$ on:

both half-cycles (full-wave), or
only one half-cycle (half-wave).

A single diode in series means only one polarity of the input will produce current.

Step-by-Step Reasoning

When the input polarity forward-biases the diode, current flows through $R$ and there is a non-zero $V_{OUT}$ .
When the input polarity reverse-biases the diode, current is (ideally) zero, so $V_{OUT} = 0$ .
Therefore only one half-cycle appears across $R$ .

Hence the rectification is half-wave rectification.

Key Takeaways

A single diode in series with a load gives half-wave rectification.
Full-wave rectification requires a bridge rectifier (4 diodes) or a centre-tapped transformer with 2 diodes.

Common Mistakes

Saying “full-wave” just because the output is always positive; you must check whether both halves are used.
Confusing rectification with smoothing (which requires a capacitor).

Things to Be Careful About

In ideal analysis (as used later in the question), assume the diode has zero p.d. when conducting and blocks completely when reverse-biased.

Techniques used

identify the function of a diode in an a.c. circuitrelate conduction direction to which half-cycles appear at the output

(ii)

Calculate $V_0$ .

$V_0$ = ______ $\text{V}$

DifficultyMedium-Easy

Worked solution

Working

For a sinusoid,

V_{\text{rms}} = \frac{V_0}{\sqrt{2}}

So

V_0 = 6.0\sqrt{2} = 8.49\ \text{V} \approx 8.5\ \text{V}

Answer

$V_0 = 8.5\ \text{V}$

Final answer

8.5 V

Detailed explanation

Background Concept

For a sinusoidal voltage

V(t) = V_0\sin(\omega t)

$V_0$ is the peak (amplitude). The r.m.s. value is defined as the square root of the mean value of $V^2$ over one full period. For a pure sine wave this leads to the standard result:

V_{\text{rms}} = \frac{V_0}{\sqrt{2}}

This is used because many electrical power equations (e.g. $P=V_{\text{rms}}^2/R$ ) work with r.m.s. values for a.c.

Understanding the Question

The graph shows a sinusoidal input voltage $V_{IN}$ with amplitude $V_0$ . You are told $V_{IN}$ has r.m.s. value $6.0\ \text{V}$ , and you must find the amplitude $V_0$ .

Approach

Use the sinusoidal relationship $V_{\text{rms}} = V_0/\sqrt{2}$ and rearrange to find $V_0$ .

Step-by-Step Reasoning

Start with

V_{\text{rms}} = \frac{V_0}{\sqrt{2}}

Rearrange:

V_0 = V_{\text{rms}}\sqrt{2}

Substitute $V_{\text{rms}} = 6.0\ \text{V}$ :

V_0 = 6.0\sqrt{2} = 8.49\ \text{V}

To 2 significant figures (matching 6.0 V):

V_0 \approx 8.5\ \text{V}

Key Takeaways

For a sine wave, peak and r.m.s. are related by a factor of $\sqrt{2}$ .
Keep significant figures consistent with the data given.

Common Mistakes

Using $V_0 = V_{\text{rms}}/\sqrt{2}$ (inverting the relationship).
Rounding too early (e.g. using $\sqrt{2}=1.4$ can be okay, but keep enough precision).

Things to Be Careful About

This part is about the input $V_{IN}$ , which is a full sinusoid, not the rectified output.
Quote the final answer with unit and sensible s.f.

Techniques used

use the relationship between peak and r.m.s. values for a sine waverearrange and substitute into an equation

(b)

Resistor $R$ has resistance $45\ \Omega$ .

Assume that there is no p.d. across the diode when it is conducting.

8M

(i)

Determine the peak power $P_0$ in the resistor.

$P_0$ = ______ $\text{W}$

DifficultyMedium-Easy

Worked solution

Working

Peak output p.d. across $R$ is $V_0$ (ideal diode).

P_0 = \frac{V_0^2}{R} = \frac{(8.49)^2}{45} = 1.60\ \text{W}

Answer

$P_0 = 1.6\ \text{W}$

Final answer

1.6 W

Detailed explanation

Background Concept

For a resistor, instantaneous power is

P = VI

and since $V = IR$ ,

P = I^2R = \frac{V^2}{R}

If the voltage across the resistor varies with time, then the power also varies with time.

In a half-wave rectifier with an ideal diode (zero p.d. when conducting), the load resistor sees:

the positive half-cycle of the input voltage,
and approximately $0\ \text{V}$ on the negative half-cycle.
The peak output voltage across the resistor is therefore the same as the input peak $V_0$ .

Understanding the Question

You are given $R = 45\ \Omega$ and told to assume the diode has no p.d. across it when conducting (ideal). You must find the peak power $P_0$ in the resistor. Peak power occurs when the resistor voltage is at its maximum value.

Approach

Identify the peak voltage across $R$ (here it is $V_0$ ).
Use $P_0 = V_0^2/R$ .

Step-by-Step Reasoning

At the top of a conducting half-cycle, the resistor has voltage $V_0$ .
The corresponding power is

P_0 = \frac{V_0^2}{R}

Substitute $V_0 = 8.49\ \text{V}$ and $R = 45\ \Omega$ :

P_0 = \frac{(8.49)^2}{45} = 1.60\ \text{W}

So $P_0 \approx 1.6\ \text{W}$ .

Key Takeaways

Peak power in a resistor depends on the square of the peak voltage.
For an ideal diode, the peak output voltage equals the peak input voltage (during the conducting half-cycle).

Common Mistakes

Using $P_0 = V_0/R$ (missing the square).
Using r.m.s. voltage instead of peak voltage when asked for peak power.

Things to Be Careful About

The diode assumption matters: a real diode drop would slightly reduce the peak voltage and hence reduce $P_0$ .
Keep enough precision in $V_0$ before squaring.

Techniques used

use the peak output voltage across a resistor to find peak powerapply the resistive power relation P = V^2 / Rsubstitute numerical values with correct units

(ii)

On Fig. 8.3, sketch the variation of the power $P$ in the resistor with $t$ between $t = 0$ and $t = 2T$ .

Fig. 8.3

DifficultyMedium

Worked solution

diagram-bii

Answer

$P=0$ for each negative half-cycle.
For each positive half-cycle, $P$ follows a $\sin^2$ shape from $0$ up to $P_0$ and back to $0$ (peaks at $t=\tfrac{T}{4}$ and $t=\tfrac{5T}{4}$ ).

Final answer

Power is a sin^2 hump during each positive half-cycle (peak P0 at T/4 and 5T/4) and zero during each negative half-cycle.

Detailed explanation

Background Concept

If the voltage across a resistor is $V(t)$ , the instantaneous power is

P(t) = \frac{V(t)^2}{R}

So if $V(t)$ is sinusoidal, power involves the square of a sinusoid. Squaring makes all values non-negative and changes the shape:

\sin^2(\omega t) \text{ varies between } 0 \text{ and } 1.

For a half-wave rectifier, the output voltage across $R$ is:

$V_{OUT}(t) = V_0\sin(\omega t)$ during the conducting half-cycle,
$V_{OUT}(t) = 0$ during the blocked half-cycle.

Understanding the Question

You must sketch $P$ against $t$ from $0$ to $2T$ . The axes are already provided, including the labelled levels $0$ , $\tfrac{1}{2}P_0$ , and $P_0$ . The key is to show:

two identical pulses in the interval $0$ to $2T$ ,
and zero power over the half-cycles when the diode does not conduct.

Approach

Write power in terms of output voltage: $P = V_{OUT}^2/R$ .
Determine $V_{OUT}(t)$ for a half-wave rectifier.
Sketch $P(t)$ $P (t)$ :
- when $V_{OUT}=0$ , then $P=0$ ;
- when $V_{OUT}$ is sinusoidal, $P$ follows a $\sin^2$ pulse.

Step-by-Step Reasoning

During a conducting (positive) half-cycle:

V_{OUT}(t) = V_0\sin(\omega t)

So

P(t) = \frac{V_0^2\sin^2(\omega t)}{R} = P_0\sin^2(\omega t)

where

P_0 = \frac{V_0^2}{R}

This pulse:

starts at $P=0$ when $\sin(\omega t)=0$ ,
rises to $P=P_0$ when $\sin(\omega t)=1$ (quarter-period into the pulse),
falls back to $0$ at the end of that half-cycle.

During the non-conducting (negative) half-cycle, $V_{OUT}=0$ , so

P(t)=0

Over $0$ to $2T$ , you therefore draw:

a $\sin^2$ “hump” from $0$ to $T/2$ peaking at $t=T/4$ ,
a flat line at $0$ from $T/2$ to $T$ ,
the same pattern again from $T$ to $2T$ (peak at $t=5T/4$ ).

diagram-bii

Key Takeaways

Power in a resistor depends on $V^2$ , so it is always non-negative.
Half-wave rectification makes the output zero for half of each cycle.
Squaring a sine gives a $\sin^2$ pulse with maximum at $P_0$ .

Common Mistakes

Sketching negative power for the negative half-cycle (power cannot be negative in a resistor).
Drawing a sine-shaped power graph instead of a $\sin^2$ -shaped graph.
Forgetting the power is zero for an entire half-cycle (not just smaller).

Things to Be Careful About

Place the peaks at $t=\tfrac{T}{4}$ and $t=\tfrac{5T}{4}$ , not at $t=\tfrac{T}{2}$ .
Ensure the curve touches the time axis at $t=0,\ \tfrac{T}{2},\ T,\ \tfrac{3T}{2},\ 2T$ .
Use the given vertical labels ( $P_0$ and $\tfrac{1}{2}P_0$ ) to set the peak height correctly.

Techniques used

express instantaneous power in terms of instantaneous voltagesquare a sinusoidal function to obtain a sin^2 time variationapply diode conduction to set the power to zero for one half-cyclesketch a waveform with correct period and peak value

(iii)

Use the answer in (b)(ii) to explain why the mean power in the resistor is $\frac{1}{4}P_0$ .

DifficultyMedium

Worked solution

Working

When conducting,

P = P_0\sin^2(\omega t)

Mean of $\sin^2$ over a half-cycle is $\tfrac{1}{2}$ , so mean power over the conducting half-cycle is $\tfrac{1}{2}P_0$ .

Power is zero for the other half-cycle, so mean over a full cycle is

\frac{1}{2}\left(\frac{1}{2}P_0\right)=\frac{1}{4}P_0.

Answer

Mean power $= \dfrac{1}{4}P_0$ .

Final answer

Mean power = (1/4) P0.

Detailed explanation

Background Concept

Mean power over a time interval is found by averaging $P(t)$ :

P_{\text{mean}} = \frac{1}{T}\int_0^T P(t)\,dt

Graphically, this is:

area under the $P$ – $t$ graph over one period,
divided by the period.

Also, a key mathematical result for a sinusoid is:

\text{mean of } \sin^2(\omega t) \text{ over a full cycle (or over }0\text{ to }\pi\text{)} = \frac{1}{2}.

Understanding the Question

From your sketch in (b)(ii), power is present only during the conducting half-cycle and is zero for the other half-cycle. The question asks you to use that sketch to justify why the overall mean power is $\tfrac{1}{4}P_0$ .

Approach

Break the cycle into two halves:

Conducting half-cycle: $P(t)=P_0\sin^2(\omega t)$ , whose average value is $\tfrac{1}{2}P_0$ .
Non-conducting half-cycle: $P(t)=0$ .
Then average those two halves over the full period.

Step-by-Step Reasoning

During the positive (conducting) half-cycle,

P(t)=P_0\sin^2(\omega t).

Over that half-cycle, the average of $\sin^2$ is $\tfrac{1}{2}$ , so the average power during the conducting half is

P_{\text{mean, cond}} = \frac{1}{2}P_0.

However, the diode blocks for the other half-cycle, so for half the time $P=0$ .

So over a full period, you spend:

half the time at an average level $\tfrac{1}{2}P_0$ ,
half the time at $0$ .
Therefore the overall mean is

P_{\text{mean}} = \frac{1}{2}\left(\frac{1}{2}P_0\right) + \frac{1}{2}(0) = \frac{1}{4}P_0.

This matches the result required.

Key Takeaways

Mean power is linked to area under the $P$ – $t$ curve.
Squaring a sine gives a function with mean value $1/2$ over its cycle.
Half-wave rectification introduces an additional factor of $1/2$ because power is zero for half the time.

Common Mistakes

Using the mean of $\sin$ (which is zero) instead of the mean of $\sin^2$ .
Forgetting that the power is exactly zero during the blocked half-cycle.

Things to Be Careful About

The $\tfrac{1}{2}$ mean value applies to $\sin^2$ over the interval where it completes a full “arch” (e.g. $0$ to $\pi$ ).
Keep clear which average you are taking: average during conduction vs average over the whole cycle.

Techniques used

use area under a power–time graph to determine mean poweruse the mean value of sin^2 over a half-cyclerelate the half-wave conduction interval to a factor of one half

(iv)

Use the information in (b)(iii) to determine the r.m.s. value of $V_{OUT}$ .

r.m.s. voltage = ______ $\text{V}$

DifficultyMedium-Easy

Worked solution

Working

P_{\text{mean}} = \frac{V_{\text{rms}}^2}{R}

But $P_{\text{mean}}=\frac{1}{4}P_0$ and $P_0=\frac{V_0^2}{R}$ , so

\frac{V_{OUT,\text{rms}}^2}{R} = \frac{1}{4}\frac{V_0^2}{R}

Hence

V_{OUT,\text{rms}} = \frac{V_0}{2} = \frac{8.49}{2} = 4.25\ \text{V} \approx 4.2\ \text{V}

Answer

r.m.s. voltage $= 4.2\ \text{V}$

Final answer

4.2 V

Detailed explanation

Background Concept

For a resistor, mean power can be written using r.m.s. voltage:

P_{\text{mean}} = \frac{V_{\text{rms}}^2}{R}.

This is why r.m.s. values are so useful in a.c. circuits: they let you use the same form as for d.c. power.

Understanding the Question

You have already shown in (b)(iii) that the mean power in the resistor is $\tfrac{1}{4}P_0$ , where $P_0$ is the peak power. You must use that to find the r.m.s. value of the rectified output voltage $V_{OUT}$ .

Approach

Write $P_{\text{mean}}$ in terms of $V_{OUT,\text{rms}}$ using $P_{\text{mean}}=V_{\text{rms}}^2/R$ .
Substitute $P_{\text{mean}}=\tfrac{1}{4}P_0$ and $P_0 = V_0^2/R$ .
Solve for $V_{OUT,\text{rms}}$ .

Step-by-Step Reasoning

Start with the r.m.s. power relation for the output:

P_{\text{mean}} = \frac{V_{OUT,\text{rms}}^2}{R}.

From (b)(iii):

P_{\text{mean}} = \frac{1}{4}P_0.

Also from (b)(i):

P_0 = \frac{V_0^2}{R}.

Substitute into the mean power equation:

\frac{V_{OUT,\text{rms}}^2}{R} = \frac{1}{4}\left(\frac{V_0^2}{R}\right).

The $R$ cancels:

V_{OUT,\text{rms}}^2 = \frac{1}{4}V_0^2.

Take the square root:

V_{OUT,\text{rms}} = \frac{V_0}{2}.

With $V_0 = 8.49\ \text{V}$ :

V_{OUT,\text{rms}} = 4.25\ \text{V} \approx 4.2\ \text{V}.

Key Takeaways

Mean power in a resistor is linked to r.m.s. voltage by $P_{\text{mean}}=V_{\text{rms}}^2/R$ .
For a half-wave rectified sine, $P_{\text{mean}}=\tfrac{1}{4}P_0$ leads to $V_{OUT,\text{rms}} = V_0/2$ .

Common Mistakes

Using $V_{OUT,\text{rms}} = V_0/\sqrt{2}$ (that would be for an un-rectified sine wave).
Forgetting to square-root at the end (treating $V^2$ as $V$ ).

Things to Be Careful About

This $V_{OUT,\text{rms}}$ is for the rectified waveform, not the original input.
Because $P \propto V^2$ , it is consistent that halving r.m.s. voltage would quarter the power, and vice versa.

Techniques used

relate mean power in a resistor to r.m.s. voltage using P = Vrms^2 / Rsubstitute the given mean-power fraction to derive Vrmsrearrange and simplify an algebraic expression

Q9MediumQuantum Physics

(a)

State what is meant by the photoelectric effect.

DifficultyEasy

Worked solution

Answer

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation is incident on it (provided the radiation has frequency above a threshold value).

Final answer

Emission of electrons from a metal surface due to incident electromagnetic radiation (for frequency above a threshold).

Detailed explanation

Background Concept

The photoelectric effect is evidence that electromagnetic radiation transfers energy in discrete packets called photons. For a given metal, electrons are bound in the surface and require a minimum energy (the work function) to escape.

Understanding the Question

You are asked to state what the photoelectric effect means. The key points are: electrons are emitted from a metal surface, and the emission is caused by incident electromagnetic radiation (light, UV, etc.), with the important condition that the frequency must be sufficiently high.

Approach

Give a precise definition: what happens (electron emission), from where (metal surface), and what causes it (incident EM radiation), including the threshold-frequency condition.

Step-by-Step Reasoning

Identify the phenomenon: electrons leave the metal surface.
Identify the cause: incident electromagnetic radiation.
Include the condition: emission occurs only if the frequency is above a threshold value for that metal.

Key Takeaways

Photoelectric effect = electron emission due to incident EM radiation.
There is a threshold frequency for each metal.

Common Mistakes

Saying “more intense light always ejects electrons” (intensity affects current, not whether emission happens).
Missing that electrons are emitted from a metal surface.
Not mentioning the threshold frequency condition.

Things to Be Careful About

Use “electrons” (or “photoelectrons”), not “photons are emitted”.
Frequency/threshold is more fundamental than “brightness” for the onset of emission.

Techniques used

state the defining features of the effectlink electron emission to incident electromagnetic radiation frequency

(b)

The photoelectric effect is investigated in two stages using the circuit shown in Fig. 9.1.

Fig. 9.1

The polished metal plate Y is illuminated with electromagnetic radiation of frequency $f$ and constant power.

In stage 1 of the investigation, frequency $f$ is set to a constant value of $2.5 \times 10^{15}\ \text{Hz}$ . The current $I$ in the ammeter is varied by adjusting the potentiometer P. Fig. 9.2 shows the variation of $I$ with the voltmeter reading $V$ . There is a value $V_S$ of $V$ at which the current just falls to zero.

In stage 2 of the investigation, stage 1 is repeated for different values of frequency. As frequency $f$ is varied, the voltmeter reading $V_S$ at which the current just falls to zero is measured. Fig. 9.3 shows the variation of $V_S$ with $f$ .

Fig. 9.2

Fig. 9.3

6M

(i)

Explain, with reference to photons, why $V_S$ depends on the frequency of the incident electromagnetic radiation.

DifficultyMedium-Easy

Worked solution

Answer

Photon energy is $E = hf$ . Electrons require energy $\phi$ (work function) to escape, so

E_{k,\max} = hf - \phi.

The stopping potential $V_S$ is when even the most energetic electrons are stopped:

eV_S = E_{k,\max}.

Hence $V_S$ increases when $f$ increases (since $hf$ increases).

Final answer

Because photon energy $hf$ increases with $f$ , so $E_{k,\max}=hf-\phi$ increases and therefore $eV_S=E_{k,\max}$ gives larger $V_S$ for higher $f$ .

Detailed explanation

Background Concept

Light can be treated as photons with energy

E = hf,

where $h$ is the Planck constant and $f$ is the radiation frequency. In the photoelectric effect, an electron needs at least the work function $\phi$ (energy needed to escape the metal). Any extra photon energy becomes the electron’s maximum kinetic energy:

hf = \phi + E_{k,\max}.

A stopping potential $V_S$ is a reverse potential difference that just prevents the fastest photoelectrons from reaching the collector. The electrical work done in stopping an electron of charge $e$ is $eV_S$ , so at the stopping condition:

eV_S = E_{k,\max}.

Understanding the Question

In the experiment, the frequency $f$ is changed (while power is constant). The question asks why the measured stopping potential $V_S$ changes when $f$ changes, and it specifically wants the explanation in terms of photons.

Approach

Connect the chain:

photon energy depends on frequency ( $E=hf$ );
maximum electron kinetic energy depends on photon energy minus work function ( $E_{k,\max}=hf-\phi$ );
stopping potential is proportional to $E_{k,\max}$ via $eV_S$ .
Therefore, changing $f$ changes $V_S$ .

Step-by-Step Reasoning

Increase frequency $f$ .
Photon energy increases because $E=hf$ .
The metal requires the same work function $\phi$ (property of the metal surface), so the “leftover” energy for the fastest electrons increases:

E_{k,\max} = hf - \phi.

To stop these faster electrons, the reverse potential difference must do more work per electron, i.e. $eV_S$ must be larger.
Using

eV_S = E_{k,\max},

a larger $E_{k,\max}$ means a larger $V_S$ .

diagram-b-i-energy

Key Takeaways

Frequency controls photon energy.
Work function is a fixed “energy cost” for a given metal.
Stopping potential measures maximum kinetic energy: $eV_S = E_{k,\max}$ .

Common Mistakes

Saying $V_S$ depends on intensity/power (with constant frequency, intensity mainly changes the current, not the stopping potential).
Forgetting the work function term $\phi$ .
Mixing up $eV_S$ with $V_S$ (you must multiply by charge to get energy).

Things to Be Careful About

$V_S$ is linked to the maximum kinetic energy, not the average.
Use the electron charge as $e = 1.60 \times 10^{-19}\ \text{C}$ if a numerical conversion is needed.
The graph of $V_S$ vs $f$ should be linear for a clean photoelectric effect because $V_S = (h/e)f - (\phi/e)$ .

Techniques used

use photon energy equation E = hfapply the photoelectric equation hf = \phi + E_{k,\max}relate stopping potential to maximum kinetic energy using eV_S = E_{k,\max}

(ii)

State three quantitative conclusions that can be drawn from the results in Fig. 9.2 and Fig. 9.3. Use the space for any working.

DifficultyMedium

Worked solution

Working

From Fig. 9.2, stopping voltage:

V_S \approx 4.4\ \text{V} \quad (I = 0).

From Fig. 9.3, threshold frequency ( $V_S=0$ ):

f_0 \approx 1.4 \times 10^{15}\ \text{Hz}.

Gradient of Fig. 9.3:

\frac{\Delta V_S}{\Delta f} = \frac{6.6 - 0}{(3.0-1.4)\times 10^{15}} = 4.125 \times 10^{-15}\ \text{V s}.

Since $V_S = (h/e)f - \phi/e$ , gradient $= h/e$ so

h = e\left(4.125 \times 10^{-15}\right) = (1.60\times 10^{-19})(4.125\times 10^{-15}) = 6.6 \times 10^{-34}\ \text{J s}.

Answer

$V_S \approx 4.4\ \text{V}$ (for $f=2.5\times 10^{15}\ \text{Hz}$ );
$f_0 \approx 1.4\times 10^{15}\ \text{Hz}$ ;
$h \approx 6.6\times 10^{-34}\ \text{J s}$ .

Final answer

V_S ≈ 4.4 V; f0 ≈ 1.4×10^15 Hz; h ≈ 6.6×10^-34 J s.

Detailed explanation

Background Concept

Two standard photoelectric results are used here:

Stopping potential is related to the maximum kinetic energy of emitted electrons:

eV_S = E_{k,\max}.

Photoelectric equation:

E_{k,\max} = hf - \phi.

Combining gives a straight-line relationship:

V_S = \left(\frac{h}{e}\right) f - \frac{\phi}{e}.

So on a graph of $V_S$ against $f$ :

gradient $= h/e$ ,
$f$ -axis intercept (where $V_S=0$ ) is the threshold frequency $f_0 = \phi/h$ .

Understanding the Question

You have two graphs:

Fig. 9.2: current $I$ vs applied potential difference $V$ for a fixed frequency. The key quantitative point is where the current becomes zero (the stopping potential for that frequency).
Fig. 9.3: stopping potential $V_S$ vs frequency $f$ . The key quantitative points are the threshold frequency (where $V_S=0$ ) and the gradient (to obtain $h/e$ and hence $h$ ).

You are asked for three quantitative conclusions, so you should extract numerical values (with units) from these graphs, possibly using calculation.

Approach

Pick three solid numerical outcomes available from the graphs:

$V_S$ from Fig. 9.2 (the $V$ value where $I=0$ ).
$f_0$ from Fig. 9.3 (the $f$ value where $V_S=0$ ).
Use the straight-line gradient of Fig. 9.3 to calculate $h$ via gradient $= h/e$ .

Step-by-Step Reasoning

Stopping potential from Fig. 9.2
- Find the point where the line meets $I=0$ .
- Read $V \approx 4.4\ \text{V}$ .
- This is the stopping voltage for the stage-1 frequency.
Threshold frequency from Fig. 9.3
- Threshold frequency is where emitted electrons have zero maximum KE, so $V_S=0$ .
- Read the $f$ -axis intercept: $f_0 \approx 1.4 \times 10^{15}\ \text{Hz}$ .
Planck constant from gradient of Fig. 9.3
- Choose two well-separated points on the best-fit line (the question provides a clear point at $(3.0,6.6)$ and the intercept $(1.4,0)$ ).
- Calculate gradient:

\text{gradient} = \frac{\Delta V_S}{\Delta f} = \frac{6.6}{(3.0-1.4)\times 10^{15}} = 4.125\times 10^{-15}\ \text{V s}.

From $V_S = (h/e)f - \phi/e$ , gradient $= h/e$ .
Multiply by $e$ to get $h$ :

h = e \times \text{gradient} = (1.60\times 10^{-19})(4.125\times 10^{-15}) \approx 6.6\times 10^{-34}\ \text{J s}.

(Other valid quantitative conclusions could include the saturation current from Fig. 9.2, or the work function from $\phi = hf_0$ , but only three are required.)

Key Takeaways

$V_S$ is read where $I=0$ on an $I$ – $V$ graph.
$f_0$ is read where $V_S=0$ on a $V_S$ – $f$ graph.
The linear form $V_S = (h/e)f - \phi/e$ lets you obtain constants from gradient/intercept.

Common Mistakes

Using the point at $V=0$ to read $V_S$ (stopping potential is where $I=0$ , not where $V=0$ ).
Forgetting that the frequency axis in Fig. 9.3 is scaled by $10^{15}\ \text{Hz}$ .
Calculating the gradient using $\Delta f/\Delta V_S$ instead of $\Delta V_S/\Delta f$ .
Treating the gradient as $h$ rather than $h/e$ .

Things to Be Careful About

Include units: $V_S$ in volts, $f$ in hertz, gradient in $\text{V s}$ , and $h$ in $\text{J s}$ .
Use two far-apart points to reduce percentage reading error in the gradient.
Keep consistent significant figures with what can be read from the graphs (typically 2–3 s.f.).

Techniques used

read key values from intercepts on graphscalculate a gradient from two points on a straight-line graphuse the relation V_S = (h/e)f - (\phi/e) to identify constants

Q10MediumNuclear Physics

(a)

Radioactive decay is a spontaneous process.

State the meaning, in this context, of the term spontaneous.

DifficultyEasy

Worked solution

Answer

Spontaneous means the decay occurs randomly by itself and is not triggered or affected by external physical conditions (e.g. temperature/pressure).

Final answer

Decay occurs randomly by itself and is not affected by external physical conditions.

Detailed explanation

Background Concept

Radioactive decay happens because an unstable nucleus can change into a more stable nucleus by emitting radiation (alpha, beta, gamma). The key idea is that we cannot predict exactly when any particular nucleus will decay.

"Spontaneous" in this topic means:

the decay event is not caused by an external trigger, and
the probability of decay is not changed by ordinary external conditions such as temperature, pressure, chemical state, or electric/magnetic fields.

Understanding the Question

The question is asking for the meaning of the word spontaneous specifically in the context of radioactive decay. It wants the standard physics meaning (random/untriggered), not a general English meaning.

Approach

State the two essential features that exam mark schemes usually credit:

decay is random for an individual nucleus, and/or
decay is independent of external conditions.

Step-by-Step Reasoning

A nucleus decays without needing anything to “start” it, so it is not triggered by collisions, heating, shining light, etc.
You also cannot say when a particular nucleus will decay: it is random (only the statistical behaviour of many nuclei is predictable).

Key Takeaways

Spontaneous = random and not influenced by external physical conditions.
Predictable behaviour comes from large numbers of nuclei, not individual ones.

Common Mistakes

Saying only “it happens quickly” or “it happens naturally” (too vague).
Confusing spontaneous with “instantaneous”.

Things to Be Careful About

The question says “in this context”: use the nuclear-physics definition, not a general one.
It is safest to include the idea “not affected by external conditions” because it is a common marking point.

Techniques used

recall the definition of spontaneous decaystate that decay is random and unaffected by external conditions

(b)

Two radioactive isotopes X and Y each decay to form a stable isotope.
A sample initially contains only atoms of isotope X. At this time, its activity is $4A$ .
Another sample initially contains only atoms of Y. At this time, its activity is $A$ .

Fig. 10.1 shows the variation of the activity of each sample with time $t$ between $t = 0$ and $t = 6T$ .

Fig. 10.1

6M

(i)

Complete Table 10.1 to give expressions, in terms of either or both of $A$ and $T$ , for the quantities indicated for each of the samples.

Table 10.1

sample	half-life	decay constant	initial activity	initial number of nuclei
X			$4A$
Y			$A$

DifficultyMedium

Worked solution

Working

From the graph:

For X: $4A \to 2A$ at $t=T$ and $4A \to A$ at $t=2T$ so $t_{1/2,X}=T$ .
For Y: $A \to 0.5A$ at $t=4T$ so $t_{1/2,Y}=4T$ .

Decay constant:

\lambda = \frac{\ln 2}{t_{1/2}}

So

\lambda_X = \frac{\ln 2}{T},\qquad \lambda_Y = \frac{\ln 2}{4T}

Initial number of nuclei, using $A=\lambda N$ :

N_{0X}=\frac{4A}{\lambda_X}=\frac{4A}{\ln 2/T}=\frac{4AT}{\ln 2}

N_{0Y}=\frac{A}{\lambda_Y}=\frac{A}{\ln 2/(4T)}=\frac{4AT}{\ln 2}

Answer

Sample X: half-life $T$ , decay constant $\lambda_X=\dfrac{\ln 2}{T}$ , initial number $N_{0X}=\dfrac{4AT}{\ln 2}$ .
Sample Y: half-life $4T$ , decay constant $\lambda_Y=\dfrac{\ln 2}{4T}$ , initial number $N_{0Y}=\dfrac{4AT}{\ln 2}$ .

Final answer

X: t1/2 = T, λ = (ln2)/T, N0 = 4AT/(ln2). Y: t1/2 = 4T, λ = (ln2)/(4T), N0 = 4AT/(ln2).

Detailed explanation

Background Concept

Two key relationships are used in this question.

Half-life $t_{1/2}$ is the time for the activity (or number of undecayed nuclei) to halve.
Decay constant $\lambda$ is related to half-life by

\lambda = \frac{\ln 2}{t_{1/2}}.

Activity is the rate of decay:

A = \lambda N,

where $A$ is activity in $\text{s}^{-1}$ (Bq) and $N$ is the number of undecayed nuclei.

Understanding the Question

You are given two separate samples:

sample X has initial activity $4A$ and its activity drops faster,
sample Y has initial activity $A$ and drops more slowly.

From the activity-time graph you must:

read the half-life of each sample,
calculate each decay constant,
use $A=\lambda N$ to find the initial number of nuclei in each sample.

Approach

Use the graph to find the time taken for the activity to halve (that is the half-life).
Convert half-life to decay constant using $\lambda = \ln 2/t_{1/2}$ .
Rearrange $A=\lambda N$ to $N=A/\lambda$ and substitute each sample’s initial activity.

Step-by-Step Reasoning

1) Half-life of X

X starts at $4A$ .
One half-life later it should be $2A$ .
The graph indicates this occurs at $t=T$ (and at $t=2T$ it has reached $A$ , which is two halvings).
So

t_{1/2,X} = T.

2) Half-life of Y

Y starts at $A$ .
Half of this is $0.5A$ .
The graph shows $0.5A$ at $t=4T$ .
So

t_{1/2,Y} = 4T.

3) Decay constants
Use

\lambda = \frac{\ln 2}{t_{1/2}}.

So

\lambda_X = \frac{\ln 2}{T},\qquad \lambda_Y = \frac{\ln 2}{4T}.

4) Initial number of nuclei
Rearrange $A=\lambda N$ :

N = \frac{A}{\lambda}.

For X, initial activity is $4A$ :

N_{0X} = \frac{4A}{\lambda_X} = \frac{4A}{\ln 2/T} = \frac{4AT}{\ln 2}.

For Y, initial activity is $A$ :

N_{0Y} = \frac{A}{\lambda_Y} = \frac{A}{\ln 2/(4T)} = \frac{4AT}{\ln 2}.

It is not a problem that $N_{0X}$ and $N_{0Y}$ come out the same here: X has a larger initial activity because it has a larger decay constant.

Key Takeaways

Read half-life directly from the graph by finding the time to halve the activity.
Convert between $t_{1/2}$ and $\lambda$ using $\lambda = \ln 2/t_{1/2}$ .
Use $A=\lambda N$ to link activity to number of nuclei.

Common Mistakes

Using $\lambda = 1/t_{1/2}$ (missing $\ln 2$ ).
Taking X’s half-life as $2T$ because it reaches $A$ at $2T$ (but $A$ is one quarter of $4A$ , i.e. two half-lives).
Forgetting to use the initial activity when calculating $N_0$ .

Things to Be Careful About

Check whether the graph point corresponds to a half ( $1/2$ ) or a quarter ( $1/4$ ) of the initial activity.
Keep symbols clear: $A$ is used as a labelled activity unit on the vertical axis; $\lambda$ is the decay constant.

Techniques used

read half-life information from an activity-time graphuse $\lambda = \ln 2 / t_{1/2}$use $A = \lambda N$ to relate activity to number of nuclei

(ii)

Determine, in terms of $T$ , the time at which the two samples will have equal activities.

time = ______ $T$

DifficultyMedium

Worked solution

Working

For X, $t_{1/2}=T$ :

A_X = 4A\left(\frac{1}{2}\right)^{t/T}

For Y, $t_{1/2}=4T$ :

A_Y = A\left(\frac{1}{2}\right)^{t/(4T)}

Equal activities:

4\left(\frac{1}{2}\right)^{t/T} = \left(\frac{1}{2}\right)^{t/(4T)}

Write $4=2^2$ :

2^{2-t/T}=2^{-t/(4T)}

So

2-\frac{t}{T}=-\frac{t}{4T}

8T-4t=-t \;\Rightarrow\; 8T=3t \;\Rightarrow\; t=\frac{8}{3}T

Answer

t = \frac{8}{3}T \;(\approx 2.67T)

Final answer

8/3 T

Detailed explanation

Background Concept

Radioactive activity decreases exponentially. In terms of half-life, it is often convenient to write

A = A_0\left(\frac{1}{2}\right)^{t/t_{1/2}},

where $A_0$ is the initial activity and $t_{1/2}$ is the half-life.

This form works because each interval of one half-life multiplies the activity by $1/2$ .

Understanding the Question

You have two samples with different starting activities and different half-lives (found from the graph in part (b)(i)). The question asks for the time (as a multiple of $T$ ) when their activities are the same.

Given (from the graph):

Sample X: $A_{0X}=4A$ , $t_{1/2,X}=T$
Sample Y: $A_{0Y}=A$ , $t_{1/2,Y}=4T$

Unknown: the time $t$ when $A_X=A_Y$ .

Approach

Write an activity-time expression for each sample using the half-life form.
Set them equal and solve for $t$ .
Use powers of 2 to avoid messy logarithms (because half-life naturally uses factors of $1/2$ ).

Step-by-Step Reasoning

1) Write activity expressions
For X:

A_X = 4A\left(\frac{1}{2}\right)^{t/T}.

For Y:

A_Y = A\left(\frac{1}{2}\right)^{t/(4T)}.

2) Set them equal

4A\left(\frac{1}{2}\right)^{t/T} = A\left(\frac{1}{2}\right)^{t/(4T)}.

Cancel $A$ :

4\left(\frac{1}{2}\right)^{t/T} = \left(\frac{1}{2}\right)^{t/(4T)}.

3) Convert to base 2 and solve
Since $\left(\frac{1}{2}\right)^x = 2^{-x}$ and $4=2^2$ :

2^2\cdot 2^{-t/T} = 2^{-t/(4T)}

Combine the left-hand powers:

2^{2-t/T} = 2^{-t/(4T)}.

If the bases are equal, the exponents must be equal:

2-\frac{t}{T} = -\frac{t}{4T}.

Multiply by $4T$ :

8T-4t = -t \;\Rightarrow\; 8T=3t.

So

t = \frac{8}{3}T \approx 2.67T.

Key Takeaways

Use $A=A_0\left(\tfrac{1}{2}\right)^{t/t_{1/2}}$ when half-lives are given.
When equating two exponentials with the same base, equate the exponents.
A faster-decaying sample can start higher but still meet a slower-decaying one later.

Common Mistakes

Using the wrong half-lives (e.g. taking X’s half-life as $2T$ ).
Forgetting to include the different initial activities $4A$ and $A$ .
Adding exponents incorrectly when manipulating powers of 2.

Things to Be Careful About

Keep the ratio $t/t_{1/2}$ dimensionless.
Give the final answer in the requested form “___ $T$ ”; $\frac{8}{3}T$ is the clean exact form.

Techniques used

write activities as exponential/half-life decay functionsequate two exponential expressions and solve using logarithmsuse powers of 2 to handle half-life relationships

(c)

A radiation detector is placed near to one of the samples in (b).

Explain why the count rate measured by the detector is less than the activity of the sample.