(a) Two metal cuboids P and Q are in thermal contact with each other. (i) P and Q are in thermal equilibrium. State what is meant by the term thermal equilibrium. (ii) Data for P and Q are given in Table 3.1. Table 3.1 P Q --- --- --- specific heat capacity / $\text{J kg}^{-1}\ \text{K}^{-1}$ 390 910 mass / $\text{kg}$ 0.54 0.37 P and Q are initially both at the same temperature. P is supplied with $24\ \text{kJ}$ of thermal energy. After some time, P and Q are once again…

(a)(i) Same temperature; no net heat transfer between P and Q. · (a)(ii) 44 K · (b)(i) -2.2 × 10^4 J · (b)(ii) 7.6 × 10^24 · (b)(iii) 6.5 × 10^2 m s^-1

(a) A capacitor of capacitance $C_1$ is connected in series with a second capacitor of capacitance $C_2$. Show that the combined capacitance $C$ of the two capacitors is given by $$ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} $$ (b) Three identical capacitors, each of capacitance $C$, are connected in a network as shown in Fig. 5.1. The variation of the charge $Q$ with the potential difference (p.d.) $V$ between the terminals X and Y is shown in Fig. 5.2. Show that $C$ is…

(a) 1/C = 1/C1 + 1/C2 · (b) 44 µF · (c)(i) 3.6 s · (c)(ii) 6.8 s

An electric field and a magnetic field are used to form a velocity selector. Charged particles, called ions, pass into a region of uniform electric and magnetic fields that is between parallel plates, as shown in Fig. 6.1. (a) The potential difference (p.d.) between the plates of the velocity selector is $V$. The separation of the plates is $d$ and the magnetic flux density is $B$. Show that the speed $u$ of ions that pass undeviated through the velocity selector is given by…

(a) ν = V/(Bd) · (b) 0.17 T · (c) Same speed ⇒ also undeviated because ν = V/(Bd) is independent of particle mass and charge. · (d) Magnetic field is into the page. · (e) Path curves upward toward the positive (top) plate.

Polonium-193 ($^{193}_{84}\text{Po}$) is an unstable nuclide. A nucleus of polonium-193 decays to a nucleus of lead-189 ($^{189}_{82}\text{Pb}$) by emitting an alpha-particle. (a) Radioactive decay is both random and spontaneous. State what is meant by: (i) random (ii) spontaneous. (b) Define half-life. (c) Data for the binding energy per nucleon of the particles involved in the decay of a nucleus of polonium-193 are given in Table 9.1. Table 9.1 particle binding energy per…

(a)(i) It is impossible to predict which nucleus will decay or when a particular nucleus will decay. · (a)(ii) Decay occurs without an external trigger and is unaffected by external physical conditions. · (b) Time for number of undecayed nuclei (or activity) to halve. · (c) 7.03 × 10^6 eV · (d)(i) Decay constant, \lambda · (d)(ii) 0.50 ms · (e)(i) They annihilate with electrons, producing two gamma photons (opposite directions). · (e)(ii) Long enough for preparation/scan,…

(i) State what is meant by the luminosity of a star. (ii) Explain how standard candles are used to determine the distance to a galaxy. (b) The Sun rotates on its axis. Points X, Y and Z are on the equator of the Sun as shown in Fig. 10.1. The wavelengths of light from points X and Y are observed and recorded in Table 10.1. Table 10.1 observed wavelength from X / $\text{nm}$ observed wavelength from Y / $\text{nm}$ --- --- 656.2877 656.2831 (i) The Sun rotates with a period…

(a)(i) Total power (energy per unit time) emitted by the star in all directions. · (a)(ii) Use an object of known luminosity, measure its flux, and apply F = L/(4πd²) to find the distance. · (b)(i) 6.93 × 10^8 m · (b)(ii) Observed wavelength from Z is smaller than the emitted wavelength (blueshift) because Z is moving towards the observer. · (b)(iii) 5.8 × 10^3 K

9702/42 Feb/Mar 2025 — Physics Solved Past Paper

Q1MediumMotion in a Circle

A steel ball is placed on the inside surface of a hollow circular cone. The ball moves in a horizontal circle at constant speed, as shown in Fig. 1.1.

Fig. 1.1

The angle of the side of the cone to the horizontal is $52^{\circ}$ . There is no friction between the ball and the cone.

(a)

Fig. 1.2 shows a cross-section through the cone and the steel ball.

Fig. 1.2

On Fig. 1.2, draw labelled arrows to show the two forces acting on the ball.

1M Similar

DifficultyEasy

Worked solution

diagram-a1

Answer

Forces on the ball:

weight $mg$ vertically downwards
normal contact force $N$ from the cone, perpendicular to the cone surface

Final answer

Weight mg downward; normal reaction N perpendicular to the cone surface.

Detailed explanation

Background Concept

A force diagram (free-body diagram) shows all the real forces acting on an object.

For a ball touching a surface:

its weight $mg$ acts vertically downwards (due to gravity)
the surface exerts a normal contact force $N$ on the ball, perpendicular to the surface
friction would act along the surface, but only if friction is present.

Understanding the Question

The ball is on the inside of a smooth (frictionless) conical surface and moves in a horizontal circle. The question asks for the two forces acting on the ball.

Because it is explicitly stated that there is no friction, there cannot be a frictional force along the surface.

Approach

List the forces:

gravity on the ball
contact force from the cone
Then draw them with correct directions from the ball.

Step-by-Step Reasoning

diagram-a1

Draw $mg$ vertically downward from the centre of the ball.
Draw the normal reaction $N$ perpendicular to the sloping cone surface, pointing away from the surface into the space inside the cone (this is the direction the surface pushes the ball).
Do not draw any friction force because the surface is smooth.

Key Takeaways

With no friction, the only contact force is the normal reaction.
The normal reaction is always perpendicular to the surface.

Common Mistakes

Drawing a friction force even though the question states there is no friction.
Drawing $N$ vertically upwards instead of perpendicular to the cone surface.
Drawing an extra “centripetal force” as a separate force (centripetal force is provided by components of real forces).

Things to Be Careful About

The direction of $mg$ is always vertical, not perpendicular to the surface.
The normal force direction depends on the surface orientation; it must be exactly $90^{\circ}$ to the cone side in the cross-section.

Techniques used

identify real forces acting on an object in contact with a surfacedraw a free-body diagram with correct force directionsuse the no-friction condition to eliminate tangential forces

(b)

Describe how the forces acting on the ball cause its acceleration to be centripetal.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The forces are $mg$ (downwards) and the normal reaction $N$ (perpendicular to the surface).

Resolve $N$ :

its vertical component balances $mg$ so there is no vertical acceleration
its horizontal component acts towards the axis of the cone and provides the centripetal force, so the acceleration is towards the centre of the circular path (centripetal).

Final answer

Vertical component of N balances mg; horizontal component of N is towards the axis and provides the centripetal force so acceleration is centripetal.

Detailed explanation

Background Concept

For uniform circular motion at constant speed, the acceleration is not zero: it is a centripetal acceleration directed towards the centre of the circle.

Its magnitude is

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

This acceleration must be caused by a resultant force towards the centre (centripetal force):

F_{\text{centripetal}} = ma = \frac{mv^2}{r}.

Crucially, “centripetal force” is not an extra force; it is the inward resultant of the real forces.

Understanding the Question

The ball moves in a horizontal circle on the inside of a smooth cone. The only forces are $mg$ and the normal reaction $N$ .

The question asks you to describe how these forces produce an acceleration that points towards the centre of the circular path.

Approach

Take the normal reaction $N$ and split it into:

a vertical component (to balance $mg$ so the ball stays at constant height)
a horizontal inward component (to provide the required centripetal force)

Step-by-Step Reasoning

Consider the two forces:

Weight $mg$ acts vertically downward.
Normal reaction $N$ is angled (perpendicular to the cone surface), so it has both vertical and horizontal components.

Because the ball stays at the same height (horizontal circle), its vertical acceleration is zero, so the net vertical force must be zero:

the vertical component of $N$ balances $mg$ .

That leaves an unbalanced horizontal component of $N$ . This component points towards the cone axis (towards the centre of the circular path). Therefore:

the resultant force is horizontal and inward
the ball’s acceleration is also horizontal and inward (Newton’s second law)
hence the acceleration is centripetal.

Key Takeaways

Constant speed in a circle still requires acceleration: direction changes.
Vertical forces balance; the inward horizontal component provides centripetal acceleration.

Common Mistakes

Saying “there is no acceleration because speed is constant”. Speed is constant but velocity changes direction.
Treating centripetal force as a separate third force.
Claiming $mg$ provides the centripetal force (it is vertical, not horizontal).

Things to Be Careful About

Use the word “component” clearly: it is a component of $N$ that acts inward.
Mention vertical equilibrium explicitly to justify why the remaining resultant is horizontal and centripetal.

Techniques used

resolve a force into vertical and horizontal componentsapply vertical equilibrium to relate force componentsidentify the inward horizontal resultant as the centripetal force

(c)

The ball moves in a circle of radius $0.15\ \text{m}$ .

Show that the speed of the ball is $1.4\ \text{m s}^{-1}$ .

3M Similar

DifficultyMedium

Worked solution

Working

Let $N$ be the normal reaction. The cone side is at $52^{\circ}$ to the horizontal, so $N$ is at $52^{\circ}$ to the vertical.

Vertical equilibrium:

N\cos 52^{\circ} = mg

Horizontal (centripetal):

N\sin 52^{\circ} = \frac{mv^2}{r}

Divide:

\tan 52^{\circ} = \frac{v^2}{rg}

So

v = \sqrt{rg\tan 52^{\circ}}

With $r=0.15\ \text{m}$ and $g=9.81\ \text{m s}^{-2}$ :

v = \sqrt{0.15 \times 9.81 \times \tan 52^{\circ}} = 1.37\ \text{m s}^{-1} \approx 1.4\ \text{m s}^{-1}

Answer

$1.4\ \text{m s}^{-1}$

Final answer

1.4 m s^-1

Detailed explanation

Background Concept

When an object moves in a horizontal circle at constant speed, the required centripetal force is

F_c = \frac{mv^2}{r}.

Here, the only forces are weight $mg$ (vertical) and the normal reaction $N$ (perpendicular to the cone surface). Because the motion is at constant height, vertical forces must balance, while the horizontal resultant provides the centripetal force.

Understanding the Question

Given:

cone side makes $52^{\circ}$ with the horizontal
radius of horizontal circular path $r = 0.15\ \text{m}$
no friction, constant speed

We must show the speed is $1.4\ \text{m s}^{-1}$ by using force components and centripetal force.

Approach

Relate the direction of the normal force to the given $52^{\circ}$ angle.
Resolve $N$ into vertical and horizontal components.
Use vertical equilibrium ( $\Sigma F_y = 0$ ) to relate $N$ to $mg$ .
Use horizontal component as centripetal force ( $\Sigma F_{\text{in}} = mv^2/r$ ).
Eliminate $N$ to get $v$ .

Step-by-Step Reasoning

1) Geometry of the normal
The cone side is at $52^{\circ}$ to the horizontal. The normal is perpendicular to the side, so it is at $90^{\circ} - 52^{\circ} = 38^{\circ}$ to the horizontal, i.e. at $52^{\circ}$ to the vertical.

2) Resolve forces

Vertical component of $N$ is $N\cos 52^{\circ}$ (adjacent to the angle with vertical).
Horizontal inward component of $N$ is $N\sin 52^{\circ}$ .

3) Vertical equilibrium (constant height)

N\cos 52^{\circ} = mg.

This tells you how large the normal reaction must be to support the weight.

4) Horizontal component provides centripetal force

N\sin 52^{\circ} = \frac{mv^2}{r}.

5) Eliminate $N$
Divide the centripetal equation by the vertical one:

\frac{N\sin 52^{\circ}}{N\cos 52^{\circ}} = \frac{mv^2/r}{mg}

so

\tan 52^{\circ} = \frac{v^2}{rg}.

Rearrange:

v^2 = rg\tan 52^{\circ}, \qquad v = \sqrt{rg\tan 52^{\circ}}.

6) Substitute values

v = \sqrt{0.15 \times 9.81 \times \tan 52^{\circ}}.

Using $\tan 52^{\circ} \approx 1.28$ :

v \approx \sqrt{0.15 \times 9.81 \times 1.28} \approx \sqrt{1.88} \approx 1.37\ \text{m s}^{-1} \approx 1.4\ \text{m s}^{-1}.

Key Takeaways

For a frictionless conical surface, $mg$ is balanced by the vertical component of $N$ .
The inward (horizontal) component of $N$ supplies the centripetal force.
The result is $v^2 = rg\tan\theta$ where $\theta$ is the cone side angle to the horizontal.

Common Mistakes

Using $N\sin 52^{\circ} = mg$ (wrong component).
Taking $52^{\circ}$ as the angle between $N$ and the horizontal instead of between the cone side and the horizontal.
Forgetting centripetal force uses the radius of the horizontal circle ( $r=0.15\ \text{m}$ ).

Things to Be Careful About

Be consistent with the chosen angle: here it is simplest to note that $N$ is at $52^{\circ}$ to the vertical.
Don’t drop units: $g$ must be in $\text{m s}^{-2}$ and $r$ in $\text{m}$ .
Final rounding: showing $1.37$ then $1.4\ \text{m s}^{-1}$ matches the “show that” style.

Techniques used

resolve the normal reaction into vertical and horizontal componentsapply vertical equilibrium to relate the normal reaction to weightequate inward force to mv^2/r and solve for speed

(d)

Calculate the angular speed $\omega$ of the ball.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\omega = \frac{v}{r} = \frac{1.4}{0.15} = 9.33\ \text{rad s}^{-1}

Answer

$\omega = 9.3\ \text{rad s}^{-1}$

Final answer

9.3 rad s^-1

Detailed explanation

Background Concept

For motion in a circle of radius $r$ at speed $v$ , the angular speed $\omega$ is related by

v = r\omega.

This comes from the arc length relation $s = r\theta$ and differentiating with respect to time.

Understanding the Question

You have already found the speed of the ball and you are given the radius of the circle $r=0.15\ \text{m}$ . The question asks for the angular speed in $\text{rad s}^{-1}$ .

Approach

Rearrange $v=r\omega$ to $\omega=v/r$ and substitute the known values.

Step-by-Step Reasoning

Start with

v = r\omega.

Rearrange:

\omega = \frac{v}{r}.

Substitute $v = 1.4\ \text{m s}^{-1}$ and $r = 0.15\ \text{m}$ :

\omega = \frac{1.4}{0.15} = 9.33\ \text{s}^{-1}.

Radians are dimensionless, so $\text{s}^{-1}$ is equivalent to $\text{rad s}^{-1}$ .

Rounded appropriately:

\omega \approx 9.3\ \text{rad s}^{-1}.

Key Takeaways

Use $\omega = v/r$ for uniform circular motion.
Units check: $\text{m s}^{-1}/\text{m} = \text{s}^{-1} = \text{rad s}^{-1}$ .

Common Mistakes

Using $\omega = 2\pi r / v$ (wrong rearrangement).
Forgetting that $r$ must be in metres.
Giving the unit as $\text{m s}^{-1}$ instead of $\text{rad s}^{-1}$ .

Things to Be Careful About

Significant figures: if $v$ was shown as $1.4\ \text{m s}^{-1}$ (2 s.f.), then quoting $\omega$ to 2 s.f. (e.g. $9.3$ ) is appropriate.
If you used an unrounded $v$ from part (c), your $\omega$ may be slightly different; this is usually allowed as error carried forward.

Techniques used

use the relation v = r omegasubstitute numerical values with consistent unitsround to an appropriate number of significant figures

(e)

The speed of the ball is increased.

Explain why the radius of the circular path of the ball increases.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Since $N\cos 52^{\circ}=mg$ , the normal force $N$ (and hence $N\sin 52^{\circ}$ ) is fixed.
But

N\sin 52^{\circ}=\frac{mv^2}{r}

so $v^2/r$ is constant and therefore increasing $v$ requires a larger $r$ .

Final answer

With cone angle fixed, the available inward component of N is fixed; since mv^2/r must equal it, increasing v implies r must increase.

Detailed explanation

Background Concept

For an object on a frictionless conical surface moving in a horizontal circle:

Vertical equilibrium requires the vertical component of the normal reaction to equal weight.
The horizontal component of the normal reaction provides the centripetal force.

From earlier parts you can show

v^2 = rg\tan\theta

for fixed cone angle $\theta$ .

Understanding the Question

Only the speed is increased; the cone angle stays the same and there is still no friction. The question asks why the radius of the horizontal circular path must increase.

Approach

Use the fact that with no vertical acceleration,

N\cos 52^{\circ} = mg

so $N$ is determined by $mg$ and the cone angle (both unchanged). Then the available inward force $N\sin 52^{\circ}$ is fixed. Since the required centripetal force is $mv^2/r$ , increasing $v$ means $r$ must increase to keep $mv^2/r$ equal to the same available inward force.

Step-by-Step Reasoning

Vertical balance (constant height):

N\cos 52^{\circ} = mg.

Here $m$ and $g$ are constant, and the angle $52^{\circ}$ is fixed by the cone. Therefore $N$ does not change.

The inward (horizontal) component is therefore also fixed:

F_{\text{in}} = N\sin 52^{\circ}.

For circular motion, this must equal the centripetal force:

N\sin 52^{\circ} = \frac{mv^2}{r}.

Since the left-hand side is fixed, the quantity $v^2/r$ must remain constant. Therefore if $v$ increases, $r$ must increase as well.

Equivalently, from

v^2 = rg\tan 52^{\circ},

with $g\tan 52^{\circ}$ constant, $v^2 \propto r$ .

Key Takeaways

With fixed cone angle and no vertical acceleration, the normal force is fixed by balancing weight.
The available centripetal force is therefore fixed, forcing $v^2/r$ to be constant.
Increasing speed leads to a larger radius.

Common Mistakes

Saying “radius increases because centripetal force increases” without explaining where the extra force would come from (it cannot, because $N$ is fixed by vertical balance).
Claiming the ball moves up the cone due to a frictional force (there is no friction).

Things to Be Careful About

The key constraint is the absence of friction plus constant height: it makes the vertical component of force balance exactly, fixing $N$ .
Use proportional reasoning carefully: it is $v^2 \propto r$ , not $v \propto r$ .

Techniques used

use the derived relationship between speed and radius for a fixed cone anglerelate centripetal force requirement to mv^2/rargue how constraints on available inward force determine the new radius

Q2MediumGravitational FieldsElectric Fields

(a)

The magnitude of the gravitational potential on the surface of a planet of radius $R$ is $\phi$ . The planet can be considered to be an isolated sphere.

On Fig. 2.1, sketch the variation of the gravitational potential with distance $x$ from the centre of the planet for values of $x$ between $R$ and $4R$ .

Fig. 2.1

3M Similar

DifficultyMedium

Worked solution

diagram-2a

Answer

For an isolated sphere, for $x \ge R$ :

V(x) = -\frac{GM}{x}

Given magnitude at $x=R$ is $\phi$ , so $V(R) = -\phi$ .

Mark and use these points:

$x=R$ : $V=-\phi$
$x=2R$ : $V=-\dfrac{\phi}{2}$
$x=4R$ : $V=-\dfrac{\phi}{4}$

Draw a smooth curve that rises from $-\phi$ at $R$ , is concave down (flattens), and approaches $0$ from below as $x$ increases.

Final answer

Gravitational potential is negative and varies as -1/x: passes through (R, -φ), (2R, -φ/2), (4R, -φ/4) and approaches 0 from below.

Detailed explanation

Background Concept

Gravitational potential $V$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point (with no change in kinetic energy). Taking $V(\infty)=0$ , the potential due to a spherically symmetric mass $M$ at distance $x$ from its centre (for points outside the mass) is

V(x) = -\frac{GM}{x}.

It is negative because energy is released when a mass falls in from infinity; the field is attractive.

Understanding the Question

You are told that at the planet surface ( $x=R$ ) the magnitude of the gravitational potential is $\phi$ . Since gravitational potential is negative, that means

V(R) = -\phi.

You must sketch $V$ against $x$ for $R \le x \le 4R$ .

Approach

Use the outside-sphere formula $V(x)=-GM/x$ (the planet is an isolated sphere).
Use the given surface value to relate $GM$ to $\phi$ .
Find a few easy points (e.g. $2R$ , $4R$ ) and use the known asymptotic behaviour $V\to 0$ as $x\to\infty$ to shape the curve.

Step-by-Step Reasoning

From

V(R)=-\frac{GM}{R}=-\phi,

so

GM = \phi R.

Hence for $x\ge R$ :

V(x) = -\frac{GM}{x} = -\frac{\phi R}{x}.

Now evaluate at the marked $x$ -values:

At $x=R$ :

V=-\frac{\phi R}{R}=-\phi.

At $x=2R$ :

V=-\frac{\phi R}{2R}=-\frac{\phi}{2}.

At $x=4R$ :

V=-\frac{\phi R}{4R}=-\frac{\phi}{4}.

As $x$ increases further, $1/x$ decreases towards 0, so $V$ increases towards 0 but stays negative. The curve is not a straight line: it rises quickly near $R$ and then flattens (approaches 0 asymptotically).

diagram-2a

Key Takeaways

For an isolated spherical mass, outside the sphere $V\propto -1/x$ .
“Magnitude is $\phi$ ” at the surface means the plotted value is $-\phi$ .
Correct sketches come from correct key points and correct end behaviour (approach to 0 from below).

Common Mistakes

Plotting $+\phi$ at $x=R$ (forgetting gravitational potential is negative).
Drawing a straight line between points (confusing $1/x$ with a linear relationship).
Making the graph cross $V=0$ within $R$ to $4R$ (it should only approach 0).

Things to Be Careful About

The horizontal axis is distance from the centre: the first point is at $x=R$ , not $x=0$ .
The potential at infinity is defined as 0, so the curve must tend to 0, not to some other value.

Techniques used

use the gravitational potential of an isolated spherical mass outside the sphereapply boundary conditions at the planet surfaceplot key points and sketch an asymptotic curve approaching zero

(b)

A satellite is in a geostationary orbit above the Earth. At time $t = 0$ , the magnitude of the gravitational potential due to the Earth at the location of the satellite is $\phi$ .

On Fig. 2.2, sketch the variation of the gravitational potential due to the Earth at the location of the satellite for values of $t$ between $t = 0$ and $t = 24$ hours.

Fig. 2.2

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-2b

Answer

A geostationary satellite has constant orbital radius, so gravitational potential due to Earth at the satellite is constant.

Given its magnitude at $t=0$ is $\phi$ , the potential is $V=-\phi$ for all $t$ .

Sketch a horizontal line at $-\phi$ from $t=0$ to $t=24\ \text{h}$ .

Final answer

Constant: a horizontal line at gravitational potential -φ for 0 to 24 h.

Detailed explanation

Background Concept

For a point at distance $r$ from the centre of Earth (outside Earth), gravitational potential is

V = -\frac{GM}{r}.

If $r$ does not change, $V$ does not change.

A geostationary satellite moves in a circular orbit with the same period as Earth’s rotation and stays above the same point on Earth. In a circular geostationary orbit, the orbital radius is constant.

Understanding the Question

You are told that at $t=0$ the magnitude of the gravitational potential at the satellite is $\phi$ . Since potential is negative,

V(0) = -\phi.

You must sketch $V$ (due to Earth) as time goes from 0 to 24 h.

Approach

Decide whether the satellite’s distance from Earth changes with time.
Use $V=-GM/r$ : constant $r$ implies constant $V$ .
Place the line at the correct signed value (negative).

Step-by-Step Reasoning

A geostationary orbit is (approximately) circular with fixed radius, so $r$ is constant.

Therefore

V(t) = -\frac{GM}{r} = \text{constant}.

At $t=0$ , magnitude is $\phi$ , so the constant value must be $-\phi$ .

So the graph is a horizontal line at $-\phi$ from 0 to 24 h.

diagram-2b

Key Takeaways

In a circular orbit, the orbital radius is constant.
Gravitational potential depends only on distance from the mass (for a spherically symmetric Earth).
“Magnitude $\phi$ ” corresponds to potential $-\phi$ .

Common Mistakes

Drawing a sinusoidal variation (confusing potential with a quantity that depends on position around the orbit; here the distance is constant).
Plotting $+\phi$ instead of $-\phi$ .

Things to Be Careful About

The question asks for potential at the satellite due to Earth (not potential energy, which would be $mV$ ).
Time axis covers a full day; the line must extend unchanged across the entire range.

Techniques used

use the definition of a geostationary orbit to identify constant orbital radiusapply V = -GM/r to infer constant gravitational potentialsketch a constant-value time graph

(c)

The electric potential difference (p.d.) between two parallel plates is $V$ , as shown in Fig. 2.3.

Fig. 2.3

The distance between the plates is $d$ . The region between the plates is a vacuum.

On Fig. 2.4, sketch the variation of the electric potential with distance from the positive plate.

Fig. 2.4

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-2c

Answer

Between parallel plates in vacuum, the field is uniform so potential varies linearly with distance.

At distance $0$ from the positive plate: $V=+V$ .

At distance $d$ (the other plate at 0 V): $V=0$ .

Sketch a straight line decreasing uniformly from $(0,\,V)$ to $(d,\,0)$ .

Final answer

Electric potential decreases linearly from V at distance 0 to 0 at distance d.

Detailed explanation

Background Concept

Electric potential $V$ is related to electric field $E$ by the potential gradient. In one dimension,

E = -\frac{\mathrm{d}V}{\mathrm{d}x}.

Between large, parallel plates in vacuum (away from edges), the electric field is uniform:

E = \frac{V}{d}

in magnitude, where $V$ is the potential difference and $d$ is the plate separation. A constant $E$ means $\mathrm{d}V/\mathrm{d}x$ is constant, so $V$ changes linearly with distance.

Understanding the Question

The top plate is at potential $+V$ and the bottom plate is at $0\ \text{V}$ . The horizontal axis is “distance from the positive plate”, so:

at distance $0$ : you are on the positive plate (potential $+V$ )
at distance $d$ : you are on the grounded plate (potential $0$ )

You must sketch how the potential varies in between.

Approach

Use the idea that the field between parallel plates is uniform (in vacuum, ignoring edge effects).
A uniform field means a constant slope on a $V$ vs distance graph.
Apply the endpoint values at $x=0$ and $x=d$ and join them with a straight line.

Step-by-Step Reasoning

Let $x$ be the distance measured from the positive plate. Then:

$V(0) = V$
$V(d) = 0$

Uniform field implies $\mathrm{d}V/\mathrm{d}x$ constant, hence $V(x)$ is linear. The simplest linear function satisfying the endpoints is

V(x) = V\left(1 - \frac{x}{d}\right).

So the graph is a straight line sloping down from $V$ at $x=0$ to $0$ at $x=d$ .

diagram-2c

Key Takeaways

In a uniform electric field, potential changes linearly with distance.
Always anchor your sketch using the known potentials at the plates.
The sign and direction are handled by the endpoints: here it must decrease from $V$ to $0$ .

Common Mistakes

Drawing a curved line (that would imply a non-uniform field).
Starting at $0$ at the positive plate (swapping the reference).
Extending the line beyond $d$ as if the same uniform field continues inside the conductor; potential is constant inside each plate.

Things to Be Careful About

The graph is of potential, not field strength: the field would be the (negative) gradient of this straight line.
The question states vacuum: avoids complications of different permittivities; but the linear potential result still holds for a uniform field in a uniform medium.

Techniques used

use the uniform field assumption between parallel platesrelate electric field to potential gradientapply boundary conditions at each plate to sketch a linear graph

Q3MediumTemperatureThermodynamicsIdeal Gases

(a)

Two metal cuboids P and Q are in thermal contact with each other.

5M

(i)

P and Q are in thermal equilibrium.

State what is meant by the term thermal equilibrium.

2M Similar

(ii)

Data for P and Q are given in Table 3.1.

Table 3.1

	P	Q
specific heat capacity / $\text{J kg}^{-1}\ \text{K}^{-1}$	390	910
mass / $\text{kg}$	0.54	0.37

P and Q are initially both at the same temperature.

P is supplied with $24\ \text{kJ}$ of thermal energy. After some time, P and Q are once again both at the same temperature as each other.

P and Q are perfectly insulated from the surroundings.

Determine the change in temperature $\Delta T$ of Q.

$\Delta T$ = ______ $\text{K}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Thermal energy supplied to P is shared between P and Q (perfect insulation), and final temperatures are equal, so both rise by the same $\Delta T$ .

24000 = (m_P c_P + m_Q c_Q)\,\Delta T

m_P c_P = 0.54 \times 390 = 210.6\ \text{J K}^{-1}

m_Q c_Q = 0.37 \times 910 = 336.7\ \text{J K}^{-1}

\Delta T = \frac{24000}{210.6+336.7} = 43.8\ \text{K}

Answer

$\Delta T \approx 44\ \text{K}$

Final answer

44 K

Detailed explanation

Background Concept

The specific heat capacity $c$ links energy input to temperature rise:

Q = mc\Delta T

where:

$Q$ is thermal energy transferred (J),
$m$ is mass (kg),
$c$ is specific heat capacity ( $\text{J kg}^{-1}\ \text{K}^{-1}$ ),
$\Delta T$ is the temperature change (K).

If objects are perfectly insulated from the surroundings, then any energy supplied stays within the system: total energy in equals total increase in internal energy of the objects.

Understanding the Question

Two cuboids P and Q start at the same temperature and are in thermal contact.
$24\ \text{kJ}$ is supplied to P.
They are insulated from the surroundings (so no energy escapes).
After some time, P and Q are again at the same temperature (thermal equilibrium again).

The question asks for the temperature change of Q, $\Delta T$ .

Key implication: at the final equilibrium, both have the same final temperature, so they have undergone the same temperature rise $\Delta T$ .

Approach

Use energy conservation:

energy supplied $=$ increase in internal energy of P $+$ increase in internal energy of Q.
each increase is $mc\Delta T$ .
factor out $\Delta T$ to solve.

Step-by-Step Reasoning

Convert the energy input:

24\ \text{kJ} = 24000\ \text{J}

Write the energy balance (no losses):

24000 = m_P c_P \Delta T + m_Q c_Q \Delta T

Factor out $\Delta T$ :

24000 = (m_P c_P + m_Q c_Q)\Delta T

Calculate the “heat capacities” $mc$ for each block:

m_P c_P = 0.54 \times 390 = 210.6\ \text{J K}^{-1}

m_Q c_Q = 0.37 \times 910 = 336.7\ \text{J K}^{-1}

Add them and solve:

\Delta T = \frac{24000}{210.6 + 336.7} = 43.8\ \text{K}

So Q’s temperature rises by about $44\ \text{K}$ .

Key Takeaways

With insulation, use conservation of energy.
When two objects end at the same temperature, they share the same $\Delta T$ .
Combine heat capacities: total $= m_Pc_P + m_Qc_Q$ .

Common Mistakes

Using only Q’s $mc$ (as if all $24\ \text{kJ}$ went into Q).
Forgetting to convert $\text{kJ}$ to $\text{J}$ .
Assuming different temperature rises for P and Q even though they end at the same final temperature.

Things to Be Careful About

Temperature changes in K and in $^{\circ}\text{C}$ have the same numerical size, but keep units as K for $mc\Delta T$ .
Check that insulation is explicitly stated; otherwise you would have to consider losses to surroundings.

Techniques used

apply conservation of energy in an insulated systemuse the relationship Q = mc\Delta Tcombine heat capacities of objects in thermal contact

(b)

Nitrogen may be assumed to be an ideal gas. A fixed amount of nitrogen gas is contained at a constant pressure of $1.6 \times 10^{5}\ \text{Pa}$ .

The variation of the volume $V$ of the gas with the temperature $\theta$ of the gas is shown in Fig. 3.1.

Fig. 3.1

7M

(i)

The temperature of the nitrogen gas is increased from $0^{\circ}\text{C}$ to $210^{\circ}\text{C}$ .

Determine the work done on the gas.

work done = ______ $\text{J}$

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 3.1: $V_0 \approx 0.18\ \text{m}^3$ at $0^{\circ}\text{C}$ and $V_{300} \approx 0.38\ \text{m}^3$ at $300^{\circ}\text{C}$ .

Linear variation:

\text{gradient} = \frac{0.38-0.18}{300} = 6.67 \times 10^{-4}\ \text{m}^3\,^{\circ}\text{C}^{-1}

V_{210} = 0.18 + 210(6.67 \times 10^{-4}) \approx 0.32\ \text{m}^3

\Delta V = 0.32-0.18 = 0.14\ \text{m}^3

Work done by gas at constant pressure:

W_{\text{by}} = p\Delta V = (1.6 \times 10^{5})(0.14)=2.24\times 10^{4}\ \text{J}

Work done on gas:

W_{\text{on}} = -W_{\text{by}} = -2.24\times 10^{4}\ \text{J}

Answer

$-2.2 \times 10^{4}\ \text{J}$

Final answer

-2.2 × 10^4 J

Detailed explanation

Background Concept

When a gas expands against an external pressure $p$ , it does work. For a constant pressure process:

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

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

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

because during expansion the surroundings do negative work on the gas.

Understanding the Question

The gas pressure stays constant at $1.6 \times 10^{5}\ \text{Pa}$ .
Temperature increases from $0^{\circ}\text{C}$ to $210^{\circ}\text{C}$ .
From the given straight-line graph of $V$ vs $\theta$ , we must find the change in volume and hence the work.
The question specifically asks for the work done on the gas, so the sign matters.

Approach

Read (or interpolate) $V$ at $0^{\circ}\text{C}$ and at $210^{\circ}\text{C}$ from the straight line.
Compute $\Delta V$ .
Use $W_{\text{on}} = -p\Delta V$ .

diagram-b1

Step-by-Step Reasoning

Read two convenient points from the line. From the figure description, suitable readings are:

at $0^{\circ}\text{C}$ : $V_0 \approx 0.18\ \text{m}^3$
at $300^{\circ}\text{C}$ : $V_{300} \approx 0.38\ \text{m}^3$

Since the plot is a straight line, use linear interpolation to get the volume at $210^{\circ}\text{C}$ :

\text{gradient} = \frac{0.38-0.18}{300}

\text{gradient} = 6.67 \times 10^{-4}\ \text{m}^3\,^{\circ}\text{C}^{-1}

V_{210} = 0.18 + 210 \times 6.67\times 10^{-4} \approx 0.32\ \text{m}^3

Change in volume:

\Delta V = V_{210}-V_0 = 0.32-0.18 = 0.14\ \text{m}^3

Work done by gas:

W_{\text{by}} = p\Delta V = 1.6 \times 10^{5} \times 0.14 = 2.24 \times 10^{4}\ \text{J}

Work done on gas is negative of this (because gas expands):

W_{\text{on}} = -2.24 \times 10^{4}\ \text{J}

Key Takeaways

For constant pressure, the work magnitude is $p\Delta V$ .
Expansion means work is done by the gas; hence work done on the gas is negative.
Straight-line graphs allow interpolation using gradients.

Common Mistakes

Giving $+p\Delta V$ when the question asks for work done on the gas.
Using $\Delta V$ in the wrong direction (mixing up initial and final volume).
Reading volumes inaccurately from the graph or not showing a clear interpolation method.

Things to Be Careful About

Pressure is already in SI units (Pa) and volume in $\text{m}^3$ , so $p\Delta V$ comes out directly in J.
The numerical value depends slightly on graph reading; use consistent readings from the straight line.

Techniques used

read values from a straight-line graphuse constant-pressure work W = p\Delta Vapply a sign convention for work done on a gas

(ii)

Determine the number $N$ of molecules of nitrogen gas.

$N$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At $0^{\circ}\text{C}$ , $T = 273\ \text{K}$ and from the graph $V \approx 0.18\ \text{m}^3$ .

pV = NkT \Rightarrow N = \frac{pV}{kT}

N = \frac{(1.6\times 10^{5})(0.18)}{(1.38\times 10^{-23})(273)} = 7.6 \times 10^{24}

Answer

$N \approx 7.6 \times 10^{24}$

Final answer

7.6 × 10^24

Detailed explanation

Background Concept

For an ideal gas, the equation of state can be written using molecules:

pV = NkT

where:

$p$ is pressure (Pa),
$V$ is volume ( $\text{m}^3$ ),
$N$ is the number of molecules,
$k$ is the Boltzmann constant ( $1.38 \times 10^{-23}\ \text{J K}^{-1}$ ),
$T$ is thermodynamic temperature (K).

This form is useful when the question asks for the number of molecules rather than moles.

Understanding the Question

You are told nitrogen behaves as an ideal gas, at constant pressure $1.6 \times 10^{5}\ \text{Pa}$ . From the $V$ vs $\theta$ graph you can read the volume at a stated temperature. Since the amount of gas is fixed, $N$ is constant and can be found from any single state.

Approach

Choose a convenient point on the graph (e.g. $0^{\circ}\text{C}$ ).
Convert to kelvin.
Rearrange $pV = NkT$ to $N = \frac{pV}{kT}$ .

Step-by-Step Reasoning

At $0^{\circ}\text{C}$ , the absolute temperature is:

T = 0 + 273 = 273\ \text{K}

Read the corresponding volume from the graph: $V \approx 0.18\ \text{m}^3$ .
Substitute into the rearranged ideal gas equation:

N = \frac{pV}{kT}

N = \frac{(1.6\times 10^{5})(0.18)}{(1.38\times 10^{-23})(273)}

Evaluating gives $N \approx 7.6 \times 10^{24}$ molecules.

(Using the $210^{\circ}\text{C}$ point would give essentially the same result, allowing for graph reading.)

Key Takeaways

Use $pV = NkT$ when asked for number of molecules.
Always convert $^{\circ}\text{C}$ to K by adding 273.

Common Mistakes

Using $T = 0\ \text{K}$ for $0^{\circ}\text{C}$ .
Using $pV = nRT$ but forgetting to convert from $n$ (moles) to number of molecules.
Using an inconsistent volume reading from the graph.

Things to Be Careful About

Make sure the volume you use corresponds to the same temperature you use.
Quote $N$ to appropriate significant figures (usually 2\u20133 s.f. given graph-reading limits).

Techniques used

apply the ideal-gas equation in molecular form pV = NkTconvert Celsius to kelvinsubstitute values and rearrange for N

(iii)

The mass of a nitrogen molecule is $4.7 \times 10^{-26}\ \text{kg}$ .

Calculate the root-mean-square (r.m.s.) speed of a nitrogen molecule at $210^{\circ}\text{C}$ .

r.m.s. speed = ______ $\text{m s}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At $210^{\circ}\text{C}$ ,

T = 210 + 273 = 483\ \text{K}

c_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3(1.38\times 10^{-23})(483)}{4.7\times 10^{-26}}}

c_{\text{rms}} = 6.5 \times 10^{2}\ \text{m s}^{-1}

Answer

$6.5 \times 10^{2}\ \text{m s}^{-1}$

Final answer

6.5 × 10^2 m s^-1

Detailed explanation

Background Concept

For an ideal gas, the mean kinetic energy per molecule is

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

Rearranging gives the root-mean-square speed $c_{\text{rms}}$ :

c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{\frac{3kT}{m}}

where $m$ is the mass of one molecule.

Understanding the Question

You are given the mass of a nitrogen molecule ( $4.7 \times 10^{-26}\ \text{kg}$ ) and asked for the r.m.s. speed at $210^{\circ}\text{C}$ . The key is to use absolute temperature in kelvin.

Approach

Convert $210^{\circ}\text{C}$ to kelvin.
Substitute into $c_{\text{rms}} = \sqrt{\frac{3kT}{m}}$ .
Calculate and give the answer with unit $\text{m s}^{-1}$ .

Step-by-Step Reasoning

Convert temperature:

T = 210 + 273 = 483\ \text{K}

Substitute values:

c_{\text{rms}} = \sqrt{\frac{3(1.38\times 10^{-23})(483)}{4.7\times 10^{-26}}}

Work inside the square root:

3kT = 3 \times 1.38\times 10^{-23} \times 483 \approx 2.00\times 10^{-20}\ \text{J}

\frac{3kT}{m} \approx \frac{2.00\times 10^{-20}}{4.7\times 10^{-26}} \approx 4.25\times 10^{5}

Take the square root:

c_{\text{rms}} \approx \sqrt{4.25\times 10^{5}} \approx 6.5\times 10^{2}\ \text{m s}^{-1}

Key Takeaways

Molecular speed depends on absolute temperature: $c_{\text{rms}} \propto \sqrt{T}$ .
Always use kelvin in kinetic theory formulae.

Common Mistakes

Using $T = 210\ \text{K}$ instead of $483\ \text{K}$ .
Forgetting the factor of 3 in $3kT$ .
Mixing up molar mass with mass of one molecule.

Things to Be Careful About

Keep powers of ten consistent: dividing by $10^{-26}$ increases the power by 26.
Quote to 2\u20133 significant figures, reflecting the given data (especially the molecular mass).

Techniques used

convert Celsius to kelvinuse c_rms = \sqrt{3kT/m}evaluate a square-root expression with powers of ten

Q4MediumOscillationsMedical Physics

A small crystal is made to vibrate with simple harmonic motion. The variation with time $t$ of the displacement $x$ of one surface of the crystal from its equilibrium position is shown in Fig. 4.1.

Fig. 4.1

(a)

Show that the angular frequency of the vibration of the surface is $4.2 \times 10^{7}\ \text{rad s}^{-1}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From the graph, one full cycle takes

T = 0.15 \times 10^{-6}\ \text{s} = 1.5 \times 10^{-7}\ \text{s}.

\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5 \times 10^{-7}} = 4.19 \times 10^{7}\ \text{rad s}^{-1}.

Answer

\omega \approx 4.2 \times 10^{7}\ \text{rad s}^{-1}.

Final answer

4.2 × 10^7 rad s^-1

Detailed explanation

Background Concept

In simple harmonic motion (SHM), the motion repeats exactly after a time period $T$ . The angular frequency $\omega$ measures how fast the oscillation progresses in radians per second. The link between period and angular frequency is

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

because one full oscillation corresponds to an angular phase change of $2\pi\ \text{rad}$ .

Understanding the Question

You are given a displacement–time graph for a vibrating crystal surface. The question asks you to use the graph to find $\omega$ and show it equals $4.2 \times 10^{7}\ \text{rad s}^{-1}$ . The key step is identifying the period $T$ (time for one complete cycle) from the graph, and converting the microsecond scale to seconds.

Approach

Read the period $T$ from the time axis by finding the time for one full cycle (e.g. from one zero crossing with the same direction of motion to the next such zero crossing).
Convert $T$ from the graph units ( $10^{-6}\ \text{s}$ ) into seconds.
Substitute into $\omega = 2\pi/T$ .

Step-by-Step Reasoning

The graph completes one cycle at $t = 0.15$ on the axis scale, where the axis unit is $10^{-6}\ \text{s}$ .

T = 0.15 \times 10^{-6}\ \text{s} = 1.5 \times 10^{-7}\ \text{s}.

Use the definition of angular frequency:

\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5 \times 10^{-7}} \approx 4.19 \times 10^{7}\ \text{rad s}^{-1}.

Rounding appropriately gives $4.2 \times 10^{7}\ \text{rad s}^{-1}$ , as required.

Key Takeaways

The period $T$ is read directly from the time between identical points in successive cycles.
Convert axis multipliers (here $10^{-6}$ ) into SI before substituting.
Use $\omega = 2\pi/T$ .

Common Mistakes

Using $\omega = 2\pi f$ but then taking $f$ wrongly from the graph (e.g. using $f = T$ instead of $1/T$ ).
Forgetting that the time axis is in $10^{-6}\ \text{s}$ , leading to an answer smaller by a factor of $10^{6}$ .
Taking half a period (e.g. peak to trough) as $T$ .

Things to Be Careful About

The period must be in seconds; here $0.15$ on the axis means $0.15\ \mu\text{s}$ .
Quote $\omega$ in $\text{rad s}^{-1}$ (not $\text{s}^{-1}$ ).

Techniques used

read the period from a displacement–time graphuse $\omega = 2\pi/T$convert micro-units into SI units

(b)

Determine the maximum acceleration $a_0$ of the vibration of the surface.

$a_0$ = ______ $\text{m s}^{-2}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From the graph, amplitude

x_0 = 40 \times 10^{-6}\ \text{m} = 4.0 \times 10^{-5}\ \text{m}.

For SHM,

a_0 = \omega^2 x_0.

Using $\omega = 4.2 \times 10^{7}\ \text{rad s}^{-1}$ ,

a_0 = (4.2 \times 10^{7})^2 (4.0 \times 10^{-5}) = 7.1 \times 10^{10}\ \text{m s}^{-2}.

Answer

a_0 = 7.1 \times 10^{10}\ \text{m s}^{-2}.

Final answer

7.1 × 10^10 m s^-2

Detailed explanation

Background Concept

For SHM,

a = -\omega^2 x.

This means the acceleration is proportional to displacement and always directed towards the equilibrium position. The maximum magnitude of acceleration occurs at the maximum displacement $x = \pm x_0$ :

a_0 = \omega^2 x_0.

Understanding the Question

You are asked for the maximum acceleration of the vibrating surface. From the graph you can read the amplitude $x_0$ , and from part (a) you have $\omega$ . Then apply $a_0 = \omega^2 x_0$ .

Approach

Read amplitude $x_0$ (peak displacement) from the displacement axis and convert using the $10^{-6}\ \text{m}$ scale.
Use $a_0 = \omega^2 x_0$ .
Ensure units end up as $\text{m s}^{-2}$ .

Step-by-Step Reasoning

The graph peak is at $x = 40$ on a scale of $10^{-6}\ \text{m}$ :

x_0 = 40 \times 10^{-6}\ \text{m} = 4.0 \times 10^{-5}\ \text{m}.

Substitute into the SHM maximum acceleration formula:

a_0 = \omega^2 x_0 = (4.2 \times 10^{7})^2 \times (4.0 \times 10^{-5}).

Evaluate:

(4.2 \times 10^{7})^2 = 1.764 \times 10^{15}

so

a_0 = (1.764 \times 10^{15})(4.0 \times 10^{-5}) = 7.06 \times 10^{10}\ \text{m s}^{-2} \approx 7.1 \times 10^{10}\ \text{m s}^{-2}.

Key Takeaways

In SHM, $a$ is largest at the extremes and zero at equilibrium.
Max acceleration magnitude is $a_0 = \omega^2 x_0$ .

Common Mistakes

Using $a_0 = \omega x_0$ (that is for maximum speed $v_0$ ).
Forgetting the axis multiplier and using $x_0 = 40\ \text{m}$ or $40\times 10^{-6}$ without units.
Using $\omega = 4.2 \times 10^{7}\ \text{s}^{-1}$ but treating it as Hz (missing factor $2\pi$ ).

Things to Be Careful About

$\omega$ is in $\text{rad s}^{-1}$ but radians are dimensionless, so $\omega^2 x_0$ correctly gives $\text{m s}^{-2}$ .
Significant figures should match the data: typically 2 s.f. here.

Techniques used

read the amplitude from a displacement–time graphuse $a_0 = \omega^2 x_0$ for SHMcarry units through a calculation

(c)

The crystal may be modelled as a single mass of $2.4 \times 10^{-4}\ \text{kg}$ that vibrates as shown in Fig. 4.1.

Calculate the total energy $E$ of the vibrations.

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

3M Similar

DifficultyMedium

Worked solution

Working

Total energy in SHM:

E = \frac{1}{2} m \omega^2 x_0^2.

With $m = 2.4 \times 10^{-4}\ \text{kg}$ , $\omega = 4.2 \times 10^{7}\ \text{rad s}^{-1}$ , $x_0 = 40 \times 10^{-6}\ \text{m}$ ,

E = \frac{1}{2}(2.4 \times 10^{-4})(4.2 \times 10^{7})^2 (40 \times 10^{-6})^2.

E = \frac{1}{2}(2.4 \times 10^{-4})(1.764 \times 10^{15})(1.6 \times 10^{-9}) = 3.4 \times 10^{2}\ \text{J}.

Answer

E = 3.4 \times 10^{2}\ \text{J}.

Final answer

3.4 × 10^2 J

Detailed explanation

Background Concept

In SHM, energy continuously transfers between kinetic energy and potential energy, but the total energy stays constant (if there is no damping). The total energy can be written as

E = \frac{1}{2}m\omega^2 x_0^2.

This comes from noting that the maximum speed is $v_0 = \omega x_0$ , so the maximum kinetic energy is

E = \frac{1}{2} m v_0^2 = \frac{1}{2}m(\omega x_0)^2.

Understanding the Question

You are told to model the crystal as a single mass vibrating with the displacement shown. That means you can treat it exactly like a particle in SHM with mass $m$ , angular frequency $\omega$ (from part a), and amplitude $x_0$ (from the graph). The question asks for the total energy of the vibration.

Approach

Read/recall amplitude $x_0$ from the graph.
Use $\omega$ from part (a).
Substitute into $E = \tfrac{1}{2}m\omega^2 x_0^2$ and evaluate carefully in standard form.

Step-by-Step Reasoning

Amplitude from the graph: peak displacement is $40$ in units of $10^{-6}\ \text{m}$ :

x_0 = 40 \times 10^{-6}\ \text{m} = 4.0 \times 10^{-5}\ \text{m}.

Substitute the given mass and the angular frequency:

E = \frac{1}{2}(2.4 \times 10^{-4})(4.2 \times 10^{7})^2 (4.0 \times 10^{-5})^2.

Work through the powers:

(4.2 \times 10^{7})^2 = 1.764 \times 10^{15}, \quad (4.0 \times 10^{-5})^2 = 1.6 \times 10^{-9}.

So

E = \frac{1}{2}(2.4 \times 10^{-4})(1.764 \times 10^{15})(1.6 \times 10^{-9}).

Combine the last two factors first:

(1.764 \times 10^{15})(1.6 \times 10^{-9}) = 2.8224 \times 10^{6}.

Then

E = \frac{1}{2}(2.4 \times 10^{-4})(2.8224 \times 10^{6}) = \frac{1}{2}(6.77376 \times 10^{2}) = 3.38688 \times 10^{2}\ \text{J}.

Rounded:

E \approx 3.4 \times 10^{2}\ \text{J}.

Key Takeaways

Total energy in SHM depends on $m$ , $\omega$ , and $x_0$ through $E \propto \omega^2 x_0^2$ .
Using $v_0 = \omega x_0$ provides a useful check: $E = \tfrac{1}{2}mv_0^2$ .

Common Mistakes

Using $E = \tfrac{1}{2}kx_0^2$ without relating $k$ to $m\omega^2$ .
Squaring $40 \times 10^{-6}$ incorrectly (common error: $40^2 \times 10^{-6}$ instead of $40^2 \times 10^{-12}$ ).
Forgetting the factor $\tfrac{1}{2}$ .

Things to Be Careful About

Keep amplitude in metres before squaring.
Because $\omega$ is very large, $\omega^2$ is extremely large; errors in powers of ten change the result dramatically.
Quote the final answer to 2–3 significant figures, consistent with the given data.

Techniques used

use the total energy expression for SHMsubstitute amplitude and angular frequency into an energy formulacheck equivalence of alternative energy forms

(d)

The crystal generates ultrasound waves that are used to obtain diagnostic information about internal structures.

5M

(i)

The crystal is made from piezoelectric material.

Explain how the crystal is made to vibrate.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

An alternating potential difference is applied across the piezoelectric crystal.

The alternating electric field makes the crystal alternately expand and contract (inverse piezoelectric effect), so it vibrates at the frequency of the applied a.c. (large amplitude if driven near resonance).

Final answer

Apply an a.c. pd so the piezoelectric crystal alternately expands and contracts, making it vibrate (at the driving frequency, maximum near resonance).

Detailed explanation

Background Concept

Piezoelectric materials couple electrical and mechanical effects:

Direct piezoelectric effect: mechanical stress produces a potential difference.
Inverse piezoelectric effect: an applied electric field produces a change in dimensions (strain).

Ultrasound transducers use the inverse effect to generate vibrations, which launch sound waves into a medium.

Understanding the Question

The question asks how the crystal is made to vibrate. This is about the method of driving a piezoelectric transducer: what you apply electrically, and what mechanical response it causes.

Approach

State that you apply an alternating potential difference across the crystal. Explain that the alternating electric field causes repeated expansion and contraction, producing oscillations at the same frequency (often chosen to match a resonant frequency for efficiency).

Step-by-Step Reasoning

Put electrodes on opposite faces of the piezoelectric crystal and connect to an a.c. supply.
The a.c. supply produces an electric field whose direction (and magnitude) changes sinusoidally.
Because of the inverse piezoelectric effect, the crystal changes thickness/length in response to the field: when the field reverses, the strain reverses.
This alternating strain is a mechanical vibration. If the driving frequency matches (or is close to) the natural frequency of the crystal, resonance gives a larger vibration amplitude.

Key Takeaways

Ultrasound is generated by driving a piezoelectric crystal with an a.c. voltage.
The inverse piezoelectric effect converts electrical oscillations into mechanical oscillations.

Common Mistakes

Describing only the direct effect (pressure producing a voltage) when the question is about generating vibration.
Saying “apply a voltage” without specifying it must be alternating to cause continuous vibration.

Things to Be Careful About

Use the correct language: alternating electric field causes alternating expansion and contraction.
If mentioning resonance, keep it as an additional efficiency point, not the main mechanism.

Techniques used

describe the inverse piezoelectric effectlink an alternating potential difference to periodic deformationrelate mechanical vibration frequency to the driving electrical frequency

(ii)

A parallel beam of ultrasound waves is incident on a muscle-bone boundary. Data for muscle and bone are given in Table 4.1.

Table 4.1

material	density / $\text{kg m}^{-3}$	speed of sound / $\text{m s}^{-1}$
muscle	1100	1600
bone	1900	4100

Calculate the percentage of the intensity of the ultrasound beam that is transmitted at this boundary.

percentage transmitted = ______ $\%$

3M Similar

DifficultyMedium

Worked solution

Working

Acoustic impedance $Z = \rho c$ .

Z_{\text{muscle}} = 1100 \times 1600 = 1.76 \times 10^{6}\ \text{kg m}^{-2}\text{s}^{-1}

Z_{\text{bone}} = 1900 \times 4100 = 7.79 \times 10^{6}\ \text{kg m}^{-2}\text{s}^{-1}

For normal incidence, transmitted intensity fraction

\frac{I_T}{I_0} = \frac{4Z_1Z_2}{(Z_1+Z_2)^2}.

\frac{I_T}{I_0} = \frac{4(1.76)(7.79)}{(1.76+7.79)^2} = 0.601.

Percentage transmitted:

0.601 \times 100 = 60.1\% \approx 60\%.

Answer

\text{percentage transmitted} \approx 60\%.

Final answer

60%

Detailed explanation

Background Concept

When an ultrasound wave reaches a boundary between two materials, some intensity is reflected and some is transmitted. How much is reflected depends on the acoustic impedance

Z = \rho c

where $\rho$ is the density and $c$ is the speed of sound in the material.

For a plane wave at normal incidence (beam perpendicular to the boundary), the intensity reflection coefficient is

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

and the transmitted fraction is

T = 1 - R = \frac{4Z_1Z_2}{(Z_1+Z_2)^2}.

Understanding the Question

A parallel ultrasound beam hits a muscle–bone boundary. You are given $\rho$ and $c$ for each medium. You must find the percentage of intensity transmitted into the bone. This is a boundary-impedance problem: compute $Z$ for each side, then compute $T$ .

Approach

Calculate $Z_{\text{muscle}}$ and $Z_{\text{bone}}$ using $Z = \rho c$ .
Use the transmission fraction formula $T = 4Z_1Z_2/(Z_1+Z_2)^2$ .
Convert $T$ into a percentage.

Step-by-Step Reasoning

diagram-1

Compute impedances:

Z_{\text{muscle}} = (1100)(1600) = 1.76 \times 10^{6}\ \text{kg m}^{-2}\text{s}^{-1}

Z_{\text{bone}} = (1900)(4100) = 7.79 \times 10^{6}\ \text{kg m}^{-2}\text{s}^{-1}

Apply the transmission coefficient at normal incidence (take $Z_1 = Z_{\text{muscle}}$ , $Z_2 = Z_{\text{bone}}$ ):

T = \frac{4Z_1Z_2}{(Z_1 + Z_2)^2}.

Using $Z$ values in units of $10^{6}$ (the $10^{6}$ cancels):

T = \frac{4(1.76)(7.79)}{(1.76+7.79)^2}.

Calculate:

\text{numerator} = 4 \times 13.7104 = 54.8416, \quad \text{denominator} = (9.55)^2 = 91.2025,

so

T = \frac{54.8416}{91.2025} = 0.601.

Convert to a percentage:

\%\ \text{transmitted} = 0.601 \times 100 = 60.1\% \approx 60\%.

Key Takeaways

Use acoustic impedance $Z = \rho c$ to quantify mismatch between media.
For normal incidence, intensity transmission is $T = 4Z_1Z_2/(Z_1+Z_2)^2$ .
Large impedance mismatch (bone vs muscle) causes significant reflection.

Common Mistakes

Using densities alone (forgetting to multiply by sound speed).
Using the amplitude coefficient formula instead of the intensity coefficient (intensity depends on the square).
Mixing up transmitted and reflected fractions (reporting $R$ instead of $T$ ).
Not converting the fraction into a percentage.

Things to Be Careful About

Ensure normal incidence is assumed (the standard formula above); at oblique incidence it becomes more complicated.
Keep enough significant figures in intermediate steps; round only at the end.
Use consistent units when calculating $Z$ (here SI throughout).

Techniques used

calculate acoustic impedance from density and sound speedapply the intensity transmission (or reflection) coefficient at a boundaryconvert a fraction to a percentage

Q5MediumCapacitance

(a)

A capacitor of capacitance $C_1$ is connected in series with a second capacitor of capacitance $C_2$ .

Show that the combined capacitance $C$ of the two capacitors is given by

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For capacitors in series, the charge on each is the same, $Q$ .

Total p.d.

V = V_1 + V_2

Using $C = Q/V$ so $V = Q/C$ :

V = \frac{Q}{C_1} + \frac{Q}{C_2} = Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right)

But $C = Q/V$ , so $V=Q/C$ and hence

\frac{Q}{C} = Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right)

Cancel $Q$ :

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

Answer

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

Final answer

1/C = 1/C1 + 1/C2

Detailed explanation

Background Concept

Capacitance is defined by

C = \frac{Q}{V}

where $Q$ is the charge stored and $V$ is the potential difference (p.d.) across the capacitor.

For capacitors:

In series, the same charge $Q$ must appear on each capacitor because charge cannot “pile up” at the junction: whatever charge arrives at one plate induces an equal and opposite charge on the facing plate of the next capacitor.
The total p.d. across components in series is the sum of the individual p.d.s.

Understanding the Question

Two capacitors with capacitances $C_1$ and $C_2$ are connected in series. The question asks you to show (derive) a formula for the single equivalent capacitance $C$ of the series combination.

Approach

Use the fact that series capacitors carry the same charge $Q$ .
Add the potential differences: $V = V_1 + V_2$ .
Replace each $V_i$ using $V_i = Q/C_i$ .
Use the definition of equivalent capacitance $C = Q/V$ and rearrange.

Step-by-Step Reasoning

diagram-a1

Take the charge on each capacitor to be $Q$ .

The total p.d. between the ends of the series pair is

V = V_1 + V_2.

Using $C = Q/V \Rightarrow V = Q/C$ for each capacitor:

V_1 = \frac{Q}{C_1},\qquad V_2 = \frac{Q}{C_2}.

So

V = \frac{Q}{C_1} + \frac{Q}{C_2} = Q\left(\frac{1}{C_1} + \frac{1}{C_2}\right).

But the equivalent capacitor $C$ is defined by the same total charge $Q$ at total p.d. $V$ :

V = \frac{Q}{C}.

Equate the two expressions for $V$ :

\frac{Q}{C} = Q\left(\frac{1}{C_1}+\frac{1}{C_2}\right)

and cancel $Q$ (non-zero during charging), giving

\frac{1}{C} = \frac{1}{C_1}+\frac{1}{C_2}.

Key Takeaways

In series: same charge, voltages add.
Combine with $V=Q/C$ to get the series capacitance rule.

Common Mistakes

Assuming the p.d. is the same across series capacitors (that is only true for parallel).
Writing $C = C_1 + C_2$ (that is the parallel rule).
Forgetting to use $V = Q/C$ and trying to manipulate $C=Q/V$ inconsistently.

Things to Be Careful About

Be explicit that the charge is the same on series capacitors; that is the key physical step.
When cancelling $Q$ , ensure you have $Q$ as a common factor on both sides first.

Techniques used

use charge continuity for capacitors in seriessum potential differences across series componentssubstitute $V=Q/C$ and rearrange algebraically

(b)

Three identical capacitors, each of capacitance $C$ , are connected in a network as shown in Fig. 5.1.

Fig. 5.1

The variation of the charge $Q$ with the potential difference (p.d.) $V$ between the terminals X and Y is shown in Fig. 5.2.

Fig. 5.2

Show that $C$ is equal to $44\ \mu\text{F}$ .

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 5.2, gradient $= Q/V = C_{\text{eq}}$ .
Using point $(V,Q)=(6.0\ \text{V},\ 400\ \mu\text{C})$ :

C_{\text{eq}} = \frac{400\times 10^{-6}}{6.0} = 6.67\times 10^{-5}\ \text{F} = 66.7\ \mu\text{F}

In Fig. 5.1: one capacitor $C$ in parallel with two capacitors $C$ in series.

Series pair:

\frac{1}{C_s}=\frac{1}{C}+\frac{1}{C}\Rightarrow C_s=\frac{C}{2}

Parallel total:

C_{\text{eq}} = C + \frac{C}{2} = \frac{3C}{2}

So

\frac{3C}{2} = 66.7\ \mu\text{F} \Rightarrow C = 44.4\ \mu\text{F} \approx 44\ \mu\text{F}

Answer

$C = 44\ \mu\text{F}$

Final answer

44 µF

Detailed explanation

Background Concept

For a capacitor (or a capacitor network behaving like a single capacitor),

Q = CV.

So a graph of $Q$ against $V$ is a straight line through the origin with gradient $C$ .

For combinations:

Parallel: same p.d. across each branch, charges add, so

C_{\text{parallel}} = C_1 + C_2 + \cdots

Series: same charge on each, p.d.s add, so

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Understanding the Question

Three identical capacitors (each capacitance $C$ ) are arranged so that:

the top branch is a single capacitor $C$ ,
the bottom branch is two capacitors $C$ in series,
and these two branches are connected in parallel between terminals X and Y.

The $Q$ – $V$ graph for the whole network is given, so we can find the network’s equivalent capacitance $C_{\text{eq}}$ , then relate it to $C$ .

Approach

Use the graph to calculate $C_{\text{eq}} = Q/V$ .
Reduce the circuit: find the series equivalent of the two lower capacitors.
Add in parallel with the top capacitor.
Set this equal to the value from the graph and solve for $C$ .

Step-by-Step Reasoning

diagram-b1

1) Equivalent capacitance from the graph
Since $Q = C_{\text{eq}}V$ , the gradient of the $Q$ – $V$ line is $C_{\text{eq}}$ .
Using the point $(6.0\ \text{V}, 400\ \mu\text{C})$ :

C_{\text{eq}} = \frac{Q}{V} = \frac{400\times 10^{-6}\ \text{C}}{6.0\ \text{V}} = 6.67\times 10^{-5}\ \text{F} = 66.7\ \mu\text{F}.

2) Equivalent capacitance of the network in terms of $C$
Bottom branch has two identical capacitors in series:

\frac{1}{C_s} = \frac{1}{C}+\frac{1}{C} = \frac{2}{C} \Rightarrow C_s = \frac{C}{2}.

This series combination is in parallel with the top capacitor $C$ , so

C_{\text{eq}} = C + \frac{C}{2} = \frac{3C}{2}.

3) Solve for $C$

\frac{3C}{2} = 66.7\ \mu\text{F} \Rightarrow C = \frac{2}{3}(66.7)\ \mu\text{F} = 44.4\ \mu\text{F} \approx 44\ \mu\text{F}.

Key Takeaways

A $Q$ – $V$ graph gives capacitance directly: gradient $=C$ .
Reduce mixed networks by doing series first, then parallel (or vice versa depending on layout).

Common Mistakes

Using $(V/Q)$ instead of $(Q/V)$ for the gradient.
Forgetting to convert $\mu\text{C}$ to C when calculating in SI units.
Treating the bottom two capacitors as parallel (they are in series).

Things to Be Careful About

The line passes through the origin, consistent with $Q=CV$ ; if you used two arbitrary points, ensure you still compute $\Delta Q/\Delta V$ .
Keep track of units: $1\ \mu\text{F} = 10^{-6}\ \text{F}$ and $1\ \mu\text{C} = 10^{-6}\ \text{C}$ , which cancel cleanly when using $Q/V$ .

Techniques used

find equivalent capacitance from the gradient of a $Q$-$V$ graphcombine series capacitancescombine parallel capacitancesequate two expressions for equivalent capacitance and solve

(c)

The capacitor network in Fig. 5.1 is charged and then connected to a resistor of resistance $54\ \text{k}\Omega$ . The capacitor network discharges through the resistor.

4M

(i)

Determine the time constant $\tau$ of the circuit. Give a unit with your answer.

$\tau$ = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Equivalent capacitance of the network (from part (b)):

C_{\text{eq}} = \frac{Q}{V} = 66.7\ \mu\text{F} = 66.7\times 10^{-6}\ \text{F}

Time constant:

\tau = RC = (54\times 10^{3})(66.7\times 10^{-6}) = 3.60\ \text{s}

Answer

$\tau = 3.6\ \text{s}$

Final answer

3.6 s

Detailed explanation

Background Concept

For a capacitor (or capacitor network treated as an equivalent capacitance $C_{\text{eq}}$ ) discharging through a resistor $R$ , the time constant is

\tau = RC_{\text{eq}}.

The time constant is the characteristic time over which current, voltage, and charge fall exponentially.

Understanding the Question

The capacitor network from Fig. 5.1 is first charged, then connected across a resistor of resistance $54\ \text{k}\Omega$ and allowed to discharge. We must find the circuit time constant $\tau$ .

We need the equivalent capacitance between X and Y. From part (b) (or from the $Q$ – $V$ graph),

C_{\text{eq}} = 66.7\ \mu\text{F}.

Approach

Convert $R$ to ohms and $C_{\text{eq}}$ to farads.
Multiply using $\tau = RC$ .
Quote the unit (seconds).

Step-by-Step Reasoning

Convert units:

R = 54\ \text{k}\Omega = 54\times 10^{3}\ \Omega,

C_{\text{eq}} = 66.7\ \mu\text{F} = 66.7\times 10^{-6}\ \text{F}.

Now apply

\tau = RC_{\text{eq}} = (54\times 10^{3})(66.7\times 10^{-6}).

Combine powers of ten: $10^{3}\times 10^{-6} = 10^{-3}$ .
Numerically, $54\times 66.7 = 3600$ (approximately), so

\tau \approx 3600\times 10^{-3}\ \text{s} = 3.60\ \text{s}.

Key Takeaways

Use the equivalent capacitance of the whole network when finding the discharge time constant.
$\tau=RC$ always has units of seconds.

Common Mistakes

Using $C = 44\ \mu\text{F}$ (capacitance of one capacitor) instead of the network’s $C_{\text{eq}} = 66.7\ \mu\text{F}$ .
Forgetting to convert $\text{k}\Omega$ to $\Omega$ or $\mu\text{F}$ to $\text{F}$ .

Things to Be Careful About

Time constant uses $C$ in farads; if you keep $R$ in $\text{k}\Omega$ and $C$ in $\mu\text{F}$ , remember that $(\text{k}\Omega)(\mu\text{F}) = 10^{-3}\ \text{s}$ .
Quote a sensible number of significant figures consistent with the data given (here $3.6\ \text{s}$ or $3.60\ \text{s}$ are both reasonable).

Techniques used

identify the equivalent capacitance of a capacitor networkapply the time constant relation $\tau = RC$perform unit conversions between $\mu\text{F}$, $\text{k}\Omega$ and SI units

(ii)

Determine the time taken for the discharge current to reduce to $15\%$ of the initial discharge current.

time = ______ $\text{s}$

2M Similar

DifficultyMedium

Worked solution

Working

For discharge,

I = I_0 e^{-t/\tau}

Given $I = 0.15 I_0$ :

0.15 = e^{-t/\tau}

Take ln:

\ln(0.15) = -\frac{t}{\tau}

So

t = -\tau \ln(0.15)

With $\tau = 3.60\ \text{s}$ :

t = -3.60\ln(0.15) = 6.83\ \text{s} \approx 6.8\ \text{s}

Answer

$6.8\ \text{s}$

Final answer

6.8 s

Detailed explanation

Background Concept

During discharge of a capacitor through a resistor, the current (and voltage and charge) decreases exponentially:

I = I_0 e^{-t/\tau}

where $I_0$ is the initial current at $t=0$ and $\tau = RC$ is the time constant.

To find the time for a quantity to fall to a certain fraction, you usually rearrange by taking natural logs.

Understanding the Question

The discharge current starts at $I_0$ and we want the time $t$ when it has fallen to $15\%$ of $I_0$ , i.e.

I = 0.15 I_0.

From part (c)(i), $\tau = 3.60\ \text{s}$ .

Approach

Substitute the fraction into the exponential decay equation: $I/I_0 = e^{-t/\tau}$ .
Take $\ln$ of both sides to bring down the exponent.
Solve for $t$ and substitute $\tau$ .

Step-by-Step Reasoning

Start with

I = I_0 e^{-t/\tau}.

Divide by $I_0$ :

\frac{I}{I_0} = e^{-t/\tau}.

Given $I/I_0 = 0.15$ :

0.15 = e^{-t/\tau}.

Take natural logarithms:

\ln(0.15) = \ln\left(e^{-t/\tau}\right) = -\frac{t}{\tau}.

So

t = -\tau\ln(0.15).

Now substitute $\tau = 3.60\ \text{s}$ :

t = -3.60\ln(0.15) = 3.60\ln\left(\frac{1}{0.15}\right).

Since $\ln(0.15) \approx -1.897$ ,

t \approx 3.60\times 1.897 = 6.83\ \text{s} \approx 6.8\ \text{s}.

Key Takeaways

Discharge current follows the same exponential form as voltage and charge.
Fractions like $15\%$ lead naturally to logarithms when solving for time.

Common Mistakes

Using $0.15 = e^{+t/\tau}$ (wrong sign in exponent).
Using $\log_{10}$ instead of $\ln$ without converting (the formula uses $e$ so use $\ln$ ).
Using $\tau$ from a single capacitor rather than the network time constant.

Things to Be Careful About

$\ln(0.15)$ is negative; the minus sign in $t=-\tau\ln(0.15)$ ensures $t$ is positive.
Keep enough significant figures in $\tau$ and in the logarithm to avoid rounding too early; round only at the end.

Techniques used

use the exponential decay equation for discharge currentrearrange an exponential using natural logarithmssubstitute the time constant to obtain a time

Q6MediumElectric FieldsMagnetic Fields

An electric field and a magnetic field are used to form a velocity selector. Charged particles, called ions, pass into a region of uniform electric and magnetic fields that is between parallel plates, as shown in Fig. 6.1.

Fig. 6.1

(a)

The potential difference (p.d.) between the plates of the velocity selector is $V$ . The separation of the plates is $d$ and the magnetic flux density is $B$ .

Show that the speed $u$ of ions that pass undeviated through the velocity selector is given by

u = \frac{V}{Bd}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Uniform electric field between parallel plates:

E = \frac{V}{d}

For undeviated motion, electric and magnetic forces balance:

qE = q u B

So

u = \frac{E}{B} = \frac{V/d}{B} = \frac{V}{Bd}

Answer

\nu = \frac{V}{Bd}

Final answer

ν = V/(Bd)

Detailed explanation

Background Concept

In a velocity selector, a charged particle moves through crossed uniform electric and magnetic fields. It experiences:

Electric force:

F_E = qE

where $q$ is the charge and $E$ is the electric field strength.

Magnetic force (for motion perpendicular to $\mathbf{B}$ ):

F_B = q u B

where $u$ is the particle speed and $B$ is the magnetic flux density.

A uniform electric field between parallel plates is related to the potential difference by:

E = \frac{V}{d}

where $V$ is the p.d. and $d$ is the plate separation.

Understanding the Question

The ions pass through the region between parallel plates where both $\mathbf{E}$ and $\mathbf{B}$ are uniform. “Pass undeviated” means the net sideways force is zero, so the magnetic force must exactly balance the electric force.

You are asked to show that the speed of ions that go straight through is

\nu = \frac{V}{Bd}.

Approach

Write the electric field between the plates as $E = V/d$ .
For an undeviated path, set the magnitudes of the electric and magnetic forces equal: $qE = quB$ .
Substitute $E = V/d$ and rearrange for $u$ .

Step-by-Step Reasoning

Start with the uniform field between the plates:

E = \frac{V}{d}.

For a straight (undeviated) path, sideways forces cancel, so:

F_E = F_B.

Using the expressions for the forces:

qE = q u B.

Cancel $q$ (this is important: the selected speed does not depend on the particle’s charge):

E = uB.

Hence:

u = u = \frac{E}{B}.

Substitute $E = V/d$ :

\nu = \frac{V/d}{B} = \frac{V}{Bd}.

Key Takeaways

In a velocity selector, the condition for no deflection is $qE = quB$ .
The selected speed depends only on the fields and plate separation: $u = V/(Bd)$ .
The particle’s charge cancels out.

Common Mistakes

Using $F_B = BIL$ (that is for a current-carrying conductor, not a single charged particle).
Forgetting $E = V/d$ for parallel plates.
Missing the fact that $\sin\theta = 1$ here because the velocity is perpendicular to $\mathbf{B}$ in a velocity selector.

Things to Be Careful About

The formula $F_B = q u B$ requires the velocity to be perpendicular to the magnetic field. Velocity selectors are designed so this is the case.
Keep symbols consistent: the question uses $u$ (speed) but the shown result uses $\nu$ ; they represent the same speed here.

Techniques used

use the uniform-field relation between electric field strength and potential differenceequate electric and magnetic forces for undeviated motionsubstitute into the force balance and rearrange algebraically

(b)

Positive ions with kinetic energy $4.1 \times 10^{-17}\ \text{J}$ and mass $3.2 \times 10^{-27}\ \text{kg}$ pass undeviated through the velocity selector when $V$ is equal to $980\ \text{V}$ and $d$ is equal to $3.6 \times 10^{-2}\ \text{m}$ .

Determine $B$ .

$B$ = ______ $\text{T}$

3M Similar

DifficultyMedium

Worked solution

Working

From kinetic energy,

E_k = \frac{1}{2} m u^2 \;\Rightarrow\; u = \sqrt{\frac{2E_k}{m}}

u = \sqrt{\frac{2(4.1\times 10^{-17})}{3.2\times 10^{-27}}} = 1.6\times 10^{5}\ \text{m s}^{-1}

Using $u = \dfrac{V}{Bd}$ ,

B = \frac{V}{ud} = \frac{980}{(1.6\times 10^5)(3.6\times 10^{-2})}

B = 1.7\times 10^{-1}\ \text{T}

Answer

B = 0.17\ \text{T}

Final answer

0.17 T

Detailed explanation

Background Concept

For a particle to pass straight through a velocity selector, the electric and magnetic forces must balance:

qE = quB.

With parallel plates,

E = \frac{V}{d}

so the selected speed is

u = u = \frac{V}{Bd}.

If the particle’s kinetic energy $E_k$ and mass $m$ are known, its speed can be found from:

E_k = \frac{1}{2} m u^2.

Understanding the Question

You are told that the ions pass undeviated when $V = 980\ \text{V}$ and $d = 3.6\times 10^{-2}\ \text{m}$ . Their kinetic energy and mass are also given, so you can calculate their speed $u$ . Then you use the velocity selector condition to find the magnetic flux density $B$ .

Approach

Calculate $u$ from $E_k = \tfrac{1}{2}mu^2$ .
Rearrange $u = V/(Bd)$ to $B = V/(ud)$ .
Substitute and compute, keeping units consistent.

Step-by-Step Reasoning

Calculate the speed from kinetic energy:

E_k = \frac{1}{2}mu^2 \Rightarrow u = \sqrt{\frac{2E_k}{m}}.

Substitute values:

u = \sqrt{\frac{2(4.1\times 10^{-17})}{3.2\times 10^{-27}}} = \sqrt{2.56\times 10^{10}} = 1.6\times 10^5\ \text{m s}^{-1}.

Now use the selector condition:

u = \frac{V}{Bd} \Rightarrow B = \frac{V}{ud}.

Substitute:

B = \frac{980}{(1.6\times 10^5)(3.6\times 10^{-2})}.

Compute the denominator:

(1.6\times 10^5)(3.6\times 10^{-2}) = 5.76\times 10^3.

Hence:

B = \frac{980}{5.76\times 10^3} \approx 1.7\times 10^{-1}\ \text{T} = 0.17\ \text{T}.

Key Takeaways

Convert kinetic energy to speed via $u = \sqrt{2E_k/m}$ .
Use $B = V/(ud)$ from the velocity selector equation.
Maintain consistent SI units throughout.

Common Mistakes

Forgetting the square root when rearranging $E_k = \tfrac{1}{2}mu^2$ .
Substituting $d$ in cm instead of m.
Rounding $u$ too early, causing noticeable error in $B$ .

Things to Be Careful About

The selected speed formula assumes perpendicular $\mathbf{u}$ and $\mathbf{B}$ .
Quote $B$ with a sensible number of significant figures (typically 2–3 here).

Techniques used

convert kinetic energy to speed using the kinetic energy relationsubstitute the selected speed into the velocity selector expressionrearrange to solve for magnetic flux density

(c)

A proton passes undeviated through the velocity selector.

An alpha particle enters the velocity selector at the same speed as the proton.

State how the expression in (a) predicts that the alpha particle also passes undeviated through the velocity selector.

1M Similar

DifficultyEasy

Worked solution

Answer

From (a),

\nu = \frac{V}{Bd}

so the speed for no deflection depends only on $V$ , $B$ and $d$ (not on mass or charge). Therefore any particle (including an $\alpha$ particle) entering with the same speed as the proton will also pass undeviated.

Final answer

Same speed ⇒ also undeviated because ν = V/(Bd) is independent of particle mass and charge.

Detailed explanation

Background Concept

In a velocity selector, the condition for straight-line motion is the balance of forces:

qE = quB \Rightarrow u = \frac{E}{B}.

Using $E = V/d$ gives:

u = \frac{V}{Bd}.

Notice $q$ cancels, and there is no $m$ anywhere in the final expression.

Understanding the Question

A proton passes undeviated, so its speed must equal the selector speed $u = V/(Bd)$ . An alpha particle enters with the same speed as the proton. The question asks you to use the expression from (a) to justify what happens to the alpha particle.

Approach

Read the expression in (a) as a “required speed” set by the apparatus. If another particle has that same speed, it will also satisfy the force-balance condition.

Step-by-Step Reasoning

From (a):

u = \frac{V}{Bd}.

The required speed depends only on the electric and magnetic field settings (through $V$ , $B$ , and $d$ ). It does not depend on the particle’s mass $m$ or charge $q$ .

Therefore, if the alpha particle enters with the same speed $u$ as the proton, then the forces still balance:

qE = quB

so the alpha particle also experiences zero net transverse force and passes undeviated.

Key Takeaways

A velocity selector selects a speed, not a particular type of particle.
Because $q$ cancels, particles with different charges are still selected if their speed matches.

Common Mistakes

Saying the alpha passes straight because it has the “same charge” (it does not: $\alpha$ has charge $+2e$ ).
Bringing in mass or kinetic energy unnecessarily; the question only needs the dependence shown in (a).

Things to Be Careful About

The correct statement is: same speed in the same $E$ and $B$ gives the same force balance, regardless of charge magnitude (as long as it is non-zero).

Techniques used

identify which quantities the derived expression depends oncompare conditions for two different charged particlesapply the undeviated condition to a new particle

(d)

By reference to Fig. 6.1 and to the forces acting on a positive ion, determine the direction of the magnetic field. Explain your reasoning.

3M Similar

DifficultyMedium

Worked solution

Answer

Electric field is from the $+$ plate to the $-$ plate, so $\mathbf{E}$ is downward and the electric force on a positive ion is downward.

For the ion to be undeviated, the magnetic force must be upward to balance this.

With ion velocity to the right, $\mathbf{F}_B = q\,\mathbf{v} \times \mathbf{B}$ must be upward, so $\mathbf{B}$ is into the page.

Final answer

Magnetic field is into the page.

Detailed explanation

Background Concept

A charged particle in electric and magnetic fields experiences the Lorentz force:

\mathbf{F} = q\mathbf{E} + q(\mathbf{v} \times \mathbf{B}).

The electric force $q\mathbf{E}$ is in the same direction as $\mathbf{E}$ for a positive charge.
The magnetic force $q(\mathbf{v} \times \mathbf{B})$ is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ , with direction given by the right-hand rule for $\mathbf{v} \times \mathbf{B}$ (for a positive charge).

For an undeviated path, the net transverse force must be zero, so the electric and magnetic forces must be equal and opposite in direction.

Understanding the Question

From Fig. 6.1, the top plate is positive and the bottom plate is negative, so the electric field points from top to bottom. The ions move horizontally to the right. You must determine the direction of the magnetic field that makes the magnetic force oppose the electric force for a positive ion.

Approach

Use plate polarities to find the direction of $\mathbf{E}$ .
For a positive ion, $\mathbf{F}_E$ is in the direction of $\mathbf{E}$ .
Undeviated means $\mathbf{F}_B$ must be opposite to $\mathbf{F}_E$ .
Use $\mathbf{F}_B = q(\mathbf{v} \times \mathbf{B})$ to infer the direction of $\mathbf{B}$ .

Step-by-Step Reasoning

diagram-6d

The electric field points from the positive plate to the negative plate, so it is downward.
A positive ion feels an electric force downward.
For the ion to travel straight, the magnetic force must be upward.

Now use the cross product direction:

Velocity $\mathbf{v}$ is to the right.
We need $\mathbf{v} \times \mathbf{B}$ to be upward.

Using the right-hand rule, with fingers pointing to the right (direction of $\mathbf{v}$ ) and thumb needing to point up (direction of $\mathbf{F}_B$ ), the magnetic field must point into the page.

Therefore, $\mathbf{B}$ is directed into the plane of the paper.

Key Takeaways

$\mathbf{E}$ goes from $+$ to $-$ .
For a positive charge, $\mathbf{F}_E$ is along $\mathbf{E}$ .
Magnetic force direction is found from $\mathbf{v} \times \mathbf{B}$ (and reverses for a negative charge).

Common Mistakes

Reversing the electric field direction (it is from $+$ to $-$ , not the other way).
Using Fleming’s left-hand rule incorrectly for moving charges (it applies to conventional current; for a single positive charge, use $\mathbf{v} \times \mathbf{B}$ ).
Forgetting the magnetic force must oppose the electric force for undeviated motion.

Things to Be Careful About

The conclusion “into/out of page” depends on the ion being positive; for a negative ion, the magnetic force direction would reverse.
Keep consistent directions: if you choose upward as positive, check that your force balance matches it.

Techniques used

deduce the direction of the electric field from plate polarityapply the direction of electric force on a positive chargeuse the right-hand rule for the magnetic force direction in v × B

(e)

The positive ions in (b) enter the velocity selector with greater kinetic energy.

On Fig. 6.1, sketch the path of these ions.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For greater kinetic energy, the ions have greater speed $u$ , so $F_B = q u B$ increases while $F_E = qE$ is unchanged.

Thus $F_B > F_E$ and the resultant force is upward (towards the $+$ plate), so the path curves upward through the field.

diagram-6e

Final answer

Path curves upward toward the positive (top) plate.

Detailed explanation

Background Concept

In the selector region:

Electric force magnitude:

F_E = qE

(and $E$ is fixed by the plates).

Magnetic force magnitude (with $\mathbf{v} \perp \mathbf{B}$ ):

F_B = quB.

The velocity selector condition for no deflection is $F_E = F_B$ . If the speed changes while $E$ and $B$ stay the same, the balance is lost and the particle deflects in the direction of the larger force.

Understanding the Question

The ions in (b) were undeviated for a particular speed. Now they enter with greater kinetic energy, so their speed is higher than the selected speed. You must sketch their path on Fig. 6.1.

From part (d), for a positive ion with motion to the right, the electric force is downward and the magnetic force is upward (because $\mathbf{B}$ is into the page).

Approach

Higher kinetic energy implies higher speed.
Decide how $F_E$ and $F_B$ change when $u$ increases.
Identify the direction of the resultant force.
Sketch a curved path bending in that direction within the field region.

Step-by-Step Reasoning

If kinetic energy increases, then speed $u$ increases because:

E_k = \frac{1}{2}mu^2.

Electric force: $F_E = qE$ stays the same (same plates, same $V$ and $d$ so same $E$ ).

Magnetic force: $F_B = quB$ increases because it is proportional to $u$ .

So for the faster ions:

F_B > F_E.

From (d), $F_B$ is upward and $F_E$ is downward, so the resultant transverse force is upward. A particle experiencing an upward transverse force while moving to the right follows a trajectory that curves upward toward the top plate.

diagram-6e

Key Takeaways

In a velocity selector, only one speed gives $F_E = F_B$ .
If $u$ increases, $F_B$ increases but $F_E$ does not, so deflection is toward the magnetic-force direction.

Common Mistakes

Drawing the path curving downward (would correspond to $F_E > F_B$ , i.e. slower ions).
Drawing a straight line despite stating the ions have higher kinetic energy.
Drawing a sharp kink rather than a smooth curve; a constant transverse force produces a smooth curvature.

Things to Be Careful About

The direction of curvature depends on the sign of charge and the direction of $\mathbf{B}$ . Here ions are positive and $\mathbf{B}$ was deduced to be into the page.
The question asks for a sketch “on Fig. 6.1”: show the curve mainly within the field region, then continuing out with an upward displacement.

Techniques used

compare magnitudes of electric and magnetic forces as speed changesdeduce the net force direction from the dominant forcesketch a trajectory consistent with the net force direction

Q7MediumMagnetic Fields

(a)

State Faraday’s law of electromagnetic induction.

2M Similar

DifficultyEasy

Worked solution

Answer

Faraday’s law: the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.

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

Final answer

Induced e.m.f. equals the negative rate of change of magnetic flux linkage: E = -d(NΦ)/dt.

Detailed explanation

Background Concept

Faraday’s law links induction to changing magnetic flux. The magnetic flux through a circuit is

\Phi = BA\cos\theta

and the flux linkage is $N\Phi$ for a coil of $N$ turns. If the flux linkage changes (because $B$ , $A$ , or $\theta$ changes, or because a conductor moves in a field), an e.m.f. is induced.

Faraday’s law states that the induced e.m.f. is proportional to how fast the flux linkage changes. The minus sign is Lenz’s law: the induced e.m.f. acts in such a direction as to oppose the change causing it.

Understanding the Question

You are asked to state Faraday’s law. For full credit you should give the clear verbal statement and/or the standard equation involving flux linkage and a time rate of change.

Approach

Write Faraday’s law in its most general form using $N\Phi$ , including the negative sign.

Step-by-Step Reasoning

Identify the relevant quantity that changes: magnetic flux linkage $N\Phi$ .
State that induced e.m.f. depends on the rate of change:

E \propto \frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

Give the full equation including Lenz’s-law sign:

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

Key Takeaways

Induction requires change of flux linkage.
The induced e.m.f. depends on the rate of that change.
The negative sign indicates opposition to the change (Lenz’s law).

Common Mistakes

Writing $E = -\mathrm{d}\Phi/\mathrm{d}t$ but forgetting the flux linkage factor $N$ when the law is asked generally.
Stating Lenz’s law only (direction opposes change) without mentioning rate of change of flux linkage.
Forgetting that Faraday’s law is about change, not just “presence of a magnetic field”.

Things to Be Careful About

Use $\Phi$ for flux and $N\Phi$ for flux linkage.
Include the minus sign if giving the equation form.
Use “rate of change” wording (not just “change”).

Techniques used

state Faraday’s law using magnetic flux linkageexpress the law as a time-derivative equation

(b)

A metal rod is accelerated uniformly from rest in a uniform magnetic field as shown in Fig. 7.1.

Fig. 7.1

The rod has length $l$ and the flux density of the magnetic field is $B$ .

An electromotive force (e.m.f.) is induced in the rod. The variation with time $t$ of the induced e.m.f. $E$ is shown in Fig. 7.2.

Fig. 7.2

7M

(i)

Explain how Fig. 7.2 shows that $E$ is proportional to the velocity $v$ of the rod.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Fig. 7.2 is a straight line through the origin, so $E \propto t$ .

For uniform acceleration from rest, $v = at$ so $v \propto t$ .

Hence $E \propto v$ .

Final answer

Since E ∝ t and v = at so v ∝ t, E is proportional to v.

Detailed explanation

Background Concept

If a quantity plotted against another gives a straight line through the origin, the relationship is proportional:

y = mx \Rightarrow y \propto x

Also, for motion with uniform acceleration from rest,

v = at

so velocity increases linearly with time.

Understanding the Question

You are given a graph of induced e.m.f. $E$ against time $t$ . The rod is said to be accelerated uniformly from rest. You must use these two facts to argue that $E$ is proportional to the rod’s velocity $v$ .

Approach

Read the shape of the $E$ – $t$ graph to infer how $E$ depends on $t$ .
Use the kinematics for uniform acceleration from rest to express $v$ in terms of $t$ .
Combine the two proportionalities.

Step-by-Step Reasoning

The graph is a straight line passing through $(0,0)$ . That means

E \propto t.

Uniform acceleration from rest means initial velocity $u=0$ and

v = u + at = at \Rightarrow v \propto t.

If $E \propto t$ and $v \propto t$ , then both are proportional to the same variable, so

E \propto v.

Key Takeaways

Straight line through origin on a graph indicates direct proportionality.
For uniform acceleration from rest, $v$ increases linearly with $t$ .
Combining proportionalities is a quick way to relate two quantities.

Common Mistakes

Saying “the graph is linear so $E$ is constant” (a linear graph does not mean constant; a horizontal line would).
Forgetting “from rest” and using $v = u + at$ without setting $u=0$ .
Claiming $E \propto t^2$ by confusing displacement ( $s \propto t^2$ ) with velocity ( $v \propto t$ ).

Things to Be Careful About

Proportionality ( $\propto$ ) is only justified here because the line passes through the origin.
Use the correct kinematics variable: velocity, not displacement.

Techniques used

interpret a straight-line graph through the origin as proportionalityuse the uniform-acceleration relation v = at to link variables

(ii)

Use Faraday’s law to show that the variation of $E$ with time $t$ is given by

E = Blat

where $a$ is the acceleration of the rod.

3M Similar

DifficultyMedium

Worked solution

Working

As the rod moves a distance $x$ , area swept out

A = lx

Flux linkage ( $N=1$ ):

\Phi = BA = Blx

Faraday’s law (magnitude):

E = \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \frac{\mathrm{d}(Blx)}{\mathrm{d}t} = Bl\frac{\mathrm{d}x}{\mathrm{d}t} = Blv

With uniform acceleration from rest, $v=at$ so

E = Blat

Answer

E = Blat

Final answer

E = Blat

Detailed explanation

Background Concept

Faraday’s law for a single conductor can be derived from the general statement

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

Here $N=1$ . If a straight rod of length $l$ moves sideways through a uniform magnetic field $B$ , it “cuts” magnetic field lines. One way to see the induced e.m.f. is to consider the area swept out by the rod: as the rod moves, the magnetic flux through that swept area changes.

Understanding the Question

A rod of length $l$ accelerates uniformly from rest in a uniform field of flux density $B$ . You must use Faraday’s law to show that the induced e.m.f. varies with time as

E = Blat.

So you need to connect $E$ to rate of change of flux, and then connect that to motion ( $v=at$ ).

Approach

Write the magnetic flux through the relevant area as $\Phi = BA$ .
Express the area $A$ in terms of the rod length $l$ and the distance moved $x$ .
Differentiate $\Phi$ with respect to time to get $E$ .
Replace $\mathrm{d}x/\mathrm{d}t$ with $v$ , and then use $v = at$ .

Step-by-Step Reasoning

diagram-bii-1

As the rod moves a distance $x$ , it sweeps out a rectangular area

A = lx.

In a uniform field perpendicular to the area, the magnetic flux is

\Phi = BA = B(lx) = Blx.

Using Faraday’s law for magnitude (the question is about the variation, so the sign is not essential here):

E = \frac{\mathrm{d}\Phi}{\mathrm{d}t}.

So

E = \frac{\mathrm{d}(Blx)}{\mathrm{d}t} = Bl\frac{\mathrm{d}x}{\mathrm{d}t}.

Recognise that $\mathrm{d}x/\mathrm{d}t = v$ :

E = Blv.

The rod starts from rest and accelerates uniformly, so

v = at.

Substitute into the expression for $E$ :

E = Bl(at) = Blat.

Key Takeaways

Flux changes because the area linked with the field changes as the rod moves.
Differentiate $\Phi = BA$ to get induced e.m.f.
For uniform acceleration from rest, $v=at$ makes $E$ increase linearly with time.

Common Mistakes

Using $A = l/t$ or other incorrect expressions for the swept area; it must be $A = lx$ .
Forgetting to differentiate: writing $E = Blx$ instead of $E = Bl\,\mathrm{d}x/\mathrm{d}t$ .
Mixing up $x$ and $v$ (displacement vs velocity).
Including the minus sign and then treating $E$ as negative without discussing direction; usually the magnitude is enough unless asked for polarity.

Things to Be Careful About

Ensure the field is perpendicular to the swept area so $\Phi = BA$ (i.e. $\cos\theta = 1$ ).
Keep symbols consistent: $x$ is displacement, $v$ is velocity.
State $N=1$ (single rod) if flux linkage form is used.

Techniques used

express magnetic flux as Φ = BA using swept areadifferentiate flux with respect to time to obtain induced e.m.f.substitute v = at for uniformly accelerated motion

(iii)

The length of the rod is $0.45\ \text{m}$ . The acceleration $a$ of the rod is $7.8\ \text{m s}^{-2}$ .

Determine the value of $B$ .

$B$ = ______ $\text{T}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 7.2, at $t = 2.0\ \text{s}$ , $E = 0.30\ \text{mV} = 3.0\times 10^{-4}\ \text{V}$ .

Using $E=Blat$ ,

B = \frac{E}{lat} = \frac{3.0\times 10^{-4}}{(0.45)(7.8)(2.0)} = 4.3\times 10^{-5}\ \text{T}

Answer

B = 4.3\times 10^{-5}\ \text{T}

Final answer

4.3 × 10^-5 T

Detailed explanation

Background Concept

Once you have an expression of the form

E = Blat,

you can determine the magnetic flux density $B$ if you know the e.m.f. $E$ at a particular time $t$ , along with $l$ and $a$ . The key practical point is to keep everything in SI units: metres, seconds, volts.

Understanding the Question

You are given $l = 0.45\ \text{m}$ and $a = 7.8\ \text{m s}^{-2}$ . From the graph of $E$ against $t$ , you must obtain a corresponding pair $(t, E)$ and use

E = Blat

to calculate $B$ .

Approach

Read a clear point from the straight line (the end point is easiest).
Convert $E$ from $\text{mV}$ to $\text{V}$ .
Rearrange for $B$ and substitute.

Step-by-Step Reasoning

From the graph, at $t = 2.0\ \text{s}$ ,

E = 0.30\ \text{mV} = 0.30\times 10^{-3}\ \text{V} = 3.0\times 10^{-4}\ \text{V}.

Rearrange $E = Blat$ :

B = \frac{E}{lat}.

Substitute values:

B = \frac{3.0\times 10^{-4}}{(0.45)(7.8)(2.0)}.

Calculate the denominator:

(0.45)(7.8)(2.0) = 7.02.

So

B = \frac{3.0\times 10^{-4}}{7.02} = 4.27\times 10^{-5}\ \text{T} \approx 4.3\times 10^{-5}\ \text{T}.

(Equivalently, you could use the gradient $E/t$ since $E/t = Bla$ for a straight line.)

Key Takeaways

Convert graph readings to SI units before substituting.
Rearranging a proportionality like $E = Blat$ is straightforward once one point is read.
A straight-line $E$ – $t$ graph allows using either a single point or the gradient.

Common Mistakes

Forgetting to convert $0.30\ \text{mV}$ to volts (leading to an answer $10^3$ too large).
Using $t=2$ but reading $E$ in volts directly as $0.30\ \text{V}$ .
Rearranging incorrectly (e.g. multiplying instead of dividing by $l$ , $a$ , or $t$ ).

Things to Be Careful About

Use $t$ in seconds, $l$ in metres, $a$ in $\text{m s}^{-2}$ , and $E$ in volts.
Quote $B$ in tesla and to an appropriate number of significant figures consistent with the graph reading (typically 2 s.f. here).

Techniques used

read a value from a straight-line graphrearrange E = Blat to solve for Bconvert millivolts to volts and use consistent SI units

Q8MediumQuantum Physics

(a)

State what is meant by a photon.

2M Similar

DifficultyEasy

Worked solution

Answer

A photon is a quantum (packet) of electromagnetic radiation.

Its energy is discrete and given by

E = hf

(where $f$ is the radiation frequency).

Final answer

A photon is a quantum (packet) of electromagnetic radiation with energy E = hf.

Detailed explanation

Background Concept

In quantum physics, electromagnetic (EM) radiation can be described as being made up of particles called photons. This is the “particle” side of wave–particle duality.

A photon carries:

Energy $E$ given by

E = hf

where $h$ is the Planck constant and $f$ is the frequency.

(Later, you also use that a photon has momentum even though it has no rest mass.)

Understanding the Question

The question asks for what “a photon” means. For 2 marks, Cambridge typically wants two distinct pieces of information: (1) it is a packet/quantum of EM radiation, and (2) its energy is quantised and related to frequency.

Approach

Give a definition-style sentence (what it is), then add the key quantitative property that characterises a photon in A-level work: $E = hf$ .

Step-by-Step Reasoning

Identify what kind of thing a photon is: a single “particle” of EM radiation (light, X-rays, etc.).
State the key quantised energy relationship used throughout the topic:

E = hf

This distinguishes the photon model from a purely classical wave model.

Key Takeaways

A photon is a quantum (packet) of EM radiation.
Photon energy is proportional to frequency: $E = hf$ .

Common Mistakes

Saying only “a particle of light” without mentioning quantisation/energy–frequency link (often loses the second mark).
Confusing photon with electron or other material particle.
Writing $E = hc/\lambda$ only; this is true but the core definition point is usually $E = hf$ .

Things to Be Careful About

Use the wording “packet” or “quantum” to indicate discreteness.
If you include an equation, ensure symbols are correct: $h$ is Planck constant, not $\hbar$ , and $f$ is frequency.

Techniques used

state the particle model of electromagnetic radiationrelate photon energy to frequency using E = hf

(b)

A laser emits red light of a single wavelength. The light is produced when electrons move from a higher energy level to a lower energy level. The difference in energy between the two levels is $1.96\ \text{eV}$ .

8M

(i)

Calculate the wavelength of the light.

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Energy difference:

E = 1.96\ \text{eV} = 1.96 \times 1.60 \times 10^{-19}\ \text{J} = 3.14 \times 10^{-19}\ \text{J}

Using $E = hc/\lambda$ :

\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.14 \times 10^{-19}} = 6.34 \times 10^{-7}\ \text{m}

Answer

$\lambda = 6.34 \times 10^{-7}\ \text{m}$

Final answer

6.34 × 10^-7 m

Detailed explanation

Background Concept

When an electron drops from a higher energy level to a lower one, it can emit a photon whose energy equals the energy difference between the levels.

Photon energy can be written as:

E = hf

and because $c = f\lambda$ , we can also write

E = \frac{hc}{\lambda}

This is often the quickest route from photon energy to wavelength.

Also, energy is sometimes given in electronvolts:

1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}

Understanding the Question

You are told the energy difference between the two electron levels is $1.96\ \text{eV}$ . The laser light has a single wavelength, meaning each emitted photon has the same energy $E = 1.96\ \text{eV}$ . The task is to find the wavelength $\lambda$ in metres.

Approach

Convert the photon energy from eV to J.
Use $E = hc/\lambda$ and rearrange to $\lambda = hc/E$ .
Substitute constants and compute.

Step-by-Step Reasoning

Convert eV to J:

E = 1.96\ \text{eV} = 1.96 \times 1.60 \times 10^{-19}\ \text{J} = 3.14 \times 10^{-19}\ \text{J}

Use the energy–wavelength form:

E = \frac{hc}{\lambda}

Rearrange:

\lambda = \frac{hc}{E}

Substitute $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.00 \times 10^8\ \text{m s}^{-1}$ :

\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.14 \times 10^{-19}} = 6.34 \times 10^{-7}\ \text{m}

This is in the red region of the visible spectrum (hundreds of nm), which is a good sense-check.

Key Takeaways

Emitted photon energy equals the energy level difference.
Convert $\text{eV} \to \text{J}$ before using SI constants.
Use $E = hc/\lambda$ for direct wavelength calculations.

Common Mistakes

Forgetting to convert eV to J (gives an answer wrong by a factor of $1.60 \times 10^{-19}$ ).
Using $E = hf$ but not connecting to wavelength via $c = f\lambda$ .
Giving wavelength in nm when the answer space requests metres.

Things to Be Careful About

Keep powers of ten consistent; this calculation is sensitive to exponent errors.
Quote the final wavelength to a sensible number of significant figures (typically 3 s.f. here).
Ensure $h$ and $c$ values are in SI units so that $\lambda$ comes out in metres.

Techniques used

convert electronvolts to joulesuse E = hc/\lambda to link photon energy and wavelengthrearrange an equation to make the required quantity the subject

(ii)

The power of the beam emitted by the laser is $1.0 \times 10^{-2}\ \text{W}$ .

Calculate the number of photons emitted per unit time by the laser.

number per unit time = ______ $\text{s}^{-1}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Energy per photon:

E = 3.14 \times 10^{-19}\ \text{J}

Power $P = 1.0 \times 10^{-2}\ \text{J s}^{-1}$ , so

\frac{N}{t} = \frac{P}{E} = \frac{1.0 \times 10^{-2}}{3.14 \times 10^{-19}} = 3.18 \times 10^{16}\ \text{s}^{-1}

Answer

$3.2 \times 10^{16}\ \text{s}^{-1}$

Final answer

3.2 × 10^16 s^-1

Detailed explanation

Background Concept

Power is the rate of energy transfer:

P = \frac{\Delta E}{\Delta t}

If a laser emits photons all of the same energy $E_{\gamma}$ , then in time $\Delta t$ the energy emitted is

\Delta E = N E_{\gamma}

So the emission rate is

\frac{N}{t} = \frac{P}{E_{\gamma}}

Understanding the Question

The laser beam has power $1.0 \times 10^{-2}\ \text{W}$ , meaning it emits $1.0 \times 10^{-2}\ \text{J}$ of energy each second. Each photon has energy equal to the level difference from part (i): $1.96\ \text{eV}$ (which is $3.14 \times 10^{-19}\ \text{J}$ ). You must find how many photons per second are needed to carry that power.

Approach

Use (or recalculate) the photon energy in joules.
Apply $N/t = P/E$ .

Step-by-Step Reasoning

Photon energy:

E_{\gamma} = 1.96\ \text{eV} = 3.14 \times 10^{-19}\ \text{J}

Photon rate:

\frac{N}{t} = \frac{P}{E_{\gamma}} = \frac{1.0 \times 10^{-2}}{3.14 \times 10^{-19}} = 3.18 \times 10^{16}\ \text{s}^{-1}

Round appropriately (often 2 s.f. is acceptable for a 1-mark answer):

\frac{N}{t} \approx 3.2 \times 10^{16}\ \text{s}^{-1}

Key Takeaways

Laser power tells you energy emitted each second.
Divide by energy per photon to get photons per second.

Common Mistakes

Using $E = 1.96$ directly as if it were joules (forgetting eV conversion).
Using wavelength directly without first getting photon energy.
Writing the unit incorrectly; it should be $\text{s}^{-1}$ .

Things to Be Careful About

Use consistent units: $P$ in watts ( $\text{J s}^{-1}$ ) and $E$ in joules.
Significant figures: power is given to 2 s.f., so the rate should not be over-precise.

Techniques used

use P = energy per unit timecompute photon emission rate using N/t = P/E

(iii)

The photons are incident normally on a surface. Half of the number of photons are absorbed by the surface, and half are reflected.

Determine the average force exerted by the beam of photons on the surface.

average force = ______ $\text{N}$

4M Similar

DifficultyMedium

Worked solution

Working

Photon momentum:

p = \frac{E}{c}

Photons per second:

\frac{N}{t} = \frac{P}{E}

Half absorbed: momentum change per photon $= p$ .

Half reflected: momentum change per photon $= 2p$ .

Total momentum change per second:

F = \frac{\Delta p}{\Delta t} = \left(\frac{N}{2t}\right)p + \left(\frac{N}{2t}\right)2p = \frac{3}{2}\left(\frac{N}{t}\right)p

Substitute $\dfrac{N}{t} = \dfrac{P}{E}$ and $p = \dfrac{E}{c}$ :

F = \frac{3}{2}\left(\frac{P}{E}\right)\left(\frac{E}{c}\right) = \frac{3P}{2c}

With $P = 1.0 \times 10^{-2}\ \text{W}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ :

F = \frac{3 \times 1.0 \times 10^{-2}}{2 \times 3.00 \times 10^{8}} = 5.0 \times 10^{-11}\ \text{N}

Answer

$5.0 \times 10^{-11}\ \text{N}$

Final answer

5.0 × 10^-11 N

Detailed explanation

Background Concept

Although photons have no rest mass, they carry momentum. For a photon:

p = \frac{E}{c}

When light hits a surface, the surface experiences a force because the photons’ momentum changes.

Force is rate of change of momentum:

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

For a beam incident normally:

Absorption: photon goes from momentum $+p$ (towards the surface) to $0$ (no photon after absorption), so

\Delta p = p

Reflection: photon reverses direction, from $+p$ to $-p$ , so

\Delta p = 2p

Understanding the Question

A laser beam (power $1.0 \times 10^{-2}\ \text{W}$ ) hits a surface at right angles. Half the photons are absorbed and half are reflected.

You must find the average force on the surface due to the photons’ momentum transfer.

Approach

Express the momentum change per photon for each case (absorbed vs reflected).
Multiply each momentum change by the number of photons per second in that category.
Add the two contributions to get total momentum change per second, which equals the force.
Use $N/t = P/E$ and $p = E/c$ to simplify so you only need $P$ and $c$ .

Step-by-Step Reasoning

Let the total photon emission/arrival rate be $N/t$ .
Split into two equal groups:

absorbed rate $= \dfrac{N}{2t}$
reflected rate $= \dfrac{N}{2t}$

Momentum change rates:

Absorbed: each photon transfers momentum $p$ to the surface, so contribution to force:

F_{\text{abs}} = \left(\frac{N}{2t}\right)p

Reflected: each photon transfers momentum $2p$ , so

F_{\text{refl}} = \left(\frac{N}{2t}\right)2p

Total force:

F = F_{\text{abs}} + F_{\text{refl}} = \left(\frac{N}{2t}\right)p + \left(\frac{N}{2t}\right)2p = \frac{3}{2}\left(\frac{N}{t}\right)p

Replace $\dfrac{N}{t}$ and $p$ :

\frac{N}{t} = \frac{P}{E}, \qquad p = \frac{E}{c}

So

F = \frac{3}{2}\left(\frac{P}{E}\right)\left(\frac{E}{c}\right) = \frac{3P}{2c}

Notice the photon energy cancels: for a given power, the force depends only on what fraction is reflected/absorbed.

Substitute values:

F = \frac{3 \times 1.0 \times 10^{-2}}{2 \times 3.00 \times 10^{8}} = 5.0 \times 10^{-11}\ \text{N}

Key Takeaways

Photon momentum: $p = E/c$ .
Force from light is due to momentum change per unit time.
Absorption gives $\Delta p = p$ ; reflection gives $\Delta p = 2p$ .
For a beam of power $P$ $P$ :
- all absorbed: $F = P/c$
- all reflected: $F = 2P/c$
- half absorbed & half reflected: $F = 1.5\,P/c$ .

Common Mistakes

Treating reflected photons as having momentum change $p$ instead of $2p$ .
Forgetting that only half the photons are in each category.
Using $p = hc/\lambda$ without consistent units or without linking to power.
Confusing energy with momentum and writing $F = P/E$ (wrong dimensions).

Things to Be Careful About

The direction is normal to the surface; since the question asks for magnitude (average force), a positive value is fine.
Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ and quote the unit of force as newtons.
Avoid double-counting: the factor $\tfrac{3}{2}$ comes from half at $p$ plus half at $2p$ .

Techniques used

use photon momentum p = E/capply change of momentum for absorption and reflectionuse force as rate of change of momentumuse P = NE to eliminate the photon energy

Q9MediumNuclear PhysicsMedical Physics

Polonium-193 ( $^{193}_{84}\text{Po}$ ) is an unstable nuclide. A nucleus of polonium-193 decays to a nucleus of lead-189 ( $^{189}_{82}\text{Pb}$ ) by emitting an alpha-particle.

(a)

Radioactive decay is both random and spontaneous.

State what is meant by:

2M

(i)

random

1M Similar

(ii)

spontaneous.

1M Similar

(b)

Define half-life.

1M Similar

(c)

Data for the binding energy per nucleon of the particles involved in the decay of a nucleus of polonium-193 are given in Table 9.1.

Table 9.1

particle	binding energy per nucleon / $\text{eV}$
$^{193}_{84}\text{Po}$	7.774
$^{189}_{82}\text{Pb}$	7.826
$^{4}_{2}\alpha$	7.074

Determine the energy, in $\text{eV}$ , released when a nucleus of polonium-193 decays into a nucleus of lead-189.

energy = ______ $\text{eV}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Total binding energy of $^{193}_{84}\text{Po}$ :

E_{\text{b,Po}} = 193 \times 7.774 = 1500.382

Total binding energy of products:

E_{\text{b,Pb}} = 189 \times 7.826 = 1479.114

E_{\text{b,}\alpha} = 4 \times 7.074 = 28.296

\Delta E = (1479.114 + 28.296 - 1500.382) \times 10^6 = 7.028 \times 10^6\ \text{eV}

Answer

7.03 \times 10^6\ \text{eV}

Final answer

7.03 × 10^6 eV

Detailed explanation

Background Concept

Binding energy is the energy required to separate a nucleus into free nucleons. The binding energy per nucleon allows you to find the total binding energy:

E_\text{b,total} = A \times (\text{binding energy per nucleon})

where $A$ is the nucleon (mass) number.

In nuclear decays and reactions, the energy released (the $Q$ -value) is related to the change in total binding energy:

If the products are more tightly bound (higher total binding energy), energy is released.

Understanding the Question

The decay is:

^{193}_{84}\text{Po} \rightarrow {}^{189}_{82}\text{Pb} + {}^{4}_{2}\alpha

You are given binding energy per nucleon values for the parent nucleus and the two products. You must determine the energy released in $\text{eV}$ .

Approach

Convert each “binding energy per nucleon” into total binding energy by multiplying by nucleon number.
Add total binding energies for the products.
Subtract the reactant total binding energy.
Express the result in $\text{eV}$ (here, multiplying by $10^6$ converts from MeV to eV, matching the expected nuclear energy scale).

Step-by-Step Reasoning

Parent nucleus total binding energy:

E_{\text{b,Po}} = 193 \times 7.774 = 1500.382

Daughter nucleus total binding energy:

E_{\text{b,Pb}} = 189 \times 7.826 = 1479.114

Alpha particle total binding energy:

E_{\text{b,}\alpha} = 4 \times 7.074 = 28.296

Total binding energy of products:

E_{\text{b,products}} = 1479.114 + 28.296 = 1507.410

Increase in binding energy (released energy):

\Delta E = (1507.410 - 1500.382) \times 10^6 = 7.028 \times 10^6\ \text{eV}

So about $7.0\ \text{MeV}$ is released, i.e. $\sim 7 \times 10^6\ \text{eV}$ .

Key Takeaways

Multiply binding energy per nucleon by $A$ to get total binding energy.
Energy released corresponds to an increase in total binding energy of the products.

Common Mistakes

Using proton number $Z$ instead of nucleon number $A$ .
Subtracting in the wrong order (giving a negative released energy).
Forgetting the alpha particle contributes to the products.
Reporting the answer without converting to $\text{eV}$ when requested.

Things to Be Careful About

Nuclear binding energies are typically in MeV per nucleon; check that your final magnitude is around a few MeV (i.e. $10^6\ \text{eV}$ ), not a few eV.
Keep consistent significant figures (typically 3 s.f.).

Techniques used

calculate total binding energy from binding energy per nucleonapply energy released = increase in total binding energyconvert from megaelectronvolts to electronvolts

(d)

A pure sample of polonium-193 contains $N_0$ nuclei. After a time $t$ the sample contains $N$ nuclei of polonium-193. The variation of $\ln(N/N_0)$ with $t$ is shown in Fig. 9.1.

Fig. 9.1

3M

(i)

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

1M Similar

(ii)

Use Fig. 9.1 to determine the half-life, in $\text{ms}$ , of polonium-193.

half-life = ______ $\text{ms}$

2M Similar

DifficultyMedium

Worked solution

Working

From the line, using $(t,\ln(N/N_0)) = (0.8\ \text{ms},-1.12)$ :

\text{gradient} = \frac{-1.12-0}{0.8-0} = -1.4\ \text{ms}^{-1}

So

\lambda = 1.4\ \text{ms}^{-1}

Half-life:

t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.4} = 0.495\ \text{ms}

Answer

0.50\ \text{ms}

Final answer

0.50 ms

Detailed explanation

Background Concept

For exponential decay,

N = N_0 e^{-\lambda t}

so

\ln\left(\frac{N}{N_0}\right) = -\lambda t

A plot of $\ln(N/N_0)$ against $t$ is a straight line with gradient $-\lambda$ .

Half-life is related to $\lambda$ by setting $N/N_0 = 1/2$ :

\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2}

and since $\ln(1/2) = -\ln 2$ ,

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

Understanding the Question

You are given a straight-line graph of $\ln(N/N_0)$ vs time $t$ (in ms). You must use the graph to find the half-life in ms.

Approach

Find the gradient of the line using two well-separated points.
Use $\text{gradient} = -\lambda$ to get $\lambda$ .
Use $t_{1/2} = \ln 2/\lambda$ .

Step-by-Step Reasoning

Choose two points on the line. The graph passes through $(0,0)$ and approximately $(0.8\ \text{ms}, -1.12)$ .
Gradient:

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{-1.12 - 0}{0.8 - 0} = -1.4\ \text{ms}^{-1}

Hence $\lambda = 1.4\ \text{ms}^{-1}$ .
Now calculate the half-life:

t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.4} = 0.495\ \text{ms}

Rounding appropriately gives $0.50\ \text{ms}$ .

Key Takeaways

Log plots turn exponentials into straight lines.
Gradient gives decay constant, then half-life follows from $\ln 2$ .

Common Mistakes

Using $t_{1/2} = 1/\lambda$ (incorrect; must include $\ln 2$ ).
Forgetting units and giving $\lambda$ in $\text{s}^{-1}$ while $t$ is in ms.
Calculating gradient as $\Delta x/\Delta y$ .

Things to Be Careful About

Use two far-apart points on the best-fit line, not two noisy plotted points.
Keep time units consistent: if gradient is in $\text{ms}^{-1}$ , then $t_{1/2}$ comes out in ms.

Techniques used

determine gradient from two points on a straight-line graphuse \ln(N/N0) = -\lambda t to find the decay constantcalculate half-life from t1/2 = ln2 / \lambda

(e)

Positron emission tomography (PET scanning) uses a radioactive tracer.

2M

(i)

State what happens to the positrons emitted by the tracer.

1M Similar

(ii)

Explain why a tracer with a half-life of approximately 2 hours is a suitable tracer to use.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

A half-life of about $2\ \text{h}$ is long enough for the tracer to be prepared and for the scan to be carried out, but short enough that it decays away soon after, reducing the radiation dose to the patient.

Final answer

Long enough for preparation/scan, short enough to reduce patient dose afterwards.

Detailed explanation

Background Concept

Half-life controls how quickly the activity falls:

A = A_0 e^{-\lambda t}

A suitable medical tracer must have:

sufficient activity for detection during the measurement time,
minimal time remaining radioactive in the patient (dose reduction).

Understanding the Question

You must explain why a tracer with half-life about 2 hours is suitable for PET. This is about practicality and safety.

Approach

State the trade-off:

not too short (otherwise activity drops too fast before/during the scan),
not too long (otherwise the patient remains radioactive for an unnecessarily long time).

Step-by-Step Reasoning

A PET scan and associated preparation/transport typically take on the order of minutes to a couple of hours.
If the half-life were much shorter (minutes), the activity might fall significantly before the scan completes, reducing count rate and image quality.
If the half-life were much longer (days), the tracer would remain in the body emitting radiation long after the scan, increasing dose without benefit.
A half-life of about 2 hours is a good compromise: adequate activity during the scan but rapid decay afterwards.

Key Takeaways

Suitable half-life balances usable signal with minimal dose.

Common Mistakes

Saying only “it is safe” without linking to half-life.
Claiming a longer half-life is always better (it increases dose).

Things to Be Careful About

You only need one clear explanation for 1 mark, but including both “long enough” and “short enough” makes the argument complete.

Techniques used

relate half-life to activity during the scanning timerelate half-life to patient radiation dose

Q10MediumAstronomy and CosmologyMotion in a Circle

(a)

4M

(i)

State what is meant by the luminosity of a star.

1M Similar

(ii)

Explain how standard candles are used to determine the distance to a galaxy.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

A standard candle has known luminosity $L$ .
Measure its apparent brightness / flux $F$ as observed from Earth.
Use

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

to calculate the distance $d$ to the galaxy.

Final answer

Use an object of known luminosity, measure its flux, and apply F = L/(4πd²) to find the distance.

Detailed explanation

Background Concept

For a source radiating equally in all directions, its total power $L$ spreads out over the surface area of a sphere of radius $d$ .

Surface area of sphere:

A = 4\pi d^2

So the power per unit area received (flux), $F$ , is:

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

A standard candle is an astronomical object whose luminosity (or absolute magnitude) is known reliably (e.g. Cepheid variables, Type Ia supernovae).

Understanding the Question

You must explain the method: identify a standard candle in the galaxy, use its known luminosity, measure how bright it appears (flux), and then compute the distance using the inverse-square relationship.

Approach

Choose/identify a standard candle in the galaxy.
Use its known luminosity $L$ .
Measure observed flux $F$ from Earth (from its apparent brightness).
Rearrange $F = L/(4\pi d^2)$ to solve for $d$ .

Step-by-Step Reasoning

From

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

rearrange:

d = \sqrt{\frac{L}{4\pi F}}

So if $L$ is known (standard candle calibration) and $F$ is measured, $d$ can be calculated.

Key Takeaways

Standard candles turn a brightness measurement into a distance measurement.
The essential physics is the inverse-square spreading of radiation.

Common Mistakes

Saying “compare brightness to get distance” without mentioning known luminosity.
Mixing up $L$ (intrinsic) and $F$ (observed).
Forgetting the $4\pi$ factor, or not using a square root when rearranging.

Things to Be Careful About

State clearly what is measured ( $F$ ) and what is known ( $L$ ).
The method assumes negligible absorption/extinction, or that it has been corrected for (not always required, but avoid claiming distance depends only on brightness with no caveats).

Techniques used

use a standard candle’s known luminositymeasure the observed flux from apparent brightnessapply the inverse-square law to determine distance

(b)

The Sun rotates on its axis. Points X, Y and Z are on the equator of the Sun as shown in Fig. 10.1.

Fig. 10.1

The wavelengths of light from points X and Y are observed and recorded in Table 10.1.

Table 10.1

observed wavelength from X / $\text{nm}$	observed wavelength from Y / $\text{nm}$
656.2877	656.2831

7M

(i)

The Sun rotates with a period of $2.07 \times 10^{6}\ \text{s}$ .

Show that the radius of the Sun is $6.93 \times 10^{8}\ \text{m}$ .

3M Similar

DifficultyMedium

Worked solution

Working

For point $Y$ , radial speed is $0$ , so emitted wavelength $\lambda \approx 656.2831\ \text{nm}$ .

Difference:

\Delta \lambda = 656.2877 - 656.2831 = 0.0046\ \text{nm}

Doppler shift:

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

v = c\frac{\Delta \lambda}{\lambda} = (3.00\times 10^8)\frac{0.0046}{656.2831} \approx 2.10\times 10^3\ \text{m s}^{-1}

Uniform circular motion:

v = \frac{2\pi r}{T} \;\Rightarrow\; r = \frac{vT}{2\pi}

r = \frac{(2.10\times 10^3)(2.07\times 10^6)}{2\pi} \approx 6.93\times 10^8\ \text{m}

Answer

$6.93 \times 10^8\ \text{m}$

Final answer

6.93 × 10^8 m

Detailed explanation

Background Concept

When a source of light moves relative to an observer, the observed wavelength changes due to the Doppler effect. For speeds much less than $c$ (true for the Sun’s rotation), the fractional wavelength shift is approximately

\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}

where $v$ is the component of the source velocity along the line of sight.

For uniform circular motion, the tangential speed at the equator is

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

where $T$ is the rotation period.

Understanding the Question

You are given observed wavelengths from two points on the Sun’s equator, $X$ (at the limb, moving away from the observer) and $Y$ (at the centre of the visible disc). You use the Doppler shift between them to find the equatorial speed $v$ , then use the period $T = 2.07\times 10^6\ \text{s}$ to find the radius $r$ .

diagram-10b-i

Approach

Treat the wavelength from $Y$ as (approximately) the emitted wavelength because the velocity component toward/away from us is zero at the disc centre.
Find $\Delta\lambda = \lambda_X - \lambda_Y$ .
Use $\Delta\lambda/\lambda = v/c$ to calculate $v$ for point $X$ .
Use $v = 2\pi r/T$ to solve for $r$ .

Step-by-Step Reasoning

Identify the unshifted wavelength. At $Y$ (centre of disc), the Sun’s rotation is sideways relative to the observer, so the radial component of velocity is zero. Hence the observed wavelength from $Y$ is essentially the emitted wavelength:

\lambda \approx \lambda_Y = 656.2831\ \text{nm}

Find the wavelength difference.

\Delta \lambda = \lambda_X - \lambda_Y = 656.2877 - 656.2831 = 0.0046\ \text{nm}

It is larger at $X$ , consistent with $X$ moving away (redshift).

Use Doppler relation to find speed at the equator.

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

So

v = c\frac{\Delta \lambda}{\lambda} = (3.00\times 10^8)\frac{0.0046}{656.2831}

v \approx 2.10\times 10^3\ \text{m s}^{-1}

Relate equatorial speed to radius using the period.

v = \frac{2\pi r}{T} \Rightarrow r = \frac{vT}{2\pi}

Substitute $T = 2.07\times 10^6\ \text{s}$ :

r = \frac{(2.10\times 10^3)(2.07\times 10^6)}{2\pi} \approx 6.93\times 10^8\ \text{m}

This matches the required value.

Key Takeaways

Doppler shift gives the radial component of velocity.
The centre of the Sun’s disc has (approximately) zero radial velocity due to rotation.
Combine Doppler speed with $v = 2\pi r/T$ to obtain the radius.

Common Mistakes

Using the wrong reference wavelength (e.g. averaging $\lambda_X$ and $\lambda_Y$ without justification).
Forgetting that Doppler shift uses the component of velocity along the line of sight.
Using $r = vT$ (missing the $2\pi$ ).
Converting $0.0046\ \text{nm}$ incorrectly (it is $4.6\times 10^{-3}\ \text{nm}$ ).

Things to Be Careful About

Keep units consistent: ratios like $\Delta\lambda/\lambda$ are unitless, so nm can be used directly (no need to convert to m if both are in nm).
Use $c = 3.00\times 10^8\ \text{m s}^{-1}$ and quote the final radius in standard form with appropriate significant figures.

Techniques used

apply the Doppler shift relation for small speedscalculate tangential speed from wavelength shiftuse v = 2 pi r / T to determine radius

(ii)

State and explain how the expected wavelength of the light observed from Z compares with the emitted wavelength.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The observed wavelength from $Z$ is less than the emitted wavelength (blueshift), because $Z$ is moving towards the observer so successive wavefronts are received closer together.

Final answer

Observed wavelength from Z is smaller than the emitted wavelength (blueshift) because Z is moving towards the observer.

Detailed explanation

Background Concept

Doppler effect for light:

If the source moves away, wavefronts are stretched and the observed wavelength increases (redshift).
If the source moves towards, wavefronts are compressed and the observed wavelength decreases (blueshift).

For small speeds compared with $c$ :

\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}

and the sign of $v$ (towards/away) determines whether $\lambda$ increases or decreases.

Understanding the Question

Point $Z$ is on the opposite limb to $X$ . From the rotation direction shown, $Z$ has a velocity component towards the observer. You must state how the observed wavelength compares to the emitted wavelength and explain using Doppler shift.

Approach

Determine whether $Z$ moves towards or away along the line of sight, then state blueshift or redshift accordingly.

Step-by-Step Reasoning

Rotation means opposite limbs have opposite radial velocity components.
Since $X$ is moving away (given by longer observed wavelength at $X$ than at $Y$ ), the opposite point $Z$ must be moving towards.
Moving towards compresses wavefront spacing, so the observed wavelength is smaller than the emitted wavelength.

Key Takeaways

Towards observer $\Rightarrow$ blueshift ( $\lambda_{\text{obs}} < \lambda_{\text{em}}$ ).
Away from observer $\Rightarrow$ redshift ( $\lambda_{\text{obs}} > \lambda_{\text{em}}$ ).

Common Mistakes

Reversing redshift/blueshift.
Comparing with the wavelength from $X$ or $Y$ instead of explicitly comparing with the emitted wavelength.

Things to Be Careful About

Use the wording “observed wavelength is less than emitted wavelength” (or “blueshift”) and include the reason: motion towards the observer causing wavefronts to be closer together.

Techniques used

identify motion towards or away from the observerapply qualitative Doppler shift reasoningcompare observed and emitted wavelengths

(iii)

The luminosity of the Sun is $3.8 \times 10^{26}\ \text{W}$ .

Use the information in (b)(i) to calculate the surface temperature of the Sun.

temperature = ______ $\text{K}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Use Stefan–Boltzmann:

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

T = \left(\frac{L}{4\pi r^2 \sigma}\right)^{1/4}

With $L = 3.8\times 10^{26}\ \text{W}$ , $r = 6.93\times 10^8\ \text{m}$ , $\sigma = 5.67\times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ :

T = \left(\frac{3.8\times 10^{26}}{4\pi(6.93\times 10^8)^2(5.67\times 10^{-8})}\right)^{1/4} \approx 5.8\times 10^3\ \text{K}

Answer

$5.8 \times 10^3\ \text{K}$

Final answer

5.8 × 10^3 K

Detailed explanation

Background Concept

If a star is approximated as a black body, the power radiated per unit area is given by the Stefan–Boltzmann law:

\frac{P}{A} = \sigma T^4

For a spherical star of radius $r$ , the surface area is $A = 4\pi r^2$ , so its luminosity is

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

where:

$L$ is luminosity (total power),
$\sigma = 5.67\times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ ,
$T$ is surface temperature.

Understanding the Question

You are given the Sun’s luminosity and (from part (b)(i)) its radius. You must use these to find the surface temperature by rearranging Stefan–Boltzmann.

Approach

Start from $L = 4\pi r^2\sigma T^4$ .
Rearrange for $T$ .
Substitute $L$ and $r$ and evaluate the fourth root.

Step-by-Step Reasoning

Start with

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

Rearrange:

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

and hence

T = \left(\frac{L}{4\pi r^2\sigma}\right)^{1/4}

Substitute values:

$L = 3.8\times 10^{26}\ \text{W}$
$r = 6.93\times 10^8\ \text{m}$
$\sigma = 5.67\times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$

Compute the denominator:

$r^2 \approx (6.93\times 10^8)^2 \approx 4.80\times 10^{17}\ \text{m}^2$
$4\pi r^2 \approx 6.0\times 10^{18}\ \text{m}^2$
$4\pi r^2\sigma \approx 6.0\times 10^{18}\times 5.67\times 10^{-8} \approx 3.4\times 10^{11}$

So

T^4 \approx \frac{3.8\times 10^{26}}{3.4\times 10^{11}} \approx 1.1\times 10^{15}

Taking the fourth root gives

T \approx 5.8\times 10^3\ \text{K}

(consistent with the known solar surface temperature about $5800\ \text{K}$ ).

Key Takeaways

Use the whole surface area $4\pi r^2$ when connecting luminosity to temperature.
Temperature comes from a fourth-root, so large changes in $L$ or $r$ produce smaller fractional changes in $T$ .

Common Mistakes

Using $L = \sigma T^4$ (missing the area factor).
Using diameter instead of radius.
Arithmetic error with indices when squaring $r$ or handling $10^x$ .

Things to Be Careful About

Keep $r$ in metres and use the correct value of $\sigma$ with units.
A fourth root is not the same as a square root; do it as two successive square roots if needed.
Quote $T$ to appropriate significant figures (typically 2 s.f. or 3 s.f. depending on given data).

Techniques used

apply the Stefan–Boltzmann law for a spherical radiatorsubstitute radius into L = 4 pi r^2 sigma T^4rearrange and evaluate a fourth root

Physics 9702/42 — February/March 2025

Answer

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