9702/11 Oct/Nov 2024 — Physics MCQ Answers

Q11MEasyPhysical Quantities and Units

What are the SI base units for the moment of a force?

Options

A $\text{kg m}^{-1}\text{ s}^{2}$
B $\text{kg s}^{-2}$
C $\text{kg m s}^{-2}$
D $\text{kg m}^{2}\text{ s}^{-2}$

Pick an answer

DifficultyEasy

Worked solution

Working

Moment $= F \times d$ .

$F$ has units $\text{N} = \text{kg m s}^{-2}$ , so

\text{moment units} = \text{N m} = (\text{kg m s}^{-2})\,\text{m} = \text{kg m}^{2}\text{ s}^{-2}.

Answer

D

Final answer

D

Detailed explanation

Background Concept

The moment of a force (also called torque) measures the turning effect of a force about a pivot.

Its magnitude is defined by

\tau = F \times d_\perp

where:

$\tau$ is the moment (torque),
$F$ is the force,
$d_\perp$ is the perpendicular distance from the pivot to the line of action of the force.

So the unit of moment is the unit of force multiplied by the unit of distance.

Understanding the Question

You are asked for the SI base units (i.e. expressed in $\text{kg}$ , $\text{m}$ , $\text{s}$ , etc.) of the moment of a force.

The options are already written in base-unit form, so the task is to identify which one matches $F \times d$ .

Approach

Start from the definition: moment $= F \times d$ .
Write the SI base units for force: $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
Multiply by $\text{m}$ and simplify powers of $\text{m}$ .
Match with the given options.

Step-by-Step Reasoning

From the definition,

\text{moment} = \text{force} \times \text{distance}.

Force in SI units:

1\ \text{N} = 1\ \text{kg m s}^{-2}.

Moment has units $\text{N m}$ , so

\text{N m} = (\text{kg m s}^{-2})\,\text{m} = \text{kg m}^2\text{ s}^{-2}.

This corresponds to option D.

Key Takeaways

Moment (torque) is $F \times d_\perp$ .
$1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
Therefore $1\ \text{N m} = 1\ \text{kg m}^2\text{ s}^{-2}$ .

Common Mistakes

Forgetting to multiply by distance and giving units of force ( $\text{kg m s}^{-2}$ ) instead.
Confusing moment with pressure-like units (involving $\text{m}^{-1}$ ), which would come from dividing by an area/length rather than multiplying by a length.
Dropping a power of $\text{m}$ when simplifying.

Things to Be Careful About

The distance in the moment definition is the perpendicular distance; it still has unit $\text{m}$ regardless.
The question asks for SI base units, so the final form should be in $\text{kg}$ , $\text{m}$ , $\text{s}$ (not just $\text{N m}$ ).

Techniques used

use the definition of moment as force multiplied by perpendicular distancesubstitute the SI base units of force (newton)combine units by multiplication and simplify indices

Q21MEasyPhysical Quantities and Units

Which statement about vector quantities is correct?

Options

A Acceleration of free fall is a vector quantity because it has a constant magnitude.
B Temperature in $^\circ\text{C}$ is a vector quantity because it can be positive or negative.
C Time is a vector quantity because it can only go in the forwards direction.
D Weight is a vector quantity because it has a direction.

Pick an answer

DifficultyEasy

Worked solution

A vector has both magnitude and direction.

Weight is a force and acts towards the centre of the Earth, so it has a direction.

Answer

D

Final answer

D

Detailed explanation

Background Concept

A vector quantity is fully described only when both its magnitude and its direction are given (e.g. force, velocity, acceleration). A scalar quantity is described by magnitude alone (e.g. time, temperature, mass).

It is not enough for a quantity to be “positive or negative” to make it a vector: negative values can simply indicate a scalar measured relative to a chosen zero or reference point.

Understanding the Question

You are asked which statement about vectors is correct. Each option gives a quantity and a reason why it is (supposedly) a vector. We must choose the option whose reason matches the definition: vectors require direction.

Approach

Check each option against the definition:

Decide whether the quantity needs direction to be specified.
Check whether the reason given is a valid reason for being a vector.
Select the only statement that is correct.

Step-by-Step Reasoning

A: Acceleration of free fall, $g$ , is indeed a vector (it is an acceleration), but the reason is wrong. Being a vector is not because its magnitude is constant; it is because acceleration has a direction (downwards).
B: Temperature (in $^\circ\text{C}$ or K) is a scalar. The fact it can be positive or negative does not make it a vector; it has no direction.
C: Time is a scalar. The fact that time progresses forwards does not mean it has a spatial direction.
D: Weight is a force, $W = mg$ , and forces are vectors. Weight acts vertically downwards (towards the Earth’s centre), so it has a direction. This matches the definition.

Therefore, the correct statement is D.

Key Takeaways

Vectors require magnitude and direction.
Scalars can be positive/negative without being vectors.
Forces (including weight) are vectors.

Common Mistakes

Saying a quantity is a vector because its magnitude is constant (vectors can have changing magnitude).
Thinking “can be negative” implies a vector (many scalars can be negative).
Confusing “only goes forwards” (time) with having a direction in space.

Things to Be Careful About

For quantities like acceleration and force, always associate a direction (e.g. “downwards”) when deciding if they are vectors.
The unit or scale used (e.g. $^\circ\text{C}$ ) does not determine whether something is a vector; the definition does.

Techniques used

use the definition of a vector as a quantity with magnitude and directiondistinguish scalars from vectors based on whether direction is requiredeliminate incorrect statements by checking the stated justification

Q31MMedium-EasyPhysical Quantities and Units

The density of the material of a rectangular block is determined by measuring the mass and linear dimensions of the block. The list shows the results obtained, together with their uncertainties.

mass $= (25.0 \pm 0.1)\text{ g}$

length $= (5.00 \pm 0.01)\text{ cm}$

width $= (2.00 \pm 0.01)\text{ cm}$

height $= (1.00 \pm 0.01)\text{ cm}$

The density is calculated to be $2.50\text{ g cm}^{-3}$ .

What is the uncertainty in this result?

Options

A $\pm 0.01\text{ g cm}^{-3}$
B $\pm 0.02\text{ g cm}^{-3}$
C $\pm 0.05\text{ g cm}^{-3}$
D $\pm 0.13\text{ g cm}^{-3}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Density

\rho = \frac{m}{lwh}

For a quotient/product, fractional uncertainties add:

\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{\Delta l}{l}+\frac{\Delta w}{w}+\frac{\Delta h}{h}

\frac{\Delta \rho}{\rho}=\frac{0.1}{25.0}+\frac{0.01}{5.00}+\frac{0.01}{2.00}+\frac{0.01}{1.00} =0.004+0.002+0.005+0.01=0.021

\Delta \rho = 0.021 \times 2.50 = 0.0525\ \text{g cm}^{-3} \approx 0.05\ \text{g cm}^{-3}

Answer

C

Final answer

C

Detailed explanation

Background Concept

When a result is calculated from several measured quantities, its uncertainty comes from the uncertainties of those measurements.

For quantities that are multiplied or divided, the fractional (or percentage) uncertainties add:

Q = \frac{ab}{c} \quad \Rightarrow \quad \frac{\Delta Q}{Q} = \frac{\Delta a}{a}+\frac{\Delta b}{b}+\frac{\Delta c}{c}

This rule matches the “worst-case” way Cambridge typically expects for uncertainty propagation at AS level.

Understanding the Question

You are given measurements (with uncertainties) of mass $m$ and the three dimensions of a rectangular block: length $l$ , width $w$ , height $h$ .

The density is

\rho = \frac{m}{V} = \frac{m}{lwh}

and the calculated value is $2.50\ \text{g cm}^{-3}$ . The question asks for the uncertainty in $\rho$ , and then to match it to one of the options.

Approach

Write $\rho$ in terms of the measured quantities: $\rho = m/(lwh)$ .
Add the fractional uncertainties of $m$ , $l$ , $w$ , and $h$ .
Multiply the final fractional uncertainty by $\rho$ to get the absolute uncertainty in $\text{g cm}^{-3}$ .
Round sensibly and choose the matching option.

Step-by-Step Reasoning

Start with

\rho = \frac{m}{lwh}

So

\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{\Delta l}{l}+\frac{\Delta w}{w}+\frac{\Delta h}{h}

Now calculate each fraction:

Mass: $\Delta m/m = 0.1/25.0 = 0.004$ (0.4%)
Length: $\Delta l/l = 0.01/5.00 = 0.002$ (0.2%)
Width: $\Delta w/w = 0.01/2.00 = 0.005$ (0.5%)
Height: $\Delta h/h = 0.01/1.00 = 0.01$ (1%)

Add them:

\frac{\Delta \rho}{\rho}=0.004+0.002+0.005+0.01=0.021

So the fractional uncertainty is $2.1\%$ .

Convert to absolute uncertainty using $\rho = 2.50\ \text{g cm}^{-3}$ :

\Delta \rho = 0.021 \times 2.50 = 0.0525\ \text{g cm}^{-3}

Round the uncertainty to 1 significant figure (typical convention):

\Delta \rho \approx 0.05\ \text{g cm}^{-3}

This matches option C.

Key Takeaways

For products/quotients, add fractional uncertainties.
Convert back to an absolute uncertainty by multiplying by the calculated value.
The largest fractional contribution often comes from the measurement with the largest relative uncertainty (here, the height).

Common Mistakes

Adding absolute uncertainties ( $0.1 + 0.01 + \dots$ ) instead of fractional/percentage uncertainties.
Using volume uncertainty incorrectly, e.g. finding $\Delta V$ by adding $\Delta l + \Delta w + \Delta h$ (units and method are wrong).
Forgetting the height has a 1% uncertainty, which dominates the total.
Rounding too early, which can shift the final option choice.

Things to Be Careful About

Always divide uncertainty by the measured value in the same units (here all are consistent in g and cm, so no conversion is needed).
Use the correct rule: powers would multiply fractional uncertainty by the power; here each dimension is to power 1.
Round the uncertainty appropriately (usually 1 s.f. unless stated otherwise), then match to the option list.

Techniques used

identify the derived quantity as a quotient of measured quantitiesadd fractional uncertainties for products and quotientsconvert fractional uncertainty to absolute uncertaintyround an uncertainty to an appropriate number of significant figures

Q41MMedium-EasyPhysical Quantities and Units

What is a reasonable estimate of the volume of one page from this examination paper?

Options

A $60\text{ mm}^{3}$
B $600\text{ mm}^{3}$
C $6000\text{ mm}^{3}$
D $60\,000\text{ mm}^{3}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Take a typical exam page as about A4: $210\,\text{mm} \times 297\,\text{mm}$ .

A \approx 210 \times 297 \approx 6.2 \times 10^{4}\ \text{mm}^2

Paper thickness $t \approx 0.10\,\text{mm}$ .

V \approx At \approx (6.2 \times 10^{4})(0.10) \approx 6.2 \times 10^{3}\ \text{mm}^3

Closest option is $6000\,\text{mm}^3$ .

Answer

C

Final answer

C

Detailed explanation

Background Concept

For a thin sheet, the volume can be estimated by

V = A\,t

where $A$ is the area of the sheet and $t$ is its thickness. Estimation questions reward sensible typical values (order of magnitude) rather than exact measurements.

Understanding the Question

We need an approximate volume of one page of the exam paper in $\text{mm}^3$ . A page’s volume depends mainly on:

its surface area (page height (\times) page width), and
the thickness of the paper.

The options span factors of 10, so we just need a reasonable estimate.

Approach

Use a typical exam page size (about A4): roughly $210\,\text{mm} \times 297\,\text{mm}$ .
Use a typical paper thickness: about $0.1\,\text{mm}$ (i.e. around $100\,\mu\text{m}$ ).
Calculate $V = A t$ and match to the nearest option.

Step-by-Step Reasoning

Estimate the area:

A \approx 210 \times 297\ \text{mm}^2

A \approx 6.2 \times 10^{4}\ \text{mm}^2

Use a typical thickness $t \approx 0.10\,\text{mm}$ .
Multiply to get volume:

V \approx At \approx (6.2 \times 10^{4})(0.10)\ \text{mm}^3

V \approx 6.2 \times 10^{3}\ \text{mm}^3

Compare with the options:

$60$ is far too small,
$600$ is still too small,
$6000$ matches $\sim 6 \times 10^{3}$ ,
$60000$ is too large by about a factor of 10.

So the best estimate is option C.

Key Takeaways

For thin objects, use $V = A t$ .
In MCQ estimation, focus on sensible typical values and order of magnitude.
Paper thickness is about $0.1\,\text{mm}$ , and an A4 page area is about $6 \times 10^{4}\,\text{mm}^2$ .

Common Mistakes

Using thickness $t = 1\,\text{mm}$ (ten times too thick), leading to option D.
Forgetting to use consistent units (e.g. mixing metres and millimetres).
Estimating area as $210 + 297$ instead of multiplying.

Things to Be Careful About

Thickness must be in mm to obtain $\text{mm}^3$ directly.
The exact page size is not needed; any reasonable page dimensions and thickness should give a value around $10^{3}$ to $10^{4}\,\text{mm}^3$ .

Techniques used

estimate the area from typical page dimensionsuse thickness multiplied by area to estimate volumeperform order-of-magnitude checking against the options

Q51MMedium-EasyKinematics

An object is projected from horizontal ground at a velocity of magnitude $u$ and angle $\theta$ to the horizontal. It hits the ground at a time $t$ after it is projected. Assume air resistance is negligible.

Fig. 1

Which statement does not describe the motion of this object?

Options

A The horizontal component of the object’s velocity is constant and has the value $u\cos\theta$ .
B The horizontal distance travelled by the object is $tu\cos\theta$ .
C The time taken for the object to reach maximum height is $\dfrac{t}{2}$ .
D The vertical component of the object’s velocity is constant and has the value $u\sin\theta$ .

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For projectile motion with negligible air resistance:

Horizontal acceleration is $0$ , so horizontal velocity is constant at $u\cos\theta$ .
Vertical acceleration is $-g$ , so vertical velocity changes with time (it is not constant).

Therefore the statement that does not describe the motion is D.

Answer

D

Final answer

D

Detailed explanation

Background Concept

In projectile motion with negligible air resistance, the only force on the object after launch is its weight $mg$ acting vertically downward. This means:

Horizontal acceleration $a_x = 0$ (no horizontal force), so horizontal velocity $v_x$ stays constant.
Vertical acceleration $a_y = -g$ (constant downward), so vertical velocity $v_y$ changes uniformly with time.

The key idea is that horizontal and vertical motions are independent (you can analyse them separately) but share the same time $t$ .

Understanding the Question

An object is projected from level ground with speed $u$ at an angle $\theta$ above the horizontal, and it lands back on the ground after a total time $t$ . You are asked which option is NOT a correct description of its motion.

So we check each statement against the standard results for a projectile that lands at the same vertical level as it started.

diagram-1

Approach

Resolve the initial velocity into components: $u\cos\theta$ horizontally and $u\sin\theta$ vertically.
Use the fact that gravity provides constant vertical acceleration $-g$ and there is zero horizontal acceleration.
Use symmetry for the vertical motion (launch and landing at same height): time up to maximum height equals time down from maximum height, so the time to maximum height is half the total flight time.
Identify the statement that contradicts these facts.

Step-by-Step Reasoning

A: “The horizontal component of the object’s velocity is constant and has the value $u\cos\theta$ ."

True, because $a_x = 0$ , so $v_x$ remains equal to its initial value $u\cos\theta$ .

B: “The horizontal distance travelled by the object is $tu\cos\theta$ ."

True, because horizontal velocity is constant: $x = v_x t = (u\cos\theta)t.$

C: “The time taken for the object to reach maximum height is $\dfrac{t}{2}$ ."

True for a projectile that lands at the same height it was launched from (level ground). The vertical motion is symmetric: the upward journey to $v_y=0$ takes the same time as the downward journey back to the starting height.

D: “The vertical component of the object’s velocity is constant and has the value $u\sin\theta$ ."

False, because the vertical acceleration is $-g$ , so vertical velocity changes: $v_y = u\sin\theta - gt,$ which clearly depends on time. Only at $t=0$ is $v_y = u\sin\theta$ .

Therefore, the statement that does not describe the motion is D.

Key Takeaways

With negligible air resistance, $a_x = 0$ so $v_x$ is constant.
Gravity gives $a_y = -g$ so $v_y$ changes uniformly with time.
For launch and landing at the same height, time to maximum height is half the total time of flight.

Common Mistakes

Thinking both components of velocity stay constant; only the horizontal component does.
Forgetting that $u\sin\theta$ is the initial vertical component, not the vertical component at all times.
Assuming $x = ut$ using the full speed $u$ instead of the horizontal component $u\cos\theta$ .

Things to Be Careful About

The symmetry result (time up = time down) relies on landing at the same vertical level as launch and constant $g$ .
“Velocity component is constant” is a statement about acceleration: constant component implies zero acceleration in that direction (which is not true vertically).

Techniques used

resolve the initial velocity into horizontal and vertical componentsapply constant-acceleration kinematics separately in perpendicular directionsuse symmetry of projectile motion for launch and landing at the same height

Q61MMedium-EasyKinematics

A person, travelling on a motorway a total distance of $200\text{ km}$ , travels the first $90\text{ km}$ at an average speed of $80\text{ km h}^{-1}$ .

Which average speed must be obtained for the rest of the journey if the person is to reach the destination in a total time of $2$ hours $0$ minutes?

Options

A $110\text{ km h}^{-1}$
B $120\text{ km h}^{-1}$
C $122\text{ km h}^{-1}$
D $126\text{ km h}^{-1}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Time for first $90\ \text{km}$ :

t_1 = \frac{90}{80} = 1.125\ \text{h}

Total time $t = 2.00\ \text{h}$ , so remaining time:

t_2 = 2.00 - 1.125 = 0.875\ \text{h}

Remaining distance:

s_2 = 200 - 90 = 110\ \text{km}

Required average speed:

v_2 = \frac{110}{0.875} = 1.257\times 10^2\ \text{km h}^{-1} \approx 126\ \text{km h}^{-1}

Answer

D

Final answer

D

Detailed explanation

Background Concept

Average speed is defined by

v = \frac{\text{distance travelled}}{\text{time taken}}

For a journey made of separate sections, you can find the time for each section using $t = s/v$ . The total journey time is the sum of the section times.

Understanding the Question

The whole journey is $200\ \text{km}$ and must take exactly $2.00\ \text{h}$ . The first $90\ \text{km}$ is already completed at an average speed of $80\ \text{km h}^{-1}$ . The question asks for the average speed needed over the remaining distance so that the total time stays at $2.00\ \text{h}$ .

Known:

Total distance $= 200\ \text{km}$
First section: $s_1 = 90\ \text{km}$ at $v_1 = 80\ \text{km h}^{-1}$
Total time $t = 2.00\ \text{h}$

Unknown:

Required average speed $v_2$ for the remaining distance.

Approach

Find the time already used on the first section: $t_1 = s_1 / v_1$ .
Subtract from the total allowed time to get the remaining time: $t_2 = t - t_1$ .
Subtract the distances to get the remaining distance: $s_2 = 200 - 90$ .
Use $v_2 = s_2 / t_2$ and compare with the options.

Step-by-Step Reasoning

Time for the first $90\ \text{km}$ :

t_1 = \frac{s_1}{v_1} = \frac{90}{80} = 1.125\ \text{h}

Remaining time available out of $2.00\ \text{h}$ :

t_2 = 2.00 - 1.125 = 0.875\ \text{h}

Remaining distance:

s_2 = 200 - 90 = 110\ \text{km}

Required average speed for the remaining distance in the remaining time:

v_2 = \frac{110}{0.875} = 125.714...\ \text{km h}^{-1} \approx 126\ \text{km h}^{-1}

This matches option D.

Key Takeaways

Use $t = s/v$ to convert a known distance and average speed into a time.
For fixed total time, the required speed on the remaining section depends on the remaining time, not the total time.
Keep units consistent (here, km and h throughout).

Common Mistakes

Dividing total distance by total time to get $100\ \text{km h}^{-1}$ and assuming that is the required remaining speed.
Subtracting speeds instead of subtracting times and distances.
Converting hours to minutes incorrectly (e.g. treating $0.875\ \text{h}$ as $87.5\ \text{min}$ instead of $52.5\ \text{min}$ ).

Things to Be Careful About

When you subtract times, ensure both are in hours (or both in minutes) before subtracting.
Rounding: $125.7\ \text{km h}^{-1}$ rounds to $126\ \text{km h}^{-1}$ , which is an exact listed option.
“Average speed for the rest of the journey” means average over the remaining $110\ \text{km}$ , not an instantaneous speed.

Techniques used

use $v = \frac{s}{t}$ to convert a journey segment into a time takensubtract times to find remaining available timesubtract distances to find remaining distanceuse $v = \frac{s}{t}$ to determine the required average speed

Q71MMedium-EasyWork, Energy and PowerDynamics

A car of mass $1200\text{ kg}$ has momentum $18\,000\text{ kg m s}^{-1}$ .

What is the kinetic energy of the car?

Options

A $4.65\text{ kJ}$
B $6.57\text{ kJ}$
C $135\text{ kJ}$
D $270\text{ kJ}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

p = mv \Rightarrow v = \frac{p}{m} = \frac{18000}{1200} = 15\ \text{m s}^{-1}

E_k = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(15^2)=135000\ \text{J} = 135\ \text{kJ}

Answer

C

Final answer

C

Detailed explanation

Background Concept

Momentum and kinetic energy are both measures of motion but they are defined differently:

Momentum:

p = mv

where $p$ is momentum, $m$ is mass, and $v$ is speed.

Kinetic energy:

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

A useful link between them comes from eliminating $v$ . Since $v = p/m$ , substituting into $E_k$ gives:

E_k = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}

This is often the quickest route when $p$ and $m$ are given.

Understanding the Question

You are told:

mass $m = 1200\ \text{kg}$
momentum $p = 18000\ \text{kg m s}^{-1}$

You are asked to find the car's kinetic energy $E_k$ and choose the correct option in kJ.

Approach

Use either of these equivalent methods:

Find the speed from $p = mv$ , then substitute into $E_k = \frac{1}{2}mv^2$ .
Use the combined result $E_k = \frac{p^2}{2m}$ directly.

Then convert from joules to kilojoules using $1\ \text{kJ} = 1000\ \text{J}$ .

Step-by-Step Reasoning

From momentum:

v = \frac{p}{m} = \frac{18000}{1200} = 15\ \text{m s}^{-1}

Now kinetic energy:

E_k = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(15^2)

Compute $15^2 = 225$ :

E_k = 600 \times 225 = 135000\ \text{J}

Convert to kJ:

135000\ \text{J} = 135\ \text{kJ}

This matches option C.

(Equivalently, using $E_k = \frac{p^2}{2m}$ :
$E_k = \frac{(18000)^2}{2(1200)} = 135000\ \text{J}$ .)

Key Takeaways

Use $p = mv$ to find speed when momentum and mass are known.
When $p$ and $m$ are given, the shortcut

E_k = \frac{p^2}{2m}

is very efficient.

Always convert $\text{J}$ to $\text{kJ}$ when options are in kJ.

Common Mistakes

Using $E_k = pv$ (not a correct general formula).
Forgetting to square $p$ when using $E_k = \frac{p^2}{2m}$ .
Converting to kJ incorrectly (dividing/multiplying by $1000$ the wrong way).
Dropping units or mixing them (here all values are already in SI).

Things to Be Careful About

Momentum is a vector, but here only its magnitude is needed for kinetic energy.
Keep track of powers: $(18000)^2$ is large, so check the final energy is reasonable.
Ensure the final choice corresponds to $135\ \text{kJ}$ , not $135000\ \text{kJ}$ or $135\ \text{J}$ .

Techniques used

use the definition of momentum to relate momentum, mass and speedeliminate speed to express kinetic energy in terms of momentum and masssubstitute values and convert to appropriate energy units

Q81MMedium-EasyDynamics

A ladder is positioned on icy (frictionless) ground and is leant against a rough wall. At the instant of release it begins to slide.

Which diagram shows the directions of the forces $P$ , $W$ and $R$ acting on the ladder as it slides?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Ground is frictionless (\Rightarrow) force at the ground contact is only the normal reaction, vertically upwards (R).

At the rough wall, the normal reaction on the ladder is horizontal to the right. On release, the top of the ladder slides down the wall, so friction at the wall acts upwards. Resultant force at the wall (P) is therefore up and to the right.

Weight (W) acts vertically downwards.

Answer

B

Final answer

B

Detailed explanation

Background Concept

When an object is in contact with a surface, the contact force can be split into:

a normal reaction (perpendicular to the surface), and
a frictional force (parallel to the surface, only present if the surface is rough).

Key rules:

If a surface is frictionless, there is no tangential (parallel) force at that contact: only the normal reaction acts.
Friction acts to oppose the relative motion (or impending relative motion) between the two surfaces.
Weight (W) always acts vertically downward through the centre of mass.

Understanding the Question

A ladder touches:

The ground (icy, frictionless) at the bottom.
A vertical wall (rough) at the top.

We must choose which option correctly shows the directions of:

(R): force from the ground on the ladder,
(P): force from the wall on the ladder (resultant of normal + friction),
(W): weight of the ladder.

The statement “at the instant of release it begins to slide” tells us we should decide the friction direction from the initial sliding direction.

Approach

Use “frictionless ground” to decide the direction of (R).
At the wall, identify the normal reaction direction (perpendicular to the wall) and then decide the friction direction by considering how the ladder starts to move.
Combine the wall’s normal reaction and friction to get the resultant (P), and match to the correct diagram.

Step-by-Step Reasoning

Force at the ground (icy / frictionless)
- Ground is horizontal, so the normal reaction is perpendicular to it: vertical upward.
- Because it is frictionless, there is no horizontal force at the ground.
  Therefore (R) must be straight up.
Forces at the wall (rough vertical wall)
- The normal reaction from a vertical wall is perpendicular to the wall: horizontal.
- The wall is on the left of the ladder, so it pushes the ladder to the right.
Direction of friction at the wall
- On release, the ladder starts to slide: typically the top end moves downward along the wall (and the bottom moves outward along the ground).
- Friction opposes this relative motion, so at the wall it acts upwards on the ladder.
Resultant force (P) at the wall
- Combine the rightward normal reaction with the upward frictional force: resultant points up and to the right.
Weight (W)
- (W) acts vertically downward.

So the correct diagram must show (R) vertical up, (W) vertical down, and (P) up-right. That is option B.

Key Takeaways

Frictionless contact (\Rightarrow) only a normal reaction (perpendicular to surface).
Rough contact (\Rightarrow) normal reaction + friction, with friction opposing relative motion.
A vertical wall provides a horizontal normal reaction.

Common Mistakes

Adding a horizontal component to (R) even though the ground is frictionless.
Making friction at the wall act downward (it should oppose the top sliding down).
Drawing the wall’s normal reaction pointing left (it must push the ladder away from the wall, to the right).

Things to Be Careful About

The label (P) is the resultant force at the wall (normal + friction), so it can be diagonal.
“At the instant of release it begins to slide” indicates kinetic/impending motion, so friction direction is decided from the initial motion, not from assuming equilibrium.
Ensure each contact force is consistent with the surface orientation: normal is always perpendicular to the surface.

Techniques used

identify the normal reaction directions at each contact surfaceuse the condition of a frictionless surface to remove any tangential forcedetermine the direction of friction by opposing the relative motion at the contactcombine perpendicular contact forces to obtain the resultant contact force

Q91MMedium-EasyKinematics

A toy parachute is dropped from a bridge and falls vertically through the air.

The graph shows the distance travelled by the parachute against time.

Fig. 1

Which region of the graph shows when the parachute is at terminal velocity?

Options

A $A$
B $B$
C $C$
D $D$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Terminal velocity means constant speed ( $a = 0$ ).

On a distance–time graph, the gradient is the speed. Constant speed corresponds to a straight line with constant gradient.

Region $B$ is a straight line.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Terminal velocity occurs when the resistive (drag) force equals the weight, so the resultant force is zero.

\sum F = 0 \;\Rightarrow\; a = 0

So the object continues moving with constant velocity (constant speed for vertical motion in one direction).

For a distance–time graph:

gradient $= \dfrac{\Delta s}{\Delta t}$ is the speed,
increasing gradient means speeding up,
constant gradient means constant speed,
zero gradient (horizontal line) means speed is zero.

Understanding the Question

A toy parachute falls vertically and a graph of distance travelled $s$ against time $t$ is shown, divided into regions $A$ , $B$ , $C$ , and $D$ .

You must choose which region corresponds to the parachute moving at terminal velocity, i.e. falling at a constant speed.

Approach

Look for the part of the distance–time graph where:

the gradient (speed) is constant,
so the graph is a straight line (not curved),
and it is not horizontal (since horizontal would mean the parachute is not moving).

Step-by-Step Reasoning

In region $A$ , the curve gets steeper: the gradient is increasing, so speed is increasing (accelerating) and it is not terminal velocity.
In region $B$ , the graph is a straight line: gradient is constant, so speed is constant. This matches terminal velocity.
In region $C$ , the curve flattens: gradient is decreasing, so the speed is decreasing (decelerating), not terminal velocity.
In region $D$ , the line is horizontal: gradient is zero, meaning speed is zero (distance no longer changes), so it is not terminal velocity during a fall.

Therefore the terminal velocity region is $B$ .

Key Takeaways

Terminal velocity means constant velocity ( $a = 0$ ).
On a distance–time graph, constant velocity corresponds to a straight line (constant gradient).

Common Mistakes

Choosing the steepest part of the graph rather than the straight-line part: terminal velocity is about constant speed, not maximum steepness.
Confusing a distance–time graph with a speed–time graph.
Picking the horizontal region $D$ : a horizontal distance–time graph means the object has stopped (speed $=0$ ).

Things to Be Careful About

Always check what the axes are: here it is distance vs time, so gradient gives speed.
Terminal velocity implies no acceleration, so the graph should not be curving (curvature indicates changing gradient and hence acceleration).

Techniques used

interpret the gradient of a distance–time graph as speedidentify constant speed from a straight-line section of a graphlink terminal velocity to zero acceleration

Q101MMedium-EasyDynamics

A lift (elevator) consists of a passenger car supported by a cable that runs over a light, frictionless pulley to a counterbalance. The counterbalance falls as the passenger car rises.

Fig. 1

Some masses are shown in the table.

mass	$/\text{kg}$
passenger car	520
counterbalance	640
passenger	80

What is the magnitude of the acceleration of the car when carrying just one passenger and when the pulley is free to rotate?

Options

A $0.032\text{ m s}^{-2}$
B $0.32\text{ m s}^{-2}$
C $0.61\text{ m s}^{-2}$
D $0.65\text{ m s}^{-2}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Total mass of car + passenger:

m_2 = 520 + 80 = 600\ \text{kg}

Counterbalance mass:

m_1 = 640\ \text{kg}

For a light, frictionless pulley (Atwood machine),

a = \frac{(m_1-m_2)g}{m_1+m_2}

a = \frac{(640-600)\times 9.81}{640+600} = 0.316\ \text{m s}^{-2} \approx 0.32\ \text{m s}^{-2}

Answer

B

Final answer

B

Detailed explanation

Background Concept

This is an Atwood machine: two masses connected by a light (massless), inextensible cable over a light, frictionless pulley.

Key consequences:

The two masses have the same magnitude of acceleration $a$ (one up, one down).
The tension $T$ is the same on both sides of the cable (light cable, frictionless/light pulley).
Each mass obeys Newton’s second law: resultant force $= ma$ .

For an Atwood machine with heavier mass $m_1$ moving down and lighter mass $m_2$ moving up, the acceleration magnitude is

a = \frac{(m_1-m_2)g}{m_1+m_2}

Understanding the Question

The counterbalance (mass $640\ \text{kg}$ ) falls while the passenger car rises. The passenger car has mass $520\ \text{kg}$ and carries one passenger of mass $80\ \text{kg}$ , so the rising side has total mass $600\ \text{kg}$ .

You are asked for the magnitude of the acceleration of the car when the pulley is free to rotate (so we treat it as ideal: no frictional torque and negligible pulley mass).

Approach

Add the car and passenger masses to get the total rising mass $m_2$ .
Identify the falling mass $m_1$ (the counterbalance).
Use the Atwood machine result, or derive it quickly from Newton’s second law on each mass and eliminate the tension $T$ .
Calculate $a$ and match it to the given options.

Step-by-Step Reasoning

Let $m_1 = 640\ \text{kg}$ (counterbalance, moving down) and $m_2 = 520 + 80 = 600\ \text{kg}$ (car + passenger, moving up).

A free-body view helps:

diagram-1

Apply Newton’s second law to each mass (take the direction of motion as positive for each mass):

For $m_1$ moving down:

m_1 g - T = m_1 a

For $m_2$ moving up:

T - m_2 g = m_2 a

Add the equations to eliminate $T$ :

m_1 g - m_2 g = (m_1+m_2)a

So

a = \frac{(m_1-m_2)g}{m_1+m_2}

Substitute values:

a = \frac{(640-600)\times 9.81}{640+600}

a = \frac{40\times 9.81}{1240} = 0.316\ \text{m s}^{-2} \approx 0.32\ \text{m s}^{-2}

So the correct option is B.

Key Takeaways

Combine masses on the same side of the cable before analysing.
For an ideal Atwood machine, $a$ depends on the mass difference divided by the total mass.
Writing Newton’s second law for each mass and eliminating tension is a reliable method.

Common Mistakes

Forgetting to add the passenger mass to the car mass.
Using $a = \frac{(m_1-m_2)g}{m_1}$ or some other incorrect denominator (it must be $m_1+m_2$ ).
Mixing signs: the forces must be consistent with your chosen positive direction for each mass.
Using $g = 10\ \text{m s}^{-2}$ and then choosing the wrong option due to rounding; you should still land closest to $0.32\ \text{m s}^{-2}$ .

Things to Be Careful About

The question states a light, frictionless pulley and cable, so you assume equal tension on both sides and no energy loss.
The acceleration is small because the mass difference ( $40\ \text{kg}$ ) is small compared with the total mass ( $1240\ \text{kg}$ ).
Give the magnitude and match to the listed options; rounding to 2 s.f. is appropriate here.

Techniques used

model the system as an Atwood machine with two connected massesapply Newton's second law separately to each masssolve simultaneous equations to eliminate tensionsubstitute masses into the acceleration expression

Q111MMedium-EasyDynamics

A stationary ball of mass $m$ is hit by a bat. The ball leaves the bat with velocity $v$ .

The bat is in contact with the ball for a short time $\Delta t$ .

What is the average force of the bat on the ball?

Options

A $mv\Delta t$
B $\dfrac{mv}{\Delta t}$
C $\dfrac{1}{2}mv^{2}\Delta t$
D $\dfrac{\dfrac{1}{2}mv^{2}}{\Delta t}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Initial momentum $= 0$ (ball stationary).

Final momentum $= mv$ .

Change in momentum:

\Delta p = mv

Average force:

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{mv}{\Delta t}

Answer

B

Final answer

B

Detailed explanation

Background Concept

The (average) resultant force on an object is related to how quickly its momentum changes:

F_{\text{avg}} = \frac{\Delta p}{\Delta t}

where momentum $p = mv$ (mass (m), velocity (v)). The quantity

\text{impulse} = F_{\text{avg}}\,\Delta t

is equal to the change in momentum (\Delta p). This is especially useful for short collisions where the force varies during contact, but we are asked for the average force.

Understanding the Question

A ball of mass $m$ starts from rest, so its initial velocity is $0$ . After being hit, it leaves the bat with velocity $v$ . The bat is in contact with the ball for time $\Delta t$ . We want the average force the bat exerts on the ball during that contact time.

Approach

Write initial and final momentum of the ball.
Find the change in momentum $\Delta p$ .
Use $F_{\text{avg}} = \Delta p / \Delta t$ .
Match the resulting expression to the given options.

Step-by-Step Reasoning

Initial momentum:

p_i = m(0) = 0

Final momentum:

p_f = mv

Change in momentum:

\Delta p = p_f - p_i = mv - 0 = mv

Average force:

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{mv}{\Delta t}

This corresponds to option B.

Key Takeaways

Use impulse: $F_{\text{avg}}\Delta t = \Delta p$ .
For an object starting from rest and reaching speed $v$ , $\Delta p = mv$ .
Average force depends on how quickly the momentum changes, so divide by $\Delta t$ .

Common Mistakes

Multiplying by $\Delta t$ instead of dividing (confusing force with impulse).
Using $\tfrac{1}{2}mv^2$ (kinetic energy) instead of momentum.
Forgetting the initial momentum is zero because the ball is stationary.

Things to Be Careful About

Momentum is a vector; here the question only needs the magnitude of the average force in the direction the ball is hit.
Options involving $\tfrac{1}{2}mv^2$ relate to energy per time (power), not force.
Always check units: $mv/\Delta t$ has units $\text{kg m s}^{-2} = \text{N}$ , which is correct for force.

Techniques used

apply the impulse–momentum relationcalculate change in momentum from initial and final velocitiesdivide impulse by contact time to obtain average force

Q121MMediumDynamics

A disc of mass $M$ is moving across a horizontal frictionless surface with constant velocity $u$ . It collides with a stationary disc of mass $4M$ .

The diagram shows the view from above of the motion of the two discs before and after the collision.

Fig. 1

What is the initial velocity $u$ of the disc of mass $M$ ?

Options

A $1.1\text{ m s}^{-1}$
B $1.4\text{ m s}^{-1}$
C $3.5\text{ m s}^{-1}$
D $5.7\text{ m s}^{-1}$

Pick an answer

DifficultyMedium

Worked solution

Working

Take $x$ along the original line of motion and $y$ perpendicular.

Conservation of momentum in $y$ :

M(4.0\sin 45^\circ) - 4M(v\sin 45^\circ)=0

v = 1.0\ \text{m s}^{-1}

Conservation of momentum in $x$ :

Mu = M(4.0\cos 45^\circ) + 4M(v\cos 45^\circ)

u = \cos 45^\circ\,(4.0 + 4v) = \frac{1}{\sqrt{2}}(4.0+4.0)=4\sqrt{2}=5.66\approx 5.7\ \text{m s}^{-1}

Answer

D

Final answer

D

Detailed explanation

Background Concept

For a collision on a frictionless horizontal surface, there is no external horizontal force on the system of two discs, so total momentum in the plane is conserved.

Momentum is a vector, so we conserve it component-wise:

in the original direction of motion ( $x$ -direction)
perpendicular to it ( $y$ -direction)

For each object, momentum is $\vec{p} = m\vec{v}$ . If you know the directions of the final velocities, it is usually easiest to resolve each velocity into components using

v_x = v\cos\theta,\qquad v_y = v\sin\theta

Understanding the Question

Initially:

disc of mass $M$ moves to the right at speed $u$
disc of mass $4M$ is stationary
So the initial momentum is entirely in the $x$ -direction, and the initial $y$ -momentum is zero.

After collision:

the $M$ disc moves at $4.0\ \text{m s}^{-1}$ at $45^\circ$ above the original line
the $4M$ disc moves at speed $v$ at $45^\circ$ below the original line

You are asked to find $u$ , the initial speed.

Approach

Choose axes with $x$ along the original direction of motion and $y$ perpendicular.
Use conservation of momentum in the $y$ -direction to find the unknown speed $v$ (because initial $y$ -momentum is zero).
Substitute $v$ into conservation of momentum in the $x$ -direction to find $u$ .
Match the result to the given options.

Step-by-Step Reasoning

diagram-1

1) Resolve the final velocities into components

For the $M$ disc (speed $4.0\ \text{m s}^{-1}$ at $45^\circ$ ):

v_{Mx} = 4.0\cos 45^\circ,\qquad v_{My} = 4.0\sin 45^\circ

For the $4M$ disc (speed $v$ at $45^\circ$ below the line):

its $x$ -component is positive: $v\cos 45^\circ$
its $y$ -component is negative: $-v\sin 45^\circ$

2) Conserve momentum in the $y$ -direction

Initial $y$ -momentum is zero.

Final $y$ -momentum:

p_y = M(4.0\sin 45^\circ) + 4M(-v\sin 45^\circ)

Set equal to zero:

M(4.0\sin 45^\circ) - 4M(v\sin 45^\circ)=0

Cancel $M\sin 45^\circ$ (non-zero):

4.0 - 4v = 0 \Rightarrow v = 1.0\ \text{m s}^{-1}

3) Conserve momentum in the $x$ -direction

Initial $x$ -momentum: $Mu$ .

Final $x$ -momentum:

p_x = M(4.0\cos 45^\circ) + 4M(v\cos 45^\circ)

So

Mu = M(4.0\cos 45^\circ) + 4M(v\cos 45^\circ)

Cancel $M$ :

u = \cos 45^\circ\,(4.0 + 4v)

Substitute $v=1.0$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$ :

u = \frac{1}{\sqrt{2}}(4.0+4.0)=\frac{8.0}{\sqrt{2}}=4\sqrt{2}=5.66\approx 5.7\ \text{m s}^{-1}

This corresponds to option D.

Key Takeaways

In 2D collisions, momentum must be conserved separately in perpendicular directions.
Use the direction information (angles) to resolve velocities into components.
If initial momentum in one direction is zero, that direction often lets you eliminate unknown speeds quickly.

Common Mistakes

Treating momentum as a scalar and equating speeds instead of vector components.
Forgetting the negative sign for the downward ( $-y$ ) component of the $4M$ disc.
Using $\sin$ and $\cos$ the wrong way round for the chosen angle.
Forgetting that the second disc has mass $4M$ when writing its momentum.

Things to Be Careful About

Define your axes clearly; the diagram’s dashed line is a strong hint to take that as the $x$ -axis.
Keep track of signs: above the line is $+y$ , below is $-y$ .
Round only at the end to match the options: $u=5.66$ rounds to $5.7\ \text{m s}^{-1}$ .

Techniques used

apply conservation of momentum in two perpendicular directionsresolve velocities into horizontal and vertical componentsequate vertical components to determine an unknown speedsubstitute into the horizontal momentum equation to find the initial speed

Q131MMedium-EasyKinematics

An object is dropped from rest on the Earth from a height of $2.0\text{ m}$ .

The same object is dropped from rest on the Moon from twice the height.

The acceleration of free fall on the Moon is approximately $16\%$ of the value on the Earth.

Assume that there are no resistive forces acting on the object.

What is the ratio

speed of the object just before hitting the surface on the Moon
speed of the object just before hitting the surface on the Earth $\,?$

Options

A $1.8$
B $2.5$
C $3.1$
D $3.5$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a drop from rest,

v^2 = u^2 + 2gh \; \Rightarrow\; v = \sqrt{2gh}

Earth: $h_E = 2.0\ \text{m}$ , $g_E = g$

v_E = \sqrt{2g(2.0)}

Moon: $h_M = 4.0\ \text{m}$ , $g_M = 0.16g$

v_M = \sqrt{2(0.16g)(4.0)}

Ratio (Earth to Moon):

\frac{v_E}{v_M} = \sqrt{\frac{2g(2.0)}{2(0.16g)(4.0)}} = \sqrt{\frac{2.0}{0.64}} = \sqrt{3.125} \approx 1.77 \approx 1.8

Answer

A

Final answer

A

Detailed explanation

Background Concept

For motion under constant acceleration (free fall with no air resistance), the kinematics (SUVAT) equation

v^2 = u^2 + 2as

relates speed $v$ , initial speed $u$ , acceleration $a$ , and displacement $s$ .

For an object dropped from rest, $u = 0$ . For vertical falling:

$a = g$ (or $g_M$ on the Moon),
$s = h$ (the drop height),
so

v = \sqrt{2gh}

This shows an important proportionality:

v \propto \sqrt{gh}

Understanding the Question

The object is dropped from rest:

on Earth from $h_E = 2.0\ \text{m}$ with acceleration $g_E = g$ ,
on the Moon from twice the height, so $h_M = 4.0\ \text{m}$ , with $g_M = 0.16g$ .

We want a ratio of the impact speeds. Using the no-resistance assumption means we can treat the acceleration as constant and use SUVAT.

(From the answer options provided, the intended ratio is consistent with $v_E / v_M$ , since that gives a value around $1.8$ .)

Approach

Use $v = \sqrt{2gh}$ for each situation (Earth and Moon).
Form the ratio so that common factors (like $2$ and $g$ ) cancel.
Substitute the given height ratio and $g_M = 0.16g_E$ .

Step-by-Step Reasoning

Earth impact speed:

v_E = \sqrt{2g_E h_E} = \sqrt{2g\,(2.0)}

Moon impact speed:

v_M = \sqrt{2g_M h_M} = \sqrt{2(0.16g)\,(4.0)}

Now form the ratio (Earth to Moon):

\frac{v_E}{v_M} = \frac{\sqrt{2g(2.0)}}{\sqrt{2(0.16g)(4.0)}} = \sqrt{\frac{2g(2.0)}{2(0.16g)(4.0)}}

Cancel $2$ and $g$ :

\frac{v_E}{v_M} = \sqrt{\frac{2.0}{0.64}} = \sqrt{3.125} \approx 1.77

Rounding to the nearest option gives $1.8$ (option A).

Key Takeaways

For free fall from rest with constant $g$ , $v = \sqrt{2gh}$ .
Speed depends on the square root of both height and gravitational field strength.
Ratios are efficient because common constants cancel.

Common Mistakes

Using $v = gt$ without being given (or finding) the time $t$ .
Forgetting that “twice the height” means $h_M = 4.0\ \text{m}$ , not $2.0\ \text{m}$ .
Not taking the square root when forming the speed ratio (treating $v$ as proportional to $gh$ instead of $\sqrt{gh}$ ).

Things to Be Careful About

The equation $v^2 = u^2 + 2as$ uses displacement $s$ (here equal to the height fallen).
Use $g_M = 0.16g_E$ exactly as stated (it is not “16% less”; it is “16% of”).
When matching to MCQ options, round appropriately (here $1.77 \to 1.8$ ).

Techniques used

apply the constant-acceleration equation v^2 = u^2 + 2asset initial speed to zero for an object dropped from restform a ratio using proportionality of v to sqrt(gh)

Q141MMedium-EasyKinematics

The graph shows how velocity $v$ varies with time $t$ for a bungee jumper.

Fig. 1

At which point is the bungee jumper momentarily at rest and at which point does she have zero acceleration?

Options

	jumper at rest	jumper with zero acceleration
A	Q	P
B	Q	R
C	R	Q
D	R	R

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Momentarily at rest means $v=0$ , so this is where the graph crosses the $t$ -axis: point $R$ .

Zero acceleration means $a=\dfrac{\mathrm{d}v}{\mathrm{d}t}=0$ , so the tangent is horizontal: at the maximum point $Q$ .

Answer

C

Final answer

C

Detailed explanation

Background Concept

A velocity–time graph shows how velocity $v$ changes with time $t$ .

Two key interpretations are:

Velocity at any instant is read directly from the vertical axis.
- The object is momentarily at rest when $v=0$ (i.e. where the graph crosses the time axis).
Acceleration is the rate of change of velocity:

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

On a $v$ – $t$ graph, $a$ is the gradient (slope) of the curve.

Zero acceleration means the gradient is zero, i.e. the curve has a horizontal tangent (typically at a maximum or minimum of $v$ ).

Understanding the Question

You are given a curved $v$ – $t$ graph for a bungee jumper with labelled points $P$ , $Q$ , and $R$ .

You must identify:

the point where the jumper is momentarily at rest (so $v=0$ ), and
the point where the jumper has zero acceleration (so gradient $=0$ ).

diagram-1

Approach

Find where $v=0$ by locating the labelled point on the time axis.
Find where $a=0$ by locating the labelled point where the graph has a turning point (horizontal tangent).
Match these two points to the options table.

Step-by-Step Reasoning

Momentarily at rest:
- At rest means velocity is zero: $v=0$ .
- On the graph, $v=0$ occurs where the curve crosses the $t$ -axis.
- The labelled crossing point is $R$ .
Zero acceleration:
- Acceleration is the gradient of the $v$ – $t$ graph.
- Zero acceleration means gradient $=0$ , i.e. a horizontal tangent.
- The curve has a maximum at $Q$ , so the tangent there is horizontal.
- Hence acceleration is zero at $Q$ .

Therefore: rest at $R$ and zero acceleration at $Q$ , which corresponds to option C.

Key Takeaways

On a $v$ $v$ – $t$ $t$ graph:
- $v=0$ (at rest) occurs where the graph crosses the time axis.
- $a=\mathrm{d}v/\mathrm{d}t$ is the gradient; $a=0$ where the tangent is horizontal.

Common Mistakes

Choosing the point with the lowest or highest velocity for “at rest” (rest requires $v=0$ , not a minimum/maximum).
Thinking acceleration is zero where $v=0$ ; these are independent (you can have $v=0$ but non-zero acceleration).
Using the value of $v$ instead of the gradient to decide about acceleration.

Things to Be Careful About

“Momentarily at rest” refers to an instant where $v=0$ (a crossing of the axis), not the starting point unless it is one of the labelled points.
For curved graphs, acceleration at a point is the gradient of the tangent at that point, not the average slope over a wider interval.

Techniques used

identify when an object is momentarily at rest from where the velocity is zerodetermine acceleration from the gradient of a velocity-time graphidentify zero acceleration from a horizontal tangent on a velocity-time graph

Q151MMedium-EasyForces, Density and PressureDynamics

A solid sphere, which is less dense than water, is held completely immersed in water a few metres below the surface. The density of the water is uniform.

The sphere is released. Immediately after release, the sphere rises.

Which row describes the changes in the magnitudes of the upthrust on the sphere and the resultant force on the sphere as it rises?

Options

	upthrust on the sphere	resultant force on the sphere
A	constant	decreasing
B	constant	increasing
C	decreasing	decreasing
D	decreasing	increasing

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Upthrust on a fully immersed object in uniform-density water is

U = \rho g V

so $U$ is constant (since $\rho$ , $g$ and $V$ are constant).

As the sphere rises it speeds up, so the drag force (downwards) increases, while weight is constant. Hence the resultant upward force decreases.

Answer

A

Final answer

A

Detailed explanation

Background Concept

Upthrust (buoyant force) on an object in a fluid comes from the pressure being larger on the lower surface than on the upper surface. For a fluid of uniform density, the upthrust on a fully immersed object is given by Archimedes’ principle:

U = \rho g V

where $\rho$ is the fluid density, $g$ is gravitational field strength, and $V$ is the volume of fluid displaced (equal to the object’s volume if it is completely immersed).

Although the absolute pressure increases with depth ( $p = \rho g h$ ), the difference in pressure between the bottom and top of the object depends only on the vertical separation, so the net upthrust is independent of depth (provided $\rho$ is constant).

When an object moves through a fluid, a drag/resistive force acts opposite to its motion. As speed increases, drag generally increases, reducing the resultant (net) force in the direction of motion. This is why terminal velocity can occur when drag grows until the resultant force is zero.

Understanding the Question

A solid sphere less dense than water is held completely submerged and released. Immediately it rises, meaning initially:

U > W

The question asks, as the sphere rises, how the magnitude of:

the upthrust $U$ on it, and
the resultant force on it
change.

The key information is: “completely immersed” and “density of the water is uniform”.

Approach

Decide whether $U$ changes with height/depth for a fully immersed object in uniform-density water.
Identify forces on the rising sphere: upthrust upward, weight downward, drag downward.
As it rises, it accelerates and its speed increases (at least initially), so infer the change in drag and hence the change in the resultant force.
Match these trends to the table.

Step-by-Step Reasoning

Upthrust:

Since the sphere stays fully immersed and water density is uniform, the displaced volume $V$ is constant.
Therefore

U = \rho g V

and $U$ is constant as it rises.

Resultant force:

Forces on the sphere while rising are:

diagram-1

Weight $W = mg$ is constant.
Upthrust $U$ is constant.
Drag $D$ acts downward (opposes upward motion). As the sphere rises from rest, its speed increases, so $D$ increases.

Resultant upward force is

F_{\text{res}} = U - W - D

As $D$ increases, $F_{\text{res}}$ decreases.

So: upthrust constant; resultant force decreasing → row A.

Key Takeaways

For a fully immersed object in a uniform-density fluid, upthrust $U = \rho g V$ is constant and does not depend on depth.
As an object speeds up in a fluid, drag increases, so the resultant force in the direction of motion tends to decrease.

Common Mistakes

Thinking upthrust decreases as the object rises because pressure is lower near the surface: absolute pressure decreases, but the net upthrust remains $\rho g V$ for uniform $\rho$ .
Ignoring drag and concluding the resultant force stays constant ( $U-W$ ). In real fluids, drag increases as speed increases.
Getting the drag direction wrong (it always opposes the motion).

Things to Be Careful About

The statement “density of the water is uniform” is crucial: it allows $U$ to be treated as constant with height.
“Immediately after release” means the sphere starts from (approximately) zero speed, but the question asks what happens as it rises, i.e. once it begins to gain speed and drag grows.
The question asks about magnitudes of forces: the resultant force remains upward but its magnitude decreases as drag increases.

Techniques used

apply Archimedes' principle to determine the upthrustidentify all forces acting on a rising object in a fluidreason qualitatively how drag changes with speeddeduce how the resultant force changes as drag increases

Q161MMedium-EasyForces, Density and Pressure

A uniform bar of weight $200\text{ N}$ and length $4.0\text{ m}$ is freely hinged on a wall at one end. The bar is horizontal and is held in equilibrium by a cable attached at a distance of $0.50\text{ m}$ from the other end. The cable is at an angle of $35^\circ$ to the horizontal.

Fig. 1

What is the tension $T$ in the cable?

Options

A $140\text{ N}$
B $170\text{ N}$
C $200\text{ N}$
D $400\text{ N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Take moments about the hinge.

Distance of cable from hinge:

4.0 - 0.50 = 3.5\ \text{m}

Weight acts at centre of bar, distance from hinge:

2.0\ \text{m}

Only the vertical component of tension produces a moment about the hinge, so moment of tension:

(3.5)(T\sin 35^{\circ})

Equilibrium:

(3.5)(T\sin 35^{\circ}) = (200)(2.0)

T = \frac{400}{3.5\sin 35^{\circ}} \approx 2.0 \times 10^2\ \text{N}

Answer

C

Final answer

C

Detailed explanation

Background Concept

For a rigid body in equilibrium:

The resultant force is zero.
The resultant moment (torque) about any point is zero.

Using moments is usually simplest when there is a hinge/pivot, because any unknown hinge forces produce no moment about the hinge (their lines of action pass through the pivot).

The moment of a force about a pivot is

\tau = F \times d_\perp

where $d_\perp$ is the perpendicular distance from the pivot to the line of action of the force. Equivalently, if the force makes an angle $\theta$ to the bar, you can use the component perpendicular to the bar.

Understanding the Question

A uniform horizontal bar (length $4.0\ \text{m}$ , weight $200\ \text{N}$ ) is hinged at the wall at one end. A cable supports the bar, attached close to the free end: it is $0.50\ \text{m}$ from the far end, and makes an angle of $35^{\circ}$ to the horizontal bar.

You are asked to find the cable tension $T$ .

Key geometry:

Weight acts at the centre of a uniform bar, i.e. at $2.0\ \text{m}$ from the hinge.
The cable attachment point is $4.0 - 0.50 = 3.5\ \text{m}$ from the hinge.

Approach

Take moments about the hinge:

Clockwise moment is due to the bar's weight.
Anticlockwise moment is due to the vertical component of the cable tension.

The horizontal component of tension acts along the bar, so its line of action passes through the hinge and produces no turning effect about the hinge.

Step-by-Step Reasoning

diagram-1

Distances from the hinge:

d_W = 2.0\ \text{m}, \qquad d_T = 4.0 - 0.50 = 3.5\ \text{m}

Resolve tension into components relative to the horizontal bar:

Vertical component: $T\sin 35^{\circ}$ (perpendicular to the bar)
Horizontal component: $T\cos 35^{\circ}$ (along the bar)

Only $T\sin 35^{\circ}$ contributes to the moment about the hinge.

Equate clockwise and anticlockwise moments about the hinge:

(3.5)(T\sin 35^{\circ}) = (200)(2.0)

Solve for $T$ :

T = \frac{400}{3.5\sin 35^{\circ}} \approx 1.99\times 10^2\ \text{N} \approx 200\ \text{N}

So the correct option is C.

Key Takeaways

For equilibrium, set the sum of moments about a pivot to zero.
Choose the pivot to eliminate unknown forces (here, the hinge).
Only the component of a force perpendicular to the lever arm produces a moment.

Common Mistakes

Using the wrong distance for the cable (forgetting it is $0.50\ \text{m}$ from the far end, not from the hinge).
Using $T\cos 35^{\circ}$ as the vertical component (swapping sine and cosine).
Taking moments using the full $T$ instead of its perpendicular component.

Things to Be Careful About

The bar is horizontal, so the lever arm distances are simple horizontal distances from the hinge.
The weight acts at the centre because the bar is uniform.
In moment calculations, always check which component actually has a perpendicular distance from the pivot; components acting through the pivot give zero moment.

Techniques used

take moments about the hinge (pivot)resolve the tension into componentsset clockwise and anticlockwise moments equal for equilibrium

Q171MMedium-EasyForces, Density and Pressure

The diagrams all show a pair of equal forces acting on a metre rule.

Which diagram shows forces that provide a couple and zero resultant force?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

For a couple: forces must be equal, opposite, parallel and act along different lines so the resultant force is zero.

A: forces are same direction (\Rightarrow) resultant (\neq 0).
C: equal and opposite but same line of action (\Rightarrow) no couple.
D: forces are not parallel (\Rightarrow) not a couple and resultant (\neq 0).
B: equal and opposite, parallel and separated (\Rightarrow) couple and zero resultant.

Answer

B

Final answer

B

Detailed explanation

Background Concept

A couple is produced by two forces that:

have equal magnitude,
act in opposite directions,
are parallel,
act along different (separated) lines of action.

For a couple:

The resultant force is zero because the forces cancel.
The resultant moment (torque) is not zero because the forces act at different points, producing a turning effect.

The moment of a couple is

\tau = Fd

where $F$ is the magnitude of one force and $d$ is the perpendicular distance between their lines of action.

Understanding the Question

Each option shows two equal forces on a metre rule. We need the one that:

has zero resultant force (forces must be equal and opposite), and
produces a couple (forces must be parallel and not act along the same line).

Approach

For each diagram, check in this order:

Are the forces equal and opposite? If not, resultant force is not zero.
Are they parallel? If not, they cannot form a couple.
Do they act along different lines of action (separated)? If they are collinear, the turning effect cancels to zero.

Step-by-Step Reasoning

A: both forces point downward. They add to give a net downward force $2F$ , so resultant force is not zero. Not correct.
B: one force is upward and the other is downward, with equal magnitude. So net force is $F - F = 0$ .
The forces are also parallel (both vertical) and act at different points along the rule, so their lines of action are separated. This produces a non-zero turning effect (a couple). Correct.
C: forces are equal and opposite and parallel, but they act at the same position (same line of action). With no separation $d = 0$ , the couple moment $\tau = Fd = 0$ . So there is no couple. Not correct.
D: one force is vertical while the other is diagonal. They are not parallel, so they do not form a couple, and their vector sum will not be zero in general. Not correct.

Therefore the correct diagram is B.

Key Takeaways

A couple requires equal and opposite parallel forces with separation between their lines of action.
Zero resultant force alone is not enough: if the lines of action coincide, the turning effect is zero.

Common Mistakes

Choosing C because the forces cancel: they do cancel, but there is no separation, so no couple.
Thinking any two opposite forces form a couple: they must also be parallel and non-collinear.

Things to Be Careful About

“Separated” means separated perpendicularly between lines of action, not just “at different points” in a vague sense.
If forces are not parallel (like D), you cannot classify them as a couple even if they look like they might twist the rule.

Techniques used

identify whether two forces are equal and oppositecheck whether forces are parallel and separated to form a coupledetermine resultant force from vector additionuse the definition of a couple as a pure turning effect with zero resultant

Q181MMedium-EasyWork, Energy and Power

A ball of mass $m$ is thrown up to height $h$ in air with an initial velocity $v$ , as shown.

Fig. 1

Air resistance is negligible. The acceleration of free fall is $g$ .

What is the total work done by the gravitational force on the ball during its flight from $P$ to $Q$ ?

Options

A zero
B $\dfrac{1}{2}mv^{2}$
C $mgh$
D $2mgh$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Work done by gravity:

W_g = -\Delta E_p

From $P$ to $Q$ the ball starts and ends at the same height, so $\Delta E_p = 0$ .

Hence $W_g = 0$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

The gravitational force is a conservative force. For any conservative force, the work done depends only on the initial and final positions, not on the path taken.

For gravity near Earth (constant $g$ ), gravitational potential energy is

E_p = mgh

and the work done by gravity is related to the change in gravitational potential energy by

W_g = -\Delta E_p

So if an object ends up at the same height it started, $\Delta E_p = 0$ and the total work done by gravity over the whole journey is zero.

Understanding the Question

A ball is thrown from point $P$ on the ground, follows a curved (projectile) path, rises to a maximum height $h$ , and lands back on the ground at point $Q$ .

You are asked for the total work done by the gravitational force from $P$ to $Q$ (the entire flight), with air resistance negligible.

Key observation: points $P$ and $Q$ are at the same vertical height.

Approach

Use the fact that gravitational work depends only on the change in height between the start and end points.

Identify the initial height and final height.
Find $\Delta E_p = mg(\Delta h)$ .
Use $W_g = -\Delta E_p$ to get the total work.

Step-by-Step Reasoning

Initial height at $P$ : take $h_P = 0$ (ground level).

Final height at $Q$ : $h_Q = 0$ (also ground level).

So the change in height is

\Delta h = h_Q - h_P = 0 - 0 = 0

Hence the change in gravitational potential energy is

\Delta E_p = mg\Delta h = mg(0) = 0

Therefore total work done by gravity is

W_g = -\Delta E_p = -0 = 0

So the correct option is A (zero).

(Although gravity does negative work on the way up and positive work on the way down, these cancel over the full trip because the ball returns to its original height.)

Key Takeaways

Gravity is conservative: total work depends only on initial and final heights.
$W_g = -\Delta E_p$ and $\Delta E_p = mg\Delta h$ .
If start and finish heights are the same, the net work done by gravity is zero.

Common Mistakes

Using the maximum height $h$ and stating the work is $mgh$ or $2mgh$ without considering the return to the ground.
Confusing the magnitude of work done during ascent ( $-mgh$ ) with the total work over the whole flight.
Mixing up the sign: gravity does negative work when displacement is upward, positive work when downward.

Things to Be Careful About

The question asks for work from $P$ to $Q$ (whole journey), not just $P$ to the highest point.
Work done by gravity is linked to vertical displacement only; horizontal motion does not change $E_p$ .
Since $P$ and $Q$ are at the same level, $\Delta h = 0$ regardless of the path shape.

Techniques used

relate work done by a conservative force to change in potential energycompare initial and final heights to find change in gravitational potential energyuse conservation of mechanical energy for motion with negligible resistance

Q191MMedium-EasyDeformation of SolidsWork, Energy and Power

A spring of spring constant $30\text{ N m}^{-1}$ is suspended vertically from its top. The spring obeys Hooke’s law. Initially the spring is not compressed and not stretched. A mass of $0.50\text{ kg}$ is attached to the bottom of the spring. The mass is released from rest and falls.

Frictional effects are negligible.

In the motion that follows, what is the maximum extension of the spring?

Options

A $0.017\text{ m}$
B $0.033\text{ m}$
C $0.16\text{ m}$
D $0.33\text{ m}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

At maximum extension $v=0$ , so loss of GPE $=$ elastic energy:

mgx = \frac{1}{2}kx^2

x = \frac{2mg}{k} = \frac{2(0.50)(9.81)}{30} = 0.327\ \text{m} \approx 0.33\ \text{m}

Answer

D

Final answer

D

Detailed explanation

Background Concept

For a vertical spring that obeys Hooke’s law, the spring force is

F = kx

where $k$ is the spring constant and $x$ is the extension from the natural (unstretched) length.

When frictional effects are negligible, mechanical energy is conserved. As the mass falls, it loses gravitational potential energy (GPE) and stores energy as elastic potential energy (EPE) in the spring:

\Delta E_{\text{GPE}} = mg\,\Delta h, \qquad E_{\text{EPE}} = \frac{1}{2}kx^2

At the instant of maximum extension, the speed is momentarily zero, so there is no kinetic energy.

Understanding the Question

The spring starts at its natural length. A $0.50\ \text{kg}$ mass is attached and released from rest. The question asks for the maximum extension of the spring during the subsequent motion.

Key point: maximum extension occurs at the turning point where $v=0$ , not at the equilibrium position (where $kx = mg$ ).

Approach

Use conservation of energy between:

initial state: mass at the natural length, released from rest;
turning point: spring extended by $x$ , mass momentarily at rest.

The mass falls a distance $x$ , so it loses GPE $mgx$ . This becomes elastic energy $\tfrac{1}{2}kx^2$ .

Step-by-Step Reasoning

Let the maximum extension be $x$ .
GPE lost when the mass moves down by $x$ :

\text{GPE lost} = mgx

Elastic potential energy stored in the spring at extension $x$ :

\text{EPE gained} = \frac{1}{2}kx^2

At maximum extension, the mass is instantaneously at rest, so kinetic energy is zero. With negligible friction:

mgx = \frac{1}{2}kx^2

Solve for $x$ (take the non-zero solution):

x = \frac{2mg}{k}

Substitute values $m=0.50\ \text{kg}$ , $g=9.81\ \text{m s}^{-2}$ , $k=30\ \text{N m}^{-1}$ :

x = \frac{2(0.50)(9.81)}{30} = 0.327\ \text{m} \approx 0.33\ \text{m}

So the correct option is D.

Key Takeaways

Maximum extension in a vertical spring drop is found using energy: $mgx = \tfrac{1}{2}kx^2$ .
The maximum extension is

x_{\max} = \frac{2mg}{k}

which is twice the equilibrium extension $x_{\text{eq}} = \tfrac{mg}{k}$ .

Common Mistakes

Using the equilibrium condition $kx = mg$ and giving $x = \tfrac{mg}{k}$ (this gives the extension at equilibrium, not the maximum extension).
Including kinetic energy at the turning point (it is zero there).
Using the wrong distance for the GPE change (the mass falls by $x$ from the release point to maximum extension).

Things to Be Careful About

The spring starts unstretched, so the initial elastic energy is zero.
The turning point is defined by $v=0$ , not by zero acceleration.
Check units: $\tfrac{mg}{k}$ has units of metres because $\text{N}/(\text{N m}^{-1}) = \text{m}$ .
Rounding: $0.327\ \text{m}$ rounds to $0.33\ \text{m}$ , matching option D.

Techniques used

apply conservation of mechanical energyequate loss of gravitational potential energy to elastic potential energysolve a quadratic relationship for the extensionchoose the correct multiple-choice option from a calculated value

Q201MMedium-EasyDeformation of Solids

A wire has original length $L$ and cross-sectional area $A$ . A tensile force $F$ is applied to the wire which causes it to have extension $x$ . The wire obeys Hooke’s law.

What is an expression for the Young modulus of the material from which the wire is made?

Options

A $\text{stress} \times \dfrac{x}{L}$
B $\dfrac{F}{A \times \text{strain}}$
C $\dfrac{F \times x}{A \times L}$
D $\dfrac{\text{strain}}{\text{stress}}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Young modulus

E = \frac{\text{stress}}{\text{strain}}

\text{stress} = \frac{F}{A}, \quad \text{strain} = \frac{x}{L}

So

E = \frac{\frac{F}{A}}{\text{strain}} = \frac{F}{A \times \text{strain}}

Answer

B

Final answer

B

Detailed explanation

Background Concept

For a wire obeying Hooke’s law (within the limit of proportionality), the extension is proportional to the applied tensile force. A material property that describes how stiff the material is in tension is the Young modulus $E$ .

Young modulus is defined as

E = \frac{\text{stress}}{\text{strain}}

where

\text{stress} = \frac{F}{A}

and

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

Stress has units of $\text{Pa}$ (since $\text{N m}^{-2}$ ), and strain is dimensionless, so $E$ also has units of $\text{Pa}$ .

Understanding the Question

You are told a wire has original length $L$ and cross-sectional area $A$ . When a tensile force $F$ is applied, the wire extends by $x$ . The question asks for an expression for the Young modulus in terms of these quantities (or in terms of stress and strain), and provides four options.

So we need to start from $E = \text{stress}/\text{strain}$ and see which option matches.

Approach

Write the definition $E = \text{stress}/\text{strain}$ .
Replace stress by $F/A$ .
Replace strain by $x/L$ (or keep it as “strain” if an option uses that word).
Compare with the listed options.

Step-by-Step Reasoning

Start from the definition:

E = \frac{\text{stress}}{\text{strain}}

Substitute $\text{stress} = F/A$ :

E = \frac{\frac{F}{A}}{\text{strain}} = \frac{F}{A \times \text{strain}}

This is exactly option B.

(If you also substitute $\text{strain} = x/L$ , you get $E = \dfrac{FL}{Ax}$ , but that exact form is not one of the options; option B is the equivalent expression.)

Key Takeaways

Young modulus is defined by

E = \frac{\text{stress}}{\text{strain}}

For a wire, $\text{stress} = F/A$ and $\text{strain} = x/L$ .
In MCQs, check for equivalent algebraic forms of the same definition.

Common Mistakes

Writing $E$ as $\text{strain}/\text{stress}$ (this is the inverse, option D).
Multiplying stress and strain instead of dividing (options A and C correspond to $\text{stress} \times \text{strain}$ ).
Forgetting that strain is dimensionless, leading to incorrect unit reasoning.

Things to Be Careful About

The word “strain” already includes the ratio $x/L$ ; don’t treat it as a separate quantity with units.
Stress must be $F/A$ , not $F \times A$ .
Check whether the option uses words (stress/strain) or symbols ( $F, A, x, L$ ) and convert appropriately.

Techniques used

use the definition of Young modulus as stress divided by strainsubstitute stress as force per unit areasubstitute strain as extension divided by original lengthmatch the derived expression to the given options

Q211MMedium-EasyDeformation of Solids

A wire is stretched by a gradually increasing force. The force–extension graph for the wire is shown.

Fig. 1

Which statement must be correct?

Options

A Point $Q$ is the elastic limit.
B Point $R$ is the limit of proportionality.
C The area under the graph from $P$ to $S$ is the elastic potential energy stored in the wire.
D The area under the graph from $P$ to $S$ is the work done in stretching the wire.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Work done in stretching is

W = \int F\,\mathrm{d}x

so it equals the area under the $F$ –extension graph between the relevant extensions (here from $P$ to $S$ ).

Answer

D

Final answer

D

Detailed explanation

Background Concept

A force–extension graph plots applied force $F$ against extension $x$ .

Work done in stretching is the energy transferred by the force as the wire extends:

W = \int F\,\mathrm{d}x

Graphically, this integral is the area under the $F$ – $x$ graph between the chosen extensions.

Elastic potential energy is the energy stored recoverably in the material. If the material is stretched only within its elastic range, then the work done is stored as elastic potential energy. If plastic deformation occurs, some work done is dissipated (e.g. as internal heating), so it is not all stored.
The limit of proportionality is the point where $F \propto x$ stops being true (end of the straight-line part).
The elastic limit is the point beyond which the wire does not return to its original length when the force is removed. This can be at or beyond the limit of proportionality.

Understanding the Question

You are shown a graph that is initially a straight line (Hooke’s law region) and then becomes curved with decreasing gradient. Points $Q$ , $R$ , and $S$ are marked along the curve.

The question asks which statement is guaranteed true just from what a force–extension graph represents.

Approach

Check each option for whether it is a definition that must always apply:

Statements about named points like elastic limit or limit of proportionality depend on where the straight line ends and where plasticity begins.
Statements about the area under the graph can be tested using the definition of work done ( $\int F\,\mathrm{d}x$ ), which is always valid.

Step-by-Step Reasoning

From the graph shape, the straight-line region ends at $Q$ . That is the best candidate for the limit of proportionality, not $R$ . So B is not a “must”.
Whether $Q$ is the elastic limit cannot be concluded from the loading curve alone. The elastic limit is defined by whether the wire returns to its original length on unloading; you would need an unloading curve or information about permanent extension. So A is not a “must”.
The area under the graph from $P$ to $S$ is:

\int_{x_P}^{x_S} F\,\mathrm{d}x

This is, by definition, the work done by the stretching force in increasing the extension from $P$ to $S$ . Therefore D must be correct.

Option C says the same area is the elastic potential energy stored. That would only be true if the stretching from $P$ to $S$ is entirely elastic (no plastic deformation). Since the graph is non-linear after $Q$ , plasticity may have started, and we are not told unloading returns to zero extension. So C is not guaranteed.

Hence the only statement that must be correct is D.

Key Takeaways

The area under an $F$ – $x$ graph always gives work done: $W = \int F\,\mathrm{d}x$ .
Elastic potential energy stored equals work done only if deformation is fully elastic.
The limit of proportionality is where the graph stops being a straight line.

Common Mistakes

Assuming “area under graph” always means elastic potential energy (it does not if plastic deformation occurs).
Confusing limit of proportionality with elastic limit.
Misidentifying the limit of proportionality as a point on the curved section (it is at the end of the straight-line part).

Things to Be Careful About

“Must be correct” means true regardless of material behaviour after the linear region.
The work done from $P$ to $S$ is the area under the curve between those extensions, not necessarily from the origin.
A single loading curve cannot prove where the elastic limit is; unloading behaviour determines that.

Techniques used

interpret a force-extension graph to identify linear and non-linear regionsuse the area under a force-extension graph to determine work donedistinguish between work done and elastic potential energy stored

Q221MMedium-EasyDeformation of Solids

A spring has an unstretched length of $0.30\text{ m}$ and a spring constant of $400\text{ N m}^{-1}$ . An object is suspended from the spring and the spring is deformed within its limit of proportionality. The new length of the spring is $0.50\text{ m}$ .

What is the elastic potential energy stored in the spring?

Options

A $8.0\text{ J}$
B $16\text{ J}$
C $40\text{ J}$
D $50\text{ J}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Extension:

x = 0.50 - 0.30 = 0.20\ \text{m}

Elastic potential energy:

E = \frac{1}{2}kx^2 = \frac{1}{2}(400)(0.20)^2 = 8.0\ \text{J}

Answer

A

Final answer

A

Detailed explanation

Background Concept

For a spring that obeys Hooke's law (i.e. within the limit of proportionality), the force is proportional to the extension:

F = kx

where $k$ is the spring constant and $x$ is the extension from the natural (unstretched) length.

The elastic potential energy stored in the spring is the work done to stretch it from $x=0$ to its final extension $x$ . Since the force increases linearly with $x$ , the energy is the area under the $F$ -against- $x$ graph (a triangle):

E = \frac{1}{2}Fx = \frac{1}{2}kx^2

Understanding the Question

You are given:

Unstretched length $= 0.30\ \text{m}$
New length $= 0.50\ \text{m}$
Spring constant $k = 400\ \text{N m}^{-1}$

You must find the elastic potential energy stored when it is stretched to this new length.

Approach

Find the extension $x$ by subtracting the unstretched length from the new length.
Use $E = \frac{1}{2}kx^2$ because the spring is within its limit of proportionality (so Hooke's law applies).
Substitute values and calculate, then match to the options.

Step-by-Step Reasoning

Extension:

x = 0.50 - 0.30 = 0.20\ \text{m}

Elastic potential energy:

E = \frac{1}{2}kx^2

Substitute $k=400\ \text{N m}^{-1}$ and $x=0.20\ \text{m}$ :

E = \frac{1}{2}(400)(0.20)^2

Calculate $(0.20)^2 = 0.040$ :

E = 200 \times 0.040 = 8.0\ \text{J}

So the correct option is A.

Key Takeaways

Extension is the change in length: $x = L - L_0$ .
For a Hooke's law spring, elastic energy is $E = \frac{1}{2}kx^2$ .
Energy depends on the square of the extension, so doubling $x$ would quadruple $E$ .

Common Mistakes

Using the total length ( $0.50\ \text{m}$ ) as $x$ instead of the extension ( $0.20\ \text{m}$ ).
Forgetting the factor of $\frac{1}{2}$ and calculating $E = kx^2$ .
Mixing up $E = Fx$ (only true for constant force) instead of the triangular area for a linearly increasing force.

Things to Be Careful About

The phrase “within its limit of proportionality” is the clue that Hooke's law applies, so $E = \frac{1}{2}kx^2$ is valid.
Check units: $k$ in $\text{N m}^{-1}$ and $x$ in $\text{m}$ gives energy in $\text{N m} = \text{J}$ .
Square the extension before multiplying; errors often occur because $x^2$ is small.

Techniques used

determine the extension from the change in lengthapply the elastic potential energy formula for a Hooke's law springsubstitute values and evaluate with correct unitsidentify the correct multiple-choice option

Q231MMedium-EasyDeformation of Solids

A wire consists of a $3.0\text{ m}$ length of metal $X$ joined to a $1.0\text{ m}$ length of metal $Y$ .

The cross-sectional area of the wire is uniform.

Fig. 1

A load hung from the wire causes metal $X$ to extend by $1.5\text{ mm}$ and metal $Y$ to extend by $1.0\text{ mm}$ .

The same load is then hung from a second wire of the same cross-sectional area, consisting of a $1.0\text{ m}$ length of metal $X$ and a $3.0\text{ m}$ length of metal $Y$ .

Both wires are extended within their limit of proportionality.

What is the total extension of this second wire?

Options

A $2.5\text{ mm}$
B $3.5\text{ mm}$
C $4.8\text{ mm}$
D $5.0\text{ mm}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a wire within the limit of proportionality,

\Delta L = \frac{FL}{AE}

Same load and same cross-sectional area $\Rightarrow$ same $\frac{F}{A}$ , so for each metal $\Delta L \propto L$ .

Metal $X$ : $3.0\ \text{m} \to 1.5\ \text{mm}$ , so $1.0\ \text{m} \to \frac{1.5}{3.0} = 0.50\ \text{mm}$ .

Metal $Y$ : $1.0\ \text{m} \to 1.0\ \text{mm}$ , so $3.0\ \text{m} \to 3.0\ \text{mm}$ .

Total extension of second wire:

0.50 + 3.0 = 3.5\ \text{mm}

Answer

B

Final answer

B

Detailed explanation

Background Concept

When a wire is stretched by a tensile force (load) and it remains within its limit of proportionality, it obeys Hooke’s law behaviour: extension is proportional to force.

A more general way to express this for a uniform wire is via Young modulus $E$ :

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

Rearranging gives the extension of a wire of length $L$ and cross-sectional area $A$ under force $F$ :

\Delta L = \frac{FL}{AE}

So for the same material (same $E$ ) under the same force and with the same area, the extension is directly proportional to length $L$ .

For two sections joined in series (one after the other), each section extends by its own amount and the total extension is the sum:

\Delta L_{\text{total}} = \Delta L_X + \Delta L_Y

Understanding the Question

The first composite wire has:

$3.0\ \text{m}$ of metal $X$ that extends $1.5\ \text{mm}$ ,
$1.0\ \text{m}$ of metal $Y$ that extends $1.0\ \text{mm}$ ,
under a certain load.

A second composite wire uses the same two metals but swaps the lengths:

$1.0\ \text{m}$ of $X$ ,
$3.0\ \text{m}$ of $Y$ ,
with the same load and the same cross-sectional area.

We must find the total extension of the second wire.

The key clues are:

“same load” and “same cross-sectional area” (so the stress $F/A$ is the same in every case),
“within their limit of proportionality” (so $\Delta L \propto L$ for each metal).

Approach

Use the first wire to find the extension per metre for metal $X$ and for metal $Y$ under this load.
Scale those extensions to the new lengths in the second wire.
Add the two extensions to get the total.

Step-by-Step Reasoning

Because the load and area are unchanged, $F$ and $A$ are the same in both experiments. For any single metal,

\Delta L = \frac{FL}{AE}

so for fixed $F$ and $A$ (and fixed $E$ for that metal), $\Delta L \propto L$ .

Metal $X$

Given $3.0\ \text{m}$ of $X$ extends $1.5\ \text{mm}$ .

Extension per metre for $X$ :

\frac{1.5\ \text{mm}}{3.0\ \text{m}} = 0.50\ \text{mm m}^{-1}

So $1.0\ \text{m}$ of $X$ extends:

0.50\ \text{mm}

Metal $Y$

Given $1.0\ \text{m}$ of $Y$ extends $1.0\ \text{mm}$ .

Extension per metre for $Y$ :

\frac{1.0\ \text{mm}}{1.0\ \text{m}} = 1.0\ \text{mm m}^{-1}

So $3.0\ \text{m}$ of $Y$ extends:

3.0 \times 1.0 = 3.0\ \text{mm}

Total extension of the second wire

Add the two series extensions:

\Delta L_{\text{total}} = 0.50 + 3.0 = 3.5\ \text{mm}

This corresponds to option B.

Key Takeaways

Within the limit of proportionality, a uniform wire obeys $\Delta L = \frac{FL}{AE}$ .
For the same material under the same load and same area, extension is directly proportional to length.
For sections joined in series, total extension is the sum of the individual extensions.

Common Mistakes

Assuming the extension depends only on the total length and ignoring that different materials have different $E$ .
Adding $1.5\ \text{mm}$ and $1.0\ \text{mm}$ again (that would be for the original wire, not the swapped one).
Forgetting to scale the given extensions to the new lengths (e.g. using $1.5\ \text{mm}$ for $1.0\ \text{m}$ of $X$ ).

Things to Be Careful About

The reason $\Delta L \propto L$ here is that $F$ , $A$ , and $E$ are constant for each metal; changing any of these would break that simple scaling.
The question states both wires are within the limit of proportionality; without this, extension might not be proportional to force (and the scaling could fail).
Keep units consistent (mm with mm, m with m); here ratios make it straightforward.

Techniques used

use the Young modulus relation for extension under a tensile loadscale extension in proportion to length for the same material under the same stressadd extensions of two sections in series

Q241MMedium-EasyWaves

The graph shows the variation of the displacement with distance for a progressive wave at one instant in time.

Fig. 1

The period of the wave is $91\text{ ms}$ .

What can be determined about the wave?

Options

A It has a velocity of $0.44\text{ m s}^{-1}$ and a frequency of $11\text{ Hz}$ .
B It has a velocity of $0.55\text{ m s}^{-1}$ and a wavelength of $5.0\text{ cm}$ .
C It is longitudinal and has a frequency of $11\text{ Hz}$ .
D It is transverse and has a wavelength of $4.0\text{ cm}$ .

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

From the displacement–distance graph, one full cycle is from crest at $x=0$ to next crest at $x\approx 4.0\ \text{cm}$ , so

\lambda = 4.0\ \text{cm} = 4.0 \times 10^{-2}\ \text{m}

Given period $T = 91\ \text{ms} = 9.1\times 10^{-2}\ \text{s}$ ,

f = \frac{1}{T} = \frac{1}{9.1\times 10^{-2}} \approx 11\ \text{Hz}

v = f\lambda = 11 \times 4.0\times 10^{-2} = 0.44\ \text{m s}^{-1}

Answer

A

Final answer

A

Detailed explanation

Background Concept

A progressive wave transfers energy through space. Key quantities are:

Period $T$ : time for one complete oscillation at a point.
Frequency $f$ : number of oscillations per second, related by

f = \frac{1}{T}

Wavelength $\lambda$ : distance between two neighbouring points in phase (e.g. crest-to-crest or trough-to-trough) on a displacement–distance snapshot.
Wave speed $v$ : speed the wave pattern travels, related by the wave equation

v = f\lambda

A graph of displacement against distance at one instant lets you read $\lambda$ directly, but it does not uniquely tell you whether the wave is transverse or longitudinal without extra context.

Understanding the Question

You are given:

A displacement–distance graph at one instant.
The period $T = 91\ \text{ms}$ .

You must decide which option states something that can be determined correctly about the wave. The key is to extract $\lambda$ from the graph, then use $T$ to find $f$ and hence $v$ .

Approach

Read the wavelength $\lambda$ from the graph by finding the distance for one full cycle (same phase point to next same phase point).
Convert the period into seconds and calculate frequency using $f=1/T$ .
Compute the wave speed from $v=f\lambda$ .
Compare your computed $f$ , $\lambda$ , and $v$ with the options. Also check whether any claim about transverse/longitudinal is actually justified.

Step-by-Step Reasoning

1) Determine wavelength from the snapshot
A wavelength is the horizontal distance between two adjacent crests.

The graph shows a crest at $x=0$ and the next crest at about $x=4.0\ \text{cm}$ .
So

\lambda \approx 4.0\ \text{cm} = 4.0\times 10^{-2}\ \text{m}

(You could also use trough-to-trough, which would give the same result.)

2) Find frequency from the given period
Convert $91\ \text{ms}$ to seconds:

T = 91\ \text{ms} = 91\times 10^{-3}\ \text{s} = 9.1\times 10^{-2}\ \text{s}

Then

f = \frac{1}{T} = \frac{1}{9.1\times 10^{-2}} \approx 11\ \text{Hz}

3) Calculate wave speed

v = f\lambda = 11 \times 4.0\times 10^{-2} = 0.44\ \text{m s}^{-1}

4) Match to the options

Option A states $v=0.44\ \text{m s}^{-1}$ and $f=11\ \text{Hz}$ , which matches the calculations.
Any option claiming “transverse” or “longitudinal” is not determinable from this information alone.

Therefore the correct choice is A.

Key Takeaways

From a displacement–distance snapshot you can read wavelength.
From the given period, you can find frequency using $f=1/T$ .
Use the wave equation $v=f\lambda$ to calculate wave speed.
The wave type (transverse/longitudinal) needs additional context; it is not reliably deduced from this graph alone.

Common Mistakes

Using the distance between a crest and the next zero crossing as $\lambda$ (that is $\lambda/4$ ).
Forgetting to convert $\text{ms}$ to $\text{s}$ , giving a frequency $1000$ times too large.
Claiming the wave is transverse/longitudinal just because the graph is drawn with a vertical displacement axis.

Things to Be Careful About

Ensure you identify two points that are in the same phase (crest-to-crest, trough-to-trough, or zero crossing with the same direction of slope).
Keep units consistent: use metres in $v=f\lambda$ to obtain $v$ in $\text{m s}^{-1}$ .
For MCQs, check that the option’s numerical values are consistent with each other and with what can actually be determined from the data given.

Techniques used

read the wavelength from a displacement–distance snapshotcalculate frequency from the perioduse v = f\lambda to determine wave speedmatch calculated quantities to the given options

Q251MEasyWaves

Which group of electromagnetic waves is arranged in order from shortest wavelength to longest wavelength?

Options

A radio waves $\to$ visible light $\to$ gamma rays
B visible light $\to$ microwaves $\to$ infrared
C visible light $\to$ ultraviolet $\to$ X-rays
D X-rays $\to$ infrared $\to$ microwaves

Pick an answer

DifficultyEasy

Worked solution

Working

Shortest wavelength to longest wavelength goes:

\gamma \;\to\; \text{X-rays} \;\to\; \text{UV} \;\to\; \text{visible} \;\to\; \text{IR} \;\to\; \text{microwaves} \;\to\; \text{radio}

So $\text{X-rays}$ have shorter wavelength than $\text{infrared}$ , which has shorter wavelength than $\text{microwaves}$ .

Answer

D

Final answer

D

Detailed explanation

Background Concept

Electromagnetic (EM) waves form a continuous spectrum. A key fact is the fixed wave speed in vacuum:

c = f\lambda

Since $c$ is constant for all EM waves in vacuum, a shorter wavelength $\lambda$ means a higher frequency $f$ , and vice versa.

The EM spectrum in order of increasing wavelength (decreasing frequency/energy) is:

\gamma \;\to\; \text{X-rays} \;\to\; \text{ultraviolet} \;\to\; \text{visible} \;\to\; \text{infrared} \;\to\; \text{microwaves} \;\to\; \text{radio}

Understanding the Question

The question asks which option lists three EM wave groups arranged from shortest wavelength to longest wavelength. So we need the correct left-to-right ordering by wavelength.

Approach

Recall the standard EM spectrum order.
Check each option to see whether it matches the required direction (shortest $\to$ longest wavelength).

Step-by-Step Reasoning

From the standard order (shortest $\to$ longest):

X-rays are very short wavelength.
Infrared has longer wavelength than visible light.
Microwaves have longer wavelength than infrared.

Therefore the sequence:

\text{X-rays} \to \text{infrared} \to \text{microwaves}

is correctly arranged from shortest to longest wavelength.

So the correct option is D.

Key Takeaways

The EM spectrum order is a key recall fact.
“Shortest wavelength” corresponds to “highest frequency” because $c=f\lambda$ .

Common Mistakes

Reversing the direction: writing longest $\to$ shortest wavelength.
Mixing up infrared and microwaves (microwaves have the longer wavelength).
Thinking visible is at the short end; gamma and X-rays are much shorter.

Things to Be Careful About

The question is explicitly about wavelength, not frequency or energy; ensure the ordering matches wavelength.
Remember the relative positions: $\text{IR}$ lies between visible and microwaves, and X-rays lie between gamma and ultraviolet.

Techniques used

recall the order of the electromagnetic spectrumcompare wavelengths to identify shortest-to-longest ordering

Q261MEasyWaves

A wave has a frequency of $5\text{ GHz}$ .

What is the period of the wave?

Options

A $200\text{ ps}$
B $2\text{ ns}$
C $20\text{ ns}$
D $20\,000\mu\text{s}$

Pick an answer

DifficultyEasy

Worked solution

Working

T = \frac{1}{f}

f = 5\ \text{GHz} = 5 \times 10^{9}\ \text{Hz}

T = \frac{1}{5 \times 10^{9}} = 2.0 \times 10^{-10}\ \text{s} = 200\ \text{ps}

Answer

A

Final answer

A

Detailed explanation

Background Concept

Frequency $f$ is the number of complete oscillations (cycles) per second, measured in hertz (Hz). The period $T$ is the time taken for one complete oscillation, measured in seconds (s).

They are reciprocals:

T = \frac{1}{f}

This always applies for periodic waves.

Understanding the Question

You are given a wave frequency of $5\ \text{GHz}$ and asked for the period. So we must convert $\text{GHz}$ to $\text{Hz}$ and then use $T = 1/f$ . The multiple-choice options are in time units (ps, ns, etc.), so we should express the final time in a suitable prefix.

Approach

Convert $5\ \text{GHz}$ into $\text{Hz}$ using $1\ \text{GHz} = 10^{9}\ \text{Hz}$ .
Calculate $T = 1/f$ .
Convert the result in seconds into picoseconds (ps) or nanoseconds (ns) to match the options.

Step-by-Step Reasoning

Convert frequency:

f = 5\ \text{GHz} = 5 \times 10^{9}\ \text{Hz}

Calculate period:

T = \frac{1}{f} = \frac{1}{5 \times 10^{9}} = 0.2 \times 10^{-9}\ \text{s} = 2.0 \times 10^{-10}\ \text{s}

Convert to picoseconds. Since $1\ \text{ps} = 10^{-12}\ \text{s}$ ,

2.0 \times 10^{-10}\ \text{s} = 200 \times 10^{-12}\ \text{s} = 200\ \text{ps}

This matches option A.

Key Takeaways

Use $T = 1/f$ for any periodic wave.
Convert prefix units carefully: $\text{GHz} = 10^{9}\ \text{Hz}$ , $\text{ps} = 10^{-12}\ \text{s}$ .

Common Mistakes

Treating $\text{GHz}$ as $10^{6}\ \text{Hz}$ instead of $10^{9}\ \text{Hz}$ .
Inverting incorrectly (writing $T = f$ instead of $T = 1/f$ ).
Converting $2.0 \times 10^{-10}\ \text{s}$ to ns incorrectly (it is $0.20\ \text{ns}$ , which is $200\ \text{ps}$ ).

Things to Be Careful About

Prefixes: $\text{G} = 10^{9}$ , $\text{n} = 10^{-9}$ , $\text{p} = 10^{-12}$ .
Check that the order of magnitude makes sense: higher frequency should mean smaller period.

Techniques used

convert frequency to hertzuse the definition of period as the reciprocal of frequencyconvert seconds into an appropriate prefix unit

Q271MEasySuperposition

Three statements about two progressive waves are listed.

1 The waves have the same frequency.
2 The waves have the same amplitude.
3 The waves are emitted with a constant phase difference.

Which statements must be correct for the two waves to be coherent?

Options

A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 and 3 only

Pick an answer

DifficultyEasy

Worked solution

Working

Coherent waves have a constant phase difference, so statement 3 is required.
A constant phase difference implies the waves must have the same frequency, so statement 1 is required.
Amplitude does not affect coherence, so statement 2 is not required.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Two waves (or two sources) are coherent if they maintain a fixed (constant) phase relationship.

Equivalently, coherence requires:

the waves have the same frequency (so their phase difference does not drift with time), and
the waves have a constant phase difference.

Their amplitudes can be the same or different; amplitude affects the visibility/contrast of interference fringes, not whether the sources are coherent.

Understanding the Question

You are given three possible statements about two progressive waves:

same frequency
same amplitude
emitted with a constant phase difference

You must decide which of these statements must be true for the waves to be coherent.

Approach

Use the definition of coherence: identify the required conditions (phase relationship and frequency), and then check whether amplitude is required.

Step-by-Step Reasoning

Statement 3 (“constant phase difference”) is the key requirement for coherence.
For the phase difference to remain constant over time, the waves must oscillate at the same rate, i.e. have the same frequency. Otherwise, even if they start with some phase difference, it will change continuously. So statement 1 is also required.
Statement 2 (“same amplitude”) is not required: two coherent waves can have different amplitudes and still interfere; the resulting maxima/minima just won’t be equally strong.

Therefore, the statements that must be correct are 1 and 3 only, which corresponds to option C.

Key Takeaways

Coherence is about a stable phase relationship, not about equal amplitudes.
Same frequency is necessary to prevent the phase difference from changing with time.

Common Mistakes

Thinking coherence requires equal amplitudes (it does not).
Selecting only “constant phase difference” and forgetting that different frequencies make the phase difference vary with time.

Things to Be Careful About

“Constant phase difference” and “same frequency” are tightly linked: if frequencies differ, phase difference cannot stay constant.
Amplitude affects the intensity pattern (fringe visibility), not the coherence condition.

Techniques used

recall the definition of coherent sourcesidentify which listed conditions are required by the definitioneliminate conditions that are not necessary for coherence

Q281MMedium-EasySuperpositionWaves

Waves $P$ and $Q$ have the same amplitude. The waves meet in phase at point $X$ and interfere to give a resultant wave with intensity $I$ .

The amplitude of wave $P$ is doubled.

What is the new intensity of the resultant wave at $X$ , in terms of $I$ ?

Options

A $0.44I$
B $1.5I$
C $2.3I$
D $3.0I$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Let the original amplitude of each wave be $A$ .

In phase at $X$ :

A_\text{res} = A + A = 2A

and $I \propto A_\text{res}^2$ , so

I \propto (2A)^2 = 4A^2

Now amplitude of $P$ is doubled: amplitudes are $2A$ and $A$ .

A_\text{new} = 2A + A = 3A

\frac{I_\text{new}}{I} = \frac{(3A)^2}{(2A)^2} = \frac{9}{4} = 2.25

Answer

$2.3I$ (option C)

Final answer

C

Detailed explanation

Background Concept

When two waves overlap, the principle of superposition says the resultant displacement is the sum of the individual displacements.

At a point where the waves are in phase (phase difference $0$ ), their displacements reach maxima and minima together, so their amplitudes add.

For waves of the same type in the same medium, the intensity $I$ is proportional to the square of the amplitude $A$ :

I \propto A^2

(This comes from intensity being proportional to energy transported per unit time, and wave energy being proportional to amplitude squared.)

Understanding the Question

Initially, waves $P$ and $Q$ have the same amplitude and meet in phase at $X$ , producing a resultant intensity $I$ .

Then only wave $P$ has its amplitude doubled. We assume they still meet in phase at $X$ (only the amplitude changes, not the phase relationship). We must find the new resultant intensity at $X$ in terms of the original $I$ .

Approach

Assign a symbol $A$ to the original amplitude of each wave.
Use in-phase superposition to find the original resultant amplitude at $X$ .
Use $I \propto A^2$ to link this amplitude to the given intensity $I$ .
Repeat for the new amplitudes ( $2A$ and $A$ ), then take a ratio to express the new intensity in terms of $I$ .

Step-by-Step Reasoning

Let each original wave have amplitude $A$ .

Original situation (in phase):
Resultant amplitude is the sum:

A_\text{res} = A + A = 2A

Intensity is proportional to amplitude squared:

I \propto (2A)^2 = 4A^2

So the given $I$ corresponds to something proportional to $4A^2$ .

New situation:
Wave $P$ has doubled amplitude, so amplitudes are $2A$ (from $P$ ) and $A$ (from $Q$ ).
Since they still meet in phase at $X$ , the new resultant amplitude is

A_\text{new} = 2A + A = 3A

New intensity:

I_\text{new} \propto (3A)^2 = 9A^2

Now compare with the original using a ratio (this removes the unknown proportionality constant):

\frac{I_\text{new}}{I} = \frac{9A^2}{4A^2} = \frac{9}{4} = 2.25

So

I_\text{new} = 2.25I \approx 2.3I

This matches option C.

Key Takeaways

In-phase waves add amplitudes: $A_\text{res} = A_1 + A_2$ .
Intensity scales as amplitude squared: $I \propto A^2$ .
Ratio methods are quick and avoid needing the constant of proportionality.

Common Mistakes

Adding intensities instead of amplitudes: interference requires adding displacements/amplitudes first.
Forgetting the square relationship: doubling amplitude does not double intensity; it quadruples it.
Using $2A$ as the new resultant amplitude: after doubling $P$ , the amplitudes are $2A$ and $A$ , so the sum is $3A$ , not $2A$ .

Things to Be Careful About

The phrase “meet in phase at $X$ ” is the key clue that amplitudes add directly (no subtraction).
Keep the original intensity linked to the original resultant amplitude ( $2A$ ), not to a single wave’s amplitude ( $A$ ).
Rounding: $2.25I$ corresponds to $2.3I$ among the options.

Techniques used

apply the principle of superposition to add amplitudes for in-phase wavesuse the proportionality between intensity and the square of amplitudeform an intensity ratio to relate the new situation to the original intensity

Q291MMedium-EasyWaves

Radio waves can be polarised, but sound waves cannot be polarised.

Which statement gives the reason for this?

Options

A Radio waves are generally of a higher frequency than sound waves.
B Radio waves are transverse waves, but sound waves are longitudinal waves.
C Radio waves can travel through a vacuum, but sound waves cannot travel through a vacuum.
D Radio waves travel at a much higher speed than sound waves.

Pick an answer

DifficultyMedium-Easy

Worked solution

Polarisation requires oscillations in a plane perpendicular to the direction of travel (transverse waves). Radio waves are transverse, whereas sound waves are longitudinal.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Polarisation is the restriction of the vibrations (oscillations) of a wave to one plane. This is only possible when the wave oscillations can occur in many directions perpendicular to the direction the wave travels.

Transverse wave: oscillations are perpendicular to the direction of wave travel. There is a choice of perpendicular directions, so you can select one (polarise the wave).
Longitudinal wave: oscillations are parallel to the direction of travel (compressions and rarefactions). There is no “sideways” vibration to restrict to a plane, so longitudinal waves cannot be polarised.

Radio waves are electromagnetic waves and are transverse. Sound waves in air are longitudinal.

Understanding the Question

The question asks why radio waves can be polarised but sound waves cannot. The options give different possible differences (frequency, type of wave, ability to travel in vacuum, speed). Only one of these differences is directly connected to whether polarisation is possible.

Approach

Recall the condition for polarisation: it is only possible for transverse waves.
Decide whether radio waves and sound waves are transverse or longitudinal.
Choose the option that states this key reason.

Step-by-Step Reasoning

Polarisation means selecting one direction/plane of vibration.
That requires the wave to have vibrations perpendicular to its direction of travel, i.e. it must be transverse.
Radio waves (electromagnetic) have electric and magnetic fields oscillating perpendicular to the direction of travel, so they are transverse and can be polarised.
Sound waves in air consist of compressions/rarefactions along the direction of travel, so they are longitudinal and cannot be polarised.

Therefore the correct statement is: radio waves are transverse but sound waves are longitudinal → Option B.

Key Takeaways

Only transverse waves can be polarised.
Sound waves in air are longitudinal, so they cannot be polarised.
Electromagnetic waves (including radio) are transverse, so they can be polarised.

Common Mistakes

Choosing frequency (A): frequency does not determine whether polarisation is possible.
Choosing vacuum travel (C): ability to travel in vacuum is unrelated to polarisation.
Choosing speed (D): wave speed is not the deciding factor for polarisation.
Thinking sound can be polarised because it can be reflected or refracted: those phenomena are not polarisation.

Things to Be Careful About

“Polarisation” refers to the direction of oscillation, not the direction of travel.
Sound can be transverse in some solids, but in air (the usual meaning at AS level) sound is longitudinal; the question is using that standard context.

Techniques used

recall that polarisation is a property of transverse wavesidentify whether each type of wave is transverse or longitudinaleliminate options that describe irrelevant differences

Q301MMedium-EasySuperposition

Light of wavelength $\lambda$ is incident normally on a diffraction grating with a total number of $N$ lines in width $w$ .

A second order maximum is observed at an angle of diffraction $\theta$ .

What is $N$ ?

Options

A $\dfrac{w\lambda}{\sin\theta}$
B $\dfrac{2w\lambda}{\sin\theta}$
C $\dfrac{w\sin\theta}{2\lambda}$
D $\dfrac{w\sin\theta}{\lambda}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a grating,

d\sin\theta = n\lambda

Second order: $n=2$ so

d = \frac{2\lambda}{\sin\theta}

With $N$ lines in width $w$ , the line spacing is

d = \frac{w}{N}

So

\frac{w}{N} = \frac{2\lambda}{\sin\theta} \quad\Rightarrow\quad N = \frac{w\sin\theta}{2\lambda}

Answer

C

Final answer

C

Detailed explanation

Background Concept

A diffraction grating has many equally spaced lines (or slits). Light from adjacent lines interferes constructively at angles where the path difference between adjacent rays is an integer multiple of the wavelength.

The condition for a principal maximum is

d\sin\theta = n\lambda

where:

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

If the grating has $N$ lines spread over a total width $w$ , then the line density is $N/w$ and the spacing is

d = \frac{w}{N}.

Understanding the Question

Light of wavelength $\lambda$ is incident normally on a grating. A second-order maximum is seen at angle $\theta$ . The question asks for the total number of lines $N$ across the grating width $w$ .

So we need to connect $N$ to $w$ and $\theta$ using the grating equation for $n=2$ .

Approach

Start with the diffraction grating equation $d\sin\theta = n\lambda$ .
Substitute $n=2$ (second order).
Replace $d$ using $d=w/N$ .
Rearrange to solve for $N$ and match it to one of the options.

Step-by-Step Reasoning

From the grating equation:

d\sin\theta = n\lambda

For the second order maximum, $n=2$ :

d\sin\theta = 2\lambda quad\Rightarrow\quad d = \frac{2\lambda}{\sin\theta}

The grating spacing is also related to the total width and number of lines:

d = \frac{w}{N}

Equate the two expressions for $d$ :

\frac{w}{N} = \frac{2\lambda}{\sin\theta}

Solve for $N$ :

N = \frac{w\sin\theta}{2\lambda}

This corresponds to option C.

Key Takeaways

Use $d\sin\theta = n\lambda$ for diffraction gratings.
Convert “ $N$ lines in width $w$ ” into spacing using $d=w/N$ .
Second order means $n=2$ .

Common Mistakes

Using $d = N/w$ instead of $d=w/N$ (that swaps density and spacing).
Forgetting that it is second order and using $n=1$ .
Writing $d\sin\theta = \lambda/n$ (incorrect rearrangement of the grating condition).

Things to Be Careful About

$\theta$ is measured from the normal, so the formula uses $\sin\theta$ (not $\cos\theta$ ).
Ensure $d$ is a distance (spacing), while $N/w$ is a line density; do not confuse them.
The result should be dimensionless: $w/\lambda$ is dimensionless and $\sin\theta$ is dimensionless, so $N$ is dimensionless.

Techniques used

use the diffraction grating equationrelate grating spacing to total number of lines and grating widthrearrange an equation to make the required quantity the subject

Q311MEasyD.C. Circuits

Kirchhoff’s second law is a consequence of a basic principle.

What is this principle?

Options

A The charge flowing in an electric circuit is conserved.
B The energy in an electric circuit is conserved.
C The sum of the electric currents entering a point in an electric circuit is equal to the sum of the electric currents leaving that point.
D The sum of the potential differences in an electric circuit is equal to the sum of the products of the current and resistance.

Pick an answer

DifficultyEasy

Worked solution

Kirchhoff’s second law (sum of emfs equals sum of potential drops around a closed loop) follows from conservation of energy: a charge gains energy from sources and loses the same energy in circuit elements.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Kirchhoff’s second law (the “loop rule”) states that around any closed loop in a circuit, the algebraic sum of the emfs equals the algebraic sum of the potential differences (voltage drops).

Potential difference is energy transferred per unit charge:

V = \frac{W}{Q}

So if a charge goes around a complete loop and returns to its starting point, the net change in its energy must be zero (it can’t come back with more or less energy than it started with, assuming no time-varying magnetic fields are inducing additional emf). That statement is simply conservation of energy applied to electric circuits.

Understanding the Question

The question asks: “Kirchhoff’s second law is a consequence of a basic principle. What is this principle?” You must choose which option describes the fundamental conservation principle behind the loop rule.

Approach

Identify what Kirchhoff’s second law says (a rule about potential differences around a loop), then match it to the relevant conservation law. Since potential difference is energy per unit charge, a closed-loop sum of potential changes being zero is a conservation of energy statement.

Step-by-Step Reasoning

Kirchhoff’s second law concerns a closed loop and relates emfs and potential drops.
Emf provides energy to charge; resistors/components dissipate energy from charge.
After completing a loop, charge returns to its starting point, so total energy gained per unit charge equals total energy lost per unit charge.
Therefore the underlying principle is conservation of energy in the circuit.
This corresponds to option B.

Key Takeaways

Kirchhoff’s second law (loop rule) comes from conservation of energy.
Kirchhoff’s first law (junction rule) comes from conservation of charge.

Common Mistakes

Choosing A (conservation of charge): that supports Kirchhoff’s first law, not the second.
Choosing D: this is closer to a statement of the law (and not even generally correct as written for all loops/components), not the underlying principle.
Choosing C: that is exactly Kirchhoff’s first law (junction rule), not the principle behind the second.

Things to Be Careful About

Distinguish clearly:
- Junction/current rule (\rightarrow) conservation of charge.
- Loop/voltage rule (\rightarrow) conservation of energy.
Kirchhoff’s second law assumes no changing magnetic flux linking the loop (otherwise an induced emf must be included).

Techniques used

recall the statement of Kirchhoff's second law for a closed looplink the loop rule to conservation of energy per unit charge

Q321MMedium-EasyD.C. Circuits

The diagram shows a circuit with a light-dependent resistor (LDR).

Fig. 1

The ammeter reads zero current.

What is the resistance of the LDR?

Options

A $6.0\text{ k}\Omega$
B $18\text{ k}\Omega$
C $26\text{ k}\Omega$
D $30\text{ k}\Omega$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Ammeter reads zero $\Rightarrow$ balanced bridge, so

\frac{10\,\text{k}\Omega}{30\,\text{k}\Omega}=\frac{6.0\,\text{k}\Omega}{R_{\text{LDR}}}

R_{\text{LDR}}=6.0\,\text{k}\Omega\times\frac{30}{10}=18\,\text{k}\Omega

Answer

B

Final answer

B

Detailed explanation

Background Concept

In a bridge circuit (often called a Wheatstone bridge), there are two potential dividers in parallel across the same supply. An ammeter (or galvanometer) connects the midpoints.

If the ammeter reads zero, there is no current through it, so the two midpoint potentials are equal. This is the balanced bridge condition. For resistors arranged as:

left branch: top $R_1$ , bottom $R_2$
right branch: top $R_3$ , bottom $R_4$

Balance implies

\frac{R_1}{R_2}=\frac{R_3}{R_4}.

This comes from the fact that each side is a potential divider: equal midpoint potentials mean the same fraction of the supply voltage is dropped across the top resistors on each side.

Understanding the Question

The circuit has:

Left branch: $10\,\text{k}\Omega$ (top) and $30\,\text{k}\Omega$ (bottom)
Right branch: $6.0\,\text{k}\Omega$ (top) and an unknown $R_{\text{LDR}}$ (bottom)

The ammeter between the midpoints reads zero, so the bridge is balanced. The question asks for the LDR resistance.

Approach

Use the balanced-bridge ratio condition:

\frac{R_{\text{top,left}}}{R_{\text{bottom,left}}}=\frac{R_{\text{top,right}}}{R_{\text{bottom,right}}}

Then substitute the given resistances and solve for $R_{\text{LDR}}$ .

Step-by-Step Reasoning

Balanced bridge:

\frac{10\,\text{k}\Omega}{30\,\text{k}\Omega}=\frac{6.0\,\text{k}\Omega}{R_{\text{LDR}}}

Rearrange:

R_{\text{LDR}}=6.0\,\text{k}\Omega\times\frac{30}{10}

Compute:

R_{\text{LDR}}=18\,\text{k}\Omega.

So the correct option is B.

Key Takeaways

A zero reading in the bridge meter means the midpoint potentials are equal.
For a balanced bridge, the ratio of the top and bottom resistors is the same in both branches.
You can solve the unknown resistance with a simple proportion.

Common Mistakes

Inverting one ratio (e.g. using $30/10$ on one side but $6/R$ on the other inconsistently).
Treating the circuit as simple series/parallel and ignoring that the ammeter reading gives the special balance condition.
Dropping the $\text{k}\Omega$ factor and mixing $\Omega$ with $\text{k}\Omega$ .

Things to Be Careful About

Use corresponding positions: top with top, bottom with bottom.
Keep units consistent (here all are in $\text{k}\Omega$ , so they cancel neatly).
The supply voltage value is irrelevant for a balance-ratio question; it cancels when comparing divider fractions.

Techniques used

recognise a balanced bridge from a zero ammeter readingequate potential divider ratios in parallel branchessolve a resistance ratio to find an unknown resistance

Q331MMedium-EasyElectricityD.C. Circuits

A torch uses three lamps connected in parallel and is powered by a cell of electromotive force (e.m.f.) $3.0\text{ V}$ and negligible internal resistance. Each lamp dissipates $0.60\text{ W}$ of power.

What is the current in the cell?

Options

A $0.067\text{ A}$
B $0.20\text{ A}$
C $0.60\text{ A}$
D $0.83\text{ A}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In parallel, each lamp has $V = 3.0\ \text{V}$ .

For one lamp:

I = \frac{P}{V} = \frac{0.60}{3.0} = 0.20\ \text{A}

Total current from the cell:

I_{\text{cell}} = 3 \times 0.20 = 0.60\ \text{A}

Answer

C

Final answer

C

Detailed explanation

Background Concept

Electrical power transferred in a component is

P = VI

where $P$ is power (W), $V$ is potential difference across the component (V), and $I$ is current through it (A).

For components connected in parallel:

the potential difference across each parallel branch is the same as the supply p.d.
the total current supplied is the sum of the currents in each branch.

Understanding the Question

A cell of e.m.f. $3.0\ \text{V}$ (with negligible internal resistance) supplies three identical lamps connected in parallel. Each lamp dissipates $0.60\ \text{W}$ . The question asks for the current in the cell, i.e. the total current drawn by all three lamps together.

Approach

Because the internal resistance is negligible, the terminal p.d. of the cell is $3.0\ \text{V}$ .
Because the lamps are in parallel, each lamp has $3.0\ \text{V}$ across it.
Use $P = VI$ for one lamp to find its current.
Multiply by 3 (or sum the three equal branch currents) to get the cell current.

Step-by-Step Reasoning

Voltage across each lamp:

V_{\text{lamp}} = 3.0\ \text{V}

Current in one lamp using $P = VI$ :

I_{\text{lamp}} = \frac{P}{V} = \frac{0.60\ \text{W}}{3.0\ \text{V}} = 0.20\ \text{A}

Total current in the cell is the sum of the three parallel branch currents:

I_{\text{cell}} = I_1 + I_2 + I_3 = 3(0.20) = 0.60\ \text{A}

So the correct option is C.

Key Takeaways

In parallel, each component has the full supply potential difference.
Power can be linked to current via $P = VI$ .
The supply current in a parallel circuit equals the sum of the branch currents.

Common Mistakes

Using $V = 1.0\ \text{V}$ for each lamp by incorrectly sharing the voltage between parallel branches (that happens in series, not parallel).
Forgetting to multiply by 3 after finding the current for one lamp.
Mixing up units or rearranging $P = VI$ incorrectly (e.g. using $I = V/P$ ).

Things to Be Careful About

“Negligible internal resistance” means the terminal p.d. is essentially the e.m.f., so you can take $V = 3.0\ \text{V}$ across each lamp.
Keep consistent significant figures: $0.60/3.0$ gives $0.20\ \text{A}$ , and the total $0.60\ \text{A}$ matches the options.

Techniques used

use the power equation to find current in a componentapply the rule that components in parallel share the same potential differencesum branch currents to obtain the supply current

Q341MMedium-EasyD.C. Circuits

A cell with internal resistance is connected to a light-dependent resistor (LDR), a fixed resistor and a voltmeter, as shown.

Fig. 1

The voltmeter reading increases.

Which quantity decreases as the voltmeter reading increases?

Options

A the charge moving through the cell per unit time
B the energy transferred to the fixed resistor per unit charge
C the intensity of the light incident on the LDR
D the terminal potential difference across the cell

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Voltmeter is across the fixed resistor, so

V = IR

As $R$ is fixed, increased voltmeter reading means $I$ increases.

For a cell with internal resistance $r$ ,

V_{\text{terminal}} = \varepsilon - Ir

So as $I$ increases, $V_{\text{terminal}}$ decreases.

Answer

D

Final answer

D

Detailed explanation

Background Concept

In a circuit, a voltmeter connected in parallel across a component measures the potential difference (p.d.) across that component. For a resistor that obeys Ohm’s law,

V = IR

so the p.d. across the resistor is proportional to the current through it (provided its resistance is constant).

A real cell is modelled as an ideal source of emf $\varepsilon$ in series with an internal resistance $r$ . When current $I$ flows, some energy is “lost” inside the cell, giving an internal p.d. drop $Ir$ . The terminal p.d. across the cell is therefore

V_{\text{terminal}} = \varepsilon - Ir

So, increasing current makes the terminal p.d. smaller.

An LDR (light-dependent resistor) has a resistance that decreases when light intensity increases.

Understanding the Question

The circuit is a series loop containing: a cell with internal resistance, an LDR, and a fixed resistor. A voltmeter is connected across (in parallel with) the fixed resistor.

We are told: the voltmeter reading increases (so the p.d. across the fixed resistor increases).

We must choose which quantity decreases as that voltmeter reading increases.

Approach

Use the fact that the voltmeter reads the p.d. across the fixed resistor.
Because the resistor is fixed, use $V = IR$ to decide what happens to the current.
Use $V_{\text{terminal}} = \varepsilon - Ir$ to see what decreases when current increases.
Check the other options quickly for consistency.

Step-by-Step Reasoning

The voltmeter is across the fixed resistor, so its reading is

V_R = IR

The resistor is fixed, so $R$ is constant. Therefore an increase in $V_R$ implies an increase in current $I$ .
With internal resistance $r$ , terminal p.d. is

V_{\text{terminal}} = \varepsilon - Ir

As $I$ increases, the term $Ir$ increases, so $V_{\text{terminal}}$ must decrease.

Check the options:

A: “charge per unit time” is current $I$ → increases, not decreases.
B: “energy transferred per unit charge” is p.d. across the fixed resistor, equal to the voltmeter reading → increases, not decreases.
C: If $I$ increases in a series circuit, total resistance must have decreased; that would happen if LDR resistance decreased, which corresponds to increased light intensity, not decreased.
D: terminal p.d. decreases as shown.

So the correct choice is D.

Key Takeaways

A voltmeter across a resistor measures $V$ and for a fixed resistor $V \propto I$ .
For a cell with internal resistance, terminal p.d. falls as current increases: $V = \varepsilon - Ir$ .
For an LDR, higher light intensity means lower resistance.

Common Mistakes

Assuming the terminal p.d. across the cell increases when current increases (it decreases because the internal drop $Ir$ increases).
Mixing up “energy per unit charge” (which is p.d.) with power or energy per unit time.
Thinking the LDR resistance increases with light intensity (it decreases).

Things to Be Careful About

The voltmeter is across the fixed resistor, not across the cell, so its reading is not the terminal p.d.
The fixed resistor’s resistance is constant, so any change in $V$ directly indicates a change in $I$ via $V=IR$ .
In internal resistance questions, always write $V_{\text{terminal}} = \varepsilon - Ir$ (or equivalently $\varepsilon = V_{\text{terminal}} + Ir$ ) to keep the sign correct.

Techniques used

relate voltmeter reading across a resistor to the circuit current using Ohm's lawuse the internal resistance relation for a cell to link terminal p.d. to currentdeduce how changing the LDR resistance affects the series current

Q351MMedium-EasyD.C. Circuits

The circuit shown contains a cell with negligible internal resistance.

Fig. 1

The energy transferred per unit charge in driving charge around the complete circuit is $E$ . The potential difference (p.d.) across $X$ is $V$ .

The cell is then replaced with a different cell of the same electromotive force (e.m.f.) that has significant internal resistance.

What is the effect on $E$ and $V$ of replacing the cell?

Options

	effect on $E$	effect on $V$
A	decreases	decreases
B	decreases	increases
C	no change	decreases
D	no change	increases

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

$E$ is the energy transferred per unit charge by the cell, i.e. $E = \mathcal{E}$ .

The new cell has the same e.m.f. $\mathcal{E}$ , so $E$ has no change.

With internal resistance $r$ , the p.d. across $X$ is the terminal p.d.

V = \mathcal{E} - Ir

Since $r$ is significant and $I > 0$ , $Ir$ is significant so $V$ decreases.

Answer

C

Final answer

C

Detailed explanation

Background Concept

The electromotive force (e.m.f.) $\mathcal{E}$ of a source is defined as the energy transferred (work done) per unit charge by the source:

\mathcal{E} = \frac{W}{Q}

A real cell can be modelled as an ideal source of e.m.f. $\mathcal{E}$ in series with an internal resistance $r$ . When current $I$ flows, some energy per unit charge is dissipated inside the cell as thermal energy in $r$ . This causes the terminal potential difference (p.d.) across the external circuit to be less than the e.m.f.

The terminal p.d. $V_{\text{terminal}}$ is

V_{\text{terminal}} = \mathcal{E} - Ir

If $r$ increases (or becomes significant), for the same external resistance the current is reduced and there is a non-negligible "lost volts" term $Ir$ .

Understanding the Question

The circuit is a simple series loop: a cell in series with a component labelled $X$ .

$E$ is stated to be “energy transferred per unit charge in driving charge around the complete circuit”. That is the definition of the e.m.f. of the cell.
$V$ is the p.d. across $X$ .

Initially the cell has negligible internal resistance, so the p.d. across $X$ is essentially the full e.m.f.

Then the cell is replaced by a different cell with the same e.m.f. but significant internal resistance. You must decide how $E$ and $V$ change.

Approach

Identify what $E$ represents: compare it to the definition of e.m.f.
Decide whether changing internal resistance changes the e.m.f. (it does not, if the e.m.f. is stated to be the same).
Use the internal-resistance relation for the p.d. across the external component: $V = \mathcal{E} - Ir$ (since $X$ is the external load).
Conclude the direction of change of $V$ when $r$ becomes significant.

Step-by-Step Reasoning

By definition, the e.m.f. is the work done (energy transferred) per unit charge by the source. The question’s wording for $E$ matches this definition, so:

E = \mathcal{E}

The replacement cell is stated to have the same e.m.f. Therefore:

Effect on $E$ : no change.

With internal resistance $r$ , the e.m.f. is shared between the internal resistance and the external component $X$ :

\mathcal{E} = V + Ir

Rearrange:

V = \mathcal{E} - Ir

Since the internal resistance is now significant and there is a current ( $I > 0$ in a closed circuit), the term $Ir$ is non-zero and noticeable. Therefore $V$ is less than $\mathcal{E}$ , i.e. compared with the negligible- $r$ case, the p.d. across $X$ decreases.

So the correct option is: no change for $E$ , decrease for $V$ (\Rightarrow) option C.

Key Takeaways

e.m.f. is energy transferred per unit charge by the source: it is a property of the cell’s chemistry and is not reduced just because internal resistance is present.
Internal resistance causes “lost volts” $Ir$ , so the terminal p.d. across the external circuit is $\mathcal{E} - Ir$ .
Adding internal resistance reduces the p.d. available to the external component.

Common Mistakes

Thinking the e.m.f. decreases when internal resistance is added. Internal resistance reduces terminal p.d., not the e.m.f. (if the e.m.f. is specified unchanged).
Confusing $V$ (p.d. across $X$ ) with $\mathcal{E}$ (cell e.m.f.). With negligible internal resistance they are approximately equal, but not when $r$ is significant.
Assuming $V$ must increase because current decreases. Even though $I$ may decrease, the key point is that $V$ becomes $\mathcal{E} - Ir$ , which is less than $\mathcal{E}$ .

Things to Be Careful About

The phrase “energy transferred per unit charge in driving charge around the complete circuit” is a strong cue for e.m.f.
The question states the replacement cell has the same e.m.f.: this locks the effect on $E$ as “no change”.
$V$ is across $X$ (external component), so it is the terminal p.d., not the p.d. across the internal resistance.

Techniques used

use the definition of e.m.f. as energy transferred per unit chargemodel a real cell as an ideal source in series with internal resistanceapply the terminal p.d. relation V = \mathcal{E} - Irdeduce the change in p.d. across an external component from the change in current

Q361MMedium-EasyD.C. CircuitsElectricity

In the circuits shown, the batteries are identical and all have negligible internal resistance. All of the resistors have the same resistance. The diodes have zero resistance when conducting and infinite resistance when not conducting.

In which circuit is the current in the battery greatest?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Conducting diode: $R=0$ ; non-conducting diode: open circuit.

A: both branches conduct, so two equal resistors in parallel:

R_{\text{eq}} = \frac{R}{2}

so

I = \frac{V}{R/2} = \frac{2V}{R}

B: one diode is reverse-biased in the loop $\Rightarrow$ open circuit $\Rightarrow I=0$ .

C: one branch conducts (single $R$ ), the other is blocked $\Rightarrow R_{\text{eq}}=R \Rightarrow I=\frac{V}{R}$ .

D: one branch conducts, one blocked $\Rightarrow R_{\text{eq}}=R \Rightarrow I=\frac{V}{R}$ .

Greatest current corresponds to smallest $R_{\text{eq}}$ , which is circuit A.

Answer

A

Final answer

A

Detailed explanation

Background Concept

An ideal diode behaves like a one-way switch:

Forward-biased (conducting): it has zero resistance, so it can be replaced by a wire.
Reverse-biased (not conducting): it has infinite resistance, so it can be replaced by an open circuit (a break).

For a given battery voltage $V$ , the current drawn from the battery is

I = \frac{V}{R_{\text{eq}}}

So the circuit with the greatest current is the one with the smallest equivalent resistance seen by the battery.

Understanding the Question

All batteries are identical and have negligible internal resistance, so the battery voltage is the same in all options and does not change with current.
All resistors have the same resistance $R$ .
You are asked: in which circuit does the battery deliver the largest current?

So you only need to:

Decide which diodes conduct (and which block), given the battery polarity in the diagram.
Simplify each circuit and find $R_{\text{eq}}$ .
Compare $I = V/R_{\text{eq}}$ .

Approach

For each option:

Replace each forward-biased diode with a wire (zero resistance).
Remove each reverse-biased branch (open circuit means no current in that branch).
Combine the remaining resistors (series/parallel) to get $R_{\text{eq}}$ .
The smallest $R_{\text{eq}}$ gives the largest battery current.

Step-by-Step Reasoning

Let the battery drive conventional current from its positive terminal through the external circuit.

Circuit A
Both diodes are oriented to allow current in the same direction through each vertical branch, so both branches conduct. Each branch then behaves like a single resistor $R$ (since the diode has $0\ \Omega$ when conducting). Two equal resistors in parallel give:

R_{\text{eq}} = \left(\frac{1}{R}+\frac{1}{R}\right)^{-1} = \frac{R}{2}

So

I_A = \frac{V}{R/2} = \frac{2V}{R}

Circuit B
This is one loop, so the same current would have to pass through both diodes. Because the two diodes are oppositely oriented, for any chosen loop current direction one of them will be reverse-biased. A reverse-biased ideal diode is an open circuit, so the loop is broken and:

I_B = 0

Circuit C
There are two parallel branches across the battery. The left branch is just a resistor $R$ , so it always provides a complete path. The right branch contains a diode oriented to block the current direction that would be driven by the battery, so that entire right branch is open and carries no current. Therefore only one resistor is effectively across the battery:

R_{\text{eq}} = R \quad \Rightarrow \quad I_C = \frac{V}{R}

Circuit D
Again there are two parallel branches. The diodes are oppositely oriented, so only one branch is forward-biased for the given battery polarity; the other branch is open. That leaves a single conducting branch containing one resistor $R$ :

R_{\text{eq}} = R \quad \Rightarrow \quad I_D = \frac{V}{R}

Comparing currents:

I_A = \frac{2V}{R} \;>\; \frac{V}{R} = I_C = I_D \;>\; 0 = I_B

So the greatest battery current is in A.

Key Takeaways

Ideal diode: forward-biased $\to$ short circuit; reverse-biased $\to$ open circuit.
Battery current is largest when the equivalent resistance across the battery is smallest.
Two identical resistors in parallel halve the resistance, doubling the current for the same voltage.

Common Mistakes

Treating a reverse-biased diode as having resistance $R$ rather than being an open circuit.
Assuming both parallel branches always conduct even if one contains a reverse-biased diode.
Forgetting that in a single loop, a single reverse-biased diode stops all current.

Things to Be Careful About

Use the battery polarity shown in the diagram to decide diode bias correctly.
Once a diode is reverse-biased, that whole branch carries zero current (so do not include its resistor in the equivalent resistance).
When comparing options, you do not need a numerical value of $V$ or $R$ ; relative $R_{\text{eq}}$ is enough.

Techniques used

decide whether each diode is forward-biased or reverse-biasedreplace conducting diodes with zero resistance and non-conducting diodes with open circuitsfind the equivalent resistance of parallel branchesuse I = V/R to compare currents

Q371MMedium-EasyD.C. Circuits

A potentiometer circuit is used to determine the electromotive force (e.m.f.) $E$ of a cell. The circuit includes a second cell of e.m.f. $1.5\text{ V}$ and internal resistance $0.50\Omega$ that is connected to a uniform resistance wire $XY$ , as shown.

Fig. 1

The resistance wire $XY$ has a length of $0.96\text{ m}$ and a resistance of $0.50\Omega$ .

The movable connection $Z$ is moved along wire $XY$ . The galvanometer reading is zero when length $XZ$ is $0.64\text{ m}$ .

What is the value of e.m.f. $E$ ?

Options

A $0.50\text{ V}$
B $0.75\text{ V}$
C $1.0\text{ V}$
D $1.1\text{ V}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Total resistance in driving circuit:

R_{\text{total}} = 0.50 + 0.50 = 1.00\ \Omega

Current in wire:

I = \frac{1.5}{1.00} = 1.5\ \text{A}

p.d. across whole wire $XY$ :

V_{XY} = IR_{XY} = 1.5 \times 0.50 = 0.75\ \text{V}

Potential gradient:

\text{gradient} = \frac{0.75}{0.96}\ \text{V m}^{-1}

At balance, $E = V_{XZ}$ :

E = 0.75\times\frac{0.64}{0.96} = 0.50\ \text{V}

Answer

A

Final answer

A

Detailed explanation

Background Concept

A potentiometer measures an e.m.f. by using a null method. A steady current is passed through a uniform resistance wire, so the potential difference along the wire is proportional to length.

If the galvanometer reads zero, no current flows in the test-cell branch, so the p.d. between the contact points on the wire equals the e.m.f. of the test cell:

E = V_{\text{balance}}

For a uniform wire,

V = k\ell

where $k$ is the potential gradient (V m $^{-1}$ ) and $\ell$ is the balance length.

Understanding the Question

A 1.5 V driving cell with internal resistance $0.50\ \Omega$ sends current through the potentiometer wire $XY$ (resistance $0.50\ \Omega$ , length $0.96\ \text{m}$ ). The slider is at $Z$ such that $XZ = 0.64\ \text{m}$ and the galvanometer shows zero. We must find the unknown e.m.f. $E$ of the test cell connected between $X$ and $Z$ .

So we need the potential drop along length $XZ$ of the wire.

Approach

Find the current in the driving circuit using the driving e.m.f. and the total series resistance (internal resistance + wire resistance).
Use $V = IR$ to find the potential drop across the entire wire $XY$ .
Because the wire is uniform, scale that p.d. by the fraction of length $XZ/XY$ to get $V_{XZ}$ .
At null deflection, set $E = V_{XZ}$ .

Step-by-Step Reasoning

The driving cell has internal resistance $0.50\ \Omega$ in series with the wire resistance $0.50\ \Omega$ , so

R_{\text{total}} = 0.50 + 0.50 = 1.00\ \Omega.

The current through the wire is then

I = \frac{\varepsilon}{R_{\text{total}}} = \frac{1.5}{1.00} = 1.5\ \text{A}.

The p.d. across the entire wire $XY$ is

V_{XY} = IR_{XY} = 1.5 \times 0.50 = 0.75\ \text{V}.

Uniform wire means p.d. is proportional to length, so the p.d. from $X$ to $Z$ is

V_{XZ} = V_{XY}\left(\frac{XZ}{XY}\right) = 0.75\left(\frac{0.64}{0.96}\right) = 0.50\ \text{V}.

At balance (zero galvanometer current), the test-cell e.m.f. equals this p.d.:

E = V_{XZ} = 0.50\ \text{V}.

So the correct option is A.

Key Takeaways

In a potentiometer, null deflection implies the test cell is not supplying current, so its e.m.f. equals the measured p.d.
A uniform potentiometer wire has constant potential gradient: $V \propto \ell$ .
When the driving cell has internal resistance, include it when finding the current through the wire.

Common Mistakes

Ignoring the internal resistance of the 1.5 V cell (would give the wrong current and gradient).
Using $1.5\ \text{V}$ directly as the p.d. across the wire (it is not, because some is lost across internal resistance).
Inverting the length ratio (using $0.96/0.64$ instead of $0.64/0.96$ ).

Things to Be Careful About

The resistance of the wire and the internal resistance are in series in the driving circuit.
The balance condition uses potential difference along the wire, not the full driving e.m.f..
Proportionality with length only holds because the wire is stated to be uniform (constant resistance per unit length).

Techniques used

calculate circuit current using e.m.f. and total series resistancedetermine the p.d. across the potentiometer wire from current and resistanceuse a uniform potential gradient to relate balance length to e.m.f.apply the null (zero galvanometer) condition for a potentiometer

Q381MEasyParticle Physics

Which particle is not a fundamental particle?

Options

A charm quark
B electron
C neutrino
D neutron

Pick an answer

DifficultyEasy

Worked solution

Working

Quarks and leptons (electron, neutrino) are fundamental particles.
A neutron is a baryon made of three quarks (e.g. $udd$ ), so it is not fundamental.

Answer

D

Final answer

D

Detailed explanation

Background Concept

In the AS Level model used in Cambridge 9702, fundamental particles are those not made of smaller constituents (within the syllabus model). The key fundamental families are:

Quarks (e.g. up, down, charm, strange, etc.)
Leptons (e.g. electron, neutrino)

Particles that are not fundamental are usually hadrons, which are composite particles made of quarks:

Baryons: made of three quarks (e.g. proton $uud$ , neutron $udd$ )
Mesons: made of a quark and an antiquark

Understanding the Question

You are given four particles and asked which one is not fundamental. So we check whether each option is a quark, a lepton, or a hadron (composite).

charm quark → quark → fundamental
electron → lepton → fundamental
neutrino → lepton → fundamental
neutron → baryon (hadron) → composite → not fundamental

Approach

Classify each option as quark / lepton / hadron.
Use the rule: quarks and leptons are fundamental; hadrons are made of quarks and are not fundamental.
Select the composite particle.

Step-by-Step Reasoning

Option A: charm quark is a quark, and quarks are fundamental → not the answer.
Option B: electron is a lepton, and leptons are fundamental → not the answer.
Option C: neutrino is a lepton, and leptons are fundamental → not the answer.
Option D: neutron is a baryon (a hadron). In the quark model it is made of three quarks, typically $udd$ .
Since it has constituents, it is not fundamental.

Therefore the correct choice is D.

Key Takeaways

Fundamental (in this syllabus): quarks and leptons.
Composite: hadrons (baryons and mesons) made of quarks.
Neutron is a baryon with quark content $udd$ .

Common Mistakes

Thinking “neutral” means “fundamental”: neutron and neutrino are both neutral-sounding, but neutron is composite while neutrino is fundamental.
Mixing up neutron with neutrino.
Assuming all particles in the nucleus are fundamental: protons and neutrons are made of quarks.

Things to Be Careful About

The question is about the quark/lepton model, not about whether something is stable or whether it has charge.
Remember: neutrons are hadrons (baryons), not leptons.

Techniques used

identify leptons and quarks as fundamental particlesdistinguish hadrons as composite particles made of quarksuse quark composition to classify a particle as non-fundamental

Q391MMedium-EasyParticle Physics

The isotope fluorine-18, $^{18}_{9}\text{F}$ , undergoes $\beta^{+}$ decay to form a stable isotope.

How many neutrons are there in a nucleus of the stable isotope?

Options

A $7$
B $8$
C $9$
D $10$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In (\beta^{+}) decay, the nucleon number (A) stays the same and the proton number (Z) decreases by (1).

So (^{18}{9}\text{F} \to {}^{18}{8}\text{O}) (stable).

Number of neutrons:

N = A - Z = 18 - 8 = 10

Answer

D

Final answer

D

Detailed explanation

Background Concept

In nuclide notation (^{A}_{Z}X):

(A) is the nucleon (mass) number = number of protons + number of neutrons.
(Z) is the proton (atomic) number = number of protons.
Number of neutrons is (N = A - Z).

In (\beta^{+}) decay (positron emission), a proton in the nucleus changes into a neutron:

p \to n + e^{+} + \nu

So:

(A) stays constant (still the same total number of nucleons).
(Z) decreases by (1) (one fewer proton).

Understanding the Question

You are told (^{18}_{9}\text{F}) undergoes (\beta^{+}) decay to form a stable isotope. You must find how many neutrons are in the nucleus of that final (stable) isotope.

Given initially:

(A = 18)
(Z = 9)

After (\beta^{+}) decay:

(A) remains (18)
(Z) becomes (8)

Then use (N = A - Z) for the product nucleus.

Approach

Use the rule for (\beta^{+}) decay to update (Z) (down by 1) while keeping (A) unchanged.
Compute neutrons in the product nucleus using (N = A - Z).
Match the result to the multiple-choice options.

Step-by-Step Reasoning

Start with (^{18}_{9}\text{F}).
In (\beta^{+}) decay, (Z) decreases by 1:

Z: 9 \to 8

and (A) is unchanged:

A: 18 \to 18

So the daughter nucleus is (^{18}_{8}\text{O}), which is a stable isotope.

Find the number of neutrons in (^{18}_{8}\text{O}):

N = A - Z = 18 - 8 = 10

Option with 10 neutrons is D.

Key Takeaways

(\beta^{+}) decay converts a proton into a neutron.
(\beta^{+}) decay: (A) unchanged, (Z) decreases by 1.
Neutron number is found from (N = A - Z).

Common Mistakes

Changing (A) during (\beta^{+}) decay (it must stay constant).
Increasing (Z) instead of decreasing it (confusing (\beta^{+}) with (\beta^{-})).
Calculating neutrons using the original fluorine nucleus instead of the daughter (stable) nucleus.

Things to Be Careful About

Always apply the decay rule first (update (Z) and (A)), then calculate (N).
Remember: (\beta^{-}) decay makes (Z) go up by 1, but (\beta^{+}) makes (Z) go down by 1.

Techniques used

apply the change in proton number for $\beta^{+}$ decayuse conservation of nucleon number in nuclear decaycalculate neutron number from $N = A - Z$

Q401MEasyParticle Physics

Which statement is correct?

Options

A A baryon is a hadron and consists of 2 quarks.
B A meson is a hadron and consists of 3 quarks.
C An electron is a fundamental particle and is a lepton.
D A neutrino is a fundamental particle and is a hadron.

Pick an answer

DifficultyEasy

Worked solution

Working

Baryons are hadrons made of $3$ quarks (not $2$ ). Mesons are hadrons made of a quark and an antiquark (not $3$ quarks). Neutrinos are leptons (not hadrons). An electron is a fundamental particle and is a lepton.

Answer

C

Final answer

C

Detailed explanation

Background Concept

In the quark model, particles are grouped as:

Hadrons: particles that experience the strong interaction and are made of quarks.
- Baryons: made of three quarks ( $qqq$ ) (or three antiquarks for antibaryons).
- Mesons: made of a quark and an antiquark ( $q\bar{q}$ ).
Leptons: fundamental particles that do not take part in the strong interaction (e.g. electron, muon, neutrino).

So: baryons and mesons are both hadrons, but have different quark compositions; electrons and neutrinos are leptons.

Understanding the Question

You are asked to choose the one correct statement among four options about:

whether a particle is a hadron or a lepton, and
how many quarks it contains (where relevant).

Only one option matches the standard definitions.

Approach

Check each option against the key facts:

Baryon quark content ( $qqq$ ).
Meson quark content ( $q\bar{q}$ ).
Electron classification (fundamental lepton).
Neutrino classification (fundamental lepton).

The option that agrees with these definitions is correct.

Step-by-Step Reasoning

Option A: “A baryon is a hadron and consists of 2 quarks.”
- Baryon is indeed a hadron, but baryons consist of three quarks, not two. So A is false.
Option B: “A meson is a hadron and consists of 3 quarks.”
- Meson is indeed a hadron, but mesons consist of a quark and an antiquark (two constituents), not three quarks. So B is false.
Option C: “An electron is a fundamental particle and is a lepton.”
- The electron is a fundamental particle (not made of quarks) and it is a lepton. So C is true.
Option D: “A neutrino is a fundamental particle and is a hadron.”
- Neutrinos are fundamental, but they are leptons, not hadrons. So D is false.

Therefore the correct statement is C.

Key Takeaways

Baryon: hadron, $qqq$ .
Meson: hadron, $q\bar{q}$ .
Electron and neutrino: fundamental leptons, not hadrons.

Common Mistakes

Swapping baryon and meson quark counts (thinking baryons are $2$ quarks or mesons are $3$ quarks).
Thinking “fundamental particle” automatically means “hadron” (it does not; hadrons are composite).
Forgetting neutrinos are leptons.

Things to Be Careful About

The word hadron implies “made of quarks” (composite), so fundamental particles like electrons/neutrinos are not hadrons.
Mesons involve an antiquark as well as a quark: $q\bar{q}$ , not $qq$ .

Techniques used

recall the definitions of baryons, mesons, hadrons and leptonsuse quark content rules to distinguish baryons from mesonsclassify fundamental particles into hadrons or leptons

Physics 9702/11 — October/November 2024

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2) Conserve momentum in the $y$ -direction

3) Conserve momentum in the $x$ -direction