(a) The list below shows some SI quantities. Underline the quantity that is not an SI base quantity. charge \quad \quad current \quad \quad length \quad \quad time (b) A square solar panel with sides of length $1300\ \text{mm}$ is shown in Fig. 1.1. Light is incident normally on the solar panel. (i) The power of the light incident on the solar panel is $750\ \text{W}$. Calculate the intensity of the light. intensity = \_\_\_\_\_\_ $\text{W m}^{-2}$ (ii) The percentage…

(a) charge · (b)(i) 4.44 × 10^2 W m^-2 · (b)(ii) 3.8 % · (b)(iii) 21.3 % · (b)(iv) Greater than the original.

9702/22 May/June 2024 — Physics Solved Past Paper

Q1MediumPhysical Quantities and UnitsWavesWork, Energy and Power

(a)

The list below shows some SI quantities.

Underline the quantity that is not an SI base quantity.

charge \quad \quad current \quad \quad length \quad \quad time

1M Similar

(b)

A square solar panel with sides of length $1300\ \text{mm}$ is shown in Fig. 1.1.

Fig. 1.1

Light is incident normally on the solar panel.

9M

(i)

The power of the light incident on the solar panel is $750\ \text{W}$ .

Calculate the intensity of the light.

intensity = ______ $\text{W m}^{-2}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Side length $= 1300\ \text{mm} = 1.300\ \text{m}$

A = (1.300)^2 = 1.69\ \text{m}^2

I = \frac{P}{A} = \frac{750}{1.69} = 4.44 \times 10^2\ \text{W m}^{-2}

Answer

$4.44 \times 10^2\ \text{W m}^{-2}$

Final answer

4.44 × 10^2 W m^-2

Detailed explanation

Background Concept

Intensity $I$ is defined as the power transferred per unit area perpendicular to the direction of energy flow:

I = \frac{P}{A}

For light incident normally (perpendicular) onto a flat surface, the relevant area is simply the surface area of the panel.

Understanding the Question

You are told:

incident power $P = 750\ \text{W}$
square panel side length $1300\ \text{mm}$
light is incident normally

You must find the intensity $I$ in $\text{W m}^{-2}$ .

Approach

Convert the side length from mm to m.
Calculate the area of the square: $A = L^2$ .
Use $I = P/A$ .

Step-by-Step Reasoning

Convert units:

1300\ \text{mm} = 1300 \times 10^{-3}\ \text{m} = 1.300\ \text{m}

Area of the square panel:

A = L^2 = (1.300\ \text{m})^2 = 1.69\ \text{m}^2

Intensity:

I = \frac{750\ \text{W}}{1.69\ \text{m}^2} = 4.44 \times 10^2\ \text{W m}^{-2}

Key Takeaways

Normal incidence means you use the full panel area.
Always convert to SI units before substituting.
Intensity has units $\text{W m}^{-2}$ .

Common Mistakes

Using $1300$ as metres instead of millimetres.
Forgetting to square the side length when finding area.
Giving intensity in $\text{W mm}^{-2}$ or without units.

Things to Be Careful About

Significant figures: inputs suggest 3 s.f. is appropriate.
If light were not normal, you would need the projected area; the question explicitly avoids that by stating “normally.”

Techniques used

convert millimetres to metrescalculate area from side lengthuse intensity as power per unit area

(ii)

The percentage uncertainty in the incident power is $\pm 3\%$ .
The uncertainty in the length of each side is $\pm 5\ \text{mm}$ .

Calculate the percentage uncertainty in the intensity of the light.

percentage uncertainty = ______ $\%$

2M Similar

DifficultyMedium

Worked solution

Working

I = \frac{P}{L^2}

Percentage uncertainty in $L$ :

\frac{5}{1300} \times 100 = 0.385\%

So percentage uncertainty in $L^2$ :

2(0.385\%) = 0.770\%

Total percentage uncertainty in $I$ :

3.0\% + 0.770\% = 3.77\% \approx 3.8\%

Answer

$3.8\%$

Final answer

3.8 %

Detailed explanation

Background Concept

When calculating uncertainty in a derived quantity:

For multiplication/division, percentage uncertainties add.
For a power, the percentage uncertainty is multiplied by the power.

If

I = \frac{P}{L^2}

then (percentage uncertainty in $I$ ) = (percentage uncertainty in $P$ ) + (percentage uncertainty in $L^2$ ).

Understanding the Question

You are given:

percentage uncertainty in incident power $P$ : $\pm 3\%$
absolute uncertainty in each side length $L$ : $\pm 5\ \text{mm}$

You must find the percentage uncertainty in intensity $I$ .

Approach

Express $I$ in terms of $P$ and $L$ .
Convert the absolute uncertainty in $L$ into a percentage uncertainty.
Double it because the area depends on $L^2$ .
Add the percentage uncertainties.

Step-by-Step Reasoning

Since the area of a square is $A = L^2$ , and $I=P/A$ :

I = \frac{P}{L^2}

Percentage uncertainty in $L$ :

\%\ \text{uncertainty in }L = \frac{\Delta L}{L} \times 100

Using $\Delta L = 5\ \text{mm}$ and $L = 1300\ \text{mm}$ (same units is fine because it is a ratio):

\frac{5}{1300} \times 100 = 0.385\%

Because $L$ is squared, the percentage uncertainty in $L^2$ is doubled:

\%\ \text{uncertainty in }L^2 = 2 \times 0.385\% = 0.770\%

Combine with the $3\%$ for $P$ (division means add):

\%\ \text{uncertainty in }I = 3.0\% + 0.770\% = 3.77\% \approx 3.8\%

Key Takeaways

If $y \propto x^n$ , then percentage uncertainty in $y$ from $x$ is $n$ times that of $x$ .
For $I = P/L^2$ , the length uncertainty contributes twice.

Common Mistakes

Adding $3\%$ and $0.385\%$ without doubling for $L^2$ .
Using $\pm 5\ \text{mm}$ as if it were $\pm 5\%$ .
Subtracting uncertainties because one quantity is in the denominator (you still add percentage uncertainties).

Things to Be Careful About

Use the uncertainty for each side; since it is the same $L$ used twice (as $L^2$ ), the factor of 2 accounts for both.
Round appropriately; $3.77\%$ to $3.8\%$ is sensible.

Techniques used

identify how a derived quantity depends on measured quantitiescombine percentage uncertainties for products and powersconvert absolute uncertainty to percentage uncertainty

(iii)

The useful power output of the solar panel is $160\ \text{W}$ .

Calculate the percentage efficiency of the solar panel.

efficiency = ______ $\%$

1M Similar

(iv)

Another square solar panel is placed so that light of the same intensity is incident normally on it. The new panel has shorter sides than the original panel. The new panel has the same power output as the original panel.

State and explain whether the efficiency of the new panel is greater than, less than or the same as the efficiency of the original panel.

3M Similar

DifficultyMedium

Worked solution

Answer

Efficiency is greater.

Same intensity $I$ and smaller area $A$ means incident power $P_{\text{in}} = IA$ is smaller.
Useful output power is the same, so

\eta = \frac{P_{\text{out}}}{P_{\text{in}}}

increases when $P_{\text{in}}$ decreases.

Final answer

Greater than the original.

Detailed explanation

Background Concept

Two key ideas:

Intensity-power-area relation for normal incidence:

P_{\text{in}} = IA

Efficiency of a power-converting device:

\eta = \frac{P_{\text{out}}}{P_{\text{in}}}

If $P_{\text{out}}$ stays the same but $P_{\text{in}}$ decreases, the ratio increases.

Understanding the Question

A second square panel is used:

light has the same intensity and is incident normally
the new panel has shorter sides (so smaller area)
the new panel has the same power output as the original panel

You must decide whether its efficiency is greater/less/same, and explain.

Approach

Use $P_{\text{in}} = IA$ to compare input powers.
Use $\eta = P_{\text{out}}/P_{\text{in}}$ to compare efficiencies.

Step-by-Step Reasoning

diagram-biv-1

Same intensity $I$ for both panels.
New panel has shorter side length, so its area $A$ is smaller.
Therefore its incident power is smaller:

P_{\text{in,new}} = I A_{\text{new}} < I A_{\text{old}} = P_{\text{in,old}}

But the useful output power is stated to be the same:

P_{\text{out,new}} = P_{\text{out,old}}

Compare efficiencies:

\eta_{\text{new}} = \frac{P_{\text{out,new}}}{P_{\text{in,new}}} \quad\text{and}\quad \eta_{\text{old}} = \frac{P_{\text{out,old}}}{P_{\text{in,old}}}

Since the numerators are equal but $P_{\text{in,new}}$ is smaller, $\eta_{\text{new}}$ must be larger.

So the new panel has greater efficiency.

Key Takeaways

For normal incidence, input power depends on area: $P_{\text{in}} = IA$ .
Efficiency increases if the same output is achieved from a smaller input.

Common Mistakes

Saying efficiency is the same because “intensity is the same.” (Intensity being the same does not mean input power is the same; area matters.)
Saying efficiency is less because the panel is smaller (size alone does not decide efficiency; the ratio of output to input does).
Forgetting to mention $P_{\text{in}} = IA$ in the explanation.

Things to Be Careful About

The word normally is important: it lets you use the panel area directly (no projection factor).
The conclusion relies on comparing ratios, not on calculating any new numbers.
In real life, it may be unrealistic for a smaller panel to have the same power output at the same intensity, but you must accept the statement and deduce the efficiency consequence.

Techniques used

relate incident power to intensity and areacompare efficiencies using proportional reasoningapply efficiency definition to a changed situation

Q2MediumKinematicsDynamics

A skydiver jumps from an aircraft at time $t = 0$ and falls vertically downwards. The variation with $t$ of her velocity $v$ is shown in Fig. 2.1.

Fig. 2.1

(a)

3M

(i)

Using Fig. 2.1, state the terminal velocity of the skydiver.

terminal velocity = ______ $\text{m s}^{-1}$

1M Similar

(ii)

By drawing a suitable line on Fig. 2.1, determine the acceleration of the skydiver at time $t = 9.0\ \text{s}$ .

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Acceleration at $t=9.0\ \text{s}$ is the gradient of the tangent to the $v$ – $t$ curve.

Using two points on the tangent (example readings): $(t, v) = (7.0\ \text{s}, 28\ \text{m s}^{-1})$ and $(11.0\ \text{s}, 35\ \text{m s}^{-1})$ ,

a = \frac{\Delta v}{\Delta t} = \frac{35 - 28}{11.0 - 7.0} = \frac{7}{4.0} = 1.75\ \text{m s}^{-2}

Answer

$\boxed{a \approx 1.8\ \text{m s}^{-2}}$

Final answer

1.8 m s^-2

Detailed explanation

Background Concept

Acceleration is the rate of change of velocity:

a = \frac{\mathrm{d}v}{\mathrm{d}t}

On a velocity–time graph:

The gradient (slope) gives acceleration.
For a curved graph, the acceleration at a particular time is the gradient of the tangent at that time (instantaneous gradient).

Understanding the Question

The skydiver’s velocity changes with time, and the $v$ – $t$ graph is curved. You are asked to find the acceleration at $t=9.0\ \text{s}$ , which means you must estimate the instantaneous gradient at that time by drawing a tangent.

Approach

Locate $t=9.0\ \text{s}$ on the time axis and the corresponding point on the curve.
Draw a straight tangent line that just touches the curve at that point.
Pick two well-separated points on the tangent (not necessarily on the curve) and read their coordinates.
Use $a = \Delta v / \Delta t$ .

Step-by-Step Reasoning

At early times the gradient is large (close to $g$ ) because drag is small.
By $t=9\ \text{s}$ drag has increased, so the gradient is smaller.
After drawing the tangent at $t=9\ \text{s}$ , choose two points far apart on the tangent to reduce percentage reading error.
Example: if the tangent passes through about $(7\ \text{s}, 28\ \text{m s}^{-1})$ and $(11\ \text{s}, 35\ \text{m s}^{-1})$ then

a = \frac{35-28}{11-7} = 1.75\ \text{m s}^{-2} \approx 1.8\ \text{m s}^{-2}.

Any reasonable pair of points from a correctly drawn tangent should give a similar value.

Key Takeaways

Gradient of a $v$ – $t$ graph is acceleration.
For a curve, draw a tangent to find instantaneous acceleration.

Common Mistakes

Using two points on the curve around $t=9\ \text{s}$ (a chord) instead of a tangent, giving an average acceleration rather than instantaneous.
Choosing points too close together, making the gradient very sensitive to reading uncertainty.
Mixing up axes and calculating $\Delta t/\Delta v$ .

Things to Be Careful About

Make sure the tangent touches the curve at the correct point ( $t=9\ \text{s}$ ).
Read coordinates from the tangent, not from the original curve unless they lie on the tangent.
Quote the unit $\text{m s}^{-2}$ and an appropriate number of significant figures.

Techniques used

draw a tangent to a curve at a specified pointcalculate gradient using two well-separated pointsinterpret gradient of a velocity–time graph as acceleration

(b)

The mass of the skydiver and her equipment is $68\ \text{kg}$ . The upthrust on the skydiver is negligible.

After reaching terminal velocity, the skydiver opens her parachute at time $t_1$ . A total drag force of $1800\ \text{N}$ acts on the skydiver.

Determine the magnitude and direction of the acceleration of the skydiver at time $t_1$ .

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

3M Similar

DifficultyMedium

Worked solution

Working

Weight:

W = mg = 68 \times 9.81 = 6.67 \times 10^2\ \text{N}

Drag after opening parachute: $1800\ \text{N}$ upwards.

Resultant force (upwards):

F = 1800 - 667 = 1.13 \times 10^3\ \text{N}

Acceleration:

a = \frac{F}{m} = \frac{1.13 \times 10^3}{68} = 16.7\ \text{m s}^{-2}

Answer

acceleration $\approx 17\ \text{m s}^{-2}$

direction: upwards

Final answer

17 m s^-2 upwards

Detailed explanation

Background Concept

For vertical motion with forces:

Weight $W = mg$ acts downward.
Drag acts opposite to the direction of motion (here upward while she is moving downward).
Upthrust is stated negligible.

Newton’s second law:

\sum F = ma

The direction of $a$ is the direction of the resultant force.

Understanding the Question

At time $t_1$ the skydiver has already reached terminal velocity, so she is moving downward at constant speed just before the parachute opens. At $t_1$ the parachute opens and a drag force of $1800\ \text{N}$ acts. You must find the acceleration immediately at that moment, including its direction.

Approach

Draw up the forces: weight downward, drag upward.
Calculate weight $mg$ .
Find the resultant force (taking upward or downward as positive, but be consistent).
Use $a = F/m$ .
State the direction based on the sign/resultant.

Step-by-Step Reasoning

Mass $m = 68\ \text{kg}$ .
Weight is

W = mg = 68\times 9.81 \approx 667\ \text{N (downward)}.

Drag is $1800\ \text{N}$ upward.
Resultant force upward because drag is larger than weight:

F_{\text{res}} = 1800 - 667 = 1133\ \text{N upward}.

Then

a = \frac{F_{\text{res}}}{m} = \frac{1133}{68} \approx 16.7\ \text{m s}^{-2}.

Since the resultant is upward, the acceleration is upward. (This means she is slowing down while still moving downward.)

Key Takeaways

Terminal velocity means forces balance (before opening the parachute), but when drag changes suddenly, they no longer balance.
Acceleration direction follows the resultant force direction.

Common Mistakes

Adding forces ( $1800 + mg$ ) instead of subtracting because they act in opposite directions.
Stating the acceleration is downward because she is moving downward (velocity direction is not the same as acceleration direction).
Forgetting that upthrust is negligible (so do not include it).

Things to Be Careful About

Keep a clear sign convention (e.g. upward positive).
Use $g = 9.81\ \text{m s}^{-2}$ unless the paper specifies otherwise.
Give both magnitude and direction as requested.

Techniques used

identify forces acting and their directionscalculate weight using W = mgapply Newton's second law to find acceleration from resultant force

(c)

The parachute is fully open at time $t_2$ . At a later time $t_3$ the skydiver reaches a constant velocity of $5.7\ \text{m s}^{-1}$ .

4M

(i)

Describe and explain the variation with time of the magnitude of her acceleration between time $t_2$ and time $t_3$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

From $t_2$ to $t_3$ , the magnitude of the acceleration decreases with time and tends to zero.

Just after $t_2$ the drag is greater than weight, so there is a large upward resultant force (upward acceleration). As the skydiver slows down, drag decreases, so the resultant force decreases and hence the acceleration decreases. At $t_3$ , drag equals weight so the resultant force is zero and the acceleration is zero (new terminal velocity).

Final answer

Acceleration magnitude decreases to zero as drag falls to equal weight.

Detailed explanation

Background Concept

Drag (air resistance) increases with speed (often approximately proportional to $v$ or $v^2$ ). For vertical motion:

Weight $mg$ is constant.
Drag changes as the speed changes.

Acceleration depends on the resultant force:

a = \frac{F_{\text{res}}}{m}

A constant velocity (terminal velocity) occurs when $F_{\text{res}} = 0$ so $a=0$ .

Understanding the Question

At $t_2$ the parachute is fully open. The skydiver then slows down until at $t_3$ she moves at a constant speed of $5.7\ \text{m s}^{-1}$ . You must describe what happens to the magnitude of acceleration over this time and explain it using forces.

Approach

Consider forces just after $t_2$ : drag is very large because speed is still relatively high and parachute area is large.
As she slows, drag reduces.
Use Newton’s second law to relate changing resultant force to changing acceleration.
At $t_3$ constant velocity implies zero acceleration and balanced forces.

Step-by-Step Reasoning

Immediately after $t_2$ , the parachute is fully deployed, so for the current (still large) downward speed, the drag force is much larger than before.
Drag acts upward, opposing the motion, while weight $mg$ acts downward.
Since drag $> mg$ initially, the resultant force is upward, so acceleration is upward (she decelerates while moving downward).
As her speed decreases, drag decreases (because drag depends on speed). Weight stays constant.
Therefore the upward resultant force $F_{\text{res}} = D - mg$ becomes smaller in magnitude as $D$ falls.
Hence the magnitude of acceleration $|a| = |F_{\text{res}}|/m$ decreases with time.
At $t_3$ she reaches a constant speed: this is a new (lower) terminal velocity where drag equals weight, $D = mg$ , so $F_{\text{res}}=0$ and $a=0$ .

Key Takeaways

Changing speed changes drag; changing drag changes resultant force.
Acceleration reduces to zero as the forces become balanced at a new terminal velocity.

Common Mistakes

Saying acceleration is constant between $t_2$ and $t_3$ (it is not, because drag changes as speed changes).
Saying acceleration becomes zero because “she has stopped accelerating” without linking it to forces balancing.
Confusing the direction of velocity (down) with direction of acceleration (up during slowing down).

Things to Be Careful About

The question asks for the magnitude of acceleration: you should still make clear the direction in the explanation (upward) to show understanding.
Terminal velocity can occur at more than one speed (before and after the parachute opens).

Techniques used

relate acceleration to resultant forceexplain how drag force depends on speeddescribe approach to a new terminal velocity

(ii)

Calculate the change in momentum of the skydiver between time $t_1$ and time $t_3$ .

change in momentum = ______ $\text{N s}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Take downward as positive.

At $t_1$ , $v_1 \approx 39\ \text{m s}^{-1}$ .

At $t_3$ , $v_3 = 5.7\ \text{m s}^{-1}$ .

\Delta p = m(v_3 - v_1) = 68(5.7 - 39) = -2.26 \times 10^3\ \text{N s}

Answer

$\Delta p \approx -2.3 \times 10^3\ \text{N s}$ (i.e. a change of $2.3 \times 10^3\ \text{N s}$ upwards).

Final answer

-2.3 × 10^3 N s

Detailed explanation

Background Concept

Linear momentum is

p = mv

Change in momentum is

\Delta p = p_\text{final} - p_\text{initial} = m(v_\text{final} - v_\text{initial})

Momentum is a vector: direction matters. In one-dimensional vertical motion, we handle direction with a sign convention.

Understanding the Question

You need the change in the skydiver’s momentum between:

$t_1$ : just after reaching the original terminal velocity (from the graph), so $v_1$ is the terminal velocity from part (a).
$t_3$ : later constant velocity $5.7\ \text{m s}^{-1}$ downward.

Mass is constant at $68\ \text{kg}$ .

Approach

Choose a positive direction (e.g. downward positive).
Write initial and final velocities with correct sign.
Use $\Delta p = m(v_3 - v_1)$ .
Interpret the sign of the answer as the direction of the change in momentum.

Step-by-Step Reasoning

Choose downward as positive.
From the plateau on the $v$ – $t$ graph, terminal velocity is about $v_1 \approx 39\ \text{m s}^{-1}$ (downward, so $+39$ ).
At $t_3$ , the skydiver is still moving downward at $v_3 = 5.7\ \text{m s}^{-1}$ (so $+5.7$ ).
Compute:

\Delta p = 68(5.7 - 39) = 68(-33.3) = -2264\ \text{kg m s}^{-1}.

Since $1\ \text{kg m s}^{-1} = 1\ \text{N s}$ ,

\Delta p \approx -2.3\times 10^3\ \text{N s}.

The negative sign (with downward taken as positive) means the change in momentum is upward: the skydiver’s downward momentum has decreased.

Key Takeaways

Use $\Delta p = m\Delta v$ for constant mass.
Include direction via a sign convention; the sign tells you the direction of the change.

Common Mistakes

Using $\Delta p = mv$ with only one velocity (forgetting to subtract initial from final).
Using the wrong initial velocity (not the terminal velocity at $t_1$ ).
Giving only a positive number without stating direction when the sign is meaningful.

Things to Be Careful About

Keep the same direction convention throughout.
Quote the unit $\text{N s}$ (or $\text{kg m s}^{-1}$ ).
Use the terminal velocity value consistent with the graph reading (your final value will depend slightly on that reading).

Techniques used

use p = mv to relate momentum to velocitycompute change in momentum using \Delta p = m\Delta vuse a consistent sign convention for direction

Q3MediumElectricityDeformation of Solids

Lightning occurs when charge builds up in the atmosphere, creating a potential difference between the ground and the atmosphere.

During a lightning strike there is an average current of $3.3 \times 10^4\ \text{A}$ for a time of $2.6 \times 10^{-5}\ \text{s}$ .

(a)

Calculate the charge transferred during the lightning strike.

charge = ______ $\text{C}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Q = It

Q = (3.3 \times 10^{4})(2.6 \times 10^{-5}) = 8.58 \times 10^{-1}\ \text{C}

Answer

Q = 0.86\ \text{C}

Final answer

0.86 C

Detailed explanation

Background Concept

Electric current $I$ is defined as the rate of flow of charge $Q$ :

I = \frac{\Delta Q}{\Delta t}

If the current is approximately constant over a time interval $t$ , then the charge transferred is

Q = It

Understanding the Question

You are given an average current $I = 3.3 \times 10^{4}\ \text{A}$ that flows for a time $t = 2.6 \times 10^{-5}\ \text{s}$ during a lightning strike. The question asks for the total charge transferred.

Approach

Use $Q = It$ because current is charge per unit time. Multiply the given current by the time duration, keeping powers of ten consistent.

Step-by-Step Reasoning

Start with

Q = It

Substitute values:

Q = (3.3 \times 10^{4})(2.6 \times 10^{-5})

Multiply the numbers and the powers of ten separately:

3.3 \times 2.6 = 8.58

10^{4} \times 10^{-5} = 10^{-1}

So

Q = 8.58 \times 10^{-1}\ \text{C} = 0.858\ \text{C}

Given data are to 2 s.f., so quote $Q$ to 2 s.f.:

Q = 0.86\ \text{C}

Key Takeaways

Use $Q = It$ for charge transferred when current and time are known.
Handle standard form carefully: multiply coefficients and add powers of ten.

Common Mistakes

Using $Q = \frac{I}{t}$ (inverting the formula).
Losing the power of ten: $10^{4} \times 10^{-5} = 10^{-1}$ , not $10^{9}$ or $10^{-9}$ .
Giving an answer with no unit (must be coulombs, C).

Things to Be Careful About

“Average current” implies you can use $Q = It$ directly.
Significant figures: match the least precise given value (here 2 s.f.).

Techniques used

apply the definition of electric current as charge flow ratesubstitute values in standard formuse correct significant figures and units

(b)

The potential difference between the ground and the atmosphere is $3.0 \times 10^7\ \text{V}$ .

Calculate the average power, in GW, transferred during the lightning strike.

power = ______ $\text{GW}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

P = VI

P = (3.0 \times 10^{7})(3.3 \times 10^{4}) = 9.9 \times 10^{11}\ \text{W}

P = \frac{9.9 \times 10^{11}}{10^{9}} = 9.9 \times 10^{2}\ \text{GW}

Answer

P = 990\ \text{GW}

Final answer

990 GW

Detailed explanation

Background Concept

Electrical power $P$ is the rate at which electrical energy is transferred.

For a component (or process) with potential difference $V$ across it and current $I$ through it:

P = VI

The unit conversion is

1\ \text{GW} = 10^{9}\ \text{W}

Understanding the Question

The lightning strike produces an average current $I = 3.3 \times 10^{4}\ \text{A}$ and the potential difference is $V = 3.0 \times 10^{7}\ \text{V}$ . You must find the average power transferred and express it in gigawatts.

Approach

Use $P = VI$ with the given $V$ and $I$ to get power in watts, then divide by $10^{9}$ to convert to gigawatts.

Step-by-Step Reasoning

Compute the power:

P = VI = (3.0 \times 10^{7})(3.3 \times 10^{4})

Multiply coefficients and powers of ten:

3.0 \times 3.3 = 9.9

10^{7} \times 10^{4} = 10^{11}

So

P = 9.9 \times 10^{11}\ \text{W}

Convert to GW:

P = \frac{9.9 \times 10^{11}}{10^{9}} = 9.9 \times 10^{2}\ \text{GW} = 990\ \text{GW}

Key Takeaways

Use $P = VI$ when potential difference and current are known.
Convert between W and GW using $10^{9}$ .

Common Mistakes

Using $P = \frac{V}{I}$ or $P = IR$ (wrong relationship).
Converting to GW by multiplying by $10^{9}$ instead of dividing.
Dropping powers of ten when multiplying in standard form.

Things to Be Careful About

“Average power” matches the use of “average current”.
Quote in GW as requested, and include the unit.

Techniques used

use the electrical power equationconvert watts to gigawattscalculate using standard form

(c)

A lightning rod is attached to a tall building to conduct charge safely to the ground. The lightning rod is modelled as a uniform cylindrical copper cable of total length $95\ \text{m}$ that runs from the ground to the top of the building, as shown in Fig. 3.1.

Fig. 3.1

4M

(i)

The resistance of the lightning rod is $9.6\ \Omega$ .
The resistivity of copper is $1.7 \times 10^{-8}\ \Omega\ \text{m}$ .

Determine the radius of the lightning rod.

radius = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

R = \frac{\rho L}{A} \;\Rightarrow\; A = \frac{\rho L}{R}

A = \frac{(1.7 \times 10^{-8})(95)}{9.6} = 1.68 \times 10^{-7}\ \text{m}^2

A = \pi r^2 \;\Rightarrow\; r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{1.68 \times 10^{-7}}{\pi}}

r = 2.3 \times 10^{-4}\ \text{m}

Answer

r = 2.3 \times 10^{-4}\ \text{m}

Final answer

2.3 × 10^-4 m

Detailed explanation

Background Concept

For a uniform conductor of length $L$ and cross-sectional area $A$ , its resistance $R$ is related to the material property resistivity $\rho$ by

R = \frac{\rho L}{A}

For a cylindrical wire/rod of radius $r$ ,

A = \pi r^2

Understanding the Question

A copper lightning rod is modelled as a uniform cylinder of length $L = 95\ \text{m}$ and resistance $R = 9.6\ \Omega$ . Copper has resistivity $\rho = 1.7 \times 10^{-8}\ \Omega\ \text{m}$ . You must determine the radius $r$ .

Approach

Use $R = \rho L/A$ and rearrange to find $A$ .
Use $A = \pi r^2$ to find $r$ by taking a square root.

Step-by-Step Reasoning

A cylinder model is appropriate because the rod is described as uniform (constant radius).

c-i-cylinder

Rearrange the resistivity equation:

A = \frac{\rho L}{R}

Substitute:

A = \frac{(1.7 \times 10^{-8})(95)}{9.6}

Compute numerator:

(1.7 \times 10^{-8})(95) = 161.5 \times 10^{-8} = 1.615 \times 10^{-6}

Then

A = \frac{1.615 \times 10^{-6}}{9.6} = 1.68 \times 10^{-7}\ \text{m}^2

Now use $A = \pi r^2$ :

r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{1.68 \times 10^{-7}}{\pi}} = 2.3 \times 10^{-4}\ \text{m}

Key Takeaways

Resistance increases with length and resistivity, and decreases with cross-sectional area.
For cylindrical conductors, you often need the geometry step $A = \pi r^2$ .

Common Mistakes

Using $R = \rho A/L$ (inverting the formula).
Forgetting that $A$ depends on $r^2$ ; radius is not proportional to area.
Taking $r = A/\pi$ instead of $r = \sqrt{A/\pi}$ .

Things to Be Careful About

Units: ensure $\rho$ is in $\Omega\ \text{m}$ , $L$ in m, giving $A$ in $\text{m}^2$ and $r$ in m.
Significant figures: final $r$ should reflect the given data (typically 2 s.f.).

Techniques used

apply the resistivity equation for a uniform wirerearrange to find cross-sectional areacalculate radius from area using the circle formula

(ii)

The radius of the copper lightning rod is doubled with no change to its length.

State the effect of this change on the resistance of the lightning rod.

1M Similar

(d)

A section of the lightning rod of length $0.12\ \text{m}$ is removed for testing. A tensile stress of $1.9 \times 10^6\ \text{Pa}$ is applied, as shown in Fig. 3.2.

Fig. 3.2

The section of the rod obeys Hooke’s law. The Young modulus of copper is $1.3 \times 10^{11}\ \text{Pa}$ .

Calculate the extension of the section.

extension = ______ $\text{m}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

E = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\Delta L / L}

\Delta L = \frac{\sigma L}{E}

\Delta L = \frac{(1.9 \times 10^{6})(0.12)}{1.3 \times 10^{11}} = 1.75 \times 10^{-6}\ \text{m}

Answer

\Delta L = 1.8 \times 10^{-6}\ \text{m}

Final answer

1.8 × 10^-6 m

Detailed explanation

Background Concept

When a material obeys Hooke's law, stress is proportional to strain.

Stress $\sigma$ is force per unit area:

\sigma = \frac{F}{A}

Strain is fractional extension:

\text{strain} = \frac{\Delta L}{L}

Young modulus $E$ is defined by

E = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\Delta L / L}

This applies in the linear (elastic) region where Hooke's law holds.

Understanding the Question

A rod section of original length $L = 0.12\ \text{m}$ has a tensile stress $\sigma = 1.9 \times 10^{6}\ \text{Pa}$ applied. The Young modulus of copper is $E = 1.3 \times 10^{11}\ \text{Pa}$ . You are asked to calculate the extension $\Delta L$ .

Approach

Use $E = \sigma /(\Delta L/L)$ and rearrange to get $\Delta L = \sigma L/E$ . No area is required because stress is already given.

Step-by-Step Reasoning

d-tension

Start with

E = \frac{\sigma}{\Delta L/L}

Rearrange:

\Delta L = \frac{\sigma L}{E}

Substitute the values:

\Delta L = \frac{(1.9 \times 10^{6})(0.12)}{1.3 \times 10^{11}}

Compute the numerator:

(1.9 \times 10^{6})(0.12) = 0.228 \times 10^{6} = 2.28 \times 10^{5}

Now divide:

\Delta L = \frac{2.28 \times 10^{5}}{1.3 \times 10^{11}} = \left(\frac{2.28}{1.3}\right) \times 10^{-6} = 1.75 \times 10^{-6}\ \text{m}

To appropriate s.f. (2 s.f.):

\Delta L = 1.8 \times 10^{-6}\ \text{m}

Key Takeaways

Young modulus links stress and strain: $E = \sigma/(\Delta L/L)$ .
If stress is given, you do not need to find force or area.

Common Mistakes

Using $E = \sigma \times \text{strain}$ instead of $E = \sigma/\text{strain}$ .
Forgetting strain is $\Delta L/L$ (some write $L/\Delta L$ ).
Attempting to use $F = kx$ without knowing $k$ .

Things to Be Careful About

Units: Pa is $\text{N m}^{-2}$ ; with $L$ in m, $\Delta L$ comes out in m.
Validity: statement “obeys Hooke’s law” is the clue that Young modulus can be used directly.
Significant figures should reflect the data given.

Techniques used

use the Young modulus relationship between stress and strainrearrange to express extension in terms of stress and original lengthsubstitute values and handle powers of ten

Q4MediumDeformation of SolidsWork, Energy and PowerForces, Density and Pressure

A pinball machine uses a spring to launch a small metal ball of mass $4.5 \times 10^{-2}\ \text{kg}$ up a ramp. The spring is compressed by $8.0 \times 10^{-2}\ \text{m}$ and held in equilibrium, as shown in Fig. 4.1.

Fig. 4.1

The ramp is at an angle of $15^\circ$ to the horizontal.

(a)

The spring obeys Hooke’s law and has a spring constant of $29\ \text{N m}^{-1}$ .

Calculate the elastic potential energy in the compressed spring.

elastic potential energy = ______ $\text{J}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

E_\text{elastic} = \frac{1}{2}kx^2

E_\text{elastic} = \frac{1}{2}(29)(8.0\times10^{-2})^2 = 9.28\times10^{-2}\ \text{J}

Answer

E_\text{elastic} = 9.3\times10^{-2}\ \text{J}

Final answer

9.3 × 10^-2 J

Detailed explanation

Background Concept

A spring that obeys Hooke’s law has force proportional to extension/compression:

F = kx

The elastic potential energy stored in the spring is the work done to compress/extend it from $0$ to $x$ , which equals the area under the $F$ – $x$ graph. Since $F$ increases linearly with $x$ , the graph is a triangle.

So:

E_\text{elastic} = \text{area} = \frac{1}{2}Fx = \frac{1}{2}(kx)x = \frac{1}{2}kx^2

Understanding the Question

You are given the spring constant $k = 29\ \text{N m}^{-1}$ and compression $x = 8.0\times10^{-2}\ \text{m}$ . The question asks for the energy stored in the spring while it is held compressed.

Approach

Use

E_\text{elastic} = \frac{1}{2}kx^2

Convert nothing (already SI), substitute, and round appropriately.

Step-by-Step Reasoning

Substitute values:

E_\text{elastic} = \frac{1}{2}(29)(0.080)^2

Square the compression:

(0.080)^2 = 0.0064

Then:

E_\text{elastic} = 0.5\times 29\times 0.0064 = 0.0928\ \text{J}

To 2 s.f. (matching $8.0\times10^{-2}$ and $29$ ),

E_\text{elastic} = 9.3\times10^{-2}\ \text{J}

Key Takeaways

For a Hooke’s-law spring, stored energy is $\frac{1}{2}kx^2$ .
The factor $\frac{1}{2}$ comes from the triangular $F$ – $x$ graph.

Common Mistakes

Using $E = kx^2$ (missing the $\frac{1}{2}$ ).
Using $x = 8.0\times10^{-2}$ as $8.0$ (not converting to metres).
Giving too many significant figures (e.g. $0.0928$ without rounding).

Things to Be Careful About

Ensure $x$ is in metres and $k$ in $\text{N m}^{-1}$ so energy comes out in joules.
Quote the final answer with sensible significant figures.

Techniques used

use the elastic potential energy formula for a springsubstitute values with consistent SI unitsround the final value to appropriate significant figures

(b)

The spring is released and expands quickly back to its original length.

6M

(i)

Calculate the increase in gravitational potential energy of the ball when the spring returns to its original length.

increase in gravitational potential energy = ______ $\text{J}$

3M Similar

DifficultyMedium

Worked solution

Working

\Delta h = (8.0\times10^{-2})\sin 15^{\circ} = 2.07\times10^{-2}\ \text{m}

\Delta E_p = mg\Delta h = (4.5\times10^{-2})(9.81)(2.07\times10^{-2}) = 9.14\times10^{-3}\ \text{J}

Answer

\Delta E_p = 9.1\times10^{-3}\ \text{J}

Final answer

9.1 × 10^-3 J

Detailed explanation

Background Concept

Gravitational potential energy (GPE) increases when an object rises vertically by height $h$ :

\Delta E_p = mgh

On a slope, the object may move a distance $s$ along the ramp, but only the vertical rise matters for GPE. If the ramp makes angle $\theta$ to the horizontal, then the vertical rise is:

\Delta h = s\sin\theta

Understanding the Question

When the compressed spring returns to its original length, the ball moves up the ramp by the same distance as the compression, $s = 8.0\times10^{-2}\ \text{m}$ , along a ramp at $15^{\circ}$ . The question asks for the increase in GPE during that motion.

diagram-b1

Approach

Find the vertical height gain using $\Delta h = s\sin\theta$ .
Use $\Delta E_p = mg\Delta h$ .

Step-by-Step Reasoning

Height gain:

\Delta h = (8.0\times10^{-2})\sin 15^{\circ}

Using $\sin 15^{\circ} \approx 0.259$ :

\Delta h \approx 0.080\times 0.259 = 2.07\times10^{-2}\ \text{m}

Now calculate GPE increase:

\Delta E_p = (4.5\times10^{-2})(9.81)(2.07\times10^{-2})

\Delta E_p = 9.14\times10^{-3}\ \text{J} \approx 9.1\times10^{-3}\ \text{J}

Key Takeaways

GPE depends only on vertical height change, not the distance along the slope.
On an incline, height gained is $s\sin\theta$ .

Common Mistakes

Using $\Delta h = s\cos\theta$ instead of $s\sin\theta$ .
Using the slope distance $s$ directly in $mgs$ .
Forgetting that the ball moves by $0.080\ \text{m}$ along the ramp during the spring’s expansion.

Things to Be Careful About

Keep $s$ in metres and use $g$ consistently (typically $9.81\ \text{m s}^{-2}$ ).
Significant figures: the data are mostly 2 s.f., so $\Delta E_p$ should be 2 s.f. as well.

Techniques used

resolve motion along an incline to find vertical height gainuse trigonometry to relate slope distance to vertical risecalculate change in gravitational potential energy using $\Delta E_p = mg\Delta h$

(ii)

The ball leaves the spring when the spring reaches its original length. Assume that all the elastic potential energy of the spring is transferred to the ball.

Calculate the speed of the ball as it leaves the spring.

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

3M Similar

DifficultyMedium

Worked solution

Working

E_\text{elastic} = 9.28\times10^{-2}\ \text{J},\qquad \Delta E_p = 9.14\times10^{-3}\ \text{J}

E_k = E_\text{elastic} - \Delta E_p = (9.28\times10^{-2} - 9.14\times10^{-3}) = 8.37\times10^{-2}\ \text{J}

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

v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2(8.37\times10^{-2})}{4.5\times10^{-2}}} = 1.93\ \text{m s}^{-1}

Answer

v = 1.9\ \text{m s}^{-1}

Final answer

1.9 m s^-1

Detailed explanation

Background Concept

Energy transfers let you connect spring energy to the ball’s motion. If all the spring’s elastic potential energy becomes energy of the ball, it can appear as:

increase in gravitational potential energy (ball rises), and
kinetic energy (ball speeds up).

Elastic energy in the spring (Hooke’s law spring):

E_\text{elastic} = \frac{1}{2}kx^2

Kinetic energy of the ball:

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

Conservation of energy for this stage:

E_\text{elastic} = \Delta E_p + E_k

Understanding the Question

At the instant the spring reaches its natural length, the ball leaves the spring. During the spring’s expansion, the ball has moved up the ramp (so it has gained GPE). The question says assume all elastic potential energy is transferred to the ball, so you should account for both GPE gain and the remaining KE.

Given:

$m = 4.5\times10^{-2}\ \text{kg}$
$E_\text{elastic} = 9.28\times10^{-2}\ \text{J}$ from (a)
$\Delta E_p = 9.14\times10^{-3}\ \text{J}$ from (b)(i)

Find: $v$ as it leaves the spring.

Approach

Use energy conservation: $E_k = E_\text{elastic} - \Delta E_p$ .
Use $E_k = \frac{1}{2}mv^2$ to solve for $v$ .

Step-by-Step Reasoning

Energy available as kinetic energy after lifting the ball:

E_k = 9.28\times10^{-2} - 9.14\times10^{-3} = 8.37\times10^{-2}\ \text{J}

Now equate to kinetic energy:

\frac{1}{2}mv^2 = 8.37\times10^{-2}

Rearrange for $v$ :

v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2(8.37\times10^{-2})}{4.5\times10^{-2}}}

Compute the bracket:

\frac{2(8.37\times10^{-2})}{4.5\times10^{-2}} \approx 3.72

So:

v \approx \sqrt{3.72} = 1.93\ \text{m s}^{-1} \approx 1.9\ \text{m s}^{-1}

Key Takeaways

When an object moves up a slope, spring energy splits into GPE and KE.
Use $E_\text{elastic} = \Delta E_p + E_k$ , not $E_\text{elastic} = E_k$ unless there is no height gain.

Common Mistakes

Setting $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ and ignoring the GPE increase.
Using mass $4.5\times10^{-2}$ as $4.5$ .
Forgetting the square root when solving for $v$ .

Things to Be Careful About

The phrase “as it leaves the spring” refers to the moment the spring reaches natural length, after the ball has risen by $0.080\sin15^{\circ}$ .
Keep track of energy units (all in J) and quote $v$ to a sensible number of significant figures.

Techniques used

apply conservation of energy between spring energy and the ball’s mechanical energysubtract the gain in gravitational potential energy from initial stored energyequate kinetic energy to $\tfrac12 mv^2$ and solve for speed

(c)

The ball comes to rest on a horizontal trapdoor of negligible mass at a distance $d$ from its pivot.
A force $F$ acts vertically downwards at a distance of $2.0\ \text{cm}$ from the pivot, as shown in Fig. 4.2.

Fig. 4.2

3M

(i)

The trapdoor is in equilibrium when $F$ is $1.7\ \text{N}$ .

Calculate $d$ .

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For equilibrium about the pivot:

F(0.020) = mgd

d = \frac{F(0.020)}{mg} = \frac{(1.7)(0.020)}{(4.5\times10^{-2})(9.81)} = 7.71\times10^{-2}\ \text{m}

Answer

d = 7.7\times10^{-2}\ \text{m}

Final answer

7.7 × 10^-2 m

Detailed explanation

Background Concept

The moment (torque) of a force about a pivot is:

\tau = F \times \text{perpendicular distance from pivot}

For an object in rotational equilibrium:

\sum \tau_\text{clockwise} = \sum \tau_\text{anticlockwise}

Understanding the Question

A ball of mass $4.5\times10^{-2}\ \text{kg}$ rests on a horizontal trapdoor at distance $d$ from the pivot. Its weight $mg$ acts downward at that point.

A downward force $F = 1.7\ \text{N}$ acts at $2.0\ \text{cm} = 0.020\ \text{m}$ from the pivot on the other side.

The trapdoor is in equilibrium, so the clockwise and anticlockwise moments about the pivot balance. You must find $d$ .

Approach

Identify the two torques about the pivot: one from $F$ , one from the ball’s weight $mg$ .
Set their moments equal.
Rearrange for $d$ .

Step-by-Step Reasoning

Weight of the ball:

mg = (4.5\times10^{-2})(9.81) = 0.441\ \text{N}

Moment balance:

F(0.020) = (mg)(d)

So:

d = \frac{F(0.020)}{mg} = \frac{1.7\times 0.020}{0.441} = 7.71\times10^{-2}\ \text{m}

Rounded:

d = 7.7\times10^{-2}\ \text{m}

Key Takeaways

Use perpendicular distance from the pivot when calculating a moment.
In equilibrium, clockwise moment equals anticlockwise moment.

Common Mistakes

Leaving $2.0\ \text{cm}$ as $2.0$ (not converting to metres).
Using mass instead of weight (forgetting to multiply by $g$ ).
Taking moments about the wrong point (not the pivot).

Things to Be Careful About

The door is horizontal, so the perpendicular distance is the horizontal distance shown.
Use consistent units (metres) for all distances when computing moments.

Techniques used

take moments about the pivotset clockwise moment equal to anticlockwise moment for equilibriumcalculate weight using $mg$ and rearrange for the unknown distance

(ii)

Force $F$ is decreased from $1.7\ \text{N}$ .

State the direction of the resultant moment about the pivot on the trapdoor.

1M Similar

DifficultyEasy

Worked solution

Answer

When $F$ is decreased, the clockwise moment due to the ball’s weight is greater, so the resultant moment is clockwise.

Final answer

Clockwise

Detailed explanation

Background Concept

A force on one side of a pivot tends to rotate the object either clockwise or anticlockwise. The direction of the resultant moment is determined by which side produces the larger torque.

In equilibrium:

\tau_\text{clockwise} = \tau_\text{anticlockwise}

If one torque decreases while the other stays the same, the resultant moment is in the direction of the larger remaining torque.

Understanding the Question

At $F = 1.7\ \text{N}$ the trapdoor is balanced: the torque from $F$ about the pivot matches the torque from the ball’s weight.

Now $F$ is decreased. The ball’s weight $mg$ (and its distance $d$ ) does not change, so the torque due to the ball stays the same, but the torque due to $F$ becomes smaller.

diagram-c2

Approach

Identify which torque is reduced (the one due to $F$ ).
Conclude that the opposite torque (from the ball) becomes dominant.
State the direction of rotation caused by the ball’s weight.

Step-by-Step Reasoning

Force $F$ acts downward on the left of the pivot, producing an anticlockwise moment.
The ball’s weight acts downward on the right of the pivot, producing a clockwise moment.
At equilibrium these were equal.
Reducing $F$ reduces the anticlockwise moment.
Therefore the clockwise moment from the ball is now larger, so the resultant moment is clockwise.

Key Takeaways

Decreasing one balancing force breaks equilibrium.
The system rotates in the direction of the torque that is now larger.

Common Mistakes

Stating “anticlockwise” because $F$ is still present, without comparing magnitudes.
Mixing up which side gives clockwise vs anticlockwise rotation.

Things to Be Careful About

Always refer to moments about the pivot.
Keep the picture consistent: left-downward gives anticlockwise; right-downward gives clockwise for a horizontal bar (as drawn).

Techniques used

compare clockwise and anticlockwise moments about a pivotuse the effect of decreasing one torque to deduce the resultant rotation direction

Q5MediumD.C. CircuitsElectricity

(a)

State Kirchhoff’s second law.

1M Similar

DifficultyEasy

Worked solution

Answer

In any closed loop, the algebraic sum of the e.m.f.s is equal to the algebraic sum of the potential differences (sum of p.d.s around the loop is zero).

Final answer

In any closed loop, the algebraic sum of the e.m.f.s equals the algebraic sum of the potential differences (sum of p.d.s around the loop is zero).

Detailed explanation

Background Concept

Kirchhoff’s second law (the loop law) is a statement of energy conservation for electric circuits. When a charge goes around a complete closed loop, the total electrical energy gained per unit charge from sources (e.m.f.s) must equal the total energy lost per unit charge in circuit elements (potential drops).

Mathematically, for a chosen loop direction:

\sum (\text{emf}) - \sum (\text{potential drops}) = 0

Equivalently:

\sum (\text{p.d.s around a loop}) = 0

Understanding the Question

You are asked to state Kirchhoff’s second law in words (no calculation). The key ideas to include are: closed loop, algebraic sum, e.m.f.s and potential differences.

Approach

Give the standard wording: “sum of emfs equals sum of potential drops” or “algebraic sum of p.d.s around a closed loop is zero”. Mention “algebraic” to indicate sign depends on direction.

Step-by-Step Reasoning

Consider a complete closed loop so the charge returns to its starting point.
The net change in energy per unit charge after a complete loop must be zero.
Therefore, energy gained per unit charge from sources (emf) equals energy lost per unit charge in components (p.d. drops).

Key Takeaways

Kirchhoff’s second law applies to a closed loop.
It is an energy conservation statement: gains (emf) balance losses (p.d.).

Common Mistakes

Stating Kirchhoff’s first law (current at a junction) instead of the second.
Forgetting the idea of a closed loop.
Omitting that it is an algebraic sum (sign depends on chosen direction).

Things to Be Careful About

Use “potential difference” (or “voltage drop”) and “e.m.f.” correctly.
“Sum of p.d.s is zero” is acceptable only if it is clear you mean the algebraic sum around a loop.

Techniques used

state Kirchhoff’s loop law using algebraic sums of potential differencesrelate e.m.f. to potential drops around a closed loop

(b)

A battery of electromotive force (e.m.f.) $9.0\ \text{V}$ and negligible internal resistance is connected in series with a variable resistor $X$ and a thermistor $Y$ as shown in Fig. 5.1.

Fig. 5.1

Fig. 5.2 shows the relationship between temperature and resistance for the thermistor.

Fig. 5.2

8M

(i)

The current in the circuit is $1.1 \times 10^{-2}\ \text{A}$ . The potential difference across $Y$ is $4.0\ \text{V}$ .

Calculate the resistance of $X$ .

resistance = ______ $\Omega$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For thermistor $Y$ :

R_Y = \frac{V_Y}{I} = \frac{4.0}{1.1\times 10^{-2}} = 3.6\times 10^{2}\ \Omega

In series:

V_X = 9.0 - 4.0 = 5.0\ \text{V}

R_X = \frac{V_X}{I} = \frac{5.0}{1.1\times 10^{-2}} = 4.5\times 10^{2}\ \Omega

Answer

$4.5\times 10^{2}\ \Omega$

Final answer

4.5 × 10^2 Ω

Detailed explanation

Background Concept

In a series circuit, the same current $I$ flows through each component. The supply e.m.f. (here, with negligible internal resistance) is equal to the sum of the potential differences across the series components:

V_{\text{supply}} = V_X + V_Y

For any resistor-like component, the potential difference across it is related to the current through it by Ohm’s law form:

V = IR

(For a thermistor, $R$ changes with temperature, but at a fixed temperature you can still use $V = IR$ at that moment.)

Understanding the Question

You are told:

Supply e.m.f. $= 9.0\ \text{V}$ and internal resistance is negligible, so the full $9.0\ \text{V}$ appears across $X$ and $Y$ together.
Circuit current $I = 1.1\times 10^{-2}\ \text{A}$ .
Potential difference across thermistor $Y$ is $V_Y = 4.0\ \text{V}$ .

You must find the resistance of the variable resistor $X$ .

Approach

Use Kirchhoff’s second law to find the p.d. across $X$ : $V_X = 9.0 - 4.0$ .
Use $R_X = V_X/I$ .

(You can also find $R_Y$ first using $R_Y = V_Y/I$ , but the key target is $R_X$ .)

Step-by-Step Reasoning

Because the components are in series, the total potential difference splits:

V_X + V_Y = 9.0

So:

V_X = 9.0 - 4.0 = 5.0\ \text{V}

Now apply $V = IR$ to resistor $X$ :

R_X = \frac{V_X}{I} = \frac{5.0}{1.1\times 10^{-2}}\ \Omega

Calculate:

R_X \approx 4.5\times 10^{2}\ \Omega

Key Takeaways

In series: same current through each component, voltages add.
Use Kirchhoff’s loop law to relate supply voltage to component voltages.
Use $R = V/I$ once you know the voltage across that component.

Common Mistakes

Using $9.0\ \text{V}$ directly for $V_X$ (forgetting that $4.0\ \text{V}$ is across $Y$ ).
Using the wrong current (in series there is only one current).
Missing the unit $\Omega$ .

Things to Be Careful About

Keep $I$ in amperes: $1.1\times 10^{-2}\ \text{A}$ , not $1.1\times 10^{-2}\ \text{mA}$ .
Quote the answer to appropriate significant figures (here typically 2 s.f. to match data).

Techniques used

apply Kirchhoff’s second law to relate supply p.d. to component p.d.s in seriesuse $V = IR$ to calculate resistance from potential difference and currentsubtract potential differences in a series loop

(ii)

The temperature of $Y$ is changed to $190^\circ\text{C}$ . The resistance of $X$ remains unchanged.

Determine the new potential difference across $Y$ .

potential difference = ______ $\text{V}$

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 5.2 at $190^{\circ}\text{C}$ , take $R_Y \approx 28\ \Omega$ .

From (i), $R_X \approx 4.5\times 10^{2}\ \Omega$ .

Total resistance:

R_{\text{tot}} = R_X + R_Y \approx 4.5\times 10^{2} + 28 = 4.78\times 10^{2}\ \Omega

Current:

I = \frac{9.0}{R_{\text{tot}}} \approx \frac{9.0}{4.78\times 10^{2}} = 1.88\times 10^{-2}\ \text{A}

New p.d. across $Y$ :

V_Y = IR_Y \approx (1.88\times 10^{-2})(28) = 5.3\times 10^{-1}\ \text{V}

Answer

$0.53\ \text{V}$

Final answer

0.53 V

Detailed explanation

Background Concept

A thermistor has a resistance that depends strongly on temperature. For an NTC thermistor, resistance decreases as temperature increases.

In a series circuit, the supply voltage divides between the series components in proportion to their resistances. This is the potential divider idea:

V_Y = V_{\text{supply}}\frac{R_Y}{R_X + R_Y}

Equivalently, you can find the circuit current first:

I = \frac{V_{\text{supply}}}{R_X + R_Y}

then use $V_Y = IR_Y$ .

Understanding the Question

You change the thermistor temperature to $190^{\circ}\text{C}$ , but you keep $R_X$ the same as in part (i). The thermistor resistance at $190^{\circ}\text{C}$ must be read from the given graph, then you calculate the new voltage across $Y$ .

Known:

$V_{\text{supply}} = 9.0\ \text{V}$ .
$R_X$ unchanged from (i) (about $4.5\times 10^{2}\ \Omega$ ).
$R_Y$ at $190^{\circ}\text{C}$ from graph (a small value, a few tens of ohms).

Unknown:

New $V_Y$ .

Approach

Read $R_Y$ at $190^{\circ}\text{C}$ from Fig. 5.2.
Use series total resistance $R_X + R_Y$ to find the new current $I$ .
Use $V_Y = IR_Y$ (or directly the potential divider equation) to get the new p.d. across $Y$ .

Step-by-Step Reasoning

Graph reading: At $190^{\circ}\text{C}$ , the curve is just above its value at $200^{\circ}\text{C}$ (about $25\ \Omega$ ). A reasonable reading is around $28\ \Omega$ (your exact value may differ slightly depending on how you read the curve).
Keep $R_X$ the same: From part (i), $R_X \approx 4.5\times 10^{2}\ \Omega$ .
Total resistance in series:

R_{\text{tot}} = R_X + R_Y \approx 450 + 28 = 478\ \Omega

New current:

I = \frac{9.0}{478} \approx 1.88\times 10^{-2}\ \text{A}

Notice the current is larger than before because the thermistor resistance has dropped dramatically.

Voltage across $Y$ :

V_Y = IR_Y \approx (1.88\times 10^{-2})(28) \approx 0.53\ \text{V}

This is much smaller than the original $4.0\ \text{V}$ because now $R_Y \ll R_X$ , so most of the $9.0\ \text{V}$ is dropped across $X$ .

Key Takeaways

For an NTC thermistor, increasing temperature decreases resistance.
In a series potential divider, the larger resistance gets the larger share of the supply voltage.
You can compute the divided voltage using either $V_Y = IR_Y$ (after finding $I$ ) or the potential divider formula.

Common Mistakes

Not reading the thermistor resistance from the graph (or reading from the wrong axis).
Assuming the current stays the same when the thermistor resistance changes (in series, current changes when total resistance changes).
Using $V_Y = 9.0\ \text{V}$ (forgetting this is only the supply, not across one component).

Things to Be Careful About

Your final numerical value depends on the graph reading at $190^{\circ}\text{C}$ ; small differences are expected.
Keep consistent significant figures (typically 2 s.f. from given data and graph-reading uncertainty).
Ensure you are using $\Omega$ for resistance and $\text{V}$ for potential difference.

Techniques used

read a resistance value from a thermistor characteristic graphapply the potential divider relationship in a series circuitcalculate current using $I = V/R$use $V = IR$ to find potential difference across a component

(iii)

The resistance of $X$ is increased. The temperature of $Y$ remains at $190^\circ\text{C}$ .

By reference to the current in the circuit, state and explain the effect of this change, if any, on the potential difference across $Y$ .

3M Similar

DifficultyMedium

Worked solution

Answer

Increasing $R_X$ increases the total series resistance, so the current $I$ in the circuit decreases.

Since $R_Y$ is constant at $190^{\circ}\text{C}$ , the p.d. across $Y$ is

V_Y = IR_Y

so $V_Y$ decreases.

Final answer

The potential difference across Y decreases.

Detailed explanation

Background Concept

For components in series:

The same current $I$ flows through every component.
The total resistance is the sum: $R_{\text{tot}} = R_X + R_Y$ .
The current is set by the supply voltage and total resistance:

I = \frac{V_{\text{supply}}}{R_X + R_Y}

The potential difference across a component is then $V = IR$ .

You can also see it as a potential divider:

V_Y = V_{\text{supply}}\frac{R_Y}{R_X + R_Y}

Understanding the Question

At $190^{\circ}\text{C}$ the thermistor resistance $R_Y$ is fixed (because its temperature is fixed). You then increase the variable resistor $R_X$ . You are asked what happens to the potential difference across $Y$ , and you must explain it by referring to the current.

Approach

Link the cause-and-effect chain:

Increasing $R_X$ increases total resistance.
With fixed supply voltage, larger total resistance means smaller current.
With $R_Y$ fixed, smaller current means smaller $V_Y = IR_Y$ .

Optionally, confirm with the potential divider fraction $R_Y/(R_X+R_Y)$ getting smaller.

Step-by-Step Reasoning

Initially, at fixed temperature, $R_Y$ is constant.
Increase $R_X$ :

R_{\text{tot}} = R_X + R_Y \text{ increases}

With $V_{\text{supply}} = 9.0\ \text{V}$ fixed:

I = \frac{V_{\text{supply}}}{R_{\text{tot}}}

So $I$ decreases.

The potential difference across $Y$ is:

V_Y = IR_Y

Since $R_Y$ is constant but $I$ has decreased, $V_Y$ must decrease.

Potential divider check:

V_Y = 9.0\frac{R_Y}{R_X + R_Y}

Increasing $R_X$ increases the denominator, so the fraction (and hence $V_Y$ ) decreases.

Key Takeaways

In series circuits, changing one resistance changes the current everywhere.
If a component’s resistance is fixed, its p.d. is directly proportional to the circuit current.
Potential divider relationships give a quick qualitative check.

Common Mistakes

Saying $V_Y$ stays constant because $R_Y$ is constant (but forgetting $I$ changes).
Saying $V_Y$ increases because $R_X$ increases (mixing up which component gets a larger share of the voltage; actually the larger resistance takes more of the supply).
Not explicitly referring to current, as the question requests.

Things to Be Careful About

Make it clear that the temperature is constant, so it is $R_Y$ that remains constant.
Distinguish between p.d. across $Y$ and the supply voltage.
Use a correct causal chain: $R_X \uparrow \Rightarrow R_{\text{tot}} \uparrow \Rightarrow I \downarrow \Rightarrow V_Y \downarrow$ .

Techniques used

reason qualitatively using the series-circuit current changeapply $V = IR$ to infer how potential difference changesapply the potential divider relationship to compare before-and-after voltages

Q6Medium-EasySuperpositionWaves

Light of a single frequency is incident normally on a diffraction grating. An interference pattern of bright and dark fringes forms on the semicircular screen shown in Fig. 6.1.

Fig. 6.1

The light has wavelength $520\ \text{nm}$ .
The separation of the lines in the grating is $3.8 \times 10^{-6}\ \text{m}$ .

(a)

Determine the total number of bright fringes formed on the screen.

number of bright fringes = ______

3M Similar

DifficultyMedium-Easy

Worked solution

Working

For a grating,

d\sin\theta = n\lambda

Maximum order when $\sin\theta = 1$ :

n_{\max} = \frac{d}{\lambda} = \frac{3.8 \times 10^{-6}}{520 \times 10^{-9}} = 7.31 \Rightarrow n_{\max}=7

Total number of bright fringes (orders $-7$ to $+7$ plus central):

N = 2n_{\max}+1 = 2(7)+1 = 15

Answer

$15$

Final answer

15

Detailed explanation

Background Concept

A diffraction grating produces principal maxima (bright fringes) when waves from adjacent slits interfere constructively. The condition for a principal maximum is

d\sin\theta = n\lambda

where:

$d$ is the grating spacing (distance between adjacent lines/slits),
$\theta$ is the angle of the maximum from the normal to the grating,
$n$ is the order number ( $n = 0, 1, 2, \dots$ ),
$\lambda$ is the wavelength.

Not all orders exist. Because $|\sin\theta| \le 1$ , the largest possible order satisfies

|n|\lambda \le d.

A semicircular screen allows maxima to be observed over angles from $-90^\circ$ to $+90^\circ$ , so the limiting case is indeed $|\sin\theta| = 1$ .

Understanding the Question

You are given monochromatic light of wavelength $520\ \text{nm}$ incident normally on a grating with line separation $d = 3.8 \times 10^{-6}\ \text{m}$ . The question asks for the total number of bright fringes (principal maxima) formed on the semicircular screen.

“Total number” means counting all orders on both sides of the central maximum, i.e. negative and positive orders, plus the central ( $n=0$ ) maximum.

Approach

Use the grating equation and the requirement $|\sin\theta| \le 1$ to find the maximum possible order $n_{\max}$ .
Count the total maxima from $-n_{\max}$ to $+n_{\max}$ , including $n=0$ .

Step-by-Step Reasoning

Convert the wavelength to metres:

\lambda = 520\ \text{nm} = 520 \times 10^{-9}\ \text{m}.

The maximum order occurs when $\sin\theta = 1$ :

1 = \sin\theta = \frac{n\lambda}{d} \Rightarrow n_{\max} = \frac{d}{\lambda}.

Substitute values:

n_{\max} = \frac{3.8 \times 10^{-6}}{520 \times 10^{-9}} \approx 7.31.

The order must be an integer, so the largest allowed is

n_{\max} = 7.

The observable bright fringes correspond to $n = 0, \pm 1, \pm 2, \dots, \pm 7$ .
That is one central maximum plus 7 on each side:

N = 2n_{\max} + 1 = 2(7) + 1 = 15.

Key Takeaways

Use $d\sin\theta = n\lambda$ for a diffraction grating.
The highest possible order comes from $|\sin\theta| \le 1$ , giving $n_{\max} = \left\lfloor \frac{d}{\lambda} \right\rfloor$ .
Total number of bright fringes on both sides is $2n_{\max} + 1$ .

Common Mistakes

Forgetting to include the central maximum ( $n=0$ ), leading to $2n_{\max}$ instead of $2n_{\max}+1$ .
Rounding $7.31$ up to $8$ (not allowed because that would require $\sin\theta > 1$ ).
Using $\lambda$ in nm without converting to m while $d$ is in m.

Things to Be Careful About

The “total number” includes both positive and negative orders.
Only integer orders are physical.
The semicircular screen implies you can consider angles up to $90^\circ$ , so the limiting condition is $\sin\theta = 1$ .

Techniques used

apply the diffraction grating equationuse the maximum condition sin(\theta) \le 1 to find the highest ordercount total principal maxima on both sides including the central maximum

(b)

The light is replaced with red light of a single frequency.

3M

(i)

State whether the frequency of the red light is greater than, less than or the same as the frequency of the original light.

1M Similar

(ii)

State and explain the effect of this change on the number of bright fringes formed on the screen. A calculation is not required.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The number of bright fringes decreases.

For maxima, $d\sin\theta = n\lambda$ and since $\sin\theta \le 1$ the largest possible order satisfies $n_{\max}\lambda \le d$ . Red light has larger $\lambda$ , so $n_{\max}$ is smaller, giving fewer orders and hence fewer bright fringes.

Final answer

The number of bright fringes decreases.

Detailed explanation

Background Concept

Principal maxima from a diffraction grating satisfy

d\sin\theta = n\lambda.

Because $|\sin\theta| \le 1$ , the order number is limited by

|n| \le \frac{d}{\lambda}.

So the maximum observable order is approximately $n_{\max} = \left\lfloor \frac{d}{\lambda} \right\rfloor$ . The total number of bright fringes is then

N = 2n_{\max} + 1.

Understanding the Question

You replace the original $520\ \text{nm}$ light with red light. Red light has a longer wavelength. The question asks what happens to the number of bright fringes and to explain why, without doing a numerical calculation.

Approach

Identify how increasing $\lambda$ affects the condition for maxima and, in particular, the maximum possible order.
Connect a smaller $n_{\max}$ to a smaller total number of maxima on the screen.

Step-by-Step Reasoning

For any maximum of order $n$ :

\sin\theta = \frac{n\lambda}{d}.

For that maximum to exist physically, we must have $\sin\theta \le 1$ , so

\frac{n\lambda}{d} \le 1 \Rightarrow n\lambda \le d.

If you switch to red light, $\lambda$ increases. In the inequality $n\lambda \le d$ , a larger $\lambda$ means that fewer integer values of $n$ can satisfy the inequality.

Therefore $n_{\max}$ becomes smaller.

Since each order gives one bright fringe on each side (plus the central), the total number

N = 2n_{\max} + 1

also decreases when $n_{\max}$ decreases.

So the interference pattern contains fewer bright fringes.

Key Takeaways

Bigger wavelength $\Rightarrow$ larger diffraction angles for a given order, but fewer possible orders overall.
The number of fringes is determined by how many integer orders are allowed by $|\sin\theta| \le 1$ .

Common Mistakes

Saying the number increases because the fringes “spread out” more. (Spacing/angles may increase, but the maximum order decreases, so the count decreases.)
Forgetting that only integer orders are allowed.

Things to Be Careful About

The question asks for the number of bright fringes, not their spacing or angular separation.
No calculation is required, but you still need to refer to the grating condition and the $\sin\theta \le 1$ limit to justify the change.

Techniques used

use n\lambda \le d from the grating condition and sin(\theta) \le 1reason qualitatively how changing wavelength changes the maximum observable orderrelate the maximum order to the count of bright fringes

Q7Medium-EasyParticle Physics

A particle Q and a particle R are each composed of one quark and one antiquark.

(a)

State the name of the class (group) of particles that includes Q and R.

1M Similar

(b)

Q has a charge of $-1e$ , where $e$ is the elementary charge. R has a charge of $0$ .

Complete Table 7.1 to show a possible second quark in each of Q and R.

Table 7.1

	charge	first quark	second quark
Q	$-1e$	strange
R	$0$	anti-up

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For Q: strange has charge $-\dfrac{1}{3}e$ .

So second particle must have charge

-1e-\left(-\frac{1}{3}e\right)=-\frac{2}{3}e

So second quark is anti-up.

For R: anti-up has charge $-\dfrac{2}{3}e$ .

So second particle must have charge

0-\left(-\frac{2}{3}e\right)=+\frac{2}{3}e

So second quark is up.

Answer

Q second quark: anti-up
R second quark: up

Final answer

Q: anti-up; R: up

Detailed explanation

Background Concept

Quarks have fractional charges (in units of the elementary charge $e$ ):

$u$ has charge $+\dfrac{2}{3}e$ so $\bar{u}$ has charge $-\dfrac{2}{3}e$ .
$d$ has charge $-\dfrac{1}{3}e$ so $\bar{d}$ has charge $+\dfrac{1}{3}e$ .
$s$ has charge $-\dfrac{1}{3}e$ so $\bar{s}$ has charge $+\dfrac{1}{3}e$ .

For a meson (quark + antiquark), the total charge is just the sum of the constituent charges.

Understanding the Question

You are told:

Q is made from strange plus a second constituent, and has total charge $-1e$ .
R is made from anti-up plus a second constituent, and has total charge $0$ .

You must choose one possible “second quark” in each case so that:

the total charge matches, and
the particle is still a meson (one quark and one antiquark).

Approach

For each particle:

Write the charge of the given first constituent.
Use: (charge of second constituent) = (total charge) − (charge of first constituent).
Identify which quark or antiquark has that required charge, and check the quark–antiquark pairing condition.

Step-by-Step Reasoning

Particle Q

First constituent: strange quark $s$ has charge $-\dfrac{1}{3}e$ .
Total charge is $-1e$ .
Required charge of the second constituent:

q_2 = -1e - \left(-\frac{1}{3}e\right) = -\frac{2}{3}e

The particle with charge $-\dfrac{2}{3}e$ is the anti-up quark $\bar{u}$ .
Check composition: $s$ is a quark and $\bar{u}$ is an antiquark, so this is a valid meson.

So Q can be $s\bar{u}$ .

Particle R

First constituent: anti-up $\bar{u}$ has charge $-\dfrac{2}{3}e$ .
Total charge is $0$ .
Required charge of the second constituent:

q_2 = 0 - \left(-\frac{2}{3}e\right) = +\frac{2}{3}e

The particle with charge $+\dfrac{2}{3}e$ is the up quark $u$ .
Check composition: $\bar{u}$ is an antiquark and $u$ is a quark, so this is a valid meson.

So R can be $\bar{u}u$ .

Key Takeaways

Quark charges are fractional; antiquarks have equal magnitude and opposite sign.
Total charge of a composite particle is the algebraic sum of constituent charges.
Mesons must be exactly quark + antiquark.

Common Mistakes

Using $u$ instead of $\bar{u}$ (or vice versa), giving the wrong sign for charge.
Forgetting that Q and R must be quark–antiquark pairs: e.g. choosing two quarks.
Arithmetic slip with negatives when subtracting a negative charge.

Things to Be Careful About

Keep charges as fractions of $e$ until the end to avoid mistakes.
Ensure the “second quark” you name is actually a quark or an antiquark with the required charge (e.g. only $u$ and $\bar{u}$ have $\pm\dfrac{2}{3}e$ ).

Techniques used

use quark and antiquark chargesadd constituent charges to match total chargeensure quark–antiquark composition is maintained

Physics 9702/22 — May/June 2024

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer