9702/12 Oct/Nov 2024 — Physics MCQ Answers

Q11MEasyPhysical Quantities and Units

A physical quantity consists of a magnitude and a unit.

Which row does not show a correct combination of a quantity and its unit?

Options

	quantity	unit
A	mass	gram
B	length	metre
C	charge	ampere
D	temperature	kelvin

Pick an answer

DifficultyEasy

Worked solution

Working

Charge $Q$ is measured in coulomb (C), where

Q = It

So $1\ \text{C} = 1\ \text{A s}$ . The ampere (A) is the unit of current, not charge.

Answer

C

Final answer

C

Detailed explanation

Background Concept

A physical quantity is described by a numerical value (magnitude) and a unit. Many quantities have their own SI units:

length in metres (m)
temperature in kelvin (K)
electric current in amperes (A)

Some quantities are defined in terms of others. For example, electric charge $Q$ is defined using current $I$ and time $t$ :

Q = It

So the unit of charge must be the unit of current multiplied by the unit of time.

Understanding the Question

You are given four rows, each pairing a physical quantity with a unit. You must choose the one row where the unit does not belong to that quantity.

Approach

Check each pairing by recalling the correct unit (or using a defining equation). The key trap here is that ampere is a common electrical unit, but it is for current, not charge.

Step-by-Step Reasoning

Row B: length in metre is correct (SI base unit for length is m).
Row D: temperature in kelvin is correct (SI base unit for thermodynamic temperature is K).
Row A: mass in gram is still a valid unit of mass (even though the SI base unit is kilogram, gram is a commonly used metric unit of mass).
Row C: charge in ampere is incorrect.

Use the definition of charge:

Q = It

So the unit of $Q$ is

\text{unit of }Q = (\text{A})(\text{s}) = \text{A s}

This unit is called the coulomb (C). Therefore charge should be in coulombs, not amperes.

So the incorrect pairing is row C.

Key Takeaways

Current $I$ has unit ampere (A).
Charge $Q$ has unit coulomb (C) and $1\ \text{C} = 1\ \text{A s}$ .
Using defining equations is a reliable way to check units.

Common Mistakes

Thinking that “ampere” is a general unit for electricity and using it for charge.
Assuming a unit must be an SI base unit to be correct (e.g. treating gram as “wrong” just because kilogram is the SI base unit).

Things to Be Careful About

Distinguish carefully between related electrical quantities: charge ( $Q$ ), current ( $I$ ), and potential difference ( $V$ ) all have different units.
When unsure, write a defining equation (like $Q = It$ ) and deduce the unit from it.

Techniques used

recall the SI unit associated with each physical quantitydistinguish between charge and current using their defining relationships

Q21MMedium-EasyWork, Energy and PowerPhysical Quantities and Units

A small car travels in a town.

What is a reasonable estimate of the kinetic energy of the car?

Options

A $5 \times 10^{4}\ \text{J}$
B $5 \times 10^{7}\ \text{J}$
C $5 \times 10^{10}\ \text{J}$
D $5 \times 10^{13}\ \text{J}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Take a small car: $m \approx 1000\ \text{kg}$ and town speed $v \approx 10\ \text{m s}^{-1}$ .

E_k = \frac{1}{2}mv^2 \approx \frac{1}{2}(1000)(10^2)=5\times 10^4\ \text{J}

Answer

A

Final answer

A

Detailed explanation

Background Concept

Kinetic energy is the energy a moving object has due to its motion. For speed $v$ (in $\text{m s}^{-1}$ ) and mass $m$ (in $\text{kg}$ ), the kinetic energy is

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

Because of the $v^2$ term, doubling the speed increases $E_k$ by a factor of 4. For estimate questions, you only need a realistic order of magnitude.

Understanding the Question

You are not given any numbers, only “a small car travels in a town”. So you must:

choose a reasonable mass for a small car (about $10^3\ \text{kg}$ ),
choose a reasonable town speed (around $10\text{ to }15\ \text{m s}^{-1}$ , i.e. about $36$ to $54\ \text{km h}^{-1}$ ),
calculate $E_k$ and match it to the closest option.

Approach

Estimate $m$ for a small car.
Estimate a typical town speed $v$ .
Use $E_k = \tfrac{1}{2}mv^2$ .
Compare the result’s power of ten with the answers.

Step-by-Step Reasoning

Choose typical values:

Small car mass: $m \approx 1000\ \text{kg}$ .
Town driving speed: $v \approx 10\ \text{m s}^{-1}$ .

Now calculate:

E_k = \frac{1}{2}mv^2 \approx \frac{1}{2}(1000)(10^2)

Compute $10^2 = 100$ :

E_k \approx \frac{1}{2}(1000)(100)=\frac{1}{2}(100000)=50000\ \text{J}=5\times 10^4\ \text{J}

This matches option A.

A quick sense-check: even if you used a higher town speed, say $v=15\ \text{m s}^{-1}$ ,

E_k \approx \frac{1}{2}(1000)(15^2)=500\times 225\approx 1.1\times 10^5\ \text{J}

which is still much closer to $10^4$ – $10^5\ \text{J}$ than to $10^7\ \text{J}$ or larger.

Key Takeaways

Use $E_k = \tfrac{1}{2}mv^2$ .
For estimation, pick realistic everyday values (car: $\sim 10^3\ \text{kg}$ ; town speed: $\sim 10\ \text{m s}^{-1}$ ).
Kinetic energy for a car in town is typically around $10^5\ \text{J}$ .

Common Mistakes

Using an unrealistic speed (e.g. $100\ \text{m s}^{-1}$ ) which inflates $E_k$ by $v^2$ .
Confusing $\text{km h}^{-1}$ with $\text{m s}^{-1}$ (forgetting that $36\ \text{km h}^{-1} \approx 10\ \text{m s}^{-1}$ ).
Forgetting the factor of $\tfrac{1}{2}$ .

Things to Be Careful About

The question wants a “reasonable estimate”, so being within an order of magnitude is enough.
Check that your chosen mass and speed are plausible for a small car in a town setting.
Compare powers of ten first: options jump by $10^3$ each time, so an overestimated speed can move you to the wrong option very quickly.

Techniques used

estimate typical values for mass and speed from everyday contextapply the kinetic energy equationcompare an order-of-magnitude result with the given options

Q31MMedium-EasyPhysical Quantities and Units

A solid bar has a square cross-section. Its length is measured as $50.0 \pm 0.2\ \text{cm}$ and its width is measured as $2.00 \pm 0.01\ \text{cm}$ .

These values are used to calculate the volume of the bar.

What is the percentage uncertainty in the calculated volume?

Options

A $\pm 0.21\%$
B $\pm 0.22\%$
C $\pm 0.90\%$
D $\pm 1.4\%$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Volume $V = Lw^2$ .

Percentage uncertainty:

\frac{\Delta V}{V} = \frac{\Delta L}{L} + 2\frac{\Delta w}{w}

For $L = 50.0 \pm 0.2\ \text{cm}$ :

\frac{\Delta L}{L} = \frac{0.2}{50.0} = 0.004 = 0.40\%

For $w = 2.00 \pm 0.01\ \text{cm}$ :

\frac{\Delta w}{w} = \frac{0.01}{2.00} = 0.005 = 0.50\%

So

\%\ \Delta V = 0.40\% + 2(0.50\%) = 1.40\%

Answer

D

Final answer

D

Detailed explanation

Background Concept

When a quantity is calculated from measured values, its uncertainty depends on how those values combine.

For multiplication/division, fractional uncertainties add:

\frac{\Delta (ab)}{ab} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b}

For a power, the fractional uncertainty is multiplied by the power:

\frac{\Delta (a^n)}{a^n} \approx |n|\frac{\Delta a}{a}

These rules are standard at AS level for estimating the maximum percentage uncertainty.

Understanding the Question

The bar has a square cross-section, so its cross-sectional area is $w^2$ where $w$ is the width. The volume is therefore:

V = L \times w^2

We are given absolute uncertainties in $L$ and $w$ and asked for the percentage uncertainty in $V$ .

Approach

Write $V$ in terms of the measured quantities ( $L$ and $w$ ).
Convert each absolute uncertainty to a percentage uncertainty.
Add percentage uncertainties, remembering to double the width’s percentage uncertainty because it is squared.
Match the result to the options.

Step-by-Step Reasoning

Because the cross-section is square:

V = Lw^2

Fractional (percentage) uncertainty in $L$ :

\frac{\Delta L}{L} = \frac{0.2}{50.0} = 0.004

So percentage uncertainty in $L$ is $0.4\%$ .

Fractional (percentage) uncertainty in $w$ :

\frac{\Delta w}{w} = \frac{0.01}{2.00} = 0.005

So percentage uncertainty in $w$ is $0.5\%$ .

Now propagate through $V = Lw^2$ :

$L$ contributes $0.4\%$ .
$w^2$ contributes $2 \times 0.5\% = 1.0\%$ .

Total percentage uncertainty:

\%\Delta V = 0.4\% + 1.0\% = 1.4\%

This corresponds to option D.

Key Takeaways

For products, add fractional/percentage uncertainties.
For powers, multiply the fractional/percentage uncertainty by the power.
A squared measurement (here $w^2$ ) doubles its percentage uncertainty contribution.

Common Mistakes

Forgetting that $w$ is squared and using $\frac{\Delta V}{V}=\frac{\Delta L}{L}+\frac{\Delta w}{w}$ .
Adding absolute uncertainties instead of percentage (fractional) uncertainties.
Calculating $0.2/50.0$ as $0.04$ (factor of 10 error).

Things to Be Careful About

Use the fractional uncertainty rule for multiplication and powers; do not mix with rules for addition/subtraction.
Keep track of the power of $w$ carefully: $w^2$ means the factor of 2 is essential.
Ensure percentages are consistent before adding (either all in % or all as decimals).

Techniques used

identify the mathematical relationship for the derived quantityconvert absolute uncertainty to fractional (percentage) uncertaintyapply uncertainty propagation for products and powers by adding fractional uncertaintiesselect the correct multiple-choice option matching the calculated value

Q41MMedium-EasyKinematics

The graph shows how the acceleration of an object moving in a straight line varies with time.

Fig. 1

The object starts from rest.

Which graph shows the variation with time of the velocity of the object over the same time interval?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Acceleration is the gradient of the $v$ – $t$ graph. The given $a$ – $t$ graph is always positive, starting at $0$ , rising to a maximum, then falling back to $0$ .

So $v$ must:

start from rest ( $v=0$ at $t=0$ ),
increase throughout (since $a>0$ ),
have gradient $0$ at the start and at the end (since $a=0$ at both ends), with the steepest gradient when $a$ is maximum.

Only option A shows a velocity increasing monotonically and flattening to a horizontal tangent at the end.

Answer

A

Final answer

A

Detailed explanation

Background Concept

Acceleration and velocity are linked by

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

So:

On a velocity–time ( $v$ – $t$ ) graph, the gradient at any time equals the acceleration at that time.
On an acceleration–time ( $a$ – $t$ ) graph, the area under the curve between two times equals the change in velocity over that interval:

\Delta v = \int a\,\mathrm{d}t

If $a$ stays positive, velocity must keep increasing (it cannot decrease without negative acceleration).

Understanding the Question

You are given an $a$ – $t$ graph for an object starting from rest. The acceleration:

is $0$ at the start,
increases smoothly to a single maximum,
then decreases smoothly back to $0$ ,
is always above the time axis (so $a>0$ throughout).

You must choose which candidate $v$ – $t$ graph has a gradient that matches this acceleration behaviour.

Approach

Use $a = \mathrm{d}v/\mathrm{d}t$ to translate features of the $a$ – $t$ graph into features of the gradient of the $v$ – $t$ graph.
Use the sign of $a$ to decide whether $v$ is increasing or decreasing.
Check the start and end conditions: starting from rest gives $v(0)=0$ , and $a=0$ at the end means the $v$ – $t$ curve must end with zero gradient (horizontal tangent).

Step-by-Step Reasoning

The object starts from rest, so the velocity at $t=0$ is

v(0)=0.

Because the acceleration is positive for all times shown ( $a>0$ ), the velocity must increase throughout the interval. That rules out any option where $v$ falls back down (which would require negative acceleration at some time).
At the start, $a=0$ , so

\left.\frac{\mathrm{d}v}{\mathrm{d}t}\right|_{t=0} = 0.

This means the $v$ – $t$ graph starts with a horizontal tangent.

As $a$ rises to a maximum, the gradient of the $v$ – $t$ graph must also rise to its steepest value.
As $a$ falls back to $0$ at the end, the gradient of the $v$ – $t$ graph must decrease back to zero, so the $v$ – $t$ graph must flatten out again (another horizontal tangent at the end).

So the correct $v$ – $t$ shape is: increasing all the way, initially shallow, then steepest around mid-time, then flattening towards the end.

Among the options:

A shows exactly this behaviour.
B returns to zero velocity, implying velocity decreases later, which would require $a<0$ at some point (not true).
C oscillates and becomes negative, requiring negative acceleration at times.
D is concave up with ever-increasing gradient, which would correspond to acceleration increasing throughout, not increasing then decreasing.

Therefore the correct choice is A.

Key Takeaways

$a$ is the gradient of a $v$ – $t$ graph.
The sign of $a$ tells you whether $v$ is increasing ( $a>0$ ) or decreasing ( $a<0$ ).
If $a$ starts and ends at zero, the $v$ – $t$ graph must start and end with zero gradient (horizontal tangent).

Common Mistakes

Choosing a graph where $v$ decreases even though $a$ is always positive.
Forgetting that $a=0$ at the end implies the $v$ – $t$ graph must flatten (horizontal tangent), not keep getting steeper.
Confusing "area under $a$ – $t$ " (change in velocity) with "gradient of $a$ – $t$ ".

Things to Be Careful About

“Starts from rest” means $v=0$ at $t=0$ , not just “small”.
A positive acceleration does not mean velocity is constant; it means velocity is increasing.
Always match shape via gradient: increasing then decreasing acceleration corresponds to a $v$ – $t$ graph whose gradient increases then decreases, i.e. an S-like rise that levels off.

Techniques used

use the gradient of a velocity–time graph as accelerationuse the area under an acceleration–time graph as change in velocitymatch qualitative graph features (sign and end-behaviour) to physical constraints

Q51MMedium-EasyKinematics

A stone falls vertically from rest. Air resistance is negligible.

What is the speed of the stone when it has fallen through a distance of $0.40\ \text{m}$ ?

Options

A $2.0\ \text{m s}^{-1}$
B $2.8\ \text{m s}^{-1}$
C $3.9\ \text{m s}^{-1}$
D $7.8\ \text{m s}^{-1}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

From rest, $u=0$ , and for free fall $a=g=9.81\ \text{m s}^{-2}$ , $s=0.40\ \text{m}$ .

v^2 = u^2 + 2as

v = \sqrt{2gs} = \sqrt{2 \times 9.81 \times 0.40} = \sqrt{7.848} = 2.80\ \text{m s}^{-1}

Answer

B

Final answer

B

Detailed explanation

Background Concept

For motion with constant acceleration, the SUVAT equation

v^2 = u^2 + 2as

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

In free fall near Earth with negligible air resistance, the acceleration is approximately constant and equal to $g \approx 9.81\ \text{m s}^{-2}$ downward.

Understanding the Question

The stone is dropped from rest (so initial speed $u=0$ ) and falls vertically through $0.40\ \text{m}$ . Air resistance is negligible, so acceleration is constant at $g$ .

We are asked for the speed after falling that distance (magnitude only), then to choose the matching option.

Approach

Use the constant-acceleration equation that does not involve time, since time is not given:

Take $u=0$ (from rest).
Take $a=g$ (free fall).
Substitute $s=0.40\ \text{m}$ into $v^2 = u^2 + 2as$ .
Solve for $v$ by taking the square root and match to the options.

Step-by-Step Reasoning

Given:

$u=0$
$a=g=9.81\ \text{m s}^{-2}$
$s=0.40\ \text{m}$

Apply SUVAT:

v^2 = u^2 + 2as

Substitute:

v^2 = 0^2 + 2(9.81)(0.40) = 7.848

So:

v = \sqrt{7.848} = 2.80\ \text{m s}^{-1}

This corresponds to option B.

Key Takeaways

In free fall with negligible air resistance, acceleration is constant: $a=g$ .
If time is not given, $v^2 = u^2 + 2as$ is usually the quickest route.
Dropped from rest means $u=0$ .

Common Mistakes

Using $v = gt$ even though $t$ is not given.
Forgetting the factor of 2 in $v^2 = u^2 + 2as$ .
Taking $g=9.8\ \text{m s}^{-2}$ is fine, but arithmetic errors can lead to the wrong option.
Giving a negative value for speed (speed is a magnitude; velocity direction is not asked here).

Things to Be Careful About

Ensure $s$ is in metres and $g$ in $\text{m s}^{-2}$ so units are consistent.
After finding $v^2$ , remember to take the square root to get $v$ .
Pick the option that matches the calculated value to appropriate significant figures (here $2.8\ \text{m s}^{-1}$ ).

Techniques used

select the constant-acceleration equation relating speed and distancesubstitute free-fall acceleration and given displacementtake the square root to obtain speed

Q61MMedium-EasyKinematics

A ball is thrown horizontally off a tall building. The ground is horizontal. Air resistance is negligible.

Which statement about the motion of the ball is correct?

Options

A The acceleration of the ball is always at right angles to the path of the ball.
B The ball follows a circular path until it hits the ground.
C The ball has constant acceleration.
D The ball’s time in the air is proportional to the velocity at which the ball is thrown.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

With negligible air resistance, the only force on the ball is its weight $mg$ (constant), so the acceleration is constant and vertically downward with magnitude $g$ .

Answer

C

Final answer

C

Detailed explanation

Background Concept

For projectile motion with air resistance negligible, the only significant force on the object after release is its weight $mg$ acting vertically downward.

By Newton's second law,

F = ma

so a constant force produces a constant acceleration. Here,

a = g \; (\text{downward})

Also, the horizontal and vertical motions are independent:

Horizontal: constant velocity (no horizontal force).
Vertical: uniformly accelerated motion with acceleration $g$ downward.

Understanding the Question

A ball is thrown horizontally from a tall building. You must choose which statement about its motion is correct given:

ground is horizontal,
air resistance is negligible.

So the path is a standard projectile (parabolic) and the acceleration is due to gravity only.

Approach

Decide what the acceleration of the ball is throughout the flight.

Identify forces acting (only weight).
Use $F=ma$ to infer acceleration.
Check each option against this model: constant downward acceleration, horizontal speed constant, vertical speed changes.

Step-by-Step Reasoning

After the ball leaves the hand, with air resistance negligible, there is no contact force and no drag force.
The only force is weight $mg$ , which is constant (since $m$ and $g$ are constant near Earth’s surface).
Therefore the resultant force is constant and vertical, so acceleration is constant and vertical:

a = g \text{ downward}

Now evaluate the options:

A: “acceleration always at right angles to the path” is false. The acceleration is always vertical downward, but the velocity (tangent to the path) is not always horizontal; it has a vertical component that increases with time. So the acceleration is not always perpendicular to the velocity/path.
B: the path is not circular; with constant horizontal velocity and constant vertical acceleration, the trajectory is a parabola.
C: true, because the acceleration is constant ( $g$ downward) throughout the motion.
D: time of flight depends on vertical motion only. For a horizontal throw, initial vertical velocity $u_y=0$ , and the time depends on height $h$ :

h = \frac{1}{2}gt^2 \;\Rightarrow\; t = \sqrt{\frac{2h}{g}}

So it is independent of the horizontal throw speed, not proportional to it.

Therefore the correct option is C.

Key Takeaways

With negligible air resistance, projectile acceleration is constant and equal to $g$ vertically downward.
Horizontal motion has constant velocity; vertical motion is uniformly accelerated.
Time of flight for a horizontal launch depends on drop height, not on horizontal speed.

Common Mistakes

Thinking the acceleration points along the path (it does not; it is always vertically downward).
Assuming the path is circular rather than parabolic.
Believing time of flight increases with horizontal launch speed; it does not when launched horizontally.

Things to Be Careful About

“Acceleration at right angles to the path” is only true for uniform circular motion; a projectile is not in circular motion.
Independence of components: changing horizontal velocity does not change the vertical fall time (when air resistance is neglected).
Use the correct vertical kinematics equation with $u_y=0$ for a horizontal throw.

Techniques used

recognise the forces acting when air resistance is negligibleidentify the direction and constancy of gravitational accelerationseparate horizontal and vertical components of motion

Q71MMedium-EasyDynamicsWork, Energy and Power

A moving object strikes a stationary object. The collision is inelastic. The objects move off together.

Assume that the two objects form an isolated system.

Which row shows the possible values of total momentum and total kinetic energy for the system before and after the collision?

Options

	total momentum before collision / $\text{kg m s}^{-1}$	total momentum after collision / $\text{kg m s}^{-1}$	total kinetic energy before collision / $\text{J}$	total kinetic energy after collision / $\text{J}$
A	6	2	90	30
B	6	6	30	90
C	6	6	90	30
D	6	6	90	90

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For an isolated system, total momentum is conserved, so momentum before = momentum after.

In an inelastic collision where objects move off together, total kinetic energy decreases.

Only row C has momentum unchanged ( $6 \to 6$ ) and kinetic energy decreased ( $90 \to 30$ ).

Answer

C

Final answer

C

Detailed explanation

Background Concept

For collisions we often consider two key system quantities:

Total momentum $p$ of the system. For a system with no external resultant force (an isolated system), momentum is conserved:

\text{total momentum before} = \text{total momentum after}.

Total kinetic energy $E_k$ . Kinetic energy is not automatically conserved in collisions. It is conserved only in elastic collisions. In an inelastic collision, some kinetic energy is transformed into other forms (internal energy, sound, deformation), so:

E_{k,\,\text{after}} < E_{k,\,\text{before}}.

If the objects move off together, the collision is (by definition) perfectly inelastic, which guarantees a loss of kinetic energy.

Understanding the Question

One object is moving and hits another that is initially stationary. The collision is stated to be inelastic and the objects stick together and move off as one. You are told to assume the two objects form an isolated system.

You must choose the table row that could represent the system’s total momentum and total kinetic energy before and after.

So we need:

same total momentum before and after,
smaller total kinetic energy after than before.

Approach

Use “isolated system” (\Rightarrow) momentum conserved, so reject any option where total momentum changes.
Use “inelastic, move off together” (\Rightarrow) kinetic energy decreases, so reject options where kinetic energy increases or stays the same.
The remaining option is the answer.

Step-by-Step Reasoning

Check momentum:
- Option A: momentum changes from $6$ to $2$ ; this violates momentum conservation for an isolated system, so A is impossible.
- Options B, C, D: momentum is $6$ before and $6$ after, so these satisfy momentum conservation.
Check kinetic energy for inelastic collision (must decrease):
- Option B: kinetic energy increases from $30$ to $90$ ; this is not possible in an inelastic collision without an external energy input.
- Option D: kinetic energy stays the same ( $90$ to $90$ ); that would correspond to an elastic collision, not an inelastic one where the objects stick together.
- Option C: kinetic energy decreases from $90$ to $30$ ; this matches an inelastic collision.

Therefore, option C is the only physically consistent row.

Key Takeaways

In an isolated system, total momentum is conserved in any collision.
In an inelastic collision, total kinetic energy decreases.
“Move off together” indicates a perfectly inelastic collision, guaranteeing $E_k$ loss.

Common Mistakes

Choosing an option where momentum changes (forgetting that “isolated system” means no external impulse).
Thinking kinetic energy is conserved in all collisions (it is conserved only in elastic collisions).
Selecting an option where kinetic energy increases, which would require an additional energy source.

Things to Be Careful About

Momentum is a vector, but the table provides magnitudes; still, for a 1D collision with consistent direction, conservation implies equal numerical totals before and after.
“Inelastic” alone means KE is not conserved; “move off together” is stronger and indicates KE must be strictly less after.
Do not confuse conservation of energy (total energy always conserved) with conservation of kinetic energy (not always conserved).

Techniques used

apply conservation of momentum to an isolated systemuse the definition of a perfectly inelastic collision to infer kinetic energy decreasescompare before-and-after values to select a consistent option

Q81MMedium-EasyDynamicsKinematics

The graph shows how a quantity $Y$ varies with a quantity $X$ for an object falling vertically at its terminal velocity towards the surface of the Earth.

Fig. 1

Which quantities could $X$ and $Y$ represent?

Options

	X	Y
A	time	acceleration
B	time	height above surface
C	distance moved	kinetic energy
D	distance moved	velocity

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

At terminal velocity, the speed is constant and the acceleration is zero.

A straight line with negative gradient that reaches $Y=0$ at a finite $X$ corresponds to a quantity decreasing uniformly to zero, e.g.

h = h_0 - vt

So $X$ can be time and $Y$ can be height above the surface.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Terminal velocity occurs when the resistive (drag) force balances the weight. The resultant force is then zero, so by Newton's second law the acceleration is zero:

F_{\text{resultant}} = 0 \Rightarrow a = 0

With $a=0$ , the velocity is constant. For constant velocity motion, displacement changes linearly with time, so any “position-like” quantity (such as height above the ground) changes as

h = h_0 - vt

where $h_0$ is the initial height and $v$ is the (constant) speed.

Understanding the Question

The object is falling vertically at terminal velocity towards the Earth's surface. The graph of $Y$ against $X$ is a straight line with negative gradient: it starts at a positive value of $Y$ when $X=0$ and reaches $Y=0$ at some positive value of $X$ .

The task is to identify which pair of physical quantities could produce that graph.

Approach

Use “terminal velocity” to decide which kinematic quantities must be constant (velocity) or zero (acceleration).
Compare that with the shape of the graph:
- a constant quantity gives a horizontal line,
- a uniformly decreasing quantity gives a straight line with negative gradient.
Check each option for consistency.

Step-by-Step Reasoning

Terminal velocity implies:
- $a = 0$ (so acceleration-time graph would be a horizontal line at $0$ ),
- $v$ is constant (so velocity-time and velocity-distance graphs would be horizontal lines).
The given graph is a straight line decreasing from a positive intercept to zero. This matches a “remaining” quantity that decreases uniformly to zero, such as height above the surface during constant-speed descent.
Check the options:
- A (time, acceleration): at terminal velocity $a=0$ at all times, so the graph should be $Y=0$ (not a decreasing line). Reject.
- B (time, height above surface): with constant downward speed, height decreases uniformly:

h = h_0 - vt

This is a straight line with negative gradient, reaching $h=0$ at the time of impact. Accept.

C (distance moved, kinetic energy): at terminal velocity, $v$ is constant so kinetic energy

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

is constant; graph would be horizontal. Reject.

D (distance moved, velocity): at terminal velocity, $v$ is constant; graph would be horizontal. Reject.

Therefore, the only consistent choice is B.

Key Takeaways

Terminal velocity means zero acceleration and constant velocity.
Constant velocity motion gives a linear position (or height) vs time graph.
Constant quantities (acceleration at $0$ , velocity, kinetic energy) produce horizontal graphs, not sloping lines.

Common Mistakes

Thinking “falling” always means acceleration $g$ ; at terminal velocity the acceleration is $0$ .
Assuming kinetic energy must increase while falling; it stops increasing once speed is constant.
Mixing up “distance moved” with “height above surface”: height decreases as distance moved increases.

Things to Be Careful About

The line reaching $Y=0$ at a finite $X$ strongly suggests a quantity that can become zero (like height above the surface at impact).
Keep the sign convention clear: the graph shows a decrease, consistent with height decreasing with time for downward motion at constant speed.

Techniques used

use the condition of terminal velocity to infer acceleration and velocity behaviourmatch the shape of a straight-line graph to a linear kinematics relationshipeliminate options by checking whether each quantity should be constant or changing

Q91MMedium-EasyDynamics

A ball of mass $2.0\ \text{kg}$ travels horizontally with a speed of $4.0\ \text{m s}^{-1}$ . The ball collides with a wall and rebounds in the opposite direction with a speed of $2.8\ \text{m s}^{-1}$ . The time of the collision is $150\ \text{ms}$ .

What is the average force exerted on the wall?

Options

A $16\ \text{N}$
B $37\ \text{N}$
C $53\ \text{N}$
D $91\ \text{N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Take initial direction as positive.

\Delta p = m(v-u)=2.0\left(-2.8-4.0\right)=-13.6\ \text{kg m s}^{-1}

\Delta t = 150\ \text{ms} = 0.150\ \text{s}

F_{\text{avg}}=\frac{|\Delta p|}{\Delta t}=\frac{13.6}{0.150}=90.7\ \text{N}\approx 91\ \text{N}

Force on wall has the same magnitude (Newton's 3rd law).

Answer

D

Final answer

D

Detailed explanation

Background Concept

When an object’s momentum changes, there must be a net force acting on it. The impulse of the force equals the change in momentum:

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

where momentum $p = mv$ , $m$ is mass and $v$ is velocity (a vector). For a collision, the force can vary during contact, but the average force over the collision time still satisfies

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

Also, by Newton’s third law, the force the ball exerts on the wall is equal in magnitude and opposite in direction to the force the wall exerts on the ball. So we can find the average force on the ball and quote the same magnitude for the wall.

Understanding the Question

A $2.0\ \text{kg}$ ball moves horizontally at $4.0\ \text{m s}^{-1}$ , hits a wall, and rebounds in the opposite direction at $2.8\ \text{m s}^{-1}$ . The collision lasts $150\ \text{ms}$ . You are asked for the average force exerted on the wall.

Key points:

“Rebounds in the opposite direction” means the final velocity has the opposite sign to the initial velocity.
Time must be converted from ms to s.
The force on the wall has the same magnitude as the force on the ball.

Approach

Choose a positive direction (take the initial motion as positive).
Write initial velocity $u$ and final velocity $v$ with signs.
Compute change in momentum $\Delta p = m(v-u)$ .
Convert collision time to seconds.
Use $F_{\text{avg}} = |\Delta p|/\Delta t$ to get the magnitude.
Match to the closest option.

Step-by-Step Reasoning

Take initial direction as positive:
$u = +4.0\ \text{m s}^{-1},\qquad v = -2.8\ \text{m s}^{-1}$
Change in momentum:
$\Delta p = m(v-u) = 2.0\left(-2.8-4.0\right)=2.0(-6.8)=-13.6\ \text{kg m s}^{-1}$
The negative sign tells us the impulse (and average force on the ball) is opposite to the initial direction.
Convert the time:
$\Delta t = 150\ \text{ms} = 150\times 10^{-3}\ \text{s} = 0.150\ \text{s}$
Average force magnitude:
$F_{\text{avg}}=\frac{|\Delta p|}{\Delta t}=\frac{13.6}{0.150}=90.7\ \text{N}\approx 91\ \text{N}$
The force exerted on the wall has the same magnitude, so the correct option is D.

Key Takeaways

Use momentum as a vector: rebound means the final velocity has the opposite sign.
Impulse relation: $F_{\text{avg}}=\Delta p/\Delta t$ .
Convert milliseconds to seconds.
Force on wall and force on ball are equal in magnitude (Newton’s third law).

Common Mistakes

Treating the rebound speed as $+2.8\ \text{m s}^{-1}$ instead of negative, giving too small a momentum change.
Forgetting to convert $150\ \text{ms}$ to $0.150\ \text{s}$ (would make the force $1000$ times too small).
Using $\Delta v = 2.8-4.0$ (subtracting speeds) rather than $-2.8-4.0$ (subtracting velocities).
Confusing “force on wall” with “force on ball” and changing the magnitude (only the direction changes).

Things to Be Careful About

Always define a positive direction first; it prevents sign errors.
Use $\text{ms}$ conversion carefully: $1\ \text{ms}=10^{-3}\ \text{s}$ .
In MCQs, round sensibly to match options: $90.7\ \text{N}$ corresponds to $91\ \text{N}$ .

Techniques used

assign a sign convention for direction of motioncalculate change in momentum using initial and final velocitiesuse impulse-momentum relation to find average forceapply Newton's third law to relate force on ball and force on wall

Q101MMediumDynamicsWork, Energy and Power

Two train carriages each of mass $5000\ \text{kg}$ roll toward one another on a horizontal frictionless track. One is travelling at a speed of $2.00\ \text{m s}^{-1}$ and the other at a speed of $1.00\ \text{m s}^{-1}$ , as shown.

Fig. 1

They collide and join together.

What is the kinetic energy lost during the collision?

Options

A $1250\ \text{J}$
B $7500\ \text{J}$
C $11\ 250\ \text{J}$
D $12\ 500\ \text{J}$

Pick an answer

DifficultyMedium

Worked solution

Working

Take right as positive.

Initial momentum:

p = 5000(2.00) + 5000(-1.00) = 5000\ \text{kg m s}^{-1}

After collision, joined mass $= 10000\ \text{kg}$ , so

v = \frac{5000}{10000} = 0.500\ \text{m s}^{-1}

Initial kinetic energy:

E_{\text{i}}=\frac12(5000)(2.00^2)+\frac12(5000)(1.00^2)=12500\ \text{J}

Final kinetic energy:

E_{\text{f}}=\frac12(10000)(0.500^2)=1250\ \text{J}

Kinetic energy lost:

\Delta E = E_{\text{i}}-E_{\text{f}}=12500-1250=11250\ \text{J}

Answer

C

Final answer

C

Detailed explanation

Background Concept

In a collision on a frictionless track, there is no external horizontal force on the system, so total linear momentum is conserved.

Kinetic energy is not necessarily conserved. If the objects join together after collision, the collision is perfectly inelastic, and some kinetic energy is converted into internal energy (deformation, heat, sound).

Key equations:

Momentum: $p = mv$ (a vector; direction matters).
Conservation of momentum (1D): $m_1u_1 + m_2u_2 = (m_1+m_2)v$ .
Kinetic energy: $E_k = \frac12 mv^2$ (a scalar; uses speed squared).

Understanding the Question

Two carriages, each of mass $5000\ \text{kg}$ , move towards each other with speeds $2.00\ \text{m s}^{-1}$ and $1.00\ \text{m s}^{-1}$ on a horizontal frictionless track.

They collide and stick together. The question asks for the kinetic energy lost:

\text{KE lost} = \text{total KE before} - \text{total KE after}.

To find KE after, we first need the common final velocity, which comes from momentum conservation.

Approach

Choose a positive direction (e.g. to the right).
Write the initial velocities with signs (one will be negative because it moves left).
Use conservation of momentum to find the final velocity $v$ of the joined carriages.
Calculate total kinetic energy before and after.
Subtract to get the kinetic energy lost and match it to the options.

Step-by-Step Reasoning

Take right as positive.

1) Initial momentum
Left carriage: $m=5000\ \text{kg}$ , $u_1=+2.00\ \text{m s}^{-1}$ .
Right carriage: $m=5000\ \text{kg}$ , $u_2=-1.00\ \text{m s}^{-1}$ (negative because it moves left).

Total initial momentum:

p = 5000(2.00) + 5000(-1.00) = 5000\ \text{kg m s}^{-1}.

2) Final velocity after sticking
Combined mass: $m_1+m_2 = 10000\ \text{kg}$ .

Momentum conservation:

5000 = 10000\,v \quad\Rightarrow\quad v = 0.500\ \text{m s}^{-1}.

(Positive, so the joined carriages move to the right.)

3) Kinetic energy before
Even though one carriage moves left, kinetic energy uses $v^2$ , so we can use the given speeds.

E_{\text{i}} = \frac12(5000)(2.00^2) + \frac12(5000)(1.00^2).

Compute:

First term: $\frac12\cdot 5000\cdot 4 = 10000\ \text{J}$
Second term: $\frac12\cdot 5000\cdot 1 = 2500\ \text{J}$

So:

E_{\text{i}} = 12500\ \text{J}.

4) Kinetic energy after

E_{\text{f}} = \frac12(10000)(0.500^2) = \frac12(10000)(0.25) = 1250\ \text{J}.

5) Kinetic energy lost

\Delta E = 12500 - 1250 = 11250\ \text{J}.

This corresponds to option C.

Key Takeaways

In 1D collisions, use momentum conservation to find the final velocity.
If objects stick together, kinetic energy decreases (perfectly inelastic).
Momentum needs signs (direction); kinetic energy does not (it depends on $v^2$ ).

Common Mistakes

Using both velocities as positive when the objects move towards each other (momentum then becomes too large).
Assuming kinetic energy is conserved just because momentum is conserved.
Using total mass incorrectly (forgetting the joined mass is $10000\ \text{kg}$ ).
Forgetting to subtract: the question asks for loss, not final kinetic energy.

Things to Be Careful About

Keep a consistent sign convention throughout the momentum calculation.
Work in SI units (already given here).
For MCQ, ensure the numerical result exactly matches one option; here $11250\ \text{J}$ matches C.

Techniques used

choose a sign convention for velocities on a straight lineapply conservation of momentum for a perfectly inelastic collisioncalculate kinetic energy before and after the collisionfind the kinetic energy lost as the difference in total kinetic energies

Q111MMedium-EasyDynamics

Which single condition must apply for an object to be in equilibrium?

Options

A The object has a constant non-zero acceleration.
B The object is stationary.
C There are no forces acting on the object.
D The resultant force acting on the object is zero.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For equilibrium, acceleration is zero.

By Newton's second law,

\sum F = ma

So with $a = 0$ , the condition is $\sum F = 0$ .

Answer

D

Final answer

D

Detailed explanation

Background Concept

An object is in (translational) equilibrium when it has no acceleration. Newton’s second law links force to acceleration:

\sum F = ma

If the acceleration is zero, then the resultant (net) force must be zero:

\sum F = 0

This does not mean there are no forces at all; it means all the forces balance so their vector sum is zero.

(For a rigid body in full equilibrium, you also need zero resultant moment, but this MCQ asks for a single condition and is focusing on the force condition.)

Understanding the Question

You are asked which ONE condition must apply for an object to be “in equilibrium”. The options include statements about acceleration, being stationary, having no forces, and having zero resultant force.

Approach

Use Newton’s laws:

Equilibrium (\Rightarrow) no acceleration.
Convert “no acceleration” into a force condition using (\sum F = ma).
Then check which option matches this necessary condition.

Step-by-Step Reasoning

In equilibrium, the velocity is constant (it may be zero or non-zero), so acceleration is:

a = 0

Apply Newton’s second law:

\sum F = ma

Substitute (a = 0):

\sum F = m(0) = 0

So the resultant force must be zero → option D.

Why the other options are not “must” conditions:

A: constant non-zero acceleration implies (\sum F \neq 0), so not equilibrium.
B: stationary is not required; an object moving at constant velocity can be in equilibrium.
C: forces can act but balance (e.g. weight and normal reaction on a book at rest).

Key Takeaways

Equilibrium (translation) means (a = 0).
(a = 0) is equivalent to (\sum F = 0) by Newton’s second law.
Balanced forces can exist even when forces are present.

Common Mistakes

Thinking equilibrium means “no forces act” rather than “forces balance”.
Thinking equilibrium requires the object to be stationary (it can move with constant velocity).
Confusing “constant velocity” with “constant non-zero acceleration”.

Things to Be Careful About

The word “resultant” is crucial: it means the vector sum of all forces.
In some contexts (rigid bodies), equilibrium also needs (\sum \tau = 0); however, this question asks for a single condition and the only universally required one listed is (\sum F = 0).

Techniques used

apply Newton's first law definition of equilibriumdistinguish between zero resultant force and zero individual forceseliminate options by considering counterexamples

Q121MMedium-EasyForces, Density and Pressure

A kite is in equilibrium at the end of a string, as shown.

Fig. 1

The kite has three forces acting on it: the weight $W$ , the tension $T$ in the string, and the force $F$ from the wind.

Which vector diagram represents the forces acting on the kite?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In equilibrium, the three forces must add to zero, so $W$ , $T$ and $F$ form a closed triangle when drawn head-to-tail.

$W$ is vertically downward. $T$ acts along the string towards the hand, and $F$ must be opposite the resultant of $W$ and $T$ to close the triangle.

Only diagram A shows a correct closed vector triangle with $W$ downward and $F$ opposite to $T$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

For an object in equilibrium, the resultant (net) force is zero:

\vec{W}+\vec{T}+\vec{F}=\vec{0}.

A standard way to show this is a vector triangle: draw the force vectors head-to-tail. If the object is in equilibrium, you return to the starting point (a closed triangle). The directions of the vectors must match the real forces:

Weight $W$ acts vertically downward.
Tension $T$ acts along the string (a tension force always pulls along the string).
Wind force $F$ acts in the direction the wind pushes the kite.

Understanding the Question

The kite is stationary, so it is in equilibrium. You are given three forces acting on it: $W$ , $T$ , and $F$ . The question asks which option (A–D) shows the correct vector diagram for these forces.

Key clues:

The diagram shows the string sloping down towards the left from the kite, so the force of tension on the kite is along that line.
Weight must be straight down.
The wind must provide a force that, together with the tension, balances the weight.

Approach

Use equilibrium: forces must form a closed head-to-tail triangle.
Fix the direction of $W$ (vertical down).
Fix the direction of $T$ (along the string).
The remaining vector $F$ must be the one that completes the triangle back to the start.
Compare with the options and choose the one that both (i) closes and (ii) has the correct directions.

Step-by-Step Reasoning

Since the kite is in equilibrium:

\vec{W}+\vec{T}+\vec{F}=\vec{0}.

$\vec{W}$ must be vertically downward in the vector diagram.
$\vec{T}$ must be along the string (tension acts along the string).
Once $\vec{W}$ and $\vec{T}$ are placed head-to-tail, the final vector $\vec{F}$ must point from the head of the second vector back to the tail of the first to make the triangle close.
Inspecting the options, only A shows the three vectors arranged head-to-tail in a closed triangle with $W$ downward and the other two vectors oriented so that one balances the horizontal component of the other while together balancing $W$ .

Therefore the correct vector diagram is A.

Key Takeaways

In equilibrium, vectors representing forces form a closed polygon (here, a triangle).
Always start by fixing directions you know for sure (e.g. weight is vertical).
Tension acts along the string and is a pulling force.

Common Mistakes

Drawing the three forces so they meet at a point rather than head-to-tail (that is not a vector-sum diagram).
Reversing the direction of the tension force (forgetting it must act along the string as a pull).
Choosing a diagram that has the right triangle shape but does not actually close head-to-tail.

Things to Be Careful About

The vector triangle is about vector addition, not the physical placement of forces on the kite.
The direction of each vector must still match the actual force direction (especially $W$ vertical and $T$ along the string).
“Closed triangle” means the final arrow ends exactly where the first arrow started (resultant zero).

Techniques used

use the condition for equilibrium that the resultant force is zerorepresent forces by a closed head-to-tail vector trianglededuce the directions of forces from the physical arrangement

Q131MMedium-EasyForces, Density and PressurePhysical Quantities and Units

Four forces act about a point $P$ , as shown.

Fig. 1

The forces act in the same plane and produce no resultant moment about point $P$ .

What is the length $XY$ ?

Options

A $6.9\ \text{m}$
B $12\ \text{m}$
C $14\ \text{m}$
D $24\ \text{m}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Clockwise moments about $P$ from the $6.0\ \text{N}$ forces:

M_{\text{cw}} = 6.0\times 7.0 + 6.0\times 3.0 = 60\ \text{N m}

For each $5.0\ \text{N}$ horizontal force, the perpendicular distance from $P$ is the vertical separation:

\text{perpendicular distance} = (\text{distance along sloping line})\sin 60^{\circ}

So anticlockwise moment is

M_{\text{acw}} = 5.0\sin 60^{\circ}\,(PX + PY) = 5.0\sin 60^{\circ}\,(XY)

Equate moments:

5.0\sin 60^{\circ}\,(XY)=60

XY = \frac{60}{5.0\sin 60^{\circ}} = 13.9\ \text{m} \approx 14\ \text{m}

Answer

C

Final answer

C

Detailed explanation

Background Concept

The moment (turning effect) of a force about a point is

\text{moment} = F \times d_\perp

where $F$ is the force and $d_\perp$ is the perpendicular distance from the point (pivot) to the line of action of the force.

For equilibrium with no resultant moment about point $P$ :

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

A common pitfall is using the distance to the point where the force is applied instead of the perpendicular distance to the line of action.

Understanding the Question

Two vertical forces of $6.0\ \text{N}$ act at horizontal distances $7.0\ \text{m}$ (left of $P$ ) and $3.0\ \text{m}$ (right of $P$ ). These produce moments easily because the forces are vertical and the given distances are horizontal (already perpendicular).

Two horizontal forces of $5.0\ \text{N}$ act at points $X$ and $Y$ on a straight sloping line through $P$ , inclined at $60^{\circ}$ to the horizontal (on the right). The unknown is the length $XY$ along this sloping line.

We are told the forces produce no resultant moment about $P$ , so we use the principle of moments.

Approach

Calculate the total clockwise moment from the two $6.0\ \text{N}$ forces about $P$ .
Work out the perpendicular distance from $P$ to each horizontal $5.0\ \text{N}$ force’s line of action. For a horizontal force, the moment arm is the vertical separation from $P$ .
Express those vertical separations in terms of distances along the sloping line using $\sin 60^{\circ}$ .
Add the two anticlockwise moments from the $5.0\ \text{N}$ forces, set equal to the clockwise total, and solve for $XY$ .

Step-by-Step Reasoning

diagram-1

1) Moments from the $6.0\ \text{N}$ vertical forces

Left force: $6.0\ \text{N}$ upward at $7.0\ \text{m}$ from $P$ gives moment magnitude $6.0\times 7.0=42\ \text{N m}$ .
Right force: $6.0\ \text{N}$ downward at $3.0\ \text{m}$ from $P$ gives moment magnitude $6.0\times 3.0=18\ \text{N m}$ .

They act in the same rotational sense (both clockwise about $P$ as drawn), so

M_{\text{cw}} = 42 + 18 = 60\ \text{N m}

2) Moments from the $5.0\ \text{N}$ horizontal forces

For a horizontal force, the perpendicular distance to its line of action is the vertical distance from $P$ to the point of application.

Points $X$ and $Y$ lie on a line inclined at $60^{\circ}$ to the horizontal, so if the distance along the sloping line from $P$ to a point is $s$ , its vertical separation is

\Delta y = s\sin 60^{\circ}

Let $PX = s_X$ and $PY = s_Y$ . Then the total anticlockwise moment from the two $5.0\ \text{N}$ forces is

M_{\text{acw}} = 5.0(s_X\sin 60^{\circ}) + 5.0(s_Y\sin 60^{\circ})

M_{\text{acw}} = 5.0\sin 60^{\circ}(s_X+s_Y)

But $s_X + s_Y$ is exactly the full distance along the line from $X$ to $Y$ , i.e.

XY = PX + PY = s_X + s_Y

So

M_{\text{acw}} = 5.0\sin 60^{\circ}(XY)

3) Equate clockwise and anticlockwise moments

5.0\sin 60^{\circ}(XY)=60

XY=\frac{60}{5.0\sin 60^{\circ}}=\frac{12}{\sin 60^{\circ}}

With $\sin 60^{\circ}=\frac{\sqrt{3}}{2}\approx 0.866$ :

XY \approx \frac{12}{0.866}=13.9\ \text{m}\approx 14\ \text{m}

So the correct option is C.

Key Takeaways

Always use perpendicular distance to the line of action when calculating moments.
For a horizontal force, the moment arm is the vertical separation from the pivot (and vice versa).
Geometry/trigonometry is often needed to find the perpendicular distance: here $\Delta y = s\sin 60^{\circ}$ .

Common Mistakes

Using $XY\cos 60^{\circ}$ instead of $XY\sin 60^{\circ}$ (wrong component for the perpendicular distance).
Taking moments of the $5.0\ \text{N}$ forces using the sloping distance directly instead of the perpendicular (vertical) distance.
Subtracting the two $5.0\ \text{N}$ moments because the forces are opposite directions; the direction of force and position together determine the sense of the moment, and here both give the same rotational sense.

Things to Be Careful About

Choose a sign convention (clockwise/anticlockwise) and apply it consistently.
Check that the distances given (7.0 m and 3.0 m) are already perpendicular to the vertical forces (they are horizontal), so no trig is needed for those.
In MCQs, round appropriately to match options: $13.9\ \text{m}$ corresponds to $14\ \text{m}$ .

Techniques used

take moments about a point and equate clockwise to anticlockwiseuse perpendicular distance to the line of action of a forceuse trigonometry to relate a sloping distance to its perpendicular componentadd moments from multiple forces with consistent sign convention

Q141MMedium-EasyForces, Density and Pressure

A rectangular block of lead of density $1.13 \times 10^{4}\ \text{kg m}^{-3}$ has sides of length $12.0\ \text{cm}$ , $15.0\ \text{cm}$ and $10.0\ \text{cm}$ .

What is the maximum pressure the block can exert when resting on a table?

Options

A $1.13\ \text{kPa}$
B $1.70\ \text{kPa}$
C $11.1\ \text{kPa}$
D $16.6\ \text{kPa}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Convert to metres: $0.12\ \text{m}$ , $0.15\ \text{m}$ , $0.10\ \text{m}$ .

V = 0.12 \times 0.15 \times 0.10 = 1.80 \times 10^{-3}\ \text{m}^3

m = \rho V = (1.13 \times 10^{4})(1.80 \times 10^{-3}) = 20.34\ \text{kg}

W = mg = 20.34 \times 9.81 \approx 2.00 \times 10^{2}\ \text{N}

Maximum pressure occurs on smallest face area:

A_{\min} = 0.12 \times 0.10 = 1.20 \times 10^{-2}\ \text{m}^2

P_{\max} = \frac{W}{A_{\min}} = \frac{2.00 \times 10^{2}}{1.20 \times 10^{-2}} \approx 1.66 \times 10^{4}\ \text{Pa} = 16.6\ \text{kPa}

Answer

D

Final answer

D

Detailed explanation

Background Concept

Pressure is defined as the normal (perpendicular) force per unit area:

P = \frac{F}{A}

For an object resting on a horizontal table, the force on the table is the weight of the object (assuming it is at rest):

W = mg

The mass $m$ can be found from the density relation:

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

So to find the maximum pressure the block can exert, we need the largest possible $P$ , which happens when $A$ is as small as possible.

Understanding the Question

You are given a rectangular block of lead with density $\rho = 1.13 \times 10^{4}\ \text{kg m}^{-3}$ and side lengths $12.0\ \text{cm}$ , $15.0\ \text{cm}$ , and $10.0\ \text{cm}$ .

The question asks for the maximum pressure when the block is resting on a table. Since pressure is $F/A$ , the weight $F$ is fixed (same block), so the pressure is maximised by placing it on the face with the smallest area.

Approach

Convert all lengths from cm to m.
Find the volume $V$ and then the mass $m = \rho V$ .
Find the weight $W = mg$ .
Compute the areas of the three faces and choose the smallest area $A_{\min}$ .
Use $P_{\max} = W/A_{\min}$ and match to the closest option.

Step-by-Step Reasoning

1) Convert dimensions to SI units

12.0\ \text{cm} = 0.120\ \text{m},\quad 15.0\ \text{cm} = 0.150\ \text{m},\quad 10.0\ \text{cm} = 0.100\ \text{m}

2) Volume of the block

V = 0.120 \times 0.150 \times 0.100 = 1.80 \times 10^{-3}\ \text{m}^3

3) Mass from density

m = \rho V = (1.13 \times 10^{4})(1.80 \times 10^{-3}) = 20.34\ \text{kg}

4) Weight (taking $g = 9.81\ \text{m s}^{-2}$ )

W = mg = 20.34 \times 9.81 \approx 1.996 \times 10^{2}\ \text{N}

So $W \approx 200\ \text{N}$ .

5) Find the smallest face area
Three possible contact areas:

A_{12\times 15} = 0.120 \times 0.150 = 0.0180\ \text{m}^2

A_{12\times 10} = 0.120 \times 0.100 = 0.0120\ \text{m}^2

A_{15\times 10} = 0.150 \times 0.100 = 0.0150\ \text{m}^2

Smallest area is $A_{\min} = 0.0120\ \text{m}^2$ .

6) Maximum pressure

P_{\max} = \frac{W}{A_{\min}} = \frac{1.996 \times 10^{2}}{0.0120} = 1.66 \times 10^{4}\ \text{Pa}

Convert to kPa ( $1\ \text{kPa} = 10^{3}\ \text{Pa}$ ):

P_{\max} = 16.6\ \text{kPa}

This corresponds to option D.

Key Takeaways

Maximum pressure for a given object occurs when it rests on the smallest contact area.
Use $m = \rho V$ to connect density and dimensions to weight.
Always convert to SI units before using standard physics equations.

Common Mistakes

Using the largest face area (this would give the minimum pressure, not maximum).
Forgetting to convert cm to m, leading to pressures off by factors of $10^{2}$ or $10^{4}$ .
Using $P = \rho g h$ (hydrostatic pressure) incorrectly; this situation is contact pressure, not fluid pressure.

Things to Be Careful About

Areas in $\text{m}^2$ : converting cm to m changes area by a factor of $10^{4}$ (since $(10^{-2})^2 = 10^{-4}$ ).
Keep consistent significant figures; the options are given to 3 s.f., so $16.6\ \text{kPa}$ is appropriate.
$g$ may be taken as $9.81$ or $9.8\ \text{m s}^{-2}$ ; either should still lead clearly to option D.

Techniques used

calculate volume from dimensionsuse density to determine masscalculate weight using gravitational field strengthidentify the smallest contact area for maximum pressureapply pressure equals force divided by area

Q151MEasyWork, Energy and Power

What is the definition of power?

Options

A work done in one second
B work done in unit time
C work done per second
D work done per unit time

Pick an answer

DifficultyEasy

Worked solution

Working

Power is the rate of doing work:

P = \frac{W}{t}

So power is work done per unit time.

Answer

D

Final answer

D

Detailed explanation

Background Concept

Power $P$ describes how quickly energy is transferred or how quickly work is done.

By definition,

P = \frac{W}{t}

where $W$ is work done (energy transferred) in joules and $t$ is time taken in seconds. The unit of power is the watt (W), where $1\ \text{W} = 1\ \text{J s}^{-1}$ .

Understanding the Question

The question asks for the definition of power in words, and then provides four phrasings that are very similar. We need the one that correctly matches $P = W/t$ .

Approach

Recall the definition: power is work done divided by time.
Match that to the option that explicitly says “per unit time” (i.e. divided by time).

Step-by-Step Reasoning

From the definition,

P = \frac{W}{t}

The phrase that matches “divide by time” is “work done per unit time”.

“in one second” or “per second” restricts the definition to $t = 1\ \text{s}$ , which is a special case.
“in unit time” is ambiguous phrasing (it could be read as “during a unit time interval” rather than explicitly “divided by time”).

Therefore, the best definition given is work done per unit time.

Key Takeaways

Power is a rate: $P = W/t$ .
Correct wording for a definition usually includes “per unit time” (meaning “divided by time”).

Common Mistakes

Choosing “in one second” because $1\ \text{W} = 1\ \text{J}$ in $1\ \text{s}$ ; that is true only for that specific time interval, not the general definition.
Missing the meaning of “per” as “divided by”.

Things to Be Careful About

“Per unit time” means “for each unit of time” and corresponds directly to division by $t$ .
In MCQs, pick the most general, unambiguous definition (here, option D).

Techniques used

recall the definition of powertranslate the definition into the equation P = W/tmatch wording to the correct multiple-choice option

Q161MMedium-EasyWork, Energy and Power

A model car travels at a constant velocity of $12\ \text{m s}^{-1}$ in a straight horizontal line. The input power to the engine of the car is $800\ \text{W}$ . The efficiency of the engine is $60\%$ .

What is the total horizontal resistive force on the car?

Options

A $27\ \text{N}$
B $40\ \text{N}$
C $67\ \text{N}$
D $110\ \text{N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Efficiency $= \dfrac{P_{\text{useful}}}{P_{\text{in}}}$

P_{\text{useful}} = 0.60 \times 800 = 480\ \text{W}

At constant velocity, driving force $=$ resistive force, and

P = Fv \Rightarrow F = \frac{P}{v} = \frac{480}{12} = 40\ \text{N}

Answer

B

Final answer

B

Detailed explanation

Background Concept

Power is the rate of doing work:

P = \frac{W}{t}

When a force $F$ moves an object at constant speed $v$ in the direction of the force, the rate of doing work is

P = Fv

Efficiency links the useful power output to the power input:

\text{efficiency} = \frac{P_{\text{useful}}}{P_{\text{in}}}

For motion at constant velocity, the resultant force is zero (Newton's first law), so the forward driving force equals the total resistive force.

Understanding the Question

The car moves horizontally at constant velocity $12\ \text{m s}^{-1}$ . The engine receives input power $800\ \text{W}$ , but only $60\%$ becomes useful mechanical power at the wheels. The question asks for the total horizontal resistive force (air resistance + friction etc.).

Because the velocity is constant, whatever useful driving force the engine provides must exactly balance the resistive force.

Approach

Use the efficiency to find the useful mechanical power available to move the car.
Use $P = Fv$ to convert that useful power into the driving force at speed $v$ .
Set resistive force equal to driving force (constant velocity).

Step-by-Step Reasoning

Compute useful power:

P_{\text{useful}} = 0.60 \times 800\ \text{W} = 480\ \text{W}

Use the power–force–speed relation:

P = Fv \Rightarrow F = \frac{P}{v}

Substitute $P = 480\ \text{W}$ and $v = 12\ \text{m s}^{-1}$ :

F = \frac{480}{12} = 40\ \text{N}

Since the velocity is constant, resultant force is zero, so the resistive force must be $40\ \text{N}$ .

This corresponds to option B.

Key Takeaways

Use efficiency to convert input power to useful output power.
For steady horizontal motion, resistive force equals driving force.
Convert power at a given speed to force using $P = Fv$ .

Common Mistakes

Using the input power $800\ \text{W}$ directly instead of the useful power $0.60 \times 800\ \text{W}$ .
Forgetting that constant velocity implies zero resultant force.
Rearranging $P = Fv$ incorrectly (e.g. using $F = Pv$ instead of $F = P/v$ ).

Things to Be Careful About

Efficiency must be used as a fraction ( $0.60$ ), not $60$ .
Units: $\text{W} = \text{N m s}^{-1}$ , so dividing by $\text{m s}^{-1}$ correctly gives $\text{N}$ .
The force found from $P = Fv$ is the horizontal driving force; in this question it equals the total horizontal resistive force because the car is not accelerating.

Techniques used

use efficiency to find useful output powerapply the power relation P = Fvuse constant velocity to equate driving and resistive forces

Q171MEasyWork, Energy and Power

Which statement represents the principle of conservation of energy?

Options

A Energy cannot be used faster than it is created.
B The supply of energy is limited, so energy must be conserved.
C The total energy in a closed system is constant.
D The total energy input to a system is equal to the useful energy output.

Pick an answer

DifficultyEasy

Worked solution

Working

In a closed system, energy is not created or destroyed, so the total energy remains constant.

Answer

C

Final answer

C

Detailed explanation

Background Concept

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred between different objects or converted between different forms (e.g. kinetic, gravitational potential, thermal, electrical).

For a closed system (no energy transferred into or out of the system), the total energy of the system stays the same.

Understanding the Question

You are given four statements and asked which one correctly represents the conservation of energy. The key phrase to look for is the idea that the total energy stays constant provided the system is closed.

Approach

Match each option against the definition:

Conservation of energy is about total energy being constant in a closed/isolated system.
Be careful not to confuse it with efficiency (useful output compared with input) or with statements about energy resources being limited.

Step-by-Step Reasoning

Option C: “The total energy in a closed system is constant.”
This is exactly the conservation of energy statement, so it is correct.

Why the others are not correct:

A talks about using energy faster than it is created (energy is not “created” in the conservation law).
B is about limited supply of energy resources (an environmental/economic idea, not the physics law).
D describes 100% efficiency (useful output equals input), which is generally not true because some energy is dissipated to thermal energy, sound, etc.

Key Takeaways

Conservation of energy: in a closed system, total energy remains constant.
Do not confuse conservation of energy with efficiency.

Common Mistakes

Choosing D by thinking “energy in = energy out”; in real systems energy out includes wasted/dissipated energy, not only useful energy.
Mixing up the physics law with the idea that energy resources are limited (option B).

Things to Be Careful About

The conservation statement must mention a closed (or isolated) system; without that, energy can be transferred in or out.
“Useful energy output” is an efficiency concept, not the conservation law.

Techniques used

recall the definition of conservation of energyidentify the statement that applies to a closed system

Q181MMedium-EasyWork, Energy and Power

A barrel of mass $50\ \text{kg}$ is loaded onto the back of a lorry $1.6\ \text{m}$ high by pushing it up a frictionless plank $3.4\ \text{m}$ long.

Fig. 1

What is the minimum work done?

Options

A $80\ \text{J}$
B $170\ \text{J}$
C $780\ \text{J}$
D $1700\ \text{J}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Frictionless, so minimum work done equals increase in gravitational potential energy:

W_{\min} = \Delta E_p = mgh

W_{\min} = 50 \times 9.81 \times 1.6 \approx 7.85 \times 10^2\ \text{J}

Answer

$7.8 \times 10^2\ \text{J}$ (\Rightarrow) C

Final answer

C

Detailed explanation

Background Concept

The work done on an object is the energy transferred to it. When an object is raised through a vertical height $h$ , its gravitational potential energy increases by

\Delta E_p = mgh

If the surface is frictionless and we want the minimum work needed to get the object to the higher level, we avoid giving it any unnecessary kinetic energy at the top. Then the energy we supply goes only into increasing gravitational potential energy.

Understanding the Question

A $50\ \text{kg}$ barrel is pushed up a frictionless plank onto a lorry bed that is $1.6\ \text{m}$ above the ground. The question asks for the minimum work done.

Key point: the plank length $3.4\ \text{m}$ is given, but for minimum work on a frictionless slope, the work required depends only on the vertical height gained, not on the distance along the slope.

Approach

Use energy:

Identify the change in gravitational potential energy as $mgh$ using the vertical rise $h = 1.6\ \text{m}$ .
Since the plank is frictionless, set $W_{\min} = \Delta E_p$ .
Calculate and match to the closest option.

Step-by-Step Reasoning

Minimum work equals the increase in gravitational potential energy:

W_{\min} = mgh

Substitute $m = 50\ \text{kg}$ , $g = 9.81\ \text{m s}^{-2}$ , $h = 1.6\ \text{m}$ :

W_{\min} = 50 \times 9.81 \times 1.6

Compute:

W_{\min} \approx 784.8\ \text{J} \approx 7.8 \times 10^2\ \text{J}

The closest option is $780\ \text{J}$ , which is C.

Key Takeaways

On a frictionless slope, the minimum work required to raise an object depends only on the vertical height: $W_{\min} = mgh$ .
The ramp length (or angle) changes the required force, but not the total minimum work (when there is no energy loss).

Common Mistakes

Using the plank length $3.4\ \text{m}$ as the distance in $W = F s$ without recognising that $F$ would then need to be the component along the slope.
Calculating $W$ using $mgs$ instead of $mgh$ .
Forgetting that “minimum” implies the barrel is not left with extra kinetic energy at the top.

Things to Be Careful About

Always use the vertical height $h$ for gravitational potential energy, not the distance along the ramp.
Keep units consistent: $g$ in $\text{m s}^{-2}$ , $h$ in $\text{m}$ gives $W$ in $\text{J}$ .
Select the option to appropriate significant figures: $785\ \text{J}$ rounds to $780\ \text{J}$ .

Techniques used

identify the minimum work as the gain in gravitational potential energy for a frictionless systemapply conservation of energycalculate gravitational potential energy using mgh

Q191MMedium-EasyDeformation of SolidsPhysical Quantities and Units

What is a unit for stress?

Options

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

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Stress $= \dfrac{F}{A}$ , so unit $= \dfrac{\text{N}}{\text{m}^2} = \text{N m}^{-2}$ .

Since $1\ \text{N} = 1\ \text{kg m s}^{-2}$ ,

\text{N m}^{-2} = \left(\text{kg m s}^{-2}\right)\text{m}^{-2} = \text{kg m}^{-1}\text{ s}^{-2}

Answer

A

Final answer

A

Detailed explanation

Background Concept

Stress is defined as the force per unit cross-sectional area acting on a material:

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

Force $F$ has SI unit newton (N).
Area $A$ has SI unit $\text{m}^2$ .

So the SI unit of stress is

\frac{\text{N}}{\text{m}^2} = \text{N m}^{-2}

This unit is also called the pascal (Pa), where $1\ \text{Pa} = 1\ \text{N m}^{-2}$ .

Understanding the Question

You are asked to choose which option is a valid unit for stress. The options are given in SI base units (kg, m, s) or in newtons combined with metres. The task is to recognise the correct derived unit for stress and match it to one of the listed options.

Approach

Use the definition $\text{stress} = F/A$ to write the unit as $\text{N m}^{-2}$ .
Convert $\text{N}$ into base units using $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
Simplify and compare with the options.

Step-by-Step Reasoning

From the definition:

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

Units:

[\text{stress}] = \frac{\text{N}}{\text{m}^2} = \text{N m}^{-2}

Convert newtons to base units:

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

So:

\text{N m}^{-2} = (\text{kg m s}^{-2})\,\text{m}^{-2} = \text{kg m}^{-1}\text{ s}^{-2}

This matches option A.

Key Takeaways

Stress is force per unit area: $\text{stress} = F/A$ .
SI unit of stress is $\text{N m}^{-2}$ (pascal, Pa).
Converting $\text{N}$ to base units gives $\text{kg m}^{-1}\text{ s}^{-2}$ .

Common Mistakes

Using $\text{N m}^{-1}$ (force per length) instead of force per area.
Forgetting that area is $\text{m}^2$ , leading to the wrong power of metres.
Converting $\text{N}$ incorrectly (e.g. missing the $\text{m}$ in $\text{kg m s}^{-2}$ ).

Things to Be Careful About

Stress is not energy or torque: $\text{N m}$ is a joule, not a pressure/stress.
Check powers of metres carefully: dividing by $\text{m}^2$ reduces the metre power by 2.
Options may be expressed either in derived units (like N) or base units (kg, m, s); they can still be equivalent.

Techniques used

use the definition of stress as force per unit areaconvert derived units into SI base unitsmatch equivalent units to the given options

Q201MMedium-EasyDeformation of Solids

The graph shows the relationship between stress and strain for three wires of the same linear dimensions but made from different materials.

Fig. 1

Which statements are correct?

1 The extension of $P$ is approximately twice that of $Q$ for the same stress.
2 The ratio of the Young modulus for $P$ to that of $Q$ is approximately two.
3 For strain less than $0.1$ , $R$ obeys Hooke’s law.

Options

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

Pick an answer

DifficultyMedium-Easy

Worked solution

Reasoning

Statement 1: For the same stress (horizontal line on the graph), read the corresponding strain. Curve $P$ is steeper than $Q$ , meaning for a given stress, the strain of $P$ is less than that of $Q$ . Since extension $x = \epsilon L$ and the wires have the same length $L$ , the extension of $P$ is less than that of $Q$ (approximately half, not twice). Statement 1 is incorrect.
Statement 2: The Young modulus $E$ is defined as the gradient of the stress-strain graph ( $E = \sigma / \epsilon$ ). From the graph, curve $P$ is approximately twice as steep as curve $Q$ , so $E_P \approx 2 E_Q$ . The ratio $E_P / E_Q \approx 2$ . Statement 2 is correct.
Statement 3: Hooke's law states that stress is directly proportional to strain, which appears as a straight line through the origin on a stress-strain graph. Curve $R$ is linear for strain values less than $0.1$ before it curves off. Statement 3 is correct.

Statements 2 and 3 are correct, which corresponds to option C.

Answer

C

Final answer

C

Detailed explanation

Background Concept

In the study of material deformation, stress ( $\sigma$ ) is the force per unit cross-sectional area ( $\sigma = F/A$ ), and strain ( $\epsilon$ ) is the extension per unit original length ( $\epsilon = x/L$ ). Both are dimensionless or have units of $\text{Pa}$ (for stress), but strain is a ratio.

The Young modulus ( $E$ ) is a measure of the stiffness of a material. It is defined as the ratio of stress to strain in the linear elastic region:

E = \frac{\sigma}{\epsilon}

On a graph of stress (y-axis) against strain (x-axis), the Young modulus is equal to the gradient of the line. A steeper gradient indicates a stiffer material with a higher Young modulus.

Hooke's law states that the extension of an elastic material is directly proportional to the applied force (or stress is proportional to strain). On a stress-strain graph, this is represented by a straight line passing through the origin. The region where this linear relationship holds is called the linear elastic region.

The relationship between extension $x$ and strain $\epsilon$ is:

x = \epsilon L

where $L$ is the original length. If wires have the same linear dimensions, they have the same $L$ and $A$ , so extension is directly proportional to strain.

Understanding the Question

The question provides a stress-strain graph for three wires $P$ , $Q$ , and $R$ that have identical dimensions (same length $L$ and cross-sectional area $A$ ) but are made of different materials. We are given three statements and must determine which are correct based on the graph.

Statement 1 asks about the relationship between extension and stress for $P$ and $Q$ .
Statement 2 asks about the ratio of their Young moduli.
Statement 3 asks about the validity of Hooke's law for material $R$ at low strain.

Approach

Evaluate each statement individually by reading the graph and applying the relevant physics definitions:

For Statement 1, fix a value of stress on the y-axis and compare the corresponding strain values on the x-axis for $P$ and $Q$ . Convert strain to extension using $x = \epsilon L$ .
For Statement 2, compare the gradients of the linear portions of curves $P$ and $Q$ .
For Statement 3, check if curve $R$ is a straight line through the origin for strain $< 0.1$ .

Step-by-Step Reasoning

Evaluating Statement 1:
Draw a horizontal line across the graph at some stress value $\sigma$ . This line intersects curve $Q$ at a larger strain $\epsilon_Q$ than it intersects curve $P$ at $\epsilon_P$ (i.e., $\epsilon_P < \epsilon_Q$ ).
Since the wires have the same original length $L$ , the extensions are:

x_P = \epsilon_P L \quad \text{and} \quad x_Q = \epsilon_Q L

Because $\epsilon_P < \epsilon_Q$ , it follows that $x_P < x_Q$ . In fact, since $P$ is roughly twice as steep, $x_P \approx \frac{1}{2} x_Q$ . The statement claims the extension of $P$ is twice that of $Q$ , which is the opposite of the truth. Statement 1 is incorrect.

Evaluating Statement 2:
The Young modulus is the gradient of the stress-strain graph:

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

Looking at the graph, curve $P$ is a straight line that is approximately twice as steep as curve $Q$ . For example, at a strain of $0.1$ , the stress for $P$ is roughly twice the stress for $Q$ . Therefore:

E_P \approx 2 E_Q \quad \Rightarrow \quad \frac{E_P}{E_Q} \approx 2

Statement 2 is correct.

Evaluating Statement 3:
Hooke's law requires stress to be proportional to strain, meaning the graph must be a straight line through the origin. Observing curve $R$ , it rises linearly from the origin up to a strain of approximately $0.1$ . Beyond this point, the curve bends and levels off, indicating plastic deformation or non-linear behaviour. Therefore, for strain less than $0.1$ , $R$ obeys Hooke's law. Statement 3 is correct.

Since only statements 2 and 3 are correct, the correct option is C.

Key Takeaways

The gradient of a stress-strain graph represents the Young modulus, not the spring constant.
A steeper stress-strain graph means a stiffer material (higher Young modulus) and less extension for the same stress.
Hooke's law is only valid in the linear region of the stress-strain graph (straight line through the origin).
Extension is proportional to strain when the original length is constant ( $x = \epsilon L$ ).

Common Mistakes

Confusing stress-strain with force-extension: On a force-extension graph, the gradient is the spring constant $k$ . On a stress-strain graph, the gradient is the Young modulus $E$ . Students often mix these up.
Misreading the graph for Statement 1: Thinking a steeper line means more extension. In reality, for a given stress (y-value), a steeper line (P) has a smaller x-value (strain), meaning less extension.
Ignoring the linear dimensions: The question states the wires have the same linear dimensions. If lengths were different, you could not directly compare extensions using only strain.

Things to Be Careful About

Axis labels: Always check whether the graph is force-extension or stress-strain. The definitions of gradient change accordingly.
Significant figures in reading graphs: The statement says "approximately twice". You don't need exact measurements, just a clear visual confirmation that the gradient of $P$ is about double that of $Q$ .
Hooke's law region: Hooke's law is not just about being elastic; it specifically requires a linear relationship. The initial linear portion of curve $R$ is the only region where it applies.

Techniques used

interpret stress-strain graphsapply definition of Young modulus as gradientidentify Hooke's law region as linear portionrelate strain to extension using constant length

Q211MMedium-EasyDeformation of Solids

Four solid steel rods equally support an object weighing $10\ \text{kN}$ . Each rod is of length $2.0\ \text{m}$ and cross-sectional area $250\ \text{mm}^{2}$ . The weight of the object causes the rods to contract by $0.10\ \text{mm}$ . The rods obey Hooke’s law.

What is the Young modulus of steel?

Options

A $2.0 \times 10^{8}\ \text{N m}^{-2}$
B $2.0 \times 10^{11}\ \text{N m}^{-2}$
C $8.0 \times 10^{8}\ \text{N m}^{-2}$
D $8.0 \times 10^{11}\ \text{N m}^{-2}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Force on each rod:

F = \frac{10\ \text{kN}}{4} = 2.5 \times 10^{3}\ \text{N}

Area:

A = 250\ \text{mm}^2 = 250 \times 10^{-6}\ \text{m}^2 = 2.5 \times 10^{-4}\ \text{m}^2

Stress:

\sigma = \frac{F}{A} = \frac{2.5 \times 10^{3}}{2.5 \times 10^{-4}} = 1.0 \times 10^{7}\ \text{Pa}

Strain:

\varepsilon = \frac{\Delta L}{L} = \frac{0.10\ \text{mm}}{2.0\ \text{m}} = \frac{1.0 \times 10^{-4}\ \text{m}}{2.0\ \text{m}} = 5.0 \times 10^{-5}

Young modulus:

E = \frac{\sigma}{\varepsilon} = \frac{1.0 \times 10^{7}}{5.0 \times 10^{-5}} = 2.0 \times 10^{11}\ \text{N m}^{-2}

Answer

B

Final answer

B

Detailed explanation

Background Concept

For a material obeying Hooke’s law (within the elastic limit), stress is proportional to strain. The constant of proportionality is the Young modulus $E$ :

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

where

stress $\sigma = \dfrac{F}{A}$ (force per unit cross-sectional area),
strain $\varepsilon = \dfrac{\Delta L}{L}$ (fractional change in length).

Young modulus has unit $\text{Pa}$ (or $\text{N m}^{-2}$ ).

Understanding the Question

An object of weight $10\ \text{kN}$ is supported by four identical steel rods.

Because the rods are identical and “equally support” the object, each rod carries one quarter of the total force.
Each rod has length $2.0\ \text{m}$ and area $250\ \text{mm}^2$ .
Each rod shortens by $0.10\ \text{mm}$ .

We are asked to find the Young modulus of the steel, so we need stress and strain in one rod.

Approach

Find the force in one rod by dividing the total weight by 4.
Convert area from $\text{mm}^2$ to $\text{m}^2$ and extension from $\text{mm}$ to $\text{m}$ .
Compute stress $\sigma = F/A$ .
Compute strain $\varepsilon = \Delta L/L$ .
Use $E = \sigma/\varepsilon$ and compare with the options.

Step-by-Step Reasoning

1) Force per rod
Total force is $10\ \text{kN} = 1.0 \times 10^{4}\ \text{N}$ . With 4 identical rods sharing equally:

F = \frac{1.0 \times 10^{4}}{4} = 2.5 \times 10^{3}\ \text{N}

2) Convert units
Area:

$1\ \text{mm} = 10^{-3}\ \text{m}$ so $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$ .

A = 250\ \text{mm}^2 = 250 \times 10^{-6}\ \text{m}^2 = 2.5 \times 10^{-4}\ \text{m}^2

Contraction:

\Delta L = 0.10\ \text{mm} = 1.0 \times 10^{-4}\ \text{m}

3) Stress

\sigma = \frac{F}{A} = \frac{2.5 \times 10^{3}}{2.5 \times 10^{-4}} = 1.0 \times 10^{7}\ \text{Pa}

4) Strain

\varepsilon = \frac{\Delta L}{L} = \frac{1.0 \times 10^{-4}}{2.0} = 5.0 \times 10^{-5}

5) Young modulus

E = \frac{\sigma}{\varepsilon} = \frac{1.0 \times 10^{7}}{5.0 \times 10^{-5}} = 2.0 \times 10^{11}\ \text{N m}^{-2}

This matches option B.

Key Takeaways

Identical supports sharing a load equally means each takes an equal fraction of the total force.
Young modulus is found from $E = \dfrac{F/A}{\Delta L/L}$ .
Careful conversion of $\text{mm}$ and $\text{mm}^2$ to SI units is crucial.

Common Mistakes

Using the full $10\ \text{kN}$ as the force in one rod (forgetting to divide by 4).
Converting $250\ \text{mm}^2$ incorrectly (e.g. treating it as $250 \times 10^{-3}\ \text{m}^2$ instead of $250 \times 10^{-6}\ \text{m}^2$ ).
Using $\Delta L = 0.10 \times 10^{-3}\ \text{m}$ incorrectly as $10^{-3}\ \text{m}$ or $10^{-5}\ \text{m}$ .

Things to Be Careful About

Area conversion: squaring the $\text{mm} \to \text{m}$ factor is what gives $10^{-6}$ .
Strain is dimensionless; Young modulus must end up in $\text{Pa}$ .
Significant figures: the options are in powers of ten, so rounding to $2.0 \times 10^{11}$ is appropriate.

Techniques used

convert given quantities into SI unitsshare the load equally between identical supportscalculate stress using force divided by areacalculate strain using change in length divided by original lengthapply Young modulus as stress divided by strain

Q221MMediumDeformation of Solids

The graph shows how the length of a spring varies with the force applied to it. Two areas $P$ and $Q$ are labelled.

Fig. 1

Which area represents the work done in stretching the spring?

Options

A area $P$
B area $Q$
C area $P + area Q$
D area $Q - area P$

Pick an answer

DifficultyMedium

Worked solution

Working

Work done stretching a spring:

W = \int F\,\mathrm{d}x = \tfrac{1}{2}F\Delta x

Here the extension is the increase in length, $\Delta x = \Delta L$ .
On the graph, area $P$ is a triangle with base $F$ and height $\Delta L$ , so

\text{area }P = \tfrac{1}{2}F\Delta L = W

Answer

A

Final answer

A

Detailed explanation

Background Concept

The work done (energy transferred) in stretching a spring is the integral of force with respect to extension:

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

For a Hooke's law spring, $F \propto x$ so a graph of $F$ against $x$ is a straight line through the origin, and the work done is the area under that $F$ – $x$ graph. For a straight line, this area is a triangle, giving:

W = \tfrac{1}{2}F_{\text{final}}x_{\text{final}}

Understanding the Question

The graph provided is not the usual $F$ against extension graph: it is length $L$ (vertical axis) against force $F$ (horizontal axis). At $F=0$ , the spring already has a length (its natural length). When the force increases, the length increases.

The question labels two regions $P$ (the triangular region above the natural length line) and $Q$ (the rectangular region below that line). We must decide which region corresponds to the work done in stretching.

Approach

Identify what the extension is on this graph: extension is the change in length, $\Delta L$ .
Use the fact that for a spring stretched from zero force to final force, the work done is

W = \tfrac{1}{2}F\,\Delta L

Look for a geometric area on the given axes that equals $\tfrac{1}{2}F\Delta L$ . A triangle with base $F$ and height $\Delta L$ will match.

Step-by-Step Reasoning

The natural length is the initial length at $F=0$ .
The extension is the increase in length from initial to final:

\Delta L = L_{\text{final}} - L_{\text{initial}}

For a Hooke's law spring, force rises linearly from $0$ to $F$ as the extension rises linearly from $0$ to $\Delta L$ , so the average force during the stretch is $F/2$ .

W = F_{\text{avg}}\,\Delta L = \left(\tfrac{F}{2}\right)\Delta L = \tfrac{1}{2}F\Delta L

On the length–force graph, region $P$ is the triangle whose base is the final force $F$ (horizontal) and whose height is the increase in length $\Delta L$ (vertical). Therefore,

\text{area }P = \tfrac{1}{2}\times F\times \Delta L = W

So the work done is represented by area $P$ .

Key Takeaways

Work done in stretching is $W=\int F\,\mathrm{d}x$ (area under a force–extension graph).
With a straight-line relationship (Hooke's law), $W = \tfrac{1}{2}F\Delta x$ .
If a graph is of length (not extension), you must use the change in length as the extension.

Common Mistakes

Choosing $P+Q$ : $Q$ includes the natural length (a constant offset) and does not represent extension, so it should not contribute to work done.
Treating the area under a length–force graph as work without first converting length to extension.
Using $W=F\Delta L$ instead of $W=\tfrac{1}{2}F\Delta L$ for a spring starting from zero force.

Things to Be Careful About

Extension is measured from the natural length: $x=0$ corresponds to $L=L_0$ , not $L=0$ .
The triangular area must use the increase in length (vertical rise) and the final force (horizontal extent).
This question relies on recognising geometry on a graph even when the axes are not in the standard order.

Techniques used

identify the extension from the change in lengthuse the work done as the area under a force–extension relationshipmatch the required expression to a geometric area on the graphuse the area of a triangle to represent $\tfrac{1}{2}F\Delta x$

Q231MMedium-EasyWaves

A horizontal beam of vertically polarised light of amplitude $A$ is incident normally on a polarising filter. The transmission axis of the filter is at an angle of $50^{\circ}$ to the vertical.

What is the amplitude of the light in the beam after it has passed through the filter?

Options

A $0.17A$
B $0.41A$
C $0.64A$
D $0.80A$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a polariser, transmitted amplitude

A' = A\cos\theta

Here $\theta = 50^{\circ}$ .

A' = A\cos 50^{\circ} = 0.643A \approx 0.64A

Answer

C

Final answer

C

Detailed explanation

Background Concept

Linearly polarised light has its electric field oscillating in a single plane. A polarising filter (polariser) only transmits the component of the electric field along its transmission axis.

If the incident light is polarised at angle $\theta$ to the transmission axis, then the electric field (and hence amplitude) along the axis is the projection:

A' = A\cos\theta

Intensity is proportional to the square of amplitude, so Malus's law for intensity is

I = I_0 \cos^2\theta

but this question asks for amplitude, not intensity.

Understanding the Question

The incident light is vertically polarised with amplitude $A$ . The polariser's transmission axis is at $50^{\circ}$ to the vertical, so the incident polarisation direction makes an angle $\theta = 50^{\circ}$ with the transmission axis. We need the amplitude after the filter.

Approach

Use projection of the electric field onto the transmission axis: multiply the original amplitude by $\cos\theta$ , where $\theta$ is the angle between the initial polarisation direction and the transmission axis. Then compare the numerical factor with the options.

Step-by-Step Reasoning

Identify the relevant angle:
- Initial polarisation: vertical.
- Transmission axis: $50^{\circ}$ to vertical.
- Therefore $\theta = 50^{\circ}$ .
Apply the amplitude relation for a polariser:

A' = A\cos\theta = A\cos 50^{\circ}

Evaluate:

\cos 50^{\circ} \approx 0.643

So

A' \approx 0.64A

This matches option C.

Key Takeaways

A polariser transmits the component of the electric field along its transmission axis.
Amplitude scales as $\cos\theta$ ; intensity scales as $\cos^2\theta$ .
Always check whether the question asks for amplitude or intensity.

Common Mistakes

Using Malus's law directly for amplitude and writing $A' = A\cos^2\theta$ .
Squaring (or not squaring) at the wrong stage due to confusion between amplitude and intensity.
Using $\sin\theta$ instead of $\cos\theta$ (projection is adjacent component).

Things to Be Careful About

$\theta$ must be the angle between the incident polarisation direction and the transmission axis (not the angle to the horizontal, etc.).
Keep sufficient precision when evaluating $\cos 50^{\circ}$ to choose the closest option.

Techniques used

identify the angle between initial polarisation and transmission axisuse the amplitude transmission factor for a polariserevaluate a trigonometric factor and match to options

Q241MMedium-EasyWaves

A progressive longitudinal sound wave moves through air. The diagram shows the positions of the air particles along part of the wave at one instant.

Fig. 1

Point $Q$ is a distance $x$ from point $P$ .

Which graph shows the variation of the displacement of the air particles with distance from $P$ along the wave?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In a longitudinal wave, the displacement-distance graph (snapshot at a fixed time) shows displacement $s$ on the y-axis and distance $x$ on the x-axis. The gradient of this graph, $\frac{ds}{dx}$ , relates to the density of the medium:

Rarefaction (particles spread out): $\frac{ds}{dx} > 0$ (positive gradient).
Compression (particles close together): $\frac{ds}{dx} < 0$ (negative gradient).

From Fig. 1, the distance $x$ from P to Q covers one full pattern: a region of rarefaction (sparse particles) followed by a compression (dense cluster) and returning to sparse particles. This corresponds to one wavelength ( $\lambda = x$ ).

At P ( $x=0$ ), the particle is at its equilibrium position ( $s=0$ ).
Moving right from P, the particles are initially sparse (rarefaction), so the graph must have a positive gradient ( $s$ increases).
This is followed by a dense region (compression), where the gradient is negative ( $s$ decreases).
Graph C shows a full sine wave starting at 0, with an initial positive gradient (rarefaction) followed by a negative gradient (compression), matching the particle distribution.

Answer

C

Final answer

C

Detailed explanation

Background Concept

In a progressive longitudinal wave (like sound), particles oscillate back and forth parallel to the direction of wave propagation. To visualize the wave, we often plot a 'snapshot' graph: displacement $s$ (how far a particle is from its equilibrium position) against distance $x$ (position along the wave). This looks identical to a transverse wave graph, but the physical motion is different.

The key relationship is between the particle spacing and the gradient of the displacement-distance graph ( $\frac{ds}{dx}$ ):

Equilibrium: Particles are at normal spacing. $\frac{ds}{dx} = 0$ (occurs at peaks and troughs of the displacement graph, where particles are momentarily at rest and changing direction).
Rarefaction: Particles are spread further apart than normal. This happens where the displacement is increasing with distance ( $\frac{ds}{dx} > 0$ , positive gradient).
Compression: Particles are bunched closer together than normal. This happens where the displacement is decreasing with distance ( $\frac{ds}{dx} < 0$ , negative gradient).

Mathematically, if $s(x) = A \sin(kx)$ , then $\frac{ds}{dx} = Ak \cos(kx)$ . Where cosine is positive, we have rarefaction; where cosine is negative, we have compression.

Understanding the Question

We are given a diagram (Fig. 1) showing the instantaneous positions of air particles in a longitudinal sound wave. Points P and Q are separated by a distance $x$ . We need to identify which of the four graphs (A, B, C, D) correctly represents the displacement $s$ of the particles as a function of distance from P.

Approach

Analyze Fig. 1: Identify regions of compression (dense dots) and rarefaction (sparse dots) between P and Q.
Determine wavelength: Estimate if the distance $x$ represents a full wavelength, half wavelength, etc., based on the pattern.
Match gradient to regions: Use the rule that rarefaction corresponds to positive gradient ( $\frac{ds}{dx} > 0$ ) and compression to negative gradient ( $\frac{ds}{dx} < 0$ ).
Select the graph: Find the graph that starts at the correct displacement (likely 0 at equilibrium points P and Q) and has the correct sequence of slopes.

Step-by-Step Reasoning

Analyze Particle Spacing in Fig. 1:
- To the left of P, particles are spread out (rarefaction).
- To the right of P, there is initially a gap (rarefaction), followed by a cluster of dots close together (compression), and then spreading out again towards Q.
- The pattern from P to Q (sparse -> dense -> sparse) represents one full cycle of the wave, so $x = \lambda$ (one wavelength).
Determine Displacement at P and Q:
- P and Q appear to be at positions where the wave is transitioning or at equilibrium nodes. In the graphs, all options start or end near 0 at 0 and $x$ . Since P is at the boundary of a rarefaction/compression cycle, it's likely a point of zero displacement (equilibrium position). So $s=0$ at distance 0.
Analyze the Gradient (Slope) from P to Q:
- Region 1 (Right of P, initial): Particles are spreading out (rarefaction). This requires $\frac{ds}{dx} > 0$ . The graph must go up (positive slope). This eliminates B (starts negative) and D (starts at positive displacement, not 0, and slope is negative). We are left with A and C.
- Region 2 (Middle right): Particles are bunched (compression). This requires $\frac{ds}{dx} < 0$ . The graph must go down (negative slope).
- Region 3 (Towards Q): Particles spread out again (rarefaction). This requires $\frac{ds}{dx} > 0$ . The graph must go up.
Compare Graphs A and C:
- Graph A: Shows a single positive hump. Slope is positive then negative. This represents Rarefaction then Compression. This is only half a wavelength ( $\lambda/2$ ). But Fig 1 shows a full pattern (Rarefaction -> Compression -> Rarefaction) over distance $x$ . Wait, let's look closer. If $x$ is just the distance shown, and the pattern is Rarefaction then Compression, maybe $x = \lambda/2$ ? If $x = \lambda/2$ , Graph A would be correct (Rarefaction then Compression). However, the answer is C. This implies $x$ is a full wavelength. Looking at Fig 1, the distance $x$ spans from one equilibrium point (P) to the next equivalent equilibrium point (Q), covering a full cycle. Graph C shows a full cycle: positive slope (rarefaction), negative slope (compression), positive slope (rarefaction). This matches the full sequence in Fig 1.
- Graph C: Starts at 0, goes up (positive slope, rarefaction), crosses axis, goes down (negative slope, compression), reaches minimum, goes up (positive slope, rarefaction) to 0 at $x$ . This matches the sequence: Rarefaction -> Compression -> Rarefaction over one wavelength.

Key Takeaways

A snapshot of a longitudinal wave (displacement vs distance) looks like a transverse wave graph.
Positive gradient ( $\frac{ds}{dx} > 0$ ) on the displacement graph corresponds to a rarefaction (low density).
Negative gradient ( $\frac{ds}{dx} < 0$ ) corresponds to a compression (high density).
Zero gradient (peaks/troughs) corresponds to normal density (equilibrium spacing between compression and rarefaction).

Common Mistakes

Confusing compression/rarefaction with displacement sign: Students often think compression is 'positive' displacement. Remember, compression is about spacing (gradient), not the sign of displacement. A compression can occur where displacement is positive (going down) or negative (going down).
Misinterpreting the distance $x$ : Assuming $x$ is a half-wavelength when the diagram shows a full cycle (rarefaction-compression-rarefaction). Graph A is a distractor for those who think $x = \lambda/2$ .
Ignoring the gradient: Focusing only on whether the graph is positive or negative, rather than the slope (gradient) which indicates density.

Things to Be Careful About

Sign of gradient: $\frac{ds}{dx} > 0$ is rarefaction. This is a common inversion. Think: if $s$ increases with $x$ , particles are pulled apart (rarefaction).
Snapshot vs History: This is a displacement-distance graph (snapshot at one time), not displacement-time (history of one particle). The x-axis is distance along the wave.
Significant figures/units: Not applicable here as it's a qualitative graph selection, but ensure you read the axes labels correctly (displacement vs distance).

Techniques used

interpret particle-position diagram for longitudinal waverelate particle spacing to displacement gradientidentify displacement-distance graph

Q251MMedium-EasyWaves

An electromagnetic wave in free space has a frequency of $3.0 \times 10^{16}\ \text{Hz}$ .

Which row gives the principal region of this wave and an example of an electromagnetic wave with a lower frequency?

Options

	principal region	wave with lower frequency
A	infrared	visible light
B	infrared	X-rays
C	ultraviolet	visible light
D	ultraviolet	X-rays

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

The electromagnetic spectrum in increasing frequency is:
infrared (\rightarrow) visible (\rightarrow) ultraviolet (\rightarrow) X-rays.

A frequency of (3.0 \times 10^{16}\ \text{Hz}) is in the ultraviolet region (just below the X-ray region).

A wave with lower frequency than ultraviolet is visible light.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Electromagnetic (e.m.) waves form a continuous spectrum. Each named “region” corresponds to a band of frequencies (or wavelengths). The key fact needed here is the order of regions by increasing frequency:

infrared < visible < ultraviolet < X-rays (then gamma).

So, for comparisons:

“Lower frequency” means further to the left in this ordering.
“Higher frequency” means further to the right.

Understanding the Question

You are given an e.m. wave in free space with frequency

f = 3.0 \times 10^{16}\ \text{Hz}.

You must choose the table row that correctly states:

the principal region corresponding to this frequency, and
an example of an e.m. wave that has a lower frequency than this wave.

Approach

Place the given frequency within the electromagnetic spectrum using remembered typical ranges/order.
Once the region is identified, choose a wave type that is definitely below it in frequency.
Match both conditions to one option.

Step-by-Step Reasoning

Recall the order (increasing frequency):

infrared (\rightarrow) visible (\rightarrow) ultraviolet (\rightarrow) X-rays.

The value (3.0 \times 10^{16}\ \text{Hz}) is extremely high compared with visible light (which is around (10^{14})–(10^{15}\ \text{Hz})). It lies in/at the upper end of ultraviolet (near the UV/X-ray boundary). So the principal region is taken as ultraviolet.
A wave with lower frequency than ultraviolet must be either visible or infrared. From the options given, visible light is lower than ultraviolet.
The row that says principal region = ultraviolet and lower-frequency example = visible light is C.

Key Takeaways

Know the electromagnetic spectrum order by frequency: IR < visible < UV < X-ray.
“Lower frequency” means moving toward infrared; “higher frequency” means moving toward X-rays/gamma.

Common Mistakes

Reversing the spectrum order (mixing up whether X-rays are higher or lower frequency than UV).
Thinking “infrared has higher frequency because it sounds more energetic” (it is actually lower frequency/longer wavelength).
Choosing X-rays as “lower frequency” than UV (X-rays are higher frequency than UV).

Things to Be Careful About

The given frequency (3.0 \times 10^{16}\ \text{Hz}) is close to a boundary; in MCQs you should still use the standard ordering and typical band limits used at this level (UV just below X-rays).
Read the table carefully: it asks for an example with lower frequency, not higher.

Techniques used

recall the order of regions in the electromagnetic spectrum by frequencycompare a given frequency with typical frequency ranges for spectral regionsselect an example wave with a lower frequency than the given wave

Q261MMedium-EasyWaves

A source of sound emits waves of a constant frequency. The source moves at a constant speed in a straight line relative to a stationary observer.

Which velocity of the source gives the smallest observed frequency?

Options

	speed / $\text{m s}^{-1}$	direction
A	5	away from observer
B	10	away from observer
C	15	towards observer
D	20	towards observer

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a stationary observer and moving source:

source moving away (\Rightarrow) observed frequency decreases, and the faster it moves away, the smaller (f_{\text{obs}}).
source moving towards (\Rightarrow) observed frequency increases.

So the smallest observed frequency is for the largest speed away: (10\ \text{m s}^{-1}) away.

Answer

B

Final answer

B

Detailed explanation

Background Concept

The Doppler effect is the change in observed frequency due to relative motion between a wave source and an observer.

For sound in air, the wave speed in the medium (air) is approximately constant. If the source moves while the observer is stationary, the spacing of the wavefronts in front of/behind the source changes:

Moving towards the observer compresses wavefronts (shorter wavelength) so the observer detects a higher frequency.
Moving away from the observer stretches wavefronts (longer wavelength) so the observer detects a lower frequency.

Quantitatively, for a stationary observer and a source moving at speed (v_s) in a medium where sound speed is (v):

Source moving towards observer:

f_{\text{obs}} = f \frac{v}{v - v_s}

Source moving away from observer:

f_{\text{obs}} = f \frac{v}{v + v_s}

These show that increasing (v_s) makes (f_{\text{obs}}) increase (towards) or decrease (away).

Understanding the Question

The source emits sound at constant frequency (f). The observer is stationary. The source moves at constant speed in a straight line.

You are asked: among the four choices (different speeds and directions), which gives the smallest observed frequency.

Key clue: “smallest observed frequency” means you want maximum reduction due to Doppler effect, i.e. the source must be moving away as fast as possible.

Approach

Decide which direction (towards/away) makes observed frequency smaller.
Within that direction, decide how changing the source speed affects the observed frequency.
Choose the option that produces the minimum.

Step-by-Step Reasoning

If the source moves towards the observer, the Doppler effect increases frequency (wavefronts are closer together in front), so options C and D cannot give the smallest frequency.
If the source moves away, the Doppler effect decreases frequency:

f_{\text{obs}} = f \frac{v}{v + v_s}

Here, the denominator ((v+v_s)) becomes larger when (v_s) is larger, so the fraction becomes smaller, so (f_{\text{obs}}) is smaller.

Compare the “away” options:
- A: (v_s = 5\ \text{m s}^{-1}) away
- B: (v_s = 10\ \text{m s}^{-1}) away

Since (10\ \text{m s}^{-1} > 5\ \text{m s}^{-1}), option B gives the smallest observed frequency.

Key Takeaways

For a stationary observer: source moving towards (\Rightarrow) frequency increases; source moving away (\Rightarrow) frequency decreases.
For “away” motion, the faster the source, the smaller the observed frequency.

Common Mistakes

Thinking that “largest speed” always gives “smallest frequency” without checking direction.
Mixing up the moving-source and moving-observer Doppler formulas.
Assuming moving towards reduces frequency (it is the opposite).

Things to Be Careful About

The question states the observer is stationary; use the moving-source case.
You do not need numerical substitution because the options differ only by speed and direction; the trend is sufficient.
For minimum observed frequency, the correct condition is: source moving away with greatest speed (among the given away options).

Techniques used

apply the Doppler effect for a moving source and stationary observercompare how observed frequency depends on source speed and directionidentify the condition for minimum observed frequency

Q271MEasyMeasurement Techniques

What could describe the time-base of a cathode-ray oscilloscope (CRO)?

Options

A the frequency per division on the screen
B the number of divisions on the screen per unit frequency
C the number of divisions on the screen per unit time
D the time per division on the screen

Pick an answer

DifficultyEasy

Worked solution

Working

On a CRO, the time-base sets the horizontal scale (time corresponding to one division).

Answer

D

Final answer

D

Detailed explanation

Background Concept

A cathode-ray oscilloscope (CRO) displays how a voltage varies with time. The screen has a grid (graticule) marked in divisions.

The vertical scale is set in $\text{V div}^{-1}$ (volts per division).
The horizontal scale is set by the time-base, in $\text{s div}^{-1}$ (seconds per division).

So the time-base tells you how much time corresponds to one square division along the horizontal axis.

Understanding the Question

You are asked which option correctly describes the CRO time-base. The options describe different “per division” or “divisions per unit” quantities involving time or frequency.

The key clue is that the time-base is about time, not frequency, and it refers to the horizontal scale.

Approach

Recall the standard way CRO scales are written:

time-base: $\text{time per division}$

Then choose the option that exactly matches that wording and unit idea.

Step-by-Step Reasoning

The CRO spot sweeps left to right at a constant rate set by the time-base circuit.
Therefore, moving one division horizontally corresponds to a fixed time interval.
This is described as:

\text{time-base} = \text{time per division}

Option D states “the time per division on the screen”, which matches.

(Options involving frequency are incorrect because the time-base is not specified as frequency; it is the reciprocal idea and, in CRO terminology, is given directly as $\text{s div}^{-1}$ .)

Key Takeaways

CRO time-base sets the horizontal time scale.
It is quoted as time per division: $\text{s div}^{-1}$ (e.g. $5.0\ \text{ms div}^{-1}$ ).

Common Mistakes

Choosing an option involving frequency instead of time.
Picking “divisions per unit time” (a sweep speed) rather than the CRO’s stated control, which is normally time per division.

Things to Be Careful About

CRO conventions: vertical is typically $\text{V div}^{-1}$ and horizontal is $\text{s div}^{-1}$ .
“Divisions per unit time” would have units $\text{div s}^{-1}$ , which is the reciprocal of the usual CRO setting.

Techniques used

interpret the meaning of the CRO time-base settingrelate the horizontal axis scale to timematch a physical definition to the correct option

Q281MEasySuperposition

What happens when two waves superpose at a point?

Options

A Their amplitudes are added together.
B Their displacements are added together.
C Their frequencies are added together.
D Their velocities are added together.

Pick an answer

DifficultyEasy

Worked solution

Working

By the principle of superposition, the resultant displacement at a point is the algebraic sum of the individual displacements.

Answer

B

Final answer

B

Detailed explanation

Background Concept

The principle of superposition states that when two (or more) waves overlap in the same region of space, the resultant displacement $y$ at any point is the algebraic sum of the displacements due to each wave at that point:

y = y_1 + y_2

Here $y$ , $y_1$ , and $y_2$ are displacements from the equilibrium position (they can be positive or negative).

Understanding the Question

The question asks what quantity “adds” when two waves overlap at a single point. It is not asking about what happens to each wave’s speed or frequency, but about what determines the instantaneous effect at that point in space.

Approach

Use the definition of superposition: at a given position and time, add the instantaneous displacements of the individual waves to get the resultant displacement.

Step-by-Step Reasoning

When two waves meet, each wave still produces its own displacement of the medium (or field) at that point.
The actual displacement observed at that point is the combined effect of both waves.
Superposition says this combination is a simple algebraic sum:

y = y_1 + y_2

Therefore, it is displacements that are added, not frequencies, speeds, or (necessarily) amplitudes.

So the correct option is B.

Key Takeaways

Superposition: resultant displacement at a point equals the sum of individual displacements.
The addition is algebraic (sign matters), which explains constructive and destructive interference.

Common Mistakes

Choosing “amplitudes are added” (A): amplitudes only add in the special case where both waves produce displacements of the same sign and are in phase at that instant.
Choosing frequency (C): frequency is set by the source; overlapping waves do not create a wave whose frequency is the sum.
Choosing velocity (D): wave speed depends on the medium (and for EM waves, the material), not on superposition.

Things to Be Careful About

“Displacement” means instantaneous value (can be $+$ or $-$ ), not maximum displacement (amplitude).
The question says “at a point”, so it is explicitly about what adds locally at that position and time: the displacements.

Techniques used

recall the principle of superpositionidentify the correct wave quantity that adds at a pointdistinguish displacement from amplitude, frequency and speed

Q291MMedium-EasySuperpositionWaves

An electromagnetic wave is diffracted as it passes through a single slit. The width of the slit is larger than the wavelength of the wave.

Which change will decrease the amount of diffraction of the wave?

Options

A Decrease the frequency of the wave.
B Decrease the time period of the wave.
C Decrease the width of the slit.
D Increase the wavelength of the wave.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Diffraction decreases when $\dfrac{a}{\lambda}$ increases (slit much wider than the wavelength).

For an electromagnetic wave, $c = f\lambda$ and $f = \dfrac{1}{T}$ .

Decreasing $T$ increases $f$ , so $\lambda = \dfrac{c}{f}$ decreases, hence $\dfrac{a}{\lambda}$ increases and diffraction decreases.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Diffraction is the spreading out of a wave when it passes through a gap (or around an obstacle). For a single slit, the amount of diffraction depends mainly on the ratio of slit width $a$ to wavelength $\lambda$ :

If $a \gg \lambda$ , the wavefront fits easily through the gap and spreads out only a little (small diffraction).
If $a$ is comparable to $\lambda$ (or smaller), spreading is much more significant (large diffraction).

For electromagnetic waves in air/vacuum, the wave speed is essentially constant:

c = f\lambda

Also, frequency $f$ and time period $T$ are related by:

f = \frac{1}{T}

So changing $f$ (or $T$ ) changes $\lambda$ .

Understanding the Question

We are told an electromagnetic wave passes through a single slit, with slit width larger than the wavelength ( $a > \lambda$ ). The question asks which change will decrease diffraction.

To decrease diffraction, we must make the condition $a \gg \lambda$ “more true”, i.e. increase $a/\lambda$ .

Approach

Check each option’s effect on wavelength (or slit width):

If $\lambda$ decreases (with $a$ fixed), then $a/\lambda$ increases (\Rightarrow) less diffraction.
If $a$ decreases (with $\lambda$ fixed), then $a/\lambda$ decreases (\Rightarrow) more diffraction.

Use $c=f\lambda$ and $f=1/T$ to connect frequency/period changes to wavelength changes.

Step-by-Step Reasoning

Option B: Decrease the time period $T$ .

From $f=1/T$ , decreasing $T$ increases $f$ .
From $c=f\lambda$ (with $c$ constant),

\lambda = \frac{c}{f}

So increasing $f$ makes $\lambda$ smaller.
3. With smaller $\lambda$ and the same slit width $a$ , the ratio $a/\lambda$ increases, so the wave diffracts less.

Therefore option B decreases the amount of diffraction.

(Checking quickly: A decreases $f$ so increases $\lambda$ (\Rightarrow) more diffraction; C decreases $a$ (\Rightarrow) more diffraction; D increases $\lambda$ (\Rightarrow) more diffraction.)

Key Takeaways

Diffraction decreases when the gap is large compared to wavelength: larger $a/\lambda$ .
For EM waves, $c$ is constant so $\lambda \propto 1/f$ .
Since $f=1/T$ , decreasing $T$ increases $f$ and decreases $\lambda$ .

Common Mistakes

Thinking “higher frequency means more diffraction”: it is actually longer wavelength (lower frequency) that increases diffraction for a fixed slit.
Mixing up period and frequency (forgetting $f=1/T$ ).
Assuming changing frequency changes wave speed; for EM waves in vacuum/air, speed stays approximately $c$ .

Things to Be Careful About

The slit width is stated to be larger than the wavelength initially, but diffraction can still change depending on whether $\lambda$ moves closer to $a$ or further below it.
Always decide diffraction changes via the ratio $a/\lambda$ , not by $a$ or $\lambda$ alone.

Techniques used

use the qualitative condition for diffraction in terms of slit width to wavelength ratiorelate frequency and time periodapply the wave equation for electromagnetic waves in vacuum/air

Q301MMedium-EasySuperposition

The diagram shows visible light incident normally on a diffraction grating.

Fig. 1

A pattern of intensity maxima forms on the screen. A line connecting the centre of the fourth order intensity maximum with the centre of the diffraction grating forms an angle of $53^{\circ}$ with the centre line. The grating has a line spacing of $2.7 \times 10^{-6}\ \text{m}$ .

What is the wavelength of the incident light?

Options

A $4.1 \times 10^{-7}\ \text{m}$
B $5.4 \times 10^{-7}\ \text{m}$
C $1.6 \times 10^{-6}\ \text{m}$
D $2.2 \times 10^{-6}\ \text{m}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Use the grating equation

d\sin\theta = n\lambda

\lambda = \frac{d\sin\theta}{n} = \frac{(2.7\times 10^{-6})\sin 53^{\circ}}{4}

\lambda \approx \frac{(2.7\times 10^{-6})(0.799)}{4} \approx 5.4\times 10^{-7}\ \text{m}

Answer

B

Final answer

B

Detailed explanation

Background Concept

A diffraction grating has many equally spaced slits. Light from adjacent slits travels to a point on a distant screen at some angle $\theta$ to the straight-through (centre-line) direction. The path difference between rays from neighbouring slits is $d\sin\theta$ , where $d$ is the grating spacing.

Constructive interference (an intensity maximum) occurs when this path difference is an integer number of wavelengths:

d\sin\theta = n\lambda

Here $n = 0,1,2,\dots$ is the order of the maximum, and $\theta$ is measured from the centre line (the direction of the $n=0$ maximum).

Understanding the Question

You are told:

the grating is illuminated normally (so the centre line is the $n=0$ direction),
the line to the fourth order maximum makes an angle $53^{\circ}$ with the centre line, so $n=4$ and $\theta=53^{\circ}$ ,
the grating spacing is $d = 2.7\times 10^{-6}\ \text{m}$ .

The question asks for the wavelength $\lambda$ of the incident light.

Approach

Use the grating equation $d\sin\theta = n\lambda$ and rearrange to $\lambda = \dfrac{d\sin\theta}{n}$ . Then substitute $d$ , $\theta$ and $n=4$ .

Step-by-Step Reasoning

Start with

d\sin\theta = n\lambda

Rearrange:

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

Substitute $d = 2.7\times 10^{-6}\ \text{m}$ , $\theta = 53^{\circ}$ , $n=4$ :

\lambda = \frac{(2.7\times 10^{-6})\sin 53^{\circ}}{4}

Compute $\sin 53^{\circ} \approx 0.799$ :

\lambda \approx \frac{(2.7\times 10^{-6})(0.799)}{4} = \frac{2.16\times 10^{-6}}{4} \approx 5.4\times 10^{-7}\ \text{m}

This matches option B.

Key Takeaways

For a diffraction grating, maxima satisfy $d\sin\theta = n\lambda$ .
The order number $n$ must match the named maximum (fourth order means $n=4$ ).
The angle $\theta$ is measured from the central ( $n=0$ ) direction.

Common Mistakes

Using $n=3$ instead of $n=4$ for the fourth order maximum.
Using $\theta$ incorrectly (e.g. measuring from the grating plane rather than from the centre line).
Forgetting to divide by $n$ when solving for $\lambda$ .

Things to Be Careful About

The grating spacing $d$ is already in metres; avoid accidental unit conversion errors.
Ensure your calculator is in degrees for $\sin 53^{\circ}$ .
A visible wavelength should be of order $10^{-7}\ \text{m}$ ; answers of order $10^{-6}\ \text{m}$ are infrared and indicate a likely order/angle mistake.

Techniques used

apply the diffraction grating equationidentify the diffraction order from the wordingrearrange to solve for wavelengthsubstitute numerical values with correct units

Q311MMedium-EasySuperpositionWaves

A horizontal glass tube, closed at one end, has a layer of dust laid inside it on its lower side. Sound is emitted from a loudspeaker that is placed near the open end of the tube.

The frequency of the sound is varied and, at one frequency, a stationary wave is formed inside the tube so that the dust forms small heaps.

The distance between four heaps of dust is $30\ \text{cm}$ .

Fig. 1

The speed of sound in the air in the tube is $330\ \text{m s}^{-1}$ .

What is the frequency of the sound emitted by the loudspeaker?

Options

A $1650\ \text{Hz}$
B $2200\ \text{Hz}$
C $3300\ \text{Hz}$
D $6600\ \text{Hz}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Dust heaps form at displacement nodes, so adjacent heaps are separated by $\lambda/2$ .

Distance spanning four heaps $= 30\ \text{cm}$ means 3 gaps:

3\left(\frac{\lambda}{2}\right)=0.30\ \text{m}

\lambda = 0.20\ \text{m}

f = \frac{v}{\lambda} = \frac{330}{0.20} = 1650\ \text{Hz}

Answer

A

Final answer

A

Detailed explanation

Background Concept

A stationary (standing) wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose. In a tube, the sound wave reflects at the ends, so the incident and reflected waves can form a stationary wave.

In a stationary wave:

Displacement nodes are points that do not move (zero displacement amplitude).
Adjacent nodes are separated by half a wavelength, $\lambda/2$ .

In a Kundt-type dust experiment, dust collects into heaps at displacement nodes because the air motion is minimal there, so the dust is not blown away.

Once the wavelength $\lambda$ is known, the frequency follows from the wave equation:

v = f\lambda

Understanding the Question

You are told that at one frequency the dust forms small heaps at regular intervals along the tube. The diagram indicates that the distance covering four heaps is $30\ \text{cm}$ .

Given:

speed of sound in the tube air: $v = 330\ \text{m s}^{-1}$
distance spanning four adjacent heaps: $0.30\ \text{m}$

Find: the frequency $f$ of the sound.

Approach

Use the physics of standing waves: heaps mark nodes, and node-to-node spacing is $\lambda/2$ .
Convert “distance between four heaps” into a number of node spacings (gaps).
Calculate $\lambda$ , then use $f = v/\lambda$ .

Step-by-Step Reasoning

diagram-1

The four heaps correspond to four consecutive nodes along the tube.
Four nodes create three equal gaps between them.
Each gap (node to next node) is $\lambda/2$ .

So:

3\left(\frac{\lambda}{2}\right)=0.30

Rearrange:

\lambda = \frac{2\times 0.30}{3} = 0.20\ \text{m}

Now apply $v=f\lambda$ :

f = \frac{v}{\lambda} = \frac{330}{0.20} = 1650\ \text{Hz}

This matches option A.

Key Takeaways

Dust heaps in this type of tube form at displacement nodes.
Adjacent nodes in a stationary wave are separated by $\lambda/2$ .
Be careful interpreting “distance spanning four heaps”: it usually means three intervals.
Use $v=f\lambda$ once $\lambda$ is found.

Common Mistakes

Treating the distance across four heaps as $4\times (\lambda/2)$ instead of $3\times (\lambda/2)$ .
Assuming heaps form at antinodes (maximum displacement) rather than nodes.
Forgetting to convert $30\ \text{cm}$ to $0.30\ \text{m}$ .

Things to Be Careful About

Wording: “distance between four heaps” is interpreted as the distance from the first to the fourth heap, i.e. three equal spacings.
Units: keep wavelength in metres when using $v$ in $\text{m s}^{-1}$ .
Stationary-wave geometry: node spacing is always $\lambda/2$ regardless of whether the tube is open/closed; the boundary condition matters for where the pattern starts/ends, not the node-to-node spacing.

Techniques used

identify dust heaps as displacement nodes in a stationary waverelate node spacing to half-wavelengthapply the wave equation to calculate frequency

Q321MMedium-EasyElectricityParticle Physics

There is a current in a resistor for a short time interval.

Which statement about the total charge that passes through the resistor is correct?

Options

A It can take any value.
B It is an integer multiple of the elementary charge.
C It is the elementary charge.
D It is the rate of flow of the current.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Charge transferred is

Q = \int I\,\mathrm{d}t \; (\approx It \text{ if } I \text{ is constant}).

Charge is quantised, so any net charge transferred must be

Q = ne

where $n$ is an integer.

Answer

B

Final answer

B

Detailed explanation

Background Concept

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

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

Rearranging gives the charge transferred in a time interval:

Q = \int I\,\mathrm{d}t.

Also, electric charge is quantised: the smallest possible magnitude of free charge is the elementary charge $e = 1.60 \times 10^{-19}\ \text{C}$ . Any net charge transferred is therefore an integer multiple of $e$ :

Q = ne, \quad n \in \mathbb{Z}.

Understanding the Question

A current flows through a resistor for a short time. The question asks what must be true about the total charge that passes through (i.e. the net charge transferred) during that time.

Approach

Use the definition of current to relate the total charge passed to the current and time.
Use charge quantisation to decide what form $Q$ can take.
Compare with the options.

Step-by-Step Reasoning

The total charge that passes is found from

Q = \int I\,\mathrm{d}t.

If the current were constant, this would reduce to $Q = It$ , but the key idea is that $Q$ is a total amount of charge.

In a metal resistor, charge is carried by electrons. Since each electron has charge magnitude $e$ , transferring $n$ electrons corresponds to a net charge transfer

Q = ne.

Even if the time is “short”, $n$ can still be extremely large, making $Q$ appear continuous, but it is still an integer multiple of $e$ in principle.

Therefore the correct statement is B.

A (“any value”) ignores quantisation.
C (“is the elementary charge”) would only be true if exactly one electron’s worth of charge passed, which is not generally the case.
D confuses charge with current: current is the rate of flow of charge, not “the rate of flow of the current”.

Key Takeaways

Total charge transferred is found using $Q = \int I\,\mathrm{d}t$ (or $Q = It$ for steady current).
Charge is quantised: $Q$ must be an integer multiple of $e$ .
Current is a rate ( $\mathrm{C\ s^{-1}}$ ), while charge is an amount ( $\mathrm{C}$ ).

Common Mistakes

Stating $Q$ can be any real number because current seems continuous (quantisation still holds).
Choosing D by misremembering the definition: it is $I = \mathrm{d}Q/\mathrm{d}t$ , not something like “rate of flow of current”.
Thinking the total charge must be $e$ (it would require exactly one charge carrier to pass, unrealistic for typical currents).

Things to Be Careful About

The question asks about total charge transferred, not the current itself.
“Short time interval” does not mean “small number of electrons”; even microseconds of current can involve huge $n$ .
Quantisation applies to net transferred charge: $Q = ne$ with integer $n$ .

Techniques used

use the definition of charge transferred as the time integral of currentapply charge quantisation in units of the elementary chargedistinguish between charge transferred and current

Q331MMedium-EasyElectricity

The potential difference $V$ across a filament lamp is slowly raised from zero to its normal operating value.

Which graph represents the variation with $V$ of the current $I$ in the lamp?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

As $V$ increases, the filament temperature increases, so its resistance increases. Hence $I$ increases but with decreasing gradient ( $I/V$ gets smaller).

Answer

A

Final answer

A

Detailed explanation

Background Concept

For a resistor obeying Ohm's law at constant temperature,

V = IR

so an $I$ – $V$ graph is a straight line through the origin with constant gradient $\left(\frac{\Delta I}{\Delta V}\right)=\frac{1}{R}$ .

A filament lamp is non-ohmic because the filament's temperature changes significantly as current passes. For a metal filament, higher temperature causes more lattice vibrations, which increases electron scattering and therefore increases the resistance $R$ .

Understanding the Question

The potential difference across a filament lamp is increased slowly from $0$ to its normal operating value. We must choose which of the given graphs shows how current $I$ varies with $V$ during this increase.

Key idea: “slowly raised” implies the filament has time to heat up as $V$ increases, so resistance changes during the process.

Approach

Start at $V=0$ : then $I=0$ , so the graph must pass through the origin.
As $V$ increases, $I$ increases, but the filament heats up.
Heating increases resistance, so the gradient $\frac{\Delta I}{\Delta V}=\frac{1}{R}$ decreases as $V$ increases.
Therefore the curve rises but becomes progressively less steep (concave down).

Step-by-Step Reasoning

At $V=0$ , there is no driving potential difference, so $I=0$ . This eliminates any option that does not start at the origin.
Initially, when the filament is cool, its resistance is relatively low, so a small increase in $V$ produces a relatively large increase in $I$ (steep initial gradient).
As $V$ is raised further, $I$ increases and power $P = VI$ increases, heating the filament.
For a metal filament, higher temperature $\Rightarrow$ higher resistance.
Since the gradient of an $I$ – $V$ graph equals $1/R$ , increasing $R$ means the gradient must decrease as $V$ increases.
The only graph that starts at the origin and rises with decreasing gradient is Option A.

Key Takeaways

A filament lamp has a non-linear $I$ – $V$ characteristic.
As $V$ increases, the filament heats up and $R$ increases.
Increasing resistance causes the $I$ – $V$ curve to become less steep (decreasing gradient).

Common Mistakes

Choosing a straight line through the origin (ohmic behaviour): that would apply to a fixed-temperature resistor, not a filament lamp.
Thinking resistance decreases with temperature: true for thermistors (NTC) but not for a metal filament.
Picking a curve with increasing gradient (concave up): that suggests resistance is decreasing with increasing $V$ , which contradicts filament heating.

Things to Be Careful About

The gradient on an $I$ – $V$ graph is $\Delta I/\Delta V$ , not $\Delta V/\Delta I$ .
“Slowly raised” matters: it indicates the filament reaches higher temperatures as $V$ increases, so the resistance change is significant.
The graph must pass through $(0,0)$ because $V=0$ implies $I=0$ for this passive component.

Techniques used

use the I–V characteristic shape for a filament lamplink increasing temperature to increasing resistanceinfer changing gradient from changing resistance

Q341MMedium-EasyElectricity

An electric current of $12\ \text{A}$ is in a wire of length $0.35\ \text{m}$ .

The average drift speed of the free electrons (charge carriers) in the wire is $5.0 \times 10^{-4}\ \text{m s}^{-1}$ .

How many free electrons are in the wire?

Options

A $1.3 \times 10^{16}$
B $5.3 \times 10^{22}$
C $1.5 \times 10^{23}$
D $4.3 \times 10^{23}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Time for electrons to drift through length $L$ :

t = \frac{L}{v} = \frac{0.35}{5.0 \times 10^{-4}} = 700\ \text{s}

Charge passing a cross-section in this time:

Q = It = 12 \times 700 = 8400\ \text{C}

Number of electrons:

N = \frac{Q}{e} = \frac{8400}{1.60 \times 10^{-19}} = 5.25 \times 10^{22} \approx 5.3 \times 10^{22}

Answer

B

Final answer

B

Detailed explanation

Background Concept

Electric current $I$ is the rate of flow of charge:

I = \frac{\Delta Q}{\Delta t} \quad \Rightarrow \quad Q = It

In a metal wire, free electrons drift with a small average drift speed $v$ . If the drift speed is steady, then in the time it takes electrons to drift through a fixed length of wire, the charge that passes any cross-section equals the total charge contained in that length (steady flow).

Each electron carries charge magnitude $e = 1.60 \times 10^{-19}\ \text{C}$ , so the number of electrons is

N = \frac{Q}{e}

Understanding the Question

You are given:

current $I = 12\ \text{A}$
length of wire segment $L = 0.35\ \text{m}$
drift speed of electrons $v = 5.0 \times 10^{-4}\ \text{m s}^{-1}$

You are asked for the number of free electrons in that length of wire.

Approach

Find the time $t$ for an electron (on average) to drift through the length $L$ using $t = L/v$ .
In that time, the amount of charge that flows past a cross-section is $Q = It$ .
Convert charge to number of electrons using $N = Q/e$ .

This avoids needing the wire’s cross-sectional area.

Step-by-Step Reasoning

Time to drift through $0.35\ \text{m}$ :

t = \frac{L}{v} = \frac{0.35}{5.0 \times 10^{-4}} = 700\ \text{s}

Charge that passes a fixed point in $700\ \text{s}$ :

Q = It = 12 \times 700 = 8400\ \text{C}

Convert charge to number of electrons:

N = \frac{Q}{e} = \frac{8400}{1.60 \times 10^{-19}} = 5.25 \times 10^{22}

Rounded to match the options:

N \approx 5.3 \times 10^{22}

So the correct option is B.

Key Takeaways

Use $Q = It$ to connect current with charge transferred.
Drift speed lets you find the transit time $t = L/v$ .
Number of carriers is $N = Q/e$ .

Common Mistakes

Using $t = v/L$ instead of $t = L/v$ .
Forgetting to divide by $e$ (or using the wrong value for $e$ ).
Confusing drift speed with electron random thermal speed (irrelevant here).

Things to Be Careful About

Units: $v$ must be in $\text{m s}^{-1}$ and $L$ in $\text{m}$ .
Significant figures: the drift speed is given to 2 s.f., so $5.3 \times 10^{22}$ is appropriate.
The reasoning relies on steady current (constant drift speed), which is implied by the question.

Techniques used

relate drift speed to transit time through a length of wireuse charge flow equation Q = Itconvert total charge to number of electrons using elementary charge

Q351MMedium-EasyD.C. Circuits

Three resistors are connected to a cell of negligible internal resistance.

Which circuit has a combined resistance of $100\ \Omega$ ?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For circuit B, the two $100\ \Omega$ resistors are in parallel:

R_{\parallel} = \left(\frac{1}{100} + \frac{1}{100}\right)^{-1} = 50\ \Omega

This is in series with $50\ \Omega$ :

R_{\text{total}} = 50 + 50 = 100\ \Omega

Answer

B

Final answer

B

Detailed explanation

Background Concept

To find the combined (equivalent) resistance of resistors:

Series: resistances add directly because the same current flows through each.

R_{\text{series}} = R_1 + R_2 + \cdots

Parallel: reciprocals add because the same potential difference is across each branch and currents add.

\frac{1}{R_{\parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots

After reducing any parallel groups to a single resistor, you can then add any series resistors.

Understanding the Question

Each option shows a cell connected to three resistors of $100\ \Omega$ , $100\ \Omega$ and $50\ \Omega$ in different arrangements. You must identify which arrangement gives a total resistance of $100\ \Omega$ .

Approach

For each circuit:

Identify which resistors are in parallel and reduce them to one equivalent resistance.
Add any resistors in series with that equivalent resistance.
Compare the total with $100\ \Omega$ .

Because this is multiple choice, you can stop once you find an option that clearly gives exactly $100\ \Omega$ .

Step-by-Step Reasoning

Option B: a $50\ \Omega$ resistor is in series with a parallel pair of two $100\ \Omega$ resistors.

First reduce the parallel pair:

\frac{1}{R_{\parallel}} = \frac{1}{100} + \frac{1}{100} = \frac{2}{100}

So,

R_{\parallel} = \frac{100}{2} = 50\ \Omega

Now add the series resistor:

R_{\text{total}} = 50\ \Omega + 50\ \Omega = 100\ \Omega

So option B matches the required combined resistance.

(Checks on others, briefly: A gives $100 + (100||50) = 100 + 33.3\approx 133\ \Omega$ ; C gives $25\ \Omega$ ; D gives $250\ \Omega$ .)

Key Takeaways

Add resistances directly only when they are in series.
For parallel, add reciprocals and then invert.
In mixed networks, reduce parallel sections first, then combine in series.

Common Mistakes

Adding parallel resistors directly (wrong: $100 + 100$ for the parallel pair).
Forgetting to invert after adding reciprocals in a parallel calculation.
Misidentifying which resistors share the same two nodes (and therefore are truly in parallel).

Things to Be Careful About

Keep track of units: all resistances remain in $\Omega$ .
A quick mental check helps: two equal resistors $R$ in parallel always give $R/2$ (so $100||100 = 50\ \Omega$ ), making option B easy to spot.
Ensure you only add resistances once you have correctly simplified each parallel group.

Techniques used

calculate equivalent resistance of parallel resistors using reciprocalsadd series resistances to find total circuit resistancecompare the calculated equivalent resistance with a target value

Q361MEasyD.C. Circuits

Each of Kirchhoff’s laws is a statement based on the conservation of a physical quantity.

Which quantity is conserved in each law?

Options

	Kirchhoff’s first law	Kirchhoff’s second law
A	charge	energy
B	energy	current
C	power	charge
D	resistance	power

Pick an answer

DifficultyEasy

Worked solution

Working

Kirchhoff’s first law (at a junction): total current in = total current out (\Rightarrow) conservation of charge.

Kirchhoff’s second law (around a closed loop): sum of emfs = sum of potential drops (\Rightarrow) conservation of energy.

Answer

A

Final answer

A

Detailed explanation

Background Concept

Kirchhoff’s laws describe how current and potential difference behave in electric circuits, and each law follows from a conservation principle:

Kirchhoff’s first law (junction law): at any junction, the total current entering equals the total current leaving. Since current is the rate of flow of charge, this is equivalent to conservation of charge (charge does not build up at a junction in a steady circuit).
Kirchhoff’s second law (loop law): around any closed loop, the total emf supplied equals the total potential difference lost across components. This expresses conservation of energy: energy gained per unit charge from sources equals energy lost per unit charge in components.

Understanding the Question

You are asked which physical quantity is conserved in:

Kirchhoff’s first law, and
Kirchhoff’s second law,
then to select the option (A–D) that matches both.

Approach

Link each law to what it physically means:

Junction law (\rightarrow) no net accumulation of charge (\rightarrow) charge conserved.
Loop law (\rightarrow) energy changes per unit charge around a loop balance (\rightarrow) energy conserved.
Then compare with the table of options.

Step-by-Step Reasoning

First law:
- Current (I) is (\Delta Q/\Delta t).
- If more current entered a junction than left, charge would accumulate at the junction.
- In steady conditions, charge does not continuously build up, so the flow rates must balance.
- Therefore, the law is based on conservation of charge.
Second law:
- Potential difference (V) is energy transferred per unit charge: (V = W/Q).
- EMFs provide energy per unit charge; resistors and other components dissipate energy per unit charge.
- Around a closed loop, a charge returns to its starting point, so the net change in energy per unit charge must be zero.
- Therefore, the law is based on conservation of energy.
The option that states first law: charge and second law: energy is A.

Key Takeaways

Junction law (\Rightarrow) charge conservation (currents in = currents out).
Loop law (\Rightarrow) energy conservation (sum of emfs = sum of potential drops).

Common Mistakes

Saying the second law conserves current: current is not conserved around a loop if there are junctions; the loop law is about voltages/energy per charge.
Saying the first law conserves energy: the junction law is about charge flow rates, not energy transfers.
Confusing power with energy: power can differ between components; Kirchhoff’s laws are not statements of constant power.

Things to Be Careful About

Kirchhoff’s second law is sometimes stated as “sum of potential differences around a loop is zero”; this is shorthand for energy conservation per unit charge, not that individual voltages are conserved.
Kirchhoff’s laws assume a steady-state circuit (no net charge accumulation at junctions).

Techniques used

recall Kirchhoff's first (junction) lawrecall Kirchhoff's second (loop) lawidentify the conserved quantity associated with each lawmatch the identified quantities to the correct option

Q371MMedium-EasyD.C. Circuits

A cell with electromotive force (e.m.f.) $1.50\ \text{V}$ delivers a current of $0.26\ \text{A}$ to a resistor of resistance $5.0\ \Omega$ connected between its terminals.

What is the internal resistance of the cell?

Options

A $0.04\ \Omega$
B $0.20\ \Omega$
C $0.39\ \Omega$
D $0.77\ \Omega$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Terminal p.d. across the external resistor:

V = IR = 0.26 \times 5.0 = 1.30\ \text{V}

Using $E = V + Ir$ :

r = \frac{E - V}{I} = \frac{1.50 - 1.30}{0.26} = 0.769...\ \Omega

Answer

$0.77\ \Omega$ (D)

Final answer

D

Detailed explanation

Background Concept

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

The key equation is:

E = V + Ir

where:

$E$ is the e.m.f. of the cell (energy supplied per unit charge),
$V$ is the terminal p.d. across the external resistor(s),
$Ir$ is the “lost volts” inside the cell.

Also, for the external resistor $R$ , Ohm’s law gives:

V = IR

Understanding the Question

You are told:

the cell e.m.f. is $E = 1.50\ \text{V}$ ,
the current delivered is $I = 0.26\ \text{A}$ ,
the external resistor connected across the terminals has resistance $R = 5.0\ \Omega$ .

You must find the internal resistance $r$ .

Approach

Find the terminal p.d. $V$ across the external $5.0\ \Omega$ resistor using $V = IR$ .
Use the internal resistance model $E = V + Ir$ and rearrange for $r$ :

r = \frac{E - V}{I}

Calculate and choose the matching option.

Step-by-Step Reasoning

Terminal p.d. across the external resistor:

V = IR = 0.26 \times 5.0 = 1.30\ \text{V}

This is less than $1.50\ \text{V}$ , which makes sense because internal resistance causes a voltage drop inside the cell.

Lost volts across the internal resistance:

E - V = 1.50 - 1.30 = 0.20\ \text{V}

Use $Ir = E - V$ to find $r$ :

r = \frac{E - V}{I} = \frac{0.20}{0.26} = 0.769...\ \Omega

Rounded to two significant figures (matching the options):

r = 0.77\ \Omega

So the correct option is D.

Key Takeaways

The terminal p.d. is the p.d. across the external load: $V = IR$ .
Internal resistance causes a “lost volts” drop: $E - V = Ir$ .
Combine them via $E = V + Ir$ and rearrange to obtain $r$ .

Common Mistakes

Using $E = IR$ with the external resistor: $E$ is not generally equal to the terminal p.d. when internal resistance is present.
Subtracting in the wrong order (using $V - E$ ), which would give a negative $r$ .
Rounding too early (e.g. rounding $0.20/0.26$ too aggressively) and missing the closest option.

Things to Be Careful About

Make sure $V$ refers to the p.d. across the external resistor, not across the internal resistance.
Keep enough significant figures during intermediate steps; round at the end to match the options.
Units: volts, amps, and ohms must be consistent (they are already in SI here).

Techniques used

calculate terminal potential difference using V = IRapply the e.m.f. relation E = V + Irrearrange to determine internal resistance

Q381MEasyParticle Physics

What is not a quark flavour?

Options

A charm
B meson
C strange
D up

Pick an answer

DifficultyEasy

Worked solution

Working

Quark flavours are $u, d, s, c, t, b$ (up, down, strange, charm, top, bottom).
A meson is a hadron made of a quark and an antiquark, not a quark flavour.

Answer

B

Final answer

B

Detailed explanation

Background Concept

In the quark model, quark flavours are the different types of quark. There are six flavours:

up ( $u$ ), down ( $d$ )
strange ( $s$ ), charm ( $c$ )
top ( $t$ ), bottom ( $b$ )

A hadron is a composite particle made of quarks held together by the strong interaction. Hadrons are divided into:

baryons: three quarks (e.g. proton $uud$ , neutron $udd$ )
mesons: a quark + an antiquark (e.g. pion $u\bar d$ )

So “meson” is a type of composite particle, not a flavour.

Understanding the Question

You are given four words and asked which one is not the name of a quark flavour. Three of the options are actual flavour names; one is a category of particle.

Approach

Recall the list of quark flavours.
Check each option against that list.
Identify any option that refers to something else (such as a hadron type).

Step-by-Step Reasoning

Option A: charm → this is a quark flavour ( $c$ ).
Option C: strange → this is a quark flavour ( $s$ ).
Option D: up → this is a quark flavour ( $u$ ).
Option B: meson → this is not a flavour; it is a hadron consisting of a quark and an antiquark.

Therefore the option that is not a quark flavour is B.

Key Takeaways

Learn the six quark flavours: $u, d, s, c, t, b$ .
“Meson” describes a composite particle (quark–antiquark), not a fundamental flavour.

Common Mistakes

Confusing a type of hadron (meson/baryon) with a quark flavour.
Thinking “meson” is a fundamental particle rather than a quark–antiquark bound state.

Things to Be Careful About

Cambridge often tests vocabulary: flavour names are specific (up, down, strange, charm, top, bottom), whereas terms like meson/baryon/lepton/hadron are particle categories.
Make sure you can separate “what it’s made of” (quark content) from “what it’s called” (flavour name vs composite particle name).

Techniques used

recall the six quark flavoursdistinguish fundamental particles from composite hadronsidentify the option that is not a flavour name

Q391MMedium-EasyParticle Physics

How many down quarks are in a nucleus of hydrogen-3, $^{3}_{1}\text{H}$ ?

Options

A 2
B 3
C 4
D 5

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For $^{3}_{1}\text{H}$ nucleus: protons $=1$ , neutrons $=3-1=2$ .

Proton $= uud$ (\Rightarrow) down quarks $=1$ .

Neutron $= udd$ (\Rightarrow) down quarks $=2$ .

Total down quarks $= 1\times 1 + 2\times 2 = 5$ .

Answer

D

Final answer

D

Detailed explanation

Background Concept

A nucleus is made of nucleons: protons and neutrons. In the quark model:

A proton has quark content $uud$ (two up quarks and one down quark).
A neutron has quark content $udd$ (one up quark and two down quarks).

So, to find how many down quarks are in a nucleus, you:

work out how many protons and neutrons are present, then
count the down quarks contributed by each nucleon type.

Understanding the Question

The nucleus of hydrogen-3, $^{3}_{1}\text{H}$ (tritium), has:

mass (nucleon) number $A = 3$ (\Rightarrow) total nucleons = 3,
proton (atomic) number $Z = 1$ (\Rightarrow) protons = 1.

The question asks for the total number of down quarks inside those nucleons.

Approach

Use nuclide notation $^{A}_{Z}X$ $_{Z}^{A} X$ :
- protons $= Z$
- neutrons $= A - Z$
Use quark compositions:
- each proton contributes 1 down quark
- each neutron contributes 2 down quarks
Add the totals.

Step-by-Step Reasoning

For $^{3}_{1}\text{H}$ :

Number of protons:
$Z = 1$
Number of neutrons:
$A - Z = 3 - 1 = 2$

Now count down quarks.

Each proton is $uud$ (\Rightarrow) 1 down quark.
Contribution from 1 proton:
$1 \times 1 = 1$
Each neutron is $udd$ (\Rightarrow) 2 down quarks.
Contribution from 2 neutrons:
$2 \times 2 = 4$

Total down quarks in the nucleus:

1 + 4 = 5

So the correct option is D.

Key Takeaways

From $^{A}_{Z}X$ : protons $= Z$ , neutrons $= A-Z$ .
Proton quark content $uud$ (1 down); neutron quark content $udd$ (2 down).
Count contributions nucleon-by-nucleon.

Common Mistakes

Using $A=3$ as the number of neutrons (forgetting that $A$ is total nucleons).
Swapping quark compositions (writing proton as $udd$ and neutron as $uud$ ).
Counting total quarks (always 3 per nucleon) instead of specifically down quarks.

Things to Be Careful About

The question asks about the nucleus, so electrons in the atom are irrelevant.
Read $^{3}_{1}\text{H}$ correctly: $Z=1$ even though $A=3$ .
Make sure you are counting down quarks only, not up quarks or total quarks.

Techniques used

identify the nucleon composition from nuclide notationuse quark compositions of protons and neutronscount a specific quark type across all nucleons

Q401MMedium-EasyParticle Physics

An isotope of boron decays to beryllium by $\beta^{+}$ emission.

Which particle is emitted in addition to the $\beta^{+}$ particle?

Options

A antineutrino
B electron
C neutrino
D neutron

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In (\beta^{+}) decay, a proton changes into a neutron:

p \rightarrow n + \beta^{+} + \nu

So, in addition to the positron (\beta^{+}), a neutrino (\nu) is emitted.

Answer

C

Final answer

C

Detailed explanation

Background Concept

In nuclear (\beta) decay, the nucleus changes its proton number (Z) by converting a proton to a neutron or vice versa, while the total nucleon number (A) stays the same.

For (\beta^{+}) (positron) emission:

A proton in the nucleus converts to a neutron.
A positron (the (\beta^{+}) particle) is emitted.
To satisfy conservation laws, a neutrino is also emitted.

A convenient way to remember the particle-level reaction is:

p \rightarrow n + e^{+} + \nu_e

where (\nu_e) is an electron neutrino.

Understanding the Question

The question states that an isotope of boron decays to beryllium by (\beta^{+}) emission. It asks which additional particle must be emitted alongside the positron.

The options include neutrino/antineutrino and unrelated particles (electron, neutron). The key is to recall what must accompany a positron in (\beta^{+}) decay.

Approach

Use the standard (\beta^{+}) decay equation and conservation laws:

Identify what happens to a proton in (\beta^{+}) decay.
Use conservation of nucleon number and charge to confirm the nuclear change.
Use lepton number conservation to decide whether the extra particle is a neutrino or antineutrino.

Step-by-Step Reasoning

In (\beta^{+}) decay, the nucleus emits a positron (e^{+}). For the nucleus to emit a positively charged lepton while overall charge is conserved, one proton must effectively turn into a neutron (reducing (Z) by 1).
The particle reaction is:

p \rightarrow n + e^{+} + \nu

Check lepton number:

A positron (e^{+}) has lepton number (-1).
Initially, there are no leptons on the left-hand side, so total lepton number starts at (0).
To end with total lepton number (0), we need a particle with lepton number (+1), i.e. a neutrino (\nu_e).

Therefore the additional emitted particle is a neutrino (not an antineutrino).

So the correct option is C.

Key Takeaways

(\beta^{+}) decay corresponds to (p \to n) inside the nucleus.
(\beta^{+}) emission always produces a neutrino (\nu_e) alongside the positron.
Neutrino vs antineutrino can be decided using lepton number conservation.

Common Mistakes

Choosing antineutrino: antineutrinos are emitted in (\beta^{-}) decay, not (\beta^{+}).
Choosing electron: an electron is the (\beta^{-}) particle, not associated with (\beta^{+}) emission.
Choosing neutron: a neutron is created inside the nucleus via conversion, but it is not emitted as the accompanying particle in standard (\beta^{+}) decay.

Things to Be Careful About

(\beta^{+}) means positron (e^{+}), not a generic “beta particle”.
Remember the pair:

\beta^-: \; n \rightarrow p + e^- + \bar{\nu}_e

\beta^+: \; p \rightarrow n + e^+ + \nu_e

Mixing these up is the most common way to lose this mark.

Techniques used

write the nuclear equation for $\beta^+$ decayapply conservation of nucleon number and chargeapply conservation of lepton number to identify the emitted lepton

Physics 9702/12 — October/November 2024

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Options

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

1) Moments from the $6.0\ \text{N}$ vertical forces

2) Moments from the $5.0\ \text{N}$ horizontal forces