9702/13 Oct/Nov 2025 — Physics MCQ Answers

Q11MEasyPhysical Quantities and Units

The table shows some physical quantities.

Which row correctly identifies the quantities as scalars or vectors?

Options

	acceleration	charge	kinetic energy	wavelength
A	scalar	vector	vector	scalar
B	vector	vector	scalar	scalar
C	scalar	scalar	scalar	vector
D	vector	scalar	scalar	scalar

Pick an answer

DifficultyEasy

Worked solution

Working

Acceleration has magnitude and direction, so it is a vector.

Charge has no direction, so it is a scalar.

Kinetic energy has no direction, so it is a scalar.

Wavelength has no direction, so it is a scalar.

Answer

D

Final answer

D

Detailed explanation

Background Concept

A scalar quantity is completely described by its magnitude (size) alone (e.g. $5\ \text{J}$ ).

A vector quantity requires both magnitude and direction to be fully specified (e.g. $5\ \text{m s}^{-2}$ to the east).

A quick test is: if changing direction changes the physical meaning, it is a vector; if direction is irrelevant, it is a scalar.

Understanding the Question

You are given four physical quantities (acceleration, charge, kinetic energy, wavelength) and four possible rows (A–D) that label each as scalar or vector. The task is to choose the row where all four classifications are correct.

Approach

Go through each quantity one by one:

Decide whether it needs a direction to be fully described.
Mark it as vector (direction needed) or scalar (direction not needed).
Compare with the options and pick the matching row.

Step-by-Step Reasoning

Acceleration: defined as rate of change of velocity. Since velocity is a vector, acceleration also has direction (it points in the direction of the change in velocity).

Therefore acceleration is a vector.
Charge: electric charge can be positive or negative, but it does not have a spatial direction associated with it.

Therefore charge is a scalar.
Kinetic energy: given by

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

Energy is not described by a direction; it is just a magnitude.

Therefore kinetic energy is a scalar.

Wavelength: a wavelength is a length (distance between successive points in phase). Length does not require direction for its description in this context.

Therefore wavelength is a scalar.

So the correct row is: acceleration vector, charge scalar, kinetic energy scalar, wavelength scalar, which corresponds to D.

Key Takeaways

Vectors require direction; scalars do not.
Acceleration is a vector because it relates to velocity (a vector).
Charge, energy, and wavelength are scalars.

Common Mistakes

Calling charge a vector because it can be positive/negative (sign is not the same as direction).
Calling wavelength a vector because waves travel in a direction (wavelength is a distance, not a direction-dependent quantity).
Confusing energy with momentum/force (momentum and force are vectors; energy is scalar).

Things to Be Careful About

A negative value (e.g. negative charge) does not automatically mean “vector”; vectors need directional information in space.
Some quantities are vectors because they are defined from vectors (e.g. acceleration from velocity), so using definitions can help when unsure.

Techniques used

classify quantities by whether direction is requiredapply definitions of scalar and vector quantitiesuse physical interpretation of common quantities

Q21MMediumPhysical Quantities and Units

Two quantities are measured.

$L = 6.8 \pm 0.1\text{ cm}$

$T = 2.42 \pm 0.08\text{ s}$

$L$ and $T$ are related to $X$ by the equation shown.

X = \frac{4\pi^2 L}{T^2}

What is the calculated value and uncertainty of $X$ ?

Options

A $45.8 \pm 0.2\text{ cm s}^{-2}$
B $45.8 \pm 0.3\text{ cm s}^{-2}$
C $46 \pm 2\text{ cm s}^{-2}$
D $46 \pm 4\text{ cm s}^{-2}$

Pick an answer

DifficultyMedium

Worked solution

Working

X = \frac{4\pi^2 L}{T^2}

T^2 = (2.42)^2 = 5.8564\ \text{s}^2

X = \frac{(4\pi^2)(6.8)}{5.8564} = \frac{(39.478)(6.8)}{5.8564} = 45.8\ \text{cm s}^{-2}

Fractional uncertainty:

\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}

\frac{\Delta X}{X} = \frac{0.1}{6.8} + 2\frac{0.08}{2.42} = 0.0147 + 0.0661 = 0.0808

\Delta X = 0.0808 \times 45.8 = 3.7 \approx 4\ \text{cm s}^{-2}

Answer

D

Final answer

D

Detailed explanation

Background Concept

When quantities are combined by multiplication/division and powers, the percentage (fractional) uncertainties add.

If

X = k\,\frac{A\,B}{C^2}

then (taking maximum uncertainty propagation, as used in many A Level questions)

\frac{\Delta X}{X} = \frac{\Delta A}{A} + \frac{\Delta B}{B} + 2\frac{\Delta C}{C}.

The factor of 2 appears because the uncertainty in a power $C^2$ is multiplied by the power.

After finding the fractional uncertainty, convert back to an absolute uncertainty using

\Delta X = X\left(\frac{\Delta X}{X}\right).

Understanding the Question

You are given measured values (with uncertainties):

$L = 6.8 \pm 0.1\ \text{cm}$
$T = 2.42 \pm 0.08\ \text{s}$

and the relationship

X = \frac{4\pi^2 L}{T^2}.

You must calculate:

the best estimate of $X$ using the central values of $L$ and $T$ ;
the uncertainty in $X$ produced by the uncertainties in $L$ and $T$ ;
then choose the matching option.

Approach

Calculate $X$ directly from the formula using $L=6.8$ and $T=2.42$ .
Use fractional-uncertainty rules: since $X \propto L$ and $X \propto T^{-2}$ , $\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}.$
Convert fractional uncertainty to absolute uncertainty in $X$ and round appropriately.

Step-by-Step Reasoning

1) Calculate $X$

Compute $T^2$ :

T^2 = (2.42)^2 = 5.8564\ \text{s}^2.

Also,

4\pi^2 \approx 39.478.

So

X = \frac{(39.478)(6.8\ \text{cm})}{5.8564\ \text{s}^2} = 45.8\ \text{cm s}^{-2}.

(Units: $\text{cm}/\text{s}^2 = \text{cm s}^{-2}$ .)

2) Fractional uncertainty in $X$

Because $X \propto L$ :

\frac{\Delta L}{L} = \frac{0.1}{6.8} = 0.0147 \ (1.47\%).

Because $X \propto T^{-2}$ :

2\frac{\Delta T}{T} = 2\left(\frac{0.08}{2.42}\right) = 2(0.0331)=0.0661 \ (6.61\%).

Add them:

\frac{\Delta X}{X} = 0.0147 + 0.0661 = 0.0808 \ (8.08\%).

3) Absolute uncertainty in $X$

\Delta X = 0.0808 \times 45.8 = 3.7\ \text{cm s}^{-2} \approx 4\ \text{cm s}^{-2}.

So

X \approx 46 \pm 4\ \text{cm s}^{-2},

which matches option D.

Key Takeaways

For products/divisions, add fractional uncertainties.
For powers, multiply the fractional uncertainty by the power (here $T^2$ gives a factor of 2).
Convert back to absolute uncertainty with $\Delta X = X(\Delta X/X)$ .
Round uncertainty (typically 1 s.f.) and round the value to the same decimal place.

Common Mistakes

Forgetting the factor of 2 from $T^2$ .
Subtracting uncertainties because of division (you still add fractional uncertainties).
Mixing absolute and percentage uncertainties (you must use fractional/percentage first here).
Giving $45.8 \pm 3.7$ without appropriate rounding to match the options.

Things to Be Careful About

The constant $4\pi^2$ is exact, so it contributes no uncertainty.
Use consistent significant figures: once the uncertainty rounds to $\pm 4$ , the value should be quoted as $46$ (not $45.8$ ) to match the precision.
Ensure units follow from the formula: $L$ in $\text{cm}$ and $T$ in $\text{s}$ gives $X$ in $\text{cm s}^{-2}$ .

Techniques used

substitute measured values into a given formulapropagate uncertainties using fractional uncertainties for products and powersconvert a fractional uncertainty into an absolute uncertaintyround the uncertainty to a suitable number of significant figures and match the value accordingly

Q31MEasyPhysical Quantities and Units

What are the SI base units of the watt?

Options

A $\text{J s}^{-1}$
B $\text{kg m}^2\text{ s}^{-3}$
C $\text{kg m}^2\text{ s}^{-1}$
D $\text{N m s}^{-1}$

Pick an answer

DifficultyEasy

Worked solution

Working

Power $P = \dfrac{\text{energy}}{\text{time}}$ , so $1\ \text{W} = 1\ \text{J s}^{-1}$ .

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

1\ \text{W} = 1\ \text{J s}^{-1} = \text{kg m}^2\text{ s}^{-3}

Answer

B

Final answer

B

Detailed explanation

Background Concept

The watt (W) is the SI unit of power. Power is the rate of transfer of energy (or the rate of doing work):

P = \frac{E}{t}

So a watt can be written in terms of joules and seconds as:

1\ \text{W} = 1\ \text{J s}^{-1}

To get SI base units, we must rewrite the joule (J) using base units (kg, m, s, A, K, mol, cd). One joule is one newton-metre:

1\ \text{J} = 1\ \text{N m}

and one newton is:

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

Understanding the Question

You are given four possible unit expressions and asked which one is the watt written purely in SI base units. That means the final unit should contain only kg, m, s (and/or other base units if needed), but not derived units like J or N.

Approach

Start from the definition $\text{W} = \text{J s}^{-1}$ .
Convert $\text{J}$ into base units using $\text{J} = \text{N m}$ and $\text{N} = \text{kg m s}^{-2}$ .
Simplify the powers of seconds.
Match the result to the options.

Step-by-Step Reasoning

From the definition of power:

1\ \text{W} = 1\ \text{J s}^{-1}

Convert joule to newton-metre:

1\ \text{J} = 1\ \text{N m}

Convert newton to base units:

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

So:

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

Now divide by seconds to get watts:

1\ \text{W} = \text{J s}^{-1} = (\text{kg m}^2\text{ s}^{-2})\,\text{s}^{-1} = \text{kg m}^2\text{ s}^{-3}

This matches option B.

Key Takeaways

$\text{W} = \text{J s}^{-1}$ because power is energy per unit time.
Convert derived units stepwise: $\text{J} \to \text{N m}$ and $\text{N} \to \text{kg m s}^{-2}$ .
Combine powers carefully: $\text{s}^{-2} \times \text{s}^{-1} = \text{s}^{-3}$ .

Common Mistakes

Stopping at $\text{J s}^{-1}$ (option A) even though it is not in SI base units.
Converting $\text{N}$ incorrectly (e.g. missing the $\text{s}^{-2}$ ).
Making an index error when combining seconds, giving $\text{s}^{-1}$ or $\text{s}^{-2}$ instead of $\text{s}^{-3}$ .

Things to Be Careful About

“SI base units” means only base symbols (kg, m, s, A, K, mol, cd) should remain.
Check that multiplication by $\text{s}^{-1}$ reduces the time unit correctly: dividing by time always decreases the power of seconds by 1 (more negative).

Techniques used

use the definition of power as energy per unit timeexpress the joule in SI base unitssimplify units using index laws

Q41MMedium-EasyPhysical Quantities and Units

The diagram shows two forces of $6.0\text{ N}$ acting on an object. The angle between the lines of action of the two forces is $60^\circ$ .

Fig. 1

What is the magnitude of the resultant force?

Options

A $6.0\text{ N}$
B $7.9\text{ N}$
C $10\text{ N}$
D $12\text{ N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Using cosine rule for two forces $6.0\,\text{N}$ at $60^\circ$ :

R^2 = 6.0^2 + 6.0^2 + 2(6.0)(6.0)\cos 60^\circ

R^2 = 36 + 36 + 72\times 0.5 = 108

R = \sqrt{108} = 10.4\,\text{N} \approx 10\,\text{N}

Answer

C

Final answer

C

Detailed explanation

Background Concept

A force is a vector, so when two forces act at a point the overall (resultant) force is found by vector addition, not by simply adding magnitudes unless they act in the same direction.

For two vectors $\vec{F}_1$ and $\vec{F}_2$ with an angle $\theta$ between them, the magnitude $R$ of the resultant $\vec{R} = \vec{F}_1 + \vec{F}_2$ can be found using the cosine rule on the vector triangle:

R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta

This works because the two force vectors form two sides of a triangle, with the included angle equal to the angle between their directions.

Understanding the Question

Two forces, each of magnitude $6.0\,\text{N}$ , act from the same point. The angle between their lines of action is $60^\circ$ . You are asked for the magnitude (size) of the resultant force.

So the known values are:

$F_1 = 6.0\,\text{N}$
$F_2 = 6.0\,\text{N}$
included angle $\theta = 60^\circ$
unknown: $R$

Approach

Treat the two forces as two sides of a triangle (or equivalently, use the parallelogram method). Because you know both magnitudes and the included angle, the quickest method is the cosine rule to find the third side, which is the resultant magnitude.

Step-by-Step Reasoning

diagram-1

Write the cosine-rule relationship for the resultant magnitude:

R^2 = 6.0^2 + 6.0^2 + 2(6.0)(6.0)\cos 60^\circ

Evaluate $\cos 60^\circ = 0.5$ and simplify:

R^2 = 36 + 36 + 72(0.5) = 72 + 36 = 108

Take the square root:

R = \sqrt{108} = 10.4\,\text{N}

Match to the given options: $10.4\,\text{N}$ rounds to $10\,\text{N}$ , so the correct choice is C.

Key Takeaways

Forces add as vectors, so the angle between them matters.
When two vector magnitudes and their included angle are given, the cosine rule is the most direct method.
For equal forces, the resultant lies between them and is larger than each individual force when the angle is less than $180^\circ$ .

Common Mistakes

Adding magnitudes directly to get $12\,\text{N}$ (only valid if the forces are parallel in the same direction).
Using $\sin$ instead of $\cos$ in the cosine rule.
Using the wrong angle (must be the included angle between the two force directions).
Rounding too early (e.g. rounding $\sqrt{108}$ incorrectly).

Things to Be Careful About

Keep the included angle as $60^\circ$ exactly; do not halve/double it.
Check that your answer is sensible: with a $60^\circ$ angle, the resultant should be between $6\,\text{N}$ and $12\,\text{N}$ , so an answer near $10\,\text{N}$ is reasonable.
For MCQ, choose the closest listed option consistent with appropriate rounding.

Techniques used

add two vectors using a vector triangleapply the cosine rule to find the magnitude of a resultantuse the included angle between vectors in a magnitude calculation

Q51MMedium-EasyKinematics

A goods train passes through a station at a constant speed of $10\text{ m s}^{-1}$ at time $t = 0$ . An express train is at rest at the station. The express train leaves the station with a uniform acceleration of $0.5\text{ m s}^{-2}$ just as the goods train goes past. Both trains move in the same direction on straight, parallel tracks.

At which time $t$ does the express train overtake the goods train?

Options

A $6\text{ s}$
B $10\text{ s}$
C $20\text{ s}$
D $40\text{ s}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Goods train (constant speed):

x_g = vt = 10t

Express train (from rest, uniform acceleration):

x_e = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.5)t^2 = 0.25t^2

Overtake when $x_e = x_g$ :

0.25t^2 = 10t

t(0.25t - 10) = 0 \Rightarrow t = 40\ \text{s} \; (t \neq 0)

Answer

D

Final answer

D

Detailed explanation

Background Concept

Overtaking means both objects are at the same position at the same time (measured from the same origin).

For motion with:

constant speed $v$ , displacement is

s = vt

uniform acceleration $a$ , displacement is

s = ut + \frac{1}{2}at^2

where $u$ is the initial speed at $t = 0$ .

Understanding the Question

At $t=0$ :

the goods train passes the station moving at constant speed $10\ \text{m s}^{-1}$ .
the express train is at the station (so it starts at the same position) and begins moving from rest with uniform acceleration $0.5\ \text{m s}^{-2}$ .

We want the time $t$ (after $t=0$ ) when the express train catches up, i.e. when their displacements from the station are equal.

Approach

Write an expression for displacement of the goods train as a function of $t$ .
Write an expression for displacement of the express train as a function of $t$ .
Set them equal to find the overtaking time.
Reject the trivial solution $t=0$ (they are together at the start but that is not “overtaking”).

Step-by-Step Reasoning

1) Goods train (constant speed $v=10\ \text{m s}^{-1}$ ):

x_g = vt = 10t

2) Express train (starts from rest, so $u=0$ , acceleration $a=0.5\ \text{m s}^{-2}$ ):

x_e = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.5)t^2 = 0.25t^2

3) Overtaking condition $x_e = x_g$ :

0.25t^2 = 10t

Divide/factor out $t$ :

t(0.25t - 10)=0

So $t=0$ or

0.25t = 10 \Rightarrow t = 40\ \text{s}

We take $t=40\ \text{s}$ because $t=0$ is just the starting moment.

Therefore the correct option is D.

Key Takeaways

Overtaking problems are solved by setting displacements equal.
Use $s=vt$ for constant speed and $s=ut+\tfrac{1}{2}at^2$ for uniform acceleration.
If you get $t=0$ as a solution, it is usually the trivial “start together” case and should be rejected.

Common Mistakes

Using $v = u + at$ instead of a displacement equation (you need positions to decide overtaking).
Forgetting that the express train starts from rest ( $u=0$ ).
Not rejecting $t=0$ and choosing the wrong option.

Things to Be Careful About

Both trains share the same starting position at the station at $t=0$ , so displacements must be measured from that point.
Keep units consistent: $v$ in $\text{m s}^{-1}$ , $a$ in $\text{m s}^{-2}$ , so $t$ comes out in seconds.
The accelerating train’s displacement grows as $t^2$ , so it will eventually catch up even if it starts slower.

Techniques used

write displacement-time expressions for each trainuse s = ut + 1/2 at^2 for uniformly accelerated motionequate displacements to find the overtaking timesolve the resulting quadratic equation

Q61MMedium-EasyKinematicsDynamics

A ball is held above the ground and released. It falls to the ground and bounces several times.

The graph shows the variation with time $t$ of the velocity $v$ of the ball.

Fig. 2

Four points on the graph are labelled $P$ , $Q$ , $R$ and $S$ .

Which statement is not correct?

Options

A The area under line $PQ$ represents the initial height of the ball.
B The collisions of the ball with the ground are inelastic.
C The gradient of the line $RS$ represents the acceleration due to free fall.
D The maximum upwards velocity of the ball is reached at point $P$ .

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

From the graph, at point $P$ the ball has $v=0$ (it is released from rest), so it is not moving upwards at $P$ .

The maximum upwards speed occurs just after a bounce, i.e. at the most negative velocity (point $R$ ), not at $P$ .

Answer

D

Final answer

D

Detailed explanation

Background Concept

For a velocity–time ( $v$ - $t$ ) graph:

The gradient is the acceleration:

a = \frac{\Delta v}{\Delta t}

The area under the graph between two times is the displacement:

s = \int v\, dt

The sign of $v$ indicates direction (positive one way, negative the opposite).

For collisions with the ground:

If the rebound speed is smaller than the impact speed, kinetic energy has decreased, so the collision is inelastic.

Understanding the Question

A ball is released, falls, hits the ground, and bounces repeatedly. The given $v$ - $t$ graph shows straight sloping sections (constant acceleration under gravity) and vertical jumps (very short collision time where velocity changes rapidly).

Points $P, Q, R, S$ are marked. We must choose which statement (A–D) is not correct.

Approach

Check each statement using standard $v$ - $t$ graph interpretations:

Use area under a section to judge a displacement/height.
Use gradient of a sloping section to identify the acceleration.
Use the pattern of decreasing speeds after each bounce to decide if collisions are elastic/inelastic.
Use the sign and magnitude of $v$ at labelled points to see where the maximum upward velocity occurs.

Step-by-Step Reasoning

A: Line $PQ$ is the initial fall from release to first impact. The area under $PQ$ is the displacement during this time. Since the ball starts above the ground and reaches the ground, this displacement corresponds to the initial height (with sign depending on the direction chosen). So A is correct.

B: After each bounce, the magnitude of the velocity just after impact is smaller than just before impact (the vertical jumps get smaller). That means the ball loses kinetic energy in each collision, so the collisions are inelastic. So B is correct.

C: Segment $RS$ is straight, so acceleration is constant. During free flight the only significant force is weight, giving acceleration $g$ (with sign depending on the chosen positive direction). Therefore the gradient of $RS$ represents the acceleration due to free fall. So C is correct.

D: At $P$ the ball is released from rest, so $v=0$ at $P$ . The maximum upward velocity occurs immediately after a bounce, where the velocity is upward (opposite sign to the downward motion). On the graph that is at the most negative value just after a collision, e.g. at $R$ , not at $P$ . Therefore D is not correct.

So the incorrect statement is D.

Key Takeaways

Area under a $v$ - $t$ graph gives displacement.
Gradient of a $v$ - $t$ graph gives acceleration.
Repeatedly decreasing rebound speeds indicate inelastic collisions.
The maximum upward velocity is at the point with greatest magnitude in the upward direction (not necessarily the first labelled point).

Common Mistakes

Thinking the area under $v$ - $t$ is distance regardless of sign; it is displacement (sign matters).
Confusing a vertical jump on a $v$ - $t$ graph with acceleration due to gravity; it represents the collision, not free fall.
Assuming point $P$ is the maximum of something because it is the start; here it is clearly $v=0$ .

Things to Be Careful About

Decide which direction is taken as positive by looking at which way the velocity is plotted during the initial fall.
Maximum upward velocity means the largest speed in the upward direction, i.e. the most negative $v$ if downward is taken as positive.
Inelastic vs elastic: look for loss of speed/energy from one bounce to the next (decreasing amplitude).

Techniques used

interpret the area under a velocity–time graph as displacementinterpret the gradient of a velocity–time graph as accelerationuse sign of velocity on a graph to determine direction of motioninfer inelastic collisions from decreasing rebound speeds

Q71MMedium-EasyKinematics

An object is projected horizontally. The object falls a vertical distance $y$ and travels a horizontal distance $x$ before landing on the ground.

Fig. 3

A second object is projected horizontally with the same initial velocity and falls a vertical distance $4y$ before landing on the ground.

Assume that air resistance is negligible.

Which horizontal distance does the second object travel?

Options

A $x$
B $2x$
C $4x$
D $16x$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Vertical motion (initial vertical velocity $=0$ ):

y = \frac{1}{2}gt^2

So $t \propto \sqrt{y}$ .

For the second object:

4y = \frac{1}{2}gt_2^2 \Rightarrow t_2 = 2t

Horizontal motion: $x = ut$ and $x_2 = ut_2$ (same $u$ ), hence

x_2 = u(2t) = 2x

Answer

B

Final answer

B

Detailed explanation

Background Concept

In projectile motion with negligible air resistance, the horizontal and vertical motions are independent:

Horizontally there is no acceleration, so the horizontal velocity stays constant at the initial value $u$ .
Vertically the object accelerates downward at constant acceleration $g$ .

For an object projected horizontally, the initial vertical component of velocity is zero, so the vertical displacement after time $t$ is

y = \frac{1}{2}gt^2

The horizontal displacement in the same time is

x = ut

Understanding the Question

The first object falls a vertical distance $y$ and travels a horizontal distance $x$ before hitting the ground.

A second object is projected horizontally with the same initial horizontal speed $u$ , but it falls a vertical distance $4y$ before landing. The question asks how the horizontal distance changes.

Because air resistance is negligible, the only acceleration is vertical ( $g$ ), so the key is how the time of flight depends on the vertical drop.

Approach

Use vertical motion to express the time of flight $t$ in terms of the vertical drop $y$ .
Replace $y$ by $4y$ to find the new time of flight $t_2$ .
Use horizontal motion $x = ut$ to scale the horizontal distance with time.

Step-by-Step Reasoning

For the first object, vertical motion starts from rest vertically ( $u_y = 0$ ), so

y = \frac{1}{2}gt^2

Rearranging gives

t = \sqrt{\frac{2y}{g}}

So time of flight is proportional to $\sqrt{y}$ .

For the second object the drop is $4y$ :

4y = \frac{1}{2}gt_2^2

Compare with the first equation (or solve explicitly):

t_2 = \sqrt{\frac{2(4y)}{g}} = \sqrt{4\frac{2y}{g}} = 2\sqrt{\frac{2y}{g}} = 2t

So the time of flight doubles.

Horizontally, the speed is constant and the same for both objects ( $u$ ). Therefore

x = ut, \qquad x_2 = ut_2

Substitute $t_2 = 2t$ :

x_2 = u(2t) = 2(ut) = 2x

So the second object travels twice the horizontal distance.

Key Takeaways

Treat projectile motion as two independent 1D motions.
For horizontal projection, vertical displacement obeys $y = \tfrac{1}{2}gt^2$ , so $t \propto \sqrt{y}$ .
With constant horizontal speed, horizontal distance is directly proportional to time.

Common Mistakes

Assuming horizontal distance scales like vertical distance (choosing $4x$ ) instead of scaling with $\sqrt{y}$ through time.
Using $y = gt^2$ (missing the factor $\tfrac{1}{2}$ ) and then making inconsistent comparisons.
Forgetting that the horizontal speed is constant only if air resistance is negligible.

Things to Be Careful About

The initial vertical velocity is zero only because the projection is horizontal.
The factor of $4$ in vertical distance leads to a factor of $\sqrt{4} = 2$ in time, not $4$ .
Keep clear which equation applies to which direction: vertical uses constant acceleration $g$ , horizontal uses $a_x = 0$ .

Techniques used

apply vertical motion under constant acceleration to find time of flightuse proportional reasoning with kinematic equationsuse horizontal motion at constant velocity to relate distance to time

Q81MMedium-EasyDynamics

A sphere falls from rest through the air. The graph shows the variation with time of the sphere’s velocity.

Fig. 4

Which diagram shows the forces acting on the sphere when it is at the velocity corresponding to point $P$ on the graph?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

At point $P$ , the velocity–time graph is still rising, so gradient $>0$ and the sphere is accelerating downwards.

So resultant force is downwards: $W > F_\text{drag}$ .

Answer

C

Final answer

C

Detailed explanation

Background Concept

When an object falls through air, two main vertical forces act:

Weight $W = mg$ acting downwards (approximately constant).
Air resistance (drag) $F_\text{drag}$ acting upwards (increases with speed).

Newton's second law links forces and motion:

\sum F = ma

So the direction of the acceleration is the same as the direction of the resultant (net) force.

On a velocity–time graph, the acceleration is the gradient:

a = \frac{\Delta v}{\Delta t}

Terminal velocity occurs when acceleration is zero (horizontal part of the graph), meaning the resultant force is zero and the two forces balance.

Understanding the Question

The sphere starts from rest and its velocity increases then levels off. Point $P$ is on the rising part of the velocity–time curve, before it becomes horizontal.

You are asked which force diagram matches the situation at that instant (at the speed corresponding to $P$ ).

Approach

Use the velocity–time graph at $P$ to decide whether acceleration is zero or non-zero, and its direction.
Use Newton's second law to deduce the direction of the resultant force.
Compare weight and drag to match the correct diagram.

Step-by-Step Reasoning

At point $P$ , the curve is still rising with time, so the gradient is positive.
- That means $a > 0$ in the direction of increasing velocity.
- Since the sphere is falling, increasing velocity means accelerating downwards.
If the sphere accelerates downwards, the resultant force must be downwards:

W - F_\text{drag} = ma > 0

So:

W > F_\text{drag}

Therefore the correct force diagram must show both forces present, with the downward weight arrow longer than the upward drag arrow.

diagram-1

This corresponds to option C.

Key Takeaways

Acceleration is the gradient of a $v$ – $t$ graph.
Non-zero acceleration means forces are unbalanced.
For a falling object before terminal velocity: $W > F_\text{drag}$ .

Common Mistakes

Choosing the balanced-forces diagram (terminal velocity) even though $P$ is not on the horizontal section.
Thinking air resistance is always equal to weight during a fall.
Using the value of velocity at $P$ instead of the gradient at $P$ .

Things to Be Careful About

Terminal velocity is identified by zero gradient (horizontal line), not merely by “large” velocity.
Drag acts opposite to motion, so for a falling object it acts upwards.
The sign/direction convention: focus on whether speed is still increasing (gradient) to infer the net force direction.

Techniques used

interpret acceleration from the gradient of a velocity–time graphapply Newton's second law to relate resultant force direction to accelerationcompare weight and drag to determine the resultant force

Q91MEasyDynamics

In which situation is total linear momentum always conserved?

Options

A in all collisions between two objects
B in collisions between two objects moving at equal and opposite velocity
C in collisions between two objects that form an isolated system
D in collisions between two objects with the same mass

Pick an answer

DifficultyEasy

Worked solution

Total linear momentum is conserved when the system is isolated, i.e. the resultant external force is zero.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Linear momentum $p$ is defined as

p = mv

For a system of particles,

\frac{\mathrm{d}p_{\text{total}}}{\mathrm{d}t} = F_{\text{external, resultant}}

So if the resultant external force on the system is zero, then

\frac{\mathrm{d}p_{\text{total}}}{\mathrm{d}t} = 0

and therefore $p_{\text{total}}$ is constant: total linear momentum is conserved. This is true for all interactions within the system (including all types of collision: elastic or inelastic), provided the system is isolated.

Understanding the Question

The question asks which collision situation always guarantees conservation of total linear momentum. The word “always” means we need a condition that is sufficient regardless of the details of the collision (masses, speeds, whether the collision is elastic/inelastic, etc.).

Approach

Use the conservation condition: momentum is conserved when there is no resultant external force on the two-object system. Then check each option to see whether it guarantees an isolated system.

Step-by-Step Reasoning

Option C: “two objects that form an isolated system.” By definition, an isolated system has zero resultant external force. Therefore total momentum is conserved. So C is correct.

Why the others are not “always” true:

A: Not all collisions occur in isolation; external forces (friction with the ground, air resistance, a hand holding one object, etc.) can act, so total momentum of the two-object system may change.
B: Equal and opposite velocities do not guarantee isolation; external forces can still act. (This condition relates to the initial momentum possibly being zero, not to whether it is conserved.)
D: Same mass also does not guarantee isolation; external forces can still act. Mass equality is irrelevant to whether momentum is conserved.

Key Takeaways

Momentum conservation is guaranteed by zero resultant external force on the system.
“Isolated system” is the key phrase that encodes this condition.
Conditions about equal masses or special initial velocities do not ensure conservation unless the system is isolated.

Common Mistakes

Thinking momentum is conserved in all collisions without checking for external forces.
Confusing “initial total momentum is zero” (e.g. equal/opposite velocities with equal masses) with “total momentum is conserved.”
Choosing options based on symmetry (same mass, opposite velocities) rather than the isolation condition.

Things to Be Careful About

Conservation of momentum applies to the chosen system: if you exclude external agents (e.g. Earth providing friction), momentum of just the two objects need not be conserved.
In exam wording, “isolated system” or “no external force” are the reliable indicators for momentum conservation.

Techniques used

use the condition for momentum conservation (zero resultant external force)identify whether the system is isolatedeliminate options that do not guarantee isolation

Q101MMediumDynamics

An object of mass $m$ moving with velocity $9.0\text{ m s}^{-1}$ has a head-on elastic collision with a stationary object of mass $2m$ .

After the collision, both objects are moving. No external forces act on the system.

What is the velocity of the object of mass $2m$ after the collision?

Options

A $-6.0\text{ m s}^{-1}$
B $-3.0\text{ m s}^{-1}$
C $4.5\text{ m s}^{-1}$
D $6.0\text{ m s}^{-1}$

Pick an answer

DifficultyMedium

Worked solution

Working

Let the final velocities be $v_1$ for mass $m$ and $v_2$ for mass $2m$ .

Momentum:

9m = mv_1 + 2mv_2 \Rightarrow 9 = v_1 + 2v_2

Kinetic energy (elastic):

\frac{1}{2}m(9.0)^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2

81 = v_1^2 + 2v_2^2

From $v_1 = 9 - 2v_2$ :

81 = (9-2v_2)^2 + 2v_2^2 = 81 - 36v_2 + 6v_2^2

0 = -36v_2 + 6v_2^2 \Rightarrow v_2( v_2 - 6) = 0

Since both move after collision, $v_2 \neq 0$ , so $v_2 = 6.0\ \text{m s}^{-1}$ .

Answer

D

Final answer

D

Detailed explanation

Background Concept

In a closed system with no external forces, linear momentum is conserved:

\sum p_\text{before} = \sum p_\text{after}

For 1D motion, momentum is $p = mv$ with a sign depending on direction.

For an elastic collision, kinetic energy is also conserved:

\sum E_{k,\text{before}} = \sum E_{k,\text{after}}

where $E_k = \tfrac{1}{2}mv^2$ .

In a head-on (1D) collision, these two conservation equations are enough to determine the two unknown final velocities.

Understanding the Question

Object 1: mass $m$ , initial velocity $u_1 = +9.0\ \text{m s}^{-1}$ .
Object 2: mass $2m$ , initially stationary, so $u_2 = 0$ .
Collision is head-on elastic and no external forces act, so momentum is conserved; “elastic” tells us kinetic energy is conserved.
We are asked for the final velocity of the object of mass $2m$ (call it $v_2$ ).

Approach

Write conservation of momentum for the system to relate $v_1$ and $v_2$ .
Write conservation of kinetic energy (because elastic collision) to give a second equation.
Solve the two simultaneous equations; reject any solution that contradicts “after the collision, both objects are moving”.
Match the computed value to the options.

Step-by-Step Reasoning

Let the final velocities be $v_1$ (mass $m$ ) and $v_2$ (mass $2m$ ).

1) Momentum conservation
Initial momentum:

p_\text{before} = m(9.0) + 2m(0) = 9m

Final momentum:

p_\text{after} = mv_1 + 2mv_2

Set equal:

9m = mv_1 + 2mv_2 \Rightarrow 9 = v_1 + 2v_2

So

v_1 = 9 - 2v_2

2) Kinetic energy conservation (elastic)
Initial kinetic energy:

E_{k,\text{before}} = \frac{1}{2}m(9.0)^2 = \frac{81}{2}m

Final kinetic energy:

E_{k,\text{after}} = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2

So:

\frac{81}{2}m = \frac{1}{2}mv_1^2 + mv_2^2

Divide by $\tfrac{1}{2}m$ :

81 = v_1^2 + 2v_2^2

3) Substitute from momentum equation

81 = (9-2v_2)^2 + 2v_2^2

Expand:

81 = (81 - 36v_2 + 4v_2^2) + 2v_2^2 = 81 - 36v_2 + 6v_2^2

Subtract $81$ :

0 = -36v_2 + 6v_2^2 = 6v_2(v_2 - 6)

So:

v_2 = 0 \quad \text{or} \quad v_2 = 6

But the question states both objects are moving after the collision, so $v_2 \neq 0$ .
Hence:

v_2 = +6.0\ \text{m s}^{-1}

Positive means it moves in the original direction of the $m$ mass.

This corresponds to option D.

Key Takeaways

“No external forces” (\Rightarrow) total momentum conserved.
“Elastic collision” (\Rightarrow) total kinetic energy conserved as well.
In 1D elastic collisions, use two conservation equations to solve for two unknown final velocities.

Common Mistakes

Using momentum conservation alone (insufficient to determine both final velocities).
Forgetting that elastic collisions conserve kinetic energy (not just momentum).
Sign errors: treating velocity as always positive rather than using direction.
Choosing the $v_2=0$ root even though the question explicitly says both objects move after collision.

Things to Be Careful About

Cancel the factor $m$ only after writing the full momentum/energy equations correctly.
Keep velocities as signed quantities throughout.
Check that the selected root is consistent with the physical statement given (here: both objects moving).

Techniques used

apply conservation of linear momentum in one dimensionapply conservation of kinetic energy for an elastic collisionsolve simultaneous equations for the final velocities

Q111MMedium-EasyDynamics

The graph shows how the momentum of a motorcycle changes with time.

Fig. 5

What is the resultant force on the motorcycle?

Options

A $500\text{ N}$
B $5000\text{ N}$
C $25\,000\text{ N}$
D $50\,000\text{ N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Resultant force:

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

From the graph, $p$ increases from $0$ to $5000\ \text{kg m s}^{-1}$ in $10\ \text{s}$ :

F = \frac{5000}{10} = 500\ \text{N}

Answer

A

Final answer

A

Detailed explanation

Background Concept

Momentum $p$ is defined as

p = mv

The resultant (net) force is the rate of change of momentum:

F = \frac{\mathrm{d}p}{\mathrm{d}t}

So on a momentum–time graph, the resultant force is given by the gradient (slope).

Unit check: momentum has unit $\text{kg m s}^{-1}$ . Dividing by time ( $\text{s}$ ) gives $\text{kg m s}^{-2} = \text{N}$ , so the gradient directly gives force in newtons.

Understanding the Question

You are given a straight-line graph of momentum $p$ against time $t$ for a motorcycle. The line goes from the origin to the point $t = 10\ \text{s}$ , $p = 5000\ \text{kg m s}^{-1}$ .

The question asks for the resultant force on the motorcycle, which is the rate at which its momentum is changing.

Approach

Use the relationship $F = \Delta p / \Delta t$ (gradient of a $p$ – $t$ graph).
Read two points on the straight line (best is the endpoints).
Compute the gradient and then choose the matching option.

Step-by-Step Reasoning

diagram-1

Take two convenient points on the straight line:
- At $t = 0\ \text{s}$ , $p = 0\ \text{kg m s}^{-1}$
- At $t = 10\ \text{s}$ , $p = 5000\ \text{kg m s}^{-1}$
Calculate the gradient:

\Delta p = 5000 - 0 = 5000\ \text{kg m s}^{-1}

\Delta t = 10 - 0 = 10\ \text{s}

F = \frac{\Delta p}{\Delta t} = \frac{5000}{10} = 500\ \text{N}

Compare with the options: $500\ \text{N}$ corresponds to option A.

Key Takeaways

On a momentum–time graph, gradient = resultant force.
Use $F = \Delta p / \Delta t$ for a straight line (constant force).
Always check units: $(\text{kg m s}^{-1})/(\text{s}) = \text{N}$ .

Common Mistakes

Using $F = p/t$ without taking a change (only works here because it starts from the origin; in general you need $\Delta p/\Delta t$ ).
Mixing up axes and calculating $\Delta t/\Delta p$ (the reciprocal), giving a tiny number.
Reading the momentum value incorrectly (e.g. $50\,000$ instead of $5000$ ) and picking an option 10 times too large.

Things to Be Careful About

Always take the slope as “rise over run”: $\Delta p$ (vertical) divided by $\Delta t$ (horizontal).
Use two well-separated points to reduce reading error; for a straight line, endpoints are best.
Include the unit newton ( $\text{N}$ ) or show the unit conversion implicitly via the gradient.

Techniques used

interpret a momentum–time graphcalculate the gradient of a straight-line graphapply $F = \mathrm{d}p/\mathrm{d}t$match a calculated value to a multiple-choice option

Q121MMedium-EasyDynamics

A rocket has a weight of $W$ and an initial acceleration of $a$ as the rocket leaves the ground vertically. The acceleration of free fall is $g$ .

Which expression gives the initial upward force exerted on the rocket due to the engine?

Options

A $\frac{Wa}{g} + W$
B $\frac{Wa}{g}$
C $\frac{Wa}{g} - W$
D $Wa + W$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Weight $W = mg \Rightarrow m = \frac{W}{g}$ .

Upward thrust $F$ and downward weight $W$ :

F - W = ma

F = ma + W = \frac{W}{g}a + W

Answer

A

Final answer

A

Detailed explanation

Background Concept

For motion in a straight line, Newton's second law states that the resultant force in the direction chosen equals mass (m) times acceleration (a):

\sum F = ma

Weight is the gravitational force on a mass:

W = mg

So if a question gives weight (W) rather than mass, you can replace (m) by (W/g).

Understanding the Question

The rocket rises vertically as it leaves the ground. There is an upward force from the engine (thrust) and a downward force due to weight (W). The rocket's initial acceleration is upward with magnitude (a). We need an expression for the engine thrust in terms of (W), (a), and (g).

Approach

Take upward as positive.
Write the resultant force: thrust (F) (up) minus weight (W) (down).
Apply (F - W = ma).
Replace (m) using (W = mg\Rightarrow m = W/g).
Simplify and compare with the options.

Step-by-Step Reasoning

Upward is positive. Forces:

Upward thrust = (F)
Downward weight = (W)

Newton's second law:

F - W = ma

But (W = mg), so:

m = \frac{W}{g}

Substitute into (F - W = ma):

F - W = \left(\frac{W}{g}\right)a

Add (W) to both sides:

F = \frac{Wa}{g} + W

This matches option A.

Key Takeaways

Always use resultant force: thrust is not equal to (ma) because weight is also acting.
Convert between weight and mass using (W = mg).
Choose a sign convention and stick to it.

Common Mistakes

Writing (F = ma) instead of (F - W = ma), forgetting weight contributes to the resultant.
Using (m = W) (mixing up mass and weight).
Incorrect sign: using (F + W = ma) when upward is taken as positive.

Things to Be Careful About

The acceleration (a) is upward; if you had chosen downward as positive, you would need (-a) and would still end with the same thrust after rearranging.
Option D ((Wa + W)) is dimensionally inconsistent because (Wa) has units (\text{N},\text{m s}^{-2}), not (\text{N}). The conversion (m = W/g) is essential to keep units consistent.

Techniques used

draw up the vertical force balance with thrust and weightapply Newton's second law to relate resultant force to accelerationsubstitute mass in terms of weight using m = W/gmatch the derived expression to the given multiple-choice options

Q131MMedium-EasyForces, Density and Pressure

A uniform rod of weight $28\text{ N}$ is supported at one end by force $F$ .

The rod is in equilibrium and attached to a frictionless hinge at end $X$ .

Fig. 6

What is the magnitude of $F$ ?

Options

A $9.0\text{ N}$
B $11\text{ N}$
C $18\text{ N}$
D $21\text{ N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Take moments about hinge at $X$ .

Moment of $F$ about $X$ :

\tau_F = FL

(since $F$ is perpendicular to the rod).

Moment of weight $28\,\text{N}$ about $X$ :

\tau_W = 28\left(\frac{L}{2}\cos 40^\circ\right)

(horizontal distance of midpoint from $X$ is $(L/2)\cos 40^\circ$ ).

Equilibrium: $\tau_F = \tau_W$

FL = 28\left(\frac{L}{2}\cos 40^\circ\right)

F = 14\cos 40^\circ = 10.7\,\text{N} \approx 11\,\text{N}

Answer

B

Final answer

B

Detailed explanation

Background Concept

For a rigid body in equilibrium:

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

The moment of a force about a point is

\tau = F \times d_\perp

where $d_\perp$ is the perpendicular distance from the point (pivot) to the force’s line of action. Equivalently, if the force $F$ acts at position vector $\vec{r}$ from the pivot,

\tau = rF\sin\theta

where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ .

Understanding the Question

A uniform rod (weight $28\,\text{N}$ ) is hinged at end $X$ and held in equilibrium by a force $F$ applied at the other end. The rod makes an angle of $40^\circ$ to the horizontal, and the force $F$ is perpendicular to the rod. The weight acts vertically downward at the midpoint (uniform rod).

We are asked for the magnitude of $F$ .

diagram-1

Approach

Choose the hinge at $X$ as the point to take moments about. This removes the unknown hinge reaction forces from the moment equation (they pass through $X$ so produce zero moment about $X$ ).

Then:

Calculate the anticlockwise moment due to $F$ .
Calculate the clockwise moment due to the weight $28\,\text{N}$ .
Set them equal (equilibrium) and solve for $F$ .

Step-by-Step Reasoning

Moment from $F$ about $X$

The force $F$ is applied at the end of the rod, a distance $L$ from the hinge, and is perpendicular to the rod. That means the angle between the position vector along the rod and the force is $90^\circ$ , so

\tau_F = rF\sin 90^\circ = LF.

So the magnitude of the moment from $F$ is simply $FL$ .

Moment from the weight about $X$

The weight $W = 28\,\text{N}$ acts at the midpoint, so its distance along the rod from the hinge is $L/2$ .

Because the weight is vertical, the perpendicular distance from $X$ to the vertical line of action of the weight is the horizontal component of the midpoint position:

d_\perp = \left(\frac{L}{2}\right)\cos 40^\circ.

Therefore the moment from the weight is

\tau_W = 28\left(\frac{L}{2}\cos 40^\circ\right).

Equilibrium condition

Set anticlockwise moment = clockwise moment:

FL = 28\left(\frac{L}{2}\cos 40^\circ\right).

Cancel $L$ :

F = 14\cos 40^\circ.

Numerically,

F = 14(0.766) = 10.7\,\text{N} \approx 11\,\text{N}.

So the correct option is B.

Key Takeaways

In equilibrium, the sum of moments about any point is zero.
Taking moments about the hinge is powerful because hinge forces produce no moment about that point.
The lever arm for a vertical force is the horizontal distance to its line of action.
If a force is perpendicular to the rod, its moment is simply $F \times$ (rod length).

Common Mistakes

Using $(L/2)\sin 40^\circ$ for the weight’s lever arm (that would be the vertical distance, not the perpendicular distance to a vertical line of action).
Forgetting the rod is uniform and so placing the weight at the end instead of the midpoint.
Not recognising that $F$ being perpendicular makes its moment $FL$ (and unnecessarily introducing extra trig).
Equating forces instead of moments (force balance alone is not enough here).

Things to Be Careful About

Always use the perpendicular distance to the line of action, not the distance along the rod.
Check the angle you use: the rod is at $40^\circ$ to the horizontal, so the horizontal component is multiplied by $\cos 40^\circ$ .
Quote the final answer to an appropriate precision matching the options (here $\approx 11\,\text{N}$ ).

Techniques used

apply the principle of moments about the hingeuse perpendicular distance to a force’s line of actionuse trigonometry to find a lever armequate clockwise and anticlockwise moments

Q141MMedium-EasyForces, Density and Pressure

Two forces, each of magnitude $F$ , act in opposite directions on a rod.

Fig. 7

Each force acts on the rod at a distance $d$ from the pivot $P$ .

What is the torque of this couple about $P$ ?

Options

A $0$
B $F \times d$
C $2F \times d$
D $2F \times 2d$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Moment about $P$ from left force $= Fd$ .

Moment about $P$ from right force $= Fd$ (same sense).

Total torque of the couple:

\tau = Fd + Fd = 2Fd

Answer

C

Final answer

C

Detailed explanation

Background Concept

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

\tau = F \times r_\perp

where $r_\perp$ is the perpendicular distance from the pivot to the line of action of the force.

A couple is formed by two equal, opposite, parallel forces with different lines of action. A couple produces a turning effect but no resultant force. The torque (moment) of a couple is

\tau_{\text{couple}} = F \times s

where $s$ is the perpendicular separation between the two lines of action.

Understanding the Question

Two forces of magnitude $F$ act vertically on a horizontal rod, one upwards on the left and one downwards on the right. The pivot $P$ is at the centre, and each force is applied a distance $d$ from $P$ . You are asked for the torque of the couple about $P$ .

Approach

Find the moment of each force about $P$ using $\tau = F \times d$ (here the perpendicular distance is $d$ for each force). Then add the two moments, noting that they produce rotation in the same sense (both tend to turn the rod the same way).

Equivalently, treat it as a couple: the separation between the lines of action is $2d$ , so the couple moment is $F \times 2d$ .

Step-by-Step Reasoning

diagram-14-1

Left force: Its line of action is a vertical line through the left application point. The perpendicular distance from $P$ to this line is $d$ .

\tau_\text{left} = Fd

Right force: Similarly, its perpendicular distance from $P$ is also $d$ .

\tau_\text{right} = Fd

Sense of rotation: Although the forces are opposite directions, they act on opposite sides of the pivot, so they both cause rotation in the same sense about $P$ . Therefore, the torques add:

\tau_{\text{total}} = Fd + Fd = 2Fd

So the correct option is $2F \times d$ , i.e. C.

(Quick couple method: separation between forces $= 2d$ , so $\tau = F(2d) = 2Fd$ .)

Key Takeaways

Torque about a point uses the perpendicular distance to the line of action.
In a couple, the two moments add to give $\tau = F \times (\text{separation})$ .
The moment of a couple is independent of the pivot position (as long as you sum both moments correctly).

Common Mistakes

Using $d$ as the separation between the forces instead of $2d$ .
Thinking the torques cancel because the forces are opposite; for a couple, the forces cancel but the moments add.
Using distance to the point of application instead of perpendicular distance to the line of action (important in non-right-angle cases).

Things to Be Careful About

Check the sense (clockwise/anticlockwise) of each torque about the pivot before adding.
Do not multiply by $2F$ ; the couple moment is not found by replacing the two forces with a single force $2F$ at some distance.
Here the lever arm is exactly $d$ for each force because the force is perpendicular to the rod; in general you may need $r\sin\theta$ .

Techniques used

use moment = force \times perpendicular distance from pivotadd moments from two forces with opposite senses about the same pivotuse couple moment = force \times perpendicular separation of lines of action

Q151MMedium-EasyForces, Density and PressurePhysical Quantities and Units

A man of weight $600\text{ N}$ stands with both feet flat on the ground.

What is a reasonable estimate of the pressure exerted on the ground by the weight of the man?

Options

A $1 \times 10^0\text{ Pa}$
B $1 \times 10^2\text{ Pa}$
C $1 \times 10^4\text{ Pa}$
D $1 \times 10^6\text{ Pa}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Pressure $p = \dfrac{F}{A}$ .

Estimate area of both feet on ground:

A \approx 2 \times (0.20 \text{ m} \times 0.10 \text{ m}) = 4.0 \times 10^{-2} \text{ m}^2

p \approx \frac{600}{4.0 \times 10^{-2}} = 1.5 \times 10^{4} \text{ Pa}

Closest option is $1 \times 10^{4} \text{ Pa}$ .

Answer

C

Final answer

C

Detailed explanation

Background Concept

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

p = \frac{F}{A}

$p$ is pressure in pascals (Pa), where $1\text{ Pa} = 1\text{ N m}^{-2}$ .
$F$ is the force pressing on the surface (here, the man’s weight).
$A$ is the contact area over which that force is distributed.

A key idea for “reasonable estimate” questions is that we do not need an exact foot area; we only need the correct order of magnitude.

Understanding the Question

We are told the man’s weight is $600\text{ N}$ . He is standing with both feet flat on the ground, so the ground supports a total downward force of $600\text{ N}$ .

We must estimate the pressure on the ground, so we need a reasonable estimate of the total area of contact of both feet, then compute $p = F/A$ and choose the closest power of ten from the options.

Approach

Use $p = F/A$ .
Estimate the area of one foot using typical dimensions (length (\sim 0.2\text{ m}), width (\sim 0.1\text{ m})).
Multiply by 2 for both feet.
Calculate $p$ and compare with the options.

Step-by-Step Reasoning

Take a typical footprint size:

length (\approx 0.20\text{ m})
width (\approx 0.10\text{ m})

So area of one foot:

A_\text{one} \approx 0.20 \times 0.10 = 2.0 \times 10^{-2}\ \text{m}^2

Both feet:

A \approx 2A_\text{one} = 4.0 \times 10^{-2}\ \text{m}^2

Now calculate the pressure:

p = \frac{F}{A} \approx \frac{600}{4.0 \times 10^{-2}} = 1.5 \times 10^{4}\ \text{Pa}

This is closest to $1 \times 10^{4}\ \text{Pa}$ , which corresponds to option C.

Key Takeaways

Use $p = F/A$ for pressure problems.
In estimation questions, pick sensible everyday dimensions and aim for the correct power of ten.
Remember $1\text{ Pa} = 1\text{ N m}^{-2}$ .

Common Mistakes

Forgetting to use both feet (using area of one foot only gives about twice the pressure).
Using unrealistic areas (e.g. $1\text{ m}^2$ gives $\sim 10^3\text{ Pa}$ ; far too large for feet).
Unit errors, especially mixing $\text{cm}^2$ with $\text{m}^2$ without converting.

Things to Be Careful About

The contact area is much smaller than the whole shoe outline if someone stands on heels, but the question states “both feet flat”, so using the full sole area is appropriate.
Converting areas: $1\text{ cm}^2 = 1\times 10^{-4}\text{ m}^2$ . A typical footprint of a few hundred $\text{cm}^2$ corresponds to a few $\times 10^{-2}\text{ m}^2$ .

Techniques used

apply the definition of pressure as force per unit areaestimate a typical contact area from everyday dimensionsuse order-of-magnitude comparison to select the closest option

Q161MEasyForces, Density and Pressure

The formula for the upthrust on a block of wood partially submerged in water is shown.

F = \rho gV

What do the symbols $\rho$ and $V$ represent?

Options

	$\rho$	$V$
A	density of water	volume of whole block
B	density of water	volume of block below surface of water
C	density of wood	volume of whole block
D	density of wood	volume of block below surface of water

Pick an answer

DifficultyEasy

Worked solution

Working

Upthrust equals the weight of displaced water:

F = \rho_{\text{water}} g V_{\text{displaced}}

For a partially submerged block, $V_{\text{displaced}}$ is the volume of the block below the water surface.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Upthrust (buoyant force) on an object in a fluid is given by Archimedes' principle: the upthrust equals the weight of the fluid displaced by the object.

If a volume $V$ of fluid of density $\rho$ is displaced, the mass of that displaced fluid is $m = \rho V$ , so its weight is $mg = \rho g V$ . Therefore the buoyant force is

F = \rho g V

Here, $\rho$ must be the density of the fluid providing the buoyant force (the surrounding liquid or gas), and $V$ must be the displaced volume of that fluid.

Understanding the Question

A block of wood is partially submerged in water. The question gives the standard upthrust formula

F = \rho g V

and asks what the symbols $\rho$ and $V$ represent. Since the fluid is water and the block is only partially submerged, the displaced volume is only the submerged part.

Approach

Use Archimedes' principle: upthrust is the weight of displaced fluid.
Identify which substance’s density appears in the formula (the displaced fluid, not the object).
For a partially submerged object, identify $V$ as the volume of the object below the water surface (the displaced water volume).

Step-by-Step Reasoning

The buoyant force depends on the properties of the fluid being displaced, because it is that fluid whose weight is being “replaced” by an upward force.
Therefore $\rho$ is the density of water (not the density of wood).
The volume of water displaced equals the volume of the object that is actually underwater.
- If the block is partially submerged, only the submerged part displaces water.
- So $V$ is the volume of the block below the surface of the water.
This matches option B.

Key Takeaways

In $F = \rho g V$ for upthrust, $\rho$ is the density of the surrounding fluid.
$V$ is the displaced volume of fluid, which equals the submerged volume of the object.
Partial submergence means $V$ is not the whole object volume.

Common Mistakes

Using the density of the object (wood) instead of the density of the fluid (water).
Taking $V$ as the total volume of the block even when it is only partially submerged.

Things to Be Careful About

The formula always refers to the displaced fluid: $\rho$ and $V$ belong to the displaced fluid, not to the object.
The displaced volume equals the submerged volume only when the object is in a single fluid and the part above the surface does not displace any fluid.

Techniques used

apply Archimedes' principle to relate upthrust to displaced fluididentify the relevant density as that of the surrounding fluididentify the relevant volume as the displaced (submerged) volume

Q171MMedium-EasyWork, Energy and Power

An object with a mass of $100\text{ g}$ falls a vertical distance of $10\text{ m}$ .

What is the change in the gravitational potential energy of the object?

Options

A $1\text{ J}$
B $10\text{ J}$
C $100\text{ J}$
D $10\,000\text{ J}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Mass $m = 100\text{ g} = 0.100\text{ kg}$ .

Change in gravitational potential energy:

\Delta E_p = mg\Delta h = 0.100 \times 9.8 \times 10 = 9.8\text{ J} \approx 10\text{ J}

Answer

B

Final answer

B

Detailed explanation

Background Concept

For an object in a uniform gravitational field near Earth’s surface, the change in gravitational potential energy is

\Delta E_p = mg\Delta h

where:

$m$ is the mass in $\text{kg}$ ,
$g$ is the gravitational field strength (about $9.8\,\text{m s}^{-2}$ ),
$\Delta h$ is the change in vertical height in $\text{m}$ .

If an object falls, its height decreases, so $\Delta h$ is negative and $\Delta E_p$ is negative (it loses GPE). In multiple-choice questions, they often want the magnitude of the change.

Understanding the Question

The object has mass $100\text{ g}$ and falls a vertical distance of $10\text{ m}$ . You are asked for the change in its gravitational potential energy. The options are numerical values in joules, so we calculate $mg\Delta h$ and choose the closest.

Approach

Convert the mass from grams to kilograms.
Use $\Delta E_p = mg\Delta h$ with $\Delta h = -10\text{ m}$ (falling) or use $10\text{ m}$ if taking the magnitude.
Compare the magnitude with the answer choices.

Step-by-Step Reasoning

Convert mass:

100\text{ g} = 100 \times 10^{-3}\text{ kg} = 0.100\text{ kg}

Compute the change in GPE (magnitude):

|\Delta E_p| = mg(10) = 0.100 \times 9.8 \times 10 = 9.8\text{ J}

$9.8\text{ J}$ rounds to $10\text{ J}$ , which corresponds to option B.

(If including sign: $\Delta E_p = -9.8\text{ J}$ because the object loses gravitational potential energy as it falls.)

Key Takeaways

Use $\Delta E_p = mg\Delta h$ for vertical height changes near Earth.
Always convert to SI units before substituting.
Falling means a decrease in GPE (negative change), but MCQs may effectively ask for the magnitude.

Common Mistakes

Using $m = 100\text{ kg}$ or $m = 100$ (forgetting to convert grams to kilograms).
Forgetting the factor of $g$ and doing $m\Delta h$ .
Choosing $100\text{ J}$ by multiplying $0.1 \times 10$ and missing $g$ .

Things to Be Careful About

Units: $m$ must be in $\text{kg}$ and $\Delta h$ in $\text{m}$ to get joules.
Sign convention: for a fall, $\Delta h < 0$ so $\Delta E_p < 0$ ; check whether the question/options are using magnitude.

Techniques used

convert given quantities into SI unitsapply the gravitational potential energy change equation $\Delta E_p = mg\Delta h$match the calculated value to the closest option

Q181MMedium-EasyWork, Energy and PowerDeformation of Solids

A bungee jumper jumps from a platform and is decelerated by an elastic bungee cord, as shown.

Fig. 8

When the jumper makes the jump, his initial gravitational potential energy relative to the ground is converted into his kinetic energy and into elastic potential energy in the cord.

At which part of the jump are all three types of energy non-zero?

Options

A on the platform before the jump
B on the way down before the cord has started to extend
C on the way down as he decelerates
D at the bottom of the jump when he is stationary

Pick an answer

DifficultyMedium-Easy

Worked solution

Gravitational potential energy is non-zero whenever the jumper is above the ground.

Kinetic energy is non-zero when the jumper is moving.

Elastic potential energy is non-zero only when the cord is stretched (extending).

All three are non-zero while the cord is stretching and the jumper is still moving downward (decelerating).

Answer

C

Final answer

C

Detailed explanation

Background Concept

Gravitational potential energy (GPE) is energy due to height in a gravitational field:

E_p = mgh

Kinetic energy (KE) is energy due to motion:

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

Elastic potential energy (EPE) is energy stored when an elastic cord (or spring) is stretched. It is zero when the cord is slack (not extended) and becomes non-zero once the cord starts to stretch.

In a bungee jump, energy is transferred from the gravitational store into kinetic and elastic stores as the jumper falls and the cord extends.

Understanding the Question

You are told that the jumper’s initial GPE (relative to the ground) is converted into KE and into EPE in the cord.

The question asks: at which stage of the jump are all three of these energies simultaneously non-zero?

So we need a stage where:

the jumper is still above the ground (so $E_p \neq 0$ ),
the jumper is moving (so $E_k \neq 0$ ),
the cord is stretched (so elastic energy $\neq 0$ ).

Approach

Check each option by matching it to the physical conditions:

Is the jumper moving? (controls KE)
Is the cord stretched? (controls EPE)
Is the jumper above the ground? (controls GPE)

The only stage that can have all three is when the jumper is moving and the cord has started to extend.

Step-by-Step Reasoning

A: on the platform before the jump
- Jumper is stationary $\Rightarrow v=0 \Rightarrow E_k = 0$ .
- Cord is slack (not stretched) $\Rightarrow$ EPE $=0$ .
- Not all three are non-zero.
B: on the way down before the cord has started to extend
- Jumper is moving $\Rightarrow E_k \neq 0$ .
- But cord has not started to extend $\Rightarrow$ EPE $=0$ .
- Not all three are non-zero.
C: on the way down as he decelerates
- Decelerating while moving downward means the speed is not yet zero $\Rightarrow E_k \neq 0$ .
- Deceleration is caused by the cord stretching, so the cord is extended $\Rightarrow$ EPE $\neq 0$ .
- The jumper is still above the ground $\Rightarrow E_p \neq 0$ .
- Therefore all three are non-zero here.
D: at the bottom of the jump when he is stationary
- Stationary $\Rightarrow v=0 \Rightarrow E_k = 0$ .
- Even though the cord is maximally stretched (EPE non-zero) and height above ground is still non-zero, KE is zero.

So the correct option is C.

Key Takeaways

$E_k$ is non-zero only when an object is moving.
Elastic potential energy is non-zero only when an elastic object is stretched (or compressed).
During bungee deceleration, the jumper is still moving while the cord is stretching, so both $E_k$ and elastic energy exist together (and GPE is still present).

Common Mistakes

Thinking that “decelerating” implies the speed is zero; it only means speed is decreasing.
Saying elastic potential energy exists before the cord starts to stretch (a slack cord stores no elastic energy).
Choosing the bottom point (stationary) and forgetting that KE is then zero.

Things to Be Careful About

The question defines GPE relative to the ground: unless the jumper is at ground level, $E_p$ is still non-zero.
Distinguish between “cord taut but not stretched” and “cord stretched”: elastic energy requires extension beyond the natural length.
“All three non-zero” forces you to find an interval where the jumper is both moving and stretching the cord, not a starting/ending instant.

Techniques used

identify which energy stores are present at different stages of motionlink non-zero elastic potential energy to extension of the cordlink non-zero kinetic energy to non-zero speedeliminate options by checking which energy term must be zero

Q191MEasyWork, Energy and Power

A sailboat is pushed by the wind at a constant velocity $v$ across a lake.

The force of the wind and the velocity of the sailboat are in the same direction.

Which additional information is required to determine the work done per unit time by the wind acting on the sailboat?

Options

A the force exerted by the wind
B the distance travelled by the sailboat per unit time
C the total energy input to the sailboat by the wind
D the weight of the sailboat

Pick an answer

DifficultyEasy

Worked solution

Working

Work done per unit time is power:

P = \frac{W}{t}

For a constant force in the same direction as the velocity,

P = Fv

The speed $v$ is already given, so we need $F$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

Power is the rate at which work is done (or energy is transferred):

P = \frac{W}{t}

When a force $F$ causes an object to move with velocity $v$ , the instantaneous power delivered by the force is

P = \vec{F} \cdot \vec{v} = Fv\cos\theta

where $\theta$ is the angle between the force and the velocity. If the force and velocity are in the same direction, $\theta = 0$ so $\cos\theta = 1$ and

P = Fv.

Understanding the Question

A sailboat moves at constant velocity $v$ across a lake. The wind exerts a force on the boat in the same direction as the boat’s velocity. The question asks what additional piece of information is needed to determine the work done per unit time by the wind, i.e. the power delivered by the wind.

Given: direction of $\vec{F}$ is the same as $\vec{v}$ , and the magnitude of velocity is $v$ .
Unknown: power $P$ .

Approach

Use the power relation for a force acting along the direction of motion:

P = Fv.

Then decide which quantity in this expression is missing from the information given.

Step-by-Step Reasoning

Because force and velocity are in the same direction,

P = \vec{F} \cdot \vec{v} = Fv.

The question already tells us the boat’s speed is $v$ (distance travelled per unit time is exactly the speed), so $v$ is not the missing information.

To calculate $P$ numerically we still need the value of the wind force $F$ .

Therefore the required additional information is: the force exerted by the wind (Option A).

Key Takeaways

“Work done per unit time” means power.
For a force parallel to motion, power is $P = Fv$ .
If $v$ is already known, the missing quantity is the force $F$ .

Common Mistakes

Choosing “distance travelled per unit time” (Option B): this is just the speed $v$ , which is already given.
Thinking the weight matters directly: weight acts vertically and does not determine the power delivered by a horizontal wind force.
Using $P = \frac{W}{t}$ but then trying to find $W$ from irrelevant energy information rather than using $P = Fv$ .

Things to Be Careful About

The formula $P = Fv$ only applies when the component of force along the velocity is used; more generally $P = Fv\cos\theta$ .
“Constant velocity” does not mean no forces act; it means the resultant force is zero. The wind force can still be doing work while resistive forces balance it.

Techniques used

use the definition of power as rate of doing workapply the relationship P = Fv for force parallel to velocityidentify which quantity is already provided and which is missing

Q201MMedium-EasyDeformation of Solids

A spring is fixed at one end. The length $L$ of the spring is increased by applying a tensile force $F$ to the other end.

The graph shows the variation of $L$ with $F$ .

Fig. 9

What is the elastic potential energy of the spring when $F$ is $6.0\text{ N}$ ?

Options

A $0.12\text{ J}$
B $0.24\text{ J}$
C $0.36\text{ J}$
D $0.60\text{ J}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

From the graph, at $F=0$ the spring length $L_0 = 8.0\,\text{cm}$ .

At $F=6.0\,\text{N}$ , $L = 12.0\,\text{cm}$ .

Extension:

x = L - L_0 = 4.0\,\text{cm} = 4.0 \times 10^{-2}\,\text{m}

Elastic potential energy (area under $F$ - $x$ graph):

E = \frac{1}{2}Fx = \frac{1}{2}(6.0)(4.0\times 10^{-2}) = 0.12\,\text{J}

Answer

A

Final answer

A

Detailed explanation

Background Concept

For a spring obeying Hooke's law, the force is proportional to extension:

F = kx

The elastic potential energy stored in the spring is the work done in stretching it from $x=0$ to some extension $x$ . On a force–extension graph, this equals the area under the graph:

E = \text{area under } F\text{–}x \text{ graph}

For a Hooke's-law spring, the $F$ – $x$ graph is a straight line from the origin, so the area is a triangle:

E = \frac{1}{2}Fx

Understanding the Question

You are given a graph of spring length $L$ (in cm) against applied force $F$ (in N). At $F=6.0\,\text{N}$ you must find the elastic potential energy stored.

The key point is that energy depends on the extension $x$ , not the total length $L$ . So you must first find the natural length (length at zero force) and then calculate the extension at $6.0\,\text{N}$ .

Approach

Read the spring length at $F=0$ to get the natural length $L_0$ .
Read the spring length at $F=6.0\,\text{N}$ to get $L$ .
Compute the extension $x = L - L_0$ and convert to metres.
Use $E = \frac{1}{2}Fx$ to find the elastic potential energy.
Match the numerical result to the options.

Step-by-Step Reasoning

From the graph:
- When $F=0$ , $L_0 = 8.0\,\text{cm}$ .
- When $F=6.0\,\text{N}$ , $L = 12.0\,\text{cm}$ .
Extension is the increase in length:

x = L - L_0 = 12.0 - 8.0 = 4.0\,\text{cm}

Convert to SI units:

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

Elastic potential energy is the work done in stretching. For a linear spring this is

E = \frac{1}{2}Fx

Substitute $F=6.0\,\text{N}$ and $x=4.0 \times 10^{-2}\,\text{m}$ :

E = \frac{1}{2}(6.0)(4.0\times 10^{-2}) = 0.12\,\text{J}

The option that matches $0.12\,\text{J}$ is A.

Key Takeaways

Elastic potential energy in a spring equals the work done stretching it.
For a Hooke's-law spring, $E = \frac{1}{2}Fx$ (triangle area under an $F$ – $x$ graph).
If a graph gives length $L$ rather than extension $x$ , first find $x = L - L_0$ .
Always convert cm to m when calculating energy in joules.

Common Mistakes

Using $L$ instead of extension $x$ in $E = \frac{1}{2}Fx$ .
Forgetting to subtract the natural length at $F=0$ .
Not converting $4.0\,\text{cm}$ to $0.040\,\text{m}$ , giving an answer 100 times too large.
Using $E = Fx$ (rectangle area) instead of $\frac{1}{2}Fx$ for a linear spring.

Things to Be Careful About

The graph is $L$ vs $F$ , but the energy relation requires $F$ vs $x$ ; here $x$ is found from the change in $L$ .
Units: $\text{N} \times \text{m} = \text{J}$ , so extension must be in metres.
Reading the intercept correctly: the value at $F=0$ is the spring's natural length.

Techniques used

read values from a graphdetermine extension from a length–force graphcalculate elastic potential energy as the area under a force–extension graphconvert centimetres to metresselect the correct multiple-choice option

Q211MMedium-EasyDeformation of Solids

A wire has an unstretched length of $2.00\text{ m}$ .

A stress of $1.6 \times 10^5\text{ Pa}$ is applied to the wire, and the new length of the wire is $2.10\text{ m}$ .

The wire obeys Hooke’s law.

What is the Young modulus of the wire?

Options

A $8.0 \times 10^3\text{ Pa}$
B $7.8 \times 10^4\text{ Pa}$
C $1.5 \times 10^5\text{ Pa}$
D $3.2 \times 10^6\text{ Pa}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Strain:

\text{strain} = \frac{\Delta L}{L_0} = \frac{2.10 - 2.00}{2.00} = \frac{0.10}{2.00} = 0.050

Young modulus:

E = \frac{\text{stress}}{\text{strain}} = \frac{1.6 \times 10^5}{0.050} = 3.2 \times 10^6\ \text{Pa}

Answer

D

Final answer

D

Detailed explanation

Background Concept

For a wire that obeys Hooke's law, stress is proportional to strain (as long as the wire is within its limit of proportionality).

Stress is force per unit cross-sectional area:

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

Strain is fractional extension (no units):

\text{strain} = \frac{\Delta L}{L_0}

Young modulus $E$ relates stress to strain:

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

It is a measure of stiffness: a larger $E$ means a smaller strain for the same applied stress.

Understanding the Question

You are given:

Original (unstretched) length $L_0 = 2.00\ \text{m}$
New length $L = 2.10\ \text{m}$ , so extension $\Delta L = 0.10\ \text{m}$
Applied stress $= 1.6 \times 10^5\ \text{Pa}$

You need to calculate the Young modulus and choose the matching option A–D.

Approach

Find the strain using $\Delta L / L_0$ .
Use $E = \text{stress} / \text{strain}$ .
Compare the calculated value with the listed options.

Step-by-Step Reasoning

Calculate extension:

\Delta L = 2.10 - 2.00 = 0.10\ \text{m}

Calculate strain:

\text{strain} = \frac{\Delta L}{L_0} = \frac{0.10}{2.00} = 0.050

This is dimensionless (no unit).

Calculate Young modulus:

E = \frac{\text{stress}}{\text{strain}} = \frac{1.6 \times 10^5\ \text{Pa}}{0.050}

Dividing by $0.050$ is multiplying by $20$ :

E = 1.6 \times 10^5 \times 20 = 3.2 \times 10^6\ \text{Pa}

This corresponds to option D.

Key Takeaways

Strain is a fractional extension: $\Delta L / L_0$ .
Young modulus is stress per strain: $E = \text{stress} / \text{strain}$ .
Strain has no units; Young modulus has the same units as stress (Pa).

Common Mistakes

Using $\Delta L / L$ instead of $\Delta L / L_0$ for strain.
Forgetting that strain is dimensionless and incorrectly adding units.
Subtracting lengths incorrectly (e.g. using $2.00 - 2.10$ and losing the sign).
Matching the wrong option due to an arithmetic slip when dividing by $0.05$ .

Things to Be Careful About

Keep enough significant figures during intermediate steps; here $0.050$ is important (not $0.5$ ).
Ensure the final unit is $\text{Pa}$ for Young modulus.
Hooke’s law condition signals that the linear relationship (and hence constant $E$ ) is valid for the data given.

Techniques used

use the definition of strain as extension divided by original lengthapply Young modulus equation as stress divided by strainselect the correct multiple-choice option by matching magnitude and unit

Q221MEasyDeformation of SolidsPhysical Quantities and Units

What are the SI base units of stress?

Options

A $\text{kg m s}^{-2}$
B $\text{kg m}^{-1}\text{ s}^{-2}$
C $\text{kg m}^{-2}\text{ s}^{-2}$
D $\text{kg m}^{-3}\text{ s}^{-2}$

Pick an answer

DifficultyEasy

Worked solution

Working

Stress $= \dfrac{F}{A}$ .

Unit of force $\text{N} = \text{kg m s}^{-2}$ .

So

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

Answer

B

Final answer

B

Detailed explanation

Background Concept

Stress is defined as the force distributed over an area:

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

It is a derived quantity, so its SI unit must be built from SI base units. The SI unit of force is the newton (N), and from Newton’s second law $F = ma$ we can express it in base units:

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

Understanding the Question

You are asked for the SI base units of stress (not just the named unit). The options all have the form $\text{kg m}^x\text{ s}^{-2}$ , so the task is to start from $\text{stress} = F/A$ and simplify.

Approach

Write stress as $F/A$ .
Replace $F$ by its base units ( $\text{kg m s}^{-2}$ ).
Divide by area ( $\text{m}^2$ ) and combine powers of $\text{m}$ .

Step-by-Step Reasoning

Start with the definition:

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

Base units:

Force: $F$ has unit $\text{N} = \text{kg m s}^{-2}$ .
Area: $A$ has unit $\text{m}^2$ .

So the unit of stress is:

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

Divide by $\text{m}^2$ (subtract indices of $\text{m}$ ):

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

This corresponds to option B.

Key Takeaways

Stress is force per unit area: $\text{stress} = F/A$ .
Convert derived units to base units using definitions (here $\text{N} = \text{kg m s}^{-2}$ ).
When dividing quantities with the same base unit, subtract the powers.

Common Mistakes

Using $\text{N m}^{-1}$ (that is force per length, not per area).
Forgetting that area is $\text{m}^2$ and only dividing by $\text{m}$ .
Leaving the answer as $\text{N m}^{-2}$ when asked specifically for SI base units.

Things to Be Careful About

The phrase “SI base units” means the answer must be in $\text{kg}$ , $\text{m}$ , $\text{s}$ (not N or Pa), even though stress is commonly measured in pascals, $\text{Pa} = \text{N m}^{-2}$ .
Combine powers correctly: $\text{m}/\text{m}^2 = \text{m}^{-1}$ .

Techniques used

use the definition of stress as force per unit areaexpress the newton in SI base unitssimplify unit indices when dividing by area

Q231MMedium-EasyDeformation of Solids

The diagram shows a force–extension graph for a rubber band as the band is extended and then the stretching force is decreased to zero.

Fig. 10

What can be deduced from the graph?

Options

A The rubber band does not return to its original length when the force is decreased to zero.
B The rubber band obeys Hooke’s law for the extensions shown.
C The rubber band remains elastic for the extensions shown.
D The shaded area represents the work done in extending the rubber band.

Pick an answer

DifficultyMedium-Easy

Worked solution

The unloading curve returns to the origin, so when the force is reduced to zero the extension is zero (original length recovered) (\Rightarrow) elastic behaviour.

The graph is not a straight line so Hooke’s law is not obeyed, and the shaded area is not the work done in extending (that is the area under the loading curve).

Answer

(\boxed{\text{C}})

Final answer

C

Detailed explanation

Background Concept

An object is elastic over a given range if it returns to its original dimensions when the applied force is removed. If, after unloading to zero force, there is still a non-zero extension, then the object has undergone plastic deformation.

Hooke’s law applies when force is proportional to extension:

F \propto x \quad \Rightarrow \quad F = kx

This shows up on a force–extension graph as a straight line through the origin.

The work done (energy transferred) in stretching a material from extension (0) to (x) is the area under the loading curve:

W = \int_0^x F\,\mathrm{d}x

If the loading and unloading curves are different (a hysteresis loop), the area enclosed between them represents energy dissipated (often as heating) in one loading–unloading cycle.

Understanding the Question

The graph shows two different curves: one for extending the rubber band (loading) and one for contracting it (unloading) back to zero force. The enclosed region is shaded.

You must decide which statement A–D is definitely supported by what the graph shows.

Approach

Check whether the unloading curve ends at the origin: that tells you whether the band returns to its original length when the force is removed (elastic vs plastic).
Check whether the loading curve is a straight line through the origin: that tells you whether Hooke’s law is obeyed.
Use the “area under curve” idea to decide what the shaded area represents.

Step-by-Step Reasoning

diagram-1

The unloading (contracting) curve returns to (F=0) at (x=0). Therefore, when the force is reduced to zero, the extension is zero: the rubber band returns to its original length. That is the definition of elastic behaviour, so statement C is true.
Statement A would require the unloading curve to reach (F=0) at some (x>0) (a permanent extension). The graph does not show this; it returns to (x=0). So A is not supported.
Statement B (Hooke’s law) would require a straight line relationship (F = kx). The loading curve is clearly curved, so Hooke’s law is not obeyed over the extensions shown.
Statement D is incorrect because:
- Work done in extending is the area under the loading curve (from (x=0) to the maximum extension).
- The shaded enclosed area between loading and unloading curves is the difference between work done loading and energy returned unloading, i.e. the energy dissipated.

Therefore the only correct deduction is C.

Key Takeaways

Elastic material returns to original length when force is removed (graph returns to origin).
Hooke’s law corresponds to a straight line through the origin on an (F)-(x) graph.
Area under an (F)-(x) curve is work done; area of a hysteresis loop is energy dissipated.

Common Mistakes

Assuming “rubber is not Hookean” without checking the graph shape; marks come from the graph evidence (non-linear curve).
Thinking the shaded loop area is the work done in extending; it is the energy lost in the cycle.
Confusing “elastic” with “obeys Hooke’s law”: an object can be elastic but not Hookean.

Things to Be Careful About

For statement A, you must look specifically at the extension when (F=0) on unloading.
For Hooke’s law, the key feature is linearity (straight line), not merely “passes through origin”.
Always distinguish:
- area under loading curve = work done to stretch,
- area under unloading curve = energy returned,
- enclosed area = energy dissipated.

Techniques used

interpret a force–extension graph during loading and unloadinguse the definition of elastic behaviour (returns to original length on unloading)identify Hooke’s law behaviour from linearity of the graphrelate work done to the area under a force–extension curve

Q241MEasyWavesSuperposition

Which row is correct for sound waves?

Options

	can travel in a vacuum	can be polarised	can diffract through a gap
A	no	no	yes
B	yes	no	no
C	no	yes	no
D	no	yes	yes

Pick an answer

DifficultyEasy

Worked solution

Working

Sound is a mechanical wave, so it cannot travel in a vacuum.

Sound in air is longitudinal, so it cannot be polarised.

All waves diffract when passing through a gap (especially when gap size is comparable to wavelength), so sound can diffract.

Answer

A

Final answer

A

Detailed explanation

Background Concept

Sound waves (in air and most fluids) are mechanical waves: they are vibrations of particles in the medium. This means they need matter to transmit the disturbance.

Sound in air is also longitudinal: particle oscillations are parallel to the direction the wave travels (compressions and rarefactions).

Polarisation is a property of transverse waves only: it means restricting the direction of oscillation to a particular plane/direction perpendicular to the direction of travel.

Diffraction is the spreading of a wave as it passes through a gap or around an obstacle. Any wave can diffract; it is most noticeable when the gap size is similar to the wavelength.

Understanding the Question

The table asks which statements are correct for sound waves:

Can sound travel in a vacuum?
Can sound be polarised?
Can sound diffract through a gap?

You must choose the row that matches the correct yes/no pattern for all three.

Approach

Decide each property from the nature of sound:

Mechanical wave (\Rightarrow) needs a medium (\Rightarrow) no vacuum propagation.
Longitudinal (in air) (\Rightarrow) cannot be polarised.
Waves spread at gaps (\Rightarrow) diffraction is possible.

Step-by-Step Reasoning

Vacuum travel: Sound needs particles to vibrate, so in a vacuum there are no particles to pass on the oscillation. Therefore: cannot travel in a vacuum (\Rightarrow) no.
Polarisation: To be polarised, the oscillations must be able to be confined to one direction perpendicular to travel. Longitudinal oscillations are along the direction of travel, so there is no transverse oscillation direction to “select”. Therefore: cannot be polarised (\Rightarrow) no.
Diffraction: Sound has a wavelength, so when it passes through a gap it can spread out into the region beyond the gap. Therefore: can diffract through a gap (\Rightarrow) yes.

So the correct row is no, no, yes, which is A.

Key Takeaways

Sound in air is a mechanical longitudinal wave.
Mechanical waves do not travel in a vacuum.
Only transverse waves can be polarised.
Diffraction is a general wave property (not restricted to light).

Common Mistakes

Thinking “all waves can be polarised” (false: only transverse waves).
Confusing diffraction with interference; diffraction is spreading at a gap/edge (though it can be explained using superposition).
Assuming diffraction only applies to light; it applies to sound as well.

Things to Be Careful About

The question is about sound waves (typically in air): treat them as longitudinal.
Diffraction is always possible in principle; it becomes significant when gap size (\sim \lambda).

Techniques used

identify sound as a mechanical longitudinal wavededuce whether propagation in a vacuum is possibleuse the condition for polarisation to decide if it appliesapply the wave property of diffraction through a gap

Q251MMediumWaves

A source emits a progressive sound wave in the horizontal direction. The wave travels away from the source in a direction towards the right.

The graph shows the variation of the displacement of the particles from their equilibrium positions with distance from the source at a particular instant in time. Displacements to the right of the equilibrium positions are shown as positive.

Fig. 11

Position $Y$ represents the displacement of a particle in the sound wave at a particular distance from the source at this instant.

Which statement about the motion of this particle is correct?

Options

A The particle is moving away from the source.
B The particle is moving towards the source.
C The particle is moving upwards.
D The particle is moving downwards.

Pick an answer

DifficultyMedium

Worked solution

Working

For a wave travelling to the right, displacement $s(x,t)=f(x-vt)$ , so

\frac{\partial s}{\partial t}=-v\frac{\partial s}{\partial x}.

At $Y$ the displacement–distance graph is rising, so $\partial s/\partial x>0$ .
Hence $\partial s/\partial t<0$ , meaning the particle velocity is to the left (towards the source).

Answer

B

Final answer

B

Detailed explanation

Background Concept

In a progressive longitudinal sound wave, particles of the medium oscillate back-and-forth along the same line that the wave travels.

A key idea is that the wave travels, but individual particles only oscillate about their equilibrium positions.

For a wave travelling to the right with speed $v$ , the displacement can be written as

s(x,t)=f(x-vt).

Differentiating gives the particle velocity (the speed/direction the particle itself is moving at that instant):

\frac{\partial s}{\partial t}=-v\frac{\partial s}{\partial x}.

So, for a right-travelling wave:

if the graph of $s$ against $x$ has positive slope ( $\partial s/\partial x>0$ ), then the particle velocity is negative ( $\partial s/\partial t<0$ ),
if the slope is negative, the particle velocity is positive.

Understanding the Question

You are given a snapshot graph of displacement $s$ vs distance $x$ at one instant.

The wave is travelling to the right.
Positive displacement means the particle is displaced to the right of equilibrium.
Point $Y$ lies on the rising part of the curve after a trough, still below the axis.

The question asks: at that instant, is the particle at $Y$ moving left/right (towards/away from the source) or up/down?

Because this is a sound wave in the horizontal direction, “upwards/downwards” should already feel suspicious: particle motion is horizontal.

Approach

Use the fact the wave travels to the right to connect particle velocity $\partial s/\partial t$ to the slope of the snapshot $\partial s/\partial x$ .
Read the sign of the slope at $Y$ .
Convert the sign of $\partial s/\partial t$ into a physical direction using the question’s sign convention.

Step-by-Step Reasoning

At point $Y$ , the curve is rising as $x$ increases, so

\frac{\partial s}{\partial x} > 0.

For a right-travelling wave,

\frac{\partial s}{\partial t}=-v\frac{\partial s}{\partial x}.

Since $v>0$ and $\partial s/\partial x>0$ , it follows that

\frac{\partial s}{\partial t} < 0.

A negative $\partial s/\partial t$ means the displacement $s$ is becoming more negative with time: the particle is moving in the negative direction.

The question states positive displacement is to the right, so negative velocity means the particle is moving to the left, i.e. towards the source.

Therefore, the correct statement is B.

Key Takeaways

Sound waves are longitudinal: particle motion is along the direction of wave travel.
A snapshot $s$ vs $x$ graph does not directly show particle direction of motion, but its slope does when combined with the travel direction.
For a right-travelling wave: $\partial s/\partial t=-v\,\partial s/\partial x$ (opposite signs).

Common Mistakes

Saying the particle moves right because the wave travels right (confusing wave propagation with particle motion).
Using the sign of displacement at $Y$ (below axis) to decide direction of motion; displacement sign is not the same as velocity sign.
Choosing C or D (up/down), forgetting a sound wave’s particle oscillations are along the direction of propagation.

Things to Be Careful About

The graph is displacement vs distance, not displacement vs time. Motion direction comes from $\partial s/\partial t$ , not directly from the shape alone.
Always use the given sign convention: here, “positive” means “to the right”.
The relation $\partial s/\partial t=-v\,\partial s/\partial x$ is specifically for a wave travelling to the right; travelling left would reverse the sign.

Techniques used

use the right-travelling wave form to relate particle velocity to the spatial gradientdetermine the sign of the particle velocity from the slope of the displacement–distance graphmap the sign of displacement/velocity onto physical direction using the given sign convention

Q261MMedium-EasyWaves

Two polarising filters are placed next to each other so that their planes are parallel.

The first polarising filter has its transmission axis at an angle of $50^\circ$ to the vertical.

The second polarising filter has its transmission axis at an angle of $20^\circ$ to the vertical. The angle between the transmission axes of the two polarising filters is $30^\circ$ .

A beam of vertically polarised light of intensity $8.0\text{ W m}^{-2}$ is incident normally on the first polarising filter.

What is the intensity of the light that is transmitted from the second polarising filter?

Options

A zero
B $2.5\text{ W m}^{-2}$
C $2.9\text{ W m}^{-2}$
D $6.0\text{ W m}^{-2}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For the first polariser, angle to vertical (incident polarisation) is $50^\circ$ :

I_1 = I_0 \cos^2 50^\circ = 8.0\cos^2 50^\circ = 3.30\ \text{W m}^{-2}

Angle between the two transmission axes is $30^\circ$ :

I_2 = I_1 \cos^2 30^\circ = 3.30\cos^2 30^\circ = 2.48\ \text{W m}^{-2}

Answer

$2.5\ \text{W m}^{-2}$ (B)

Final answer

B

Detailed explanation

Background Concept

A polarising filter (polariser) only transmits the component of the electric field parallel to its transmission axis. For plane-polarised light incident on a polariser, the transmitted intensity is given by Malus's law:

I = I_0 \cos^2 \theta

where:

$I_0$ is the incident intensity,
$I$ is the transmitted intensity,
$\theta$ is the angle between the incident light's plane of polarisation and the polariser's transmission axis.

When light passes through a first polariser, the transmitted light becomes polarised along the first polariser's axis. A second polariser then further reduces intensity depending on the angle between the two polariser axes.

Understanding the Question

The incident light is vertically polarised with intensity $I_0 = 8.0\ \text{W m}^{-2}$ .
First polariser axis is at $50^\circ$ to the vertical, so the incident polarisation is at $50^\circ$ to its axis.
Second polariser axis is at $20^\circ$ to the vertical, so the angle between axes is $|50^\circ-20^\circ|=30^\circ$ (also stated).
We want the intensity after the second polariser.

Approach

Use Malus's law for transmission through the first polariser using $\theta = 50^\circ$ .
After the first polariser, the light is polarised along the first axis.
Use Malus's law again for the second polariser using the angle between axes, $30^\circ$ .

Step-by-Step Reasoning

1) Through the first polariser
The incoming polarisation is vertical. The first transmission axis is $50^\circ$ to vertical, so $\theta_1 = 50^\circ$ .

I_1 = I_0 \cos^2 \theta_1 = 8.0\cos^2 50^\circ

Calculate:

$\cos 50^\circ \approx 0.643$
$\cos^2 50^\circ \approx 0.413$

So:

I_1 \approx 8.0 \times 0.413 = 3.30\ \text{W m}^{-2}

2) Through the second polariser
After the first polariser, the transmitted light is polarised along the first axis. The angle between the two axes is $\theta_2 = 30^\circ$ .

I_2 = I_1 \cos^2 \theta_2 = 3.30\cos^2 30^\circ

Since $\cos 30^\circ = 0.866$ and $\cos^2 30^\circ = 0.75$ :

I_2 = 3.30 \times 0.75 = 2.48\ \text{W m}^{-2} \approx 2.5\ \text{W m}^{-2}

This matches option B.

Key Takeaways

Use Malus's law $I = I_0\cos^2\theta$ for polarised light through a polariser.
After the first polariser, the polarisation direction becomes aligned with the first polariser's axis.
For multiple polarisers, apply Malus's law successively using the relevant angles.

Common Mistakes

Using $\theta = 30^\circ$ for the first polariser instead of $50^\circ$ (the first polariser compares to the incident vertical polarisation).
Forgetting to apply Malus's law twice (stopping after the first polariser).
Using $\cos \theta$ instead of $\cos^2 \theta$ .
Treating intensities as adding or subtracting rather than multiplying by the $\cos^2$ factors.

Things to Be Careful About

Always identify what the angle is between (polarisation direction vs transmission axis).
Keep track that the first polariser sets the new polarisation direction.
Ensure the final value is quoted with appropriate rounding to match the options.

Techniques used

determine the angle between the incident polarisation direction and the transmission axisapply Malus's law to find transmitted intensity through a polariserapply Malus's law successively through two polarisersuse the given angle between transmission axes

Q271MMedium-EasyWaves

A dolphin is swimming at a speed of $16.0\text{ m s}^{-1}$ directly towards a stationary underwater microphone.

The dolphin emits a sound of frequency $122\text{ kHz}$ . The speed of sound in water is $1480\text{ m s}^{-1}$ .

What is the frequency of sound detected by the microphone?

Options

A $1300\text{ Hz}$
B $1330\text{ Hz}$
C $121\text{ kHz}$
D $123\text{ kHz}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a source moving towards a stationary detector:

f' = f\frac{v}{v - v_s}

f' = 122\,\text{kHz}\times \frac{1480}{1480-16.0}

f' = 122\,\text{kHz}\times \frac{1480}{1464} \approx 1.23\times 10^2\,\text{kHz} \approx 123\,\text{kHz}

Answer

D

Final answer

D

Detailed explanation

Background Concept

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

For sound in a medium (water here), the wave speed in the medium is $v$ . If the source moves towards a stationary observer, successive wavefronts are emitted from positions closer and closer to the observer, so the wavelength in front of the source decreases and the observed frequency increases.

For a moving source and stationary observer:

f' = f\frac{v}{v-v_s}

where:

$f$ is the emitted frequency,
$f'$ is the detected frequency,
$v$ is the speed of sound in the medium,
$v_s$ is the speed of the source towards the observer.

Understanding the Question

The dolphin (source) swims directly towards the microphone (observer).
Microphone is stationary.
Given: $v_s = 16.0\,\text{m s}^{-1}$ , $f=122\,\text{kHz}$ , speed of sound $v=1480\,\text{m s}^{-1}$ .
Need: the detected frequency $f'$ .

Because the dolphin is moving towards the microphone, the detected frequency must be slightly higher than $122\,\text{kHz}$ .

Approach

Use the Doppler formula for a moving source and stationary observer. Substitute the given values, compute the factor $\dfrac{v}{v-v_s}$ , and then choose the option that matches the calculated frequency.

Step-by-Step Reasoning

Start with:

f' = f\frac{v}{v-v_s}

Substitute $f=122\,\text{kHz}$ , $v=1480\,\text{m s}^{-1}$ , $v_s=16.0\,\text{m s}^{-1}$ :

f' = 122\,\text{kHz}\times \frac{1480}{1480-16.0}

Compute the denominator:

1480-16.0 = 1464\,\text{m s}^{-1}

So:

f' = 122\,\text{kHz}\times \frac{1480}{1464}

The ratio is slightly bigger than 1:

\frac{1480}{1464} \approx 1.01093

Thus:

f' \approx 122\,\text{kHz}\times 1.01093 \approx 123.3\,\text{kHz}

Rounded to the nearest kHz (as in the options): $f' \approx 123\,\text{kHz}$ , which is option D.

Key Takeaways

For a source moving towards a stationary detector, frequency increases: use $f' = f\dfrac{v}{v-v_s}$ .
Always check the direction: towards means higher $f'$ , away means lower $f'$ .
A quick reasonableness check helps: here $v_s \ll v$ , so the shift is small (about 1%).

Common Mistakes

Using the wrong sign: writing $v+v_s$ in the denominator gives a smaller frequency (that would correspond to moving away).
Using the formula for a moving observer instead of a moving source.
Unit confusion: mixing Hz and kHz or choosing the 1300 Hz options, which are the wrong order of magnitude.

Things to Be Careful About

The source speed must be along the line joining source and detector (it is: “directly towards”).
Keep $v$ and $v_s$ in the same units (both are in $\text{m s}^{-1}$ here).
When matching options, round appropriately: $123.3\,\text{kHz}$ corresponds to $123\,\text{kHz}$ in the list.

Techniques used

select the Doppler-shift equation for a moving source and stationary observersubstitute given speeds with consistent unitsround the result to an appropriate number of significant figuresmatch the calculated value to the closest option

Q281MMedium-EasySuperposition

A student carries out a double-slit experiment using a laser emitting red light of wavelength of $680\text{ nm}$ . The light is incident normally on a double slit.

The diagram shows part of the pattern of bright fringes visible on a screen at a distance of $2.4\text{ m}$ from the slits. The distance across five bright fringes is measured as $34\text{ mm}$ .

Fig. 12

What is the slit separation?

Options

A $8.5 \times 10^{-3}\text{ m}$
B $2.4 \times 10^{-4}\text{ m}$
C $1.9 \times 10^{-4}\text{ m}$
D $4.8 \times 10^{-5}\text{ m}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Distance from 1st to 5th bright fringe is $34\ \text{mm}$ , i.e. $4$ fringe spacings.

x = \frac{34\ \text{mm}}{4} = 8.5\ \text{mm} = 8.5 \times 10^{-3}\ \text{m}

Double-slit:

x = \frac{\lambda D}{a} \;\Rightarrow\; a = \frac{\lambda D}{x}

a = \frac{(680\times 10^{-9})(2.4)}{8.5\times 10^{-3}} = 1.92\times 10^{-4}\ \text{m}

Answer

C

Final answer

C

Detailed explanation

Background Concept

In a double-slit experiment, the bright fringes are equally spaced (for small angles). The spacing $x$ between adjacent bright fringes on a screen a distance $D$ away is related to the slit separation $a$ and wavelength $\lambda$ by

x = \frac{\lambda D}{a}

This comes from the condition for maxima (constructive interference) $a\sin\theta = n\lambda$ and the small-angle approximations $\sin\theta \approx \tan\theta$ with $y \approx D\tan\theta$ .

Understanding the Question

You are given:

wavelength $\lambda = 680\ \text{nm}$ ,
screen distance $D = 2.4\ \text{m}$ ,
the measured distance across five bright fringes is $34\ \text{mm}$ (the diagram indicates this is from the 1st to the 5th bright fringe).

The unknown is the slit separation $a$ .

Approach

Convert the “distance across five bright fringes” into the spacing $x$ between adjacent bright fringes.
Convert all quantities to SI units.
Use $x = \lambda D / a$ and rearrange to find $a$ .
Match the numerical result to one of the options.

Step-by-Step Reasoning

1) Interpret the 34 mm measurement
From the 1st to the 5th bright fringe there are 4 equal gaps (spacings):

34\ \text{mm} = 4x \quad \Rightarrow \quad x = \frac{34}{4}\ \text{mm} = 8.5\ \text{mm}

Convert to metres:

x = 8.5\ \text{mm} = 8.5\times 10^{-3}\ \text{m}

2) Convert wavelength to metres

\lambda = 680\ \text{nm} = 680\times 10^{-9}\ \text{m}

3) Use the double-slit relation
Rearrange:

x = \frac{\lambda D}{a} \Rightarrow a = \frac{\lambda D}{x}

Substitute:

a = \frac{(680\times 10^{-9})(2.4)}{8.5\times 10^{-3}} = 1.92\times 10^{-4}\ \text{m}

Rounded to the options, this is $1.9\times 10^{-4}\ \text{m}$ , which corresponds to C.

Key Takeaways

For double-slit interference, fringe spacing satisfies $x = \lambda D/a$ .
A measurement “from the 1st to the 5th fringe” represents 4 spacings, not 5.
Always convert nm and mm into metres before substitution.

Common Mistakes

Taking $34\ \text{mm}$ as $5x$ instead of $4x$ .
Forgetting unit conversions (e.g. using $680$ instead of $680\times 10^{-9}$ ).
Rearranging incorrectly (e.g. $a = x/(\lambda D)$ instead of $a = \lambda D/x$ ).

Things to Be Careful About

The wording/diagram matters: “across five bright fringes” can be ambiguous, but the diagram here indicates from the 1st to the 5th fringe, i.e. 4 intervals.
Keep consistent SI units so the final $a$ comes out in metres.
Use appropriate significant figures to match the options (here, 2 s.f.).

Techniques used

interpret a multi-fringe measurement to find the fringe spacingapply the double-slit fringe-spacing equationconvert between nanometres, millimetres and metresrearrange an equation to solve for slit separationselect the matching multiple-choice option

Q291MMedium-EasySuperposition

The diagram shows a stationary wave on a stretched spring at an instant in time.

Fig. 13

Two particles on the spring, $P$ and $Q$ , are shown.

Which statement about the vibrations of $P$ and $Q$ is correct?

Options

A They have different frequencies.
B They have the same amplitudes.
C They have different periods.
D They are always in phase.

Pick an answer

DifficultyMedium-Easy

Worked solution

In a stationary wave, all particles oscillate with the same frequency (so same period), but with amplitudes that depend on position.

Points within the same segment between adjacent nodes oscillate in phase.

$P$ and $Q$ are on the same loop (same side of a node), so they are always in phase.

Answer

D

Final answer

D

Detailed explanation

Background Concept

A stationary wave is formed by the superposition of two progressive waves of the same frequency and amplitude travelling in opposite directions.

Key properties:

Every point on the medium oscillates with the same frequency $f$ (therefore the same period $T = 1/f$ ).
The amplitude depends on position: it is zero at nodes and maximum at antinodes.
All points between the same two adjacent nodes oscillate in phase (reach maxima, minima, and cross equilibrium together). Points in neighbouring loops (separated by a node) are in antiphase.

Understanding the Question

You are shown a snapshot of a stationary wave on a stretched spring. Two points are marked:

$P$ is at the top of a crest (an antinode position, maximum displacement in that loop at this instant).
$Q$ is on the side of the same crest, above the equilibrium line.

The question asks which statement about their vibrations is correct (frequency/period/amplitude/phase).

Approach

Use the standard stationary-wave rules:

Decide whether $P$ and $Q$ are in the same loop (same region between two nodes) or separated by a node.
Use that to decide the phase relationship.
Use stationary-wave facts to eliminate options about frequency/period/amplitude.

Step-by-Step Reasoning

diagram-1

Frequency/period: In a stationary wave, the entire pattern is produced by waves of one frequency, so every particle oscillates with the same $f$ and hence the same period $T=1/f$ .

So A (different frequencies) is false.
So C (different periods) is false.

Amplitude: Amplitude varies with position. $P$ is at the crest (an antinode position) so it has a larger amplitude than a point like $Q$ that is closer to a node than the antinode.

So B (same amplitudes) is false.

Phase: Points between the same pair of nodes are always in phase. Since $P$ and $Q$ are on the same crest (same loop), they oscillate together.

Therefore D is correct.

Key Takeaways

In stationary waves, all points have the same frequency (and period).
Amplitude depends on position: node $\to$ zero amplitude, antinode $\to$ maximum amplitude.
Phase is constant within one loop: same loop in phase, adjacent loops in antiphase.

Common Mistakes

Thinking the snapshot displacement tells you the amplitude directly (amplitude is the maximum possible displacement over time, not the displacement in one instant).
Saying points with different amplitudes must have different frequencies/periods (they do not in a stationary wave).
Mixing up “in phase” with “same displacement”: points can be in phase even when their instantaneous displacements differ.

Things to Be Careful About

Check whether the two points lie on the same side of a node. A node is where the stationary-wave curve crosses the equilibrium line and stays at zero amplitude.
Remember: in-phase means same stage of motion (maxima/minima/zero crossings at the same time), not necessarily equal displacement at every instant.

Techniques used

recall properties of stationary wavescompare amplitudes at different positions along a stationary wavedetermine relative phase of points within the same loop of a stationary waveuse the equivalence of frequency and period for all points in a stationary wave

Q301MMedium-EasySuperposition

Green laser light passes through a diffraction grating and forms an interference pattern.

The diffraction grating contains $400$ lines per mm.

The wavelength of the laser light is $550\text{ nm}$ .

What is the highest order diffraction maximum produced by the grating?

Options

A $4$
B $5$
C $8$
D $9$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Lines per metre:

400\ \text{mm}^{-1} = 400 \times 10^{3}\ \text{m}^{-1} = 4.0 \times 10^{5}\ \text{m}^{-1}

Grating spacing:

d = \frac{1}{4.0 \times 10^{5}} = 2.5 \times 10^{-6}\ \text{m}

Using

d\sin\theta = n\lambda

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

n_{\max} \le \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{550 \times 10^{-9}} \approx 4.55

So $n_{\max} = 4$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

A diffraction grating produces bright maxima (principal maxima) when light from adjacent slits arrives in phase. The condition for the $n$ th order maximum is

d\sin\theta = n\lambda

where:

$d$ is the slit spacing (grating spacing),
$\theta$ is the angle to the maximum (from the normal),
$n$ is the order number ( $n = 0, 1, 2, \dots$ ),
$\lambda$ is the wavelength.

A key physical constraint is that $\sin\theta$ cannot exceed 1, so there is a largest possible integer value of $n$ .

Understanding the Question

You are told:

grating has $400$ lines per mm (this is the line density),
wavelength $\lambda = 550\ \text{nm}$ .

You must find the highest order maximum that can exist. That means: compute $d$ from the line density, then find the largest integer $n$ such that $n\lambda \le d$ .

Approach

Convert the line density into lines per metre.
Use $d = 1/(\text{lines per metre})$ .
Use $d\sin\theta = n\lambda$ and the limit $\sin\theta \le 1$ to find $n \le d/\lambda$ .
Take the greatest integer not exceeding $d/\lambda$ and match it to the options.

Step-by-Step Reasoning

Convert $400\ \text{mm}^{-1}$ to per metre:

400\ \text{mm}^{-1} = 400 \times 10^3\ \text{m}^{-1} = 4.0 \times 10^5\ \text{m}^{-1}

So the spacing between adjacent lines is

d = \frac{1}{4.0 \times 10^5} = 2.5 \times 10^{-6}\ \text{m}

The grating equation gives

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

For a real angle $\theta$ , we must have $\sin\theta \le 1$ , so

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

Substitute $\lambda = 550\ \text{nm} = 550\times 10^{-9}\ \text{m}$ :

\frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{550 \times 10^{-9}} \approx 4.55

$n$ must be an integer, so the highest possible order is

n_{\max} = 4

This corresponds to option A.

Key Takeaways

Convert line density to spacing using $d = 1/N$ (with $N$ in lines per metre).
Highest observable order is limited by $\sin\theta \le 1$ , giving $n_{\max} = \lfloor d/\lambda \rfloor$ .

Common Mistakes

Not converting $\text{mm}^{-1}$ to $\text{m}^{-1}$ (missing the factor of $10^3$ ).
Using $d = 400\times 10^{-3}$ or similar incorrect inversions.
Rounding $4.55$ up to $5$ (cannot, because that would require $\sin\theta > 1$ ).

Things to Be Careful About

Keep powers of ten consistent: $\text{nm} = 10^{-9}\ \text{m}$ .
The maximum order is the largest integer that keeps $\sin\theta \le 1$ , so always round down.

Techniques used

convert grating line density into slit spacingapply the diffraction grating equationuse the condition sin(theta) \le 1 to find the maximum possible orderselect the nearest lower integer order

Q311MMedium-EasyElectricity

There is a potential difference $V$ across a resistor of resistance $R$ . The current in the resistor is $I$ .

Which equation gives the power $P$ dissipated by the resistor?

Options

A $P = V^2 I$
B $P = \frac{V^2}{I}$
C $P = V^2 R$
D $P = \frac{V^2}{R}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Using

P = VI

and $V = IR \Rightarrow I = \dfrac{V}{R}$ .

So

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

Answer

D

Final answer

D

Detailed explanation

Background Concept

Electrical power $P$ is the rate at which electrical energy is transferred (or dissipated as thermal energy in a resistor). For a component with potential difference $V$ across it and current $I$ through it:

P = VI

For an ohmic resistor (constant resistance $R$ ), Ohm's law applies:

V = IR

These two equations can be combined to express power in different equivalent forms, depending on which quantities ( $V$ , $I$ , $R$ ) are known.

Understanding the Question

We are told there is a potential difference $V$ across a resistor of resistance $R$ , with current $I$ in the resistor. The question asks which expression gives the power dissipated by the resistor, and the options are written using $V$ with either $I$ or $R$ .

So we want a correct formula for $P$ in terms of $V$ and $R$ (since most options involve $V^2$ and either $I$ or $R$ ).

Approach

Start from the fundamental power relationship $P = VI$ . Then use Ohm’s law $V = IR$ to eliminate the variable that you don’t want (here, eliminate $I$ to obtain an expression in $V$ and $R$ ). Finally, compare the result to the options.

Step-by-Step Reasoning

Use the definition of electrical power:

P = VI

From Ohm’s law for the resistor:

V = IR \quad \Rightarrow \quad I = \frac{V}{R}

Substitute this $I$ into the power equation:

P = V\left(\frac{V}{R}\right)

Simplify:

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

This matches option D.

Key Takeaways

Use $P = VI$ as the starting point for power in circuits.
Use $V = IR$ to rewrite power in alternative forms:

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

Check that the final expression uses the variables asked for.

Common Mistakes

Choosing $P = V^2 I$ (option A): power is not proportional to $V^2$ and $I$ together.
Mixing up substitutions, e.g. using $I = R/V$ instead of $I = V/R$ .
Writing $P = V^2 R$ (option C): this has the wrong dependence on $R$ (increasing $R$ at fixed $V$ should reduce current and hence reduce power).

Things to Be Careful About

The forms $P = I^2R$ and $P = V^2/R$ come specifically from combining $P = VI$ with Ohm’s law; they apply to an ohmic resistor.
Dimensional check can help: $V^2/R$ has units $\text{V}^2/\Omega = \text{W}$ , consistent with power.

Techniques used

use the electrical power relation P = VIapply Ohm's law V = IRsubstitute to eliminate the current

Q321MMedium-EasyParticle Physics

What is a possible charge on a particle?

Options

A $6.40 \times 10^{-20}\text{ C}$
B $4.00 \times 10^{-19}\text{ C}$
C $1.12 \times 10^{-18}\text{ C}$
D $9.11 \times 10^{-18}\text{ C}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Charge must be an integer multiple of $e = 1.60 \times 10^{-19}\ \text{C}$ .

For option C:

\frac{1.12 \times 10^{-18}}{1.60 \times 10^{-19}} = 7

This is an integer, so it is possible.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Electric charge is quantised: any isolated particle’s charge $Q$ must be an integer multiple of the elementary charge $e$ .

Q = ne

where $e = 1.60 \times 10^{-19}\ \text{C}$ and $n$ is an integer (can be positive or negative). So to test whether a charge value is possible, check whether $Q/e$ is an integer.

Understanding the Question

You are given four numerical values of charge. The question asks which one could be the charge on a particle, i.e. which one matches $Q = ne$ for some integer $n$ .

Approach

For each option, divide the given charge by $e = 1.60 \times 10^{-19}\ \text{C}$ . The only acceptable option is the one that gives an integer (to appropriate precision).

Step-by-Step Reasoning

Test option C:

\frac{Q}{e} = \frac{1.12 \times 10^{-18}}{1.60 \times 10^{-19}}

Handle numbers and powers of ten separately:

\frac{1.12}{1.60} \times 10^{(-18)-(-19)} = 0.70 \times 10^1 = 7

Since $7$ is an integer, $Q = 7e$ is a possible charge value.

So the correct option is C.

Key Takeaways

Charge values occur in discrete steps of $e$ .
A “possible” charge must satisfy $Q/e$ is an integer.

Common Mistakes

Using the wrong value of $e$ (should be $1.60 \times 10^{-19}\ \text{C}$ ).
Forgetting to check the power of ten when dividing.
Accepting non-integer multiples (e.g. $2.5e$ ) as valid.

Things to Be Careful About

Rounding: an answer should be clearly an integer given the stated significant figures.
The sign of charge is not specified here; the test is about the magnitude being an integer multiple of $e$ (the charge could also be negative).

Techniques used

use charge quantisation in integer multiples of the elementary chargedivide each option by the elementary charge to test for an integer resultselect the option that corresponds to an allowed integer multiple

Q331MMedium-EasyElectricity

The graphs show possible current–voltage ( $I$ – $V$ ) characteristics for a filament lamp and for a semiconductor diode.

Fig. 14

Which row identifies the $I$ – $V$ graphs for the lamp and for the diode?

Options

	filament lamp	semiconductor diode
A	P	R
B	P	S
C	Q	R
D	Q	S

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Filament lamp: as $V$ increases the filament heats up, $R$ increases, so the gradient $\frac{\Delta I}{\Delta V}$ decreases (curve becomes less steep) (\to) graph $P$ .
Diode: negligible current until a threshold forward voltage, then a rapid rise (\to) graph $R$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

An $I$ – $V$ characteristic shows how the current $I$ through a component depends on the potential difference $V$ across it.

The gradient at any point is

\text{gradient} = \frac{\Delta I}{\Delta V}

This is the conductance. A smaller gradient means a larger resistance.

Filament lamp (non-ohmic): as current increases, the filament temperature rises. For a metal filament, resistivity increases with temperature, so the resistance increases as $V$ and $I$ increase. That means the graph becomes less steep at higher $V$ .
Semiconductor diode: in forward bias, current is very small until a “knee” (threshold) voltage is reached, after which the current rises very rapidly. (In reverse bias, current is approximately zero for ordinary supply voltages.)

Understanding the Question

You are shown four possible $I$ – $V$ graphs labelled $P, Q, R, S$ and must decide which one matches a filament lamp and which one matches a diode.

From the description:

$P$ : starts at the origin and rises with decreasing gradient (concave down).
$Q$ : starts at the origin and rises with increasing gradient (concave up).
$R$ : zero current initially for positive $V$ , then a steep rise after a threshold.
$S$ : straight line but with non-zero current at $V=0$ (not typical for either component).

Approach

Identify the filament lamp by looking for a curve that becomes less steep as $V$ increases (resistance increasing with temperature).
Identify the diode by looking for a threshold/knee: almost no current at small forward voltages, then a sharp rise.
Match these to the option table.

Step-by-Step Reasoning

Filament lamp:

At low $V$ , the filament is cooler, so resistance is lower and current rises relatively easily.
As $V$ increases, $I$ increases and the filament heats.
Heating increases resistivity and hence resistance, so each additional volt produces a smaller increase in current.
Therefore, the $I$ – $V$ curve should bend over: gradient decreases with $V$ .
This matches graph $P$ .

Diode:

In forward bias, current is tiny for small $V$ .
Once the forward voltage reaches the knee value, current increases very rapidly.
This matches graph $R$ , which shows near-zero current until a threshold, then a steep rise.

So the correct row is filament lamp $P$ and diode $R$ , which is option A.

Key Takeaways

The gradient of an $I$ – $V$ graph indicates how the resistance is changing.
Filament lamp: resistance increases as it heats up, so the curve becomes less steep at higher $V$ .
Diode: has a forward threshold (“knee”) then a rapid current rise.

Common Mistakes

Choosing the concave-up curve ( $Q$ ) for a filament lamp: that would imply resistance decreases as it heats, which is not the usual behaviour for a metal filament.
Choosing a straight line through the origin for the filament lamp: that would be an ohmic resistor, not a filament lamp.
Choosing $S$ for the diode: a non-zero current at $V=0$ does not represent a passive diode characteristic.

Things to Be Careful About

Remember: decreasing gradient with increasing $V$ means increasing resistance.
A diode’s key feature is the threshold/knee in forward bias; it is not a straight line.
Some exam graphs only show the first quadrant; in reality, a filament lamp characteristic would be roughly symmetric for positive and negative $V$ .

Techniques used

interpret the gradient of an I–V graph as conductance (inverse resistance)identify non-ohmic behaviour from a changing gradientrecognise the diode threshold (knee) behaviour on an I–V characteristic

Q341MMediumD.C. Circuits

The diagram shows a circuit containing a battery and cells with negligible internal resistance.

Some values of current, electromotive force (e.m.f.) and resistance are shown.

One resistor is labelled $P$ .

Fig. 15

What is the resistance of resistor $P$ ?

Options

A $0.40\Omega$
B $0.80\Omega$
C $2.0\Omega$
D $2.8\Omega$

Pick an answer

DifficultyMedium

Worked solution

Working

At the top junction, the currents combine, so current in $P$ is

I_P = 3.0 + 2.0 = 5.0\ \text{A}

In the left loop, the $9.0\ \text{V}$ source is opposed by the $1.0\ \text{V}$ cell, so net e.m.f. is $8.0\ \text{V}$ . The p.d. across the $2.0\ \Omega$ resistor is

V = IR = 3.0\times 2.0 = 6.0\ \text{V}

Hence p.d. across $P$ :

V_P = 8.0 - 6.0 = 2.0\ \text{V}

So

R_P = \frac{V_P}{I_P} = \frac{2.0}{5.0} = 0.40\ \Omega

Answer

A

Final answer

A

Detailed explanation

Background Concept

In a d.c. circuit with negligible internal resistance, two key rules are used:

Kirchhoff’s first law (junction rule): the total current entering a junction equals the total current leaving it.
Kirchhoff’s second law (loop rule): around any closed loop, the sum of the e.m.f.s equals the sum of the potential drops ((IR) terms), taking account of whether sources aid or oppose each other.

Also, for a resistor:

V = IR

where (V) is the potential difference across it and (I) is the current through it.

Understanding the Question

You are given a circuit containing a (9.0\ \text{V}) battery, a (2.0\ \Omega) resistor, and cells (one labelled (1.0\ \text{V})), plus an unknown resistor (P). The currents in the left and right vertical sides are (3.0\ \text{A}) and (2.0\ \text{A}) (in the directions shown).

You must find the resistance of (P) by determining:

the current through (P), and
the potential difference across (P).

Approach

Use the given branch currents and KCL at the junction connected to (P) to find (I_P).
Choose a loop that contains known sources and known resistor(s) and (P). Apply KVL to find the p.d. across (P).
Use (R_P = V_P/I_P) and match to the multiple-choice options.

Step-by-Step Reasoning

1) Current through (P)

At the top junction, the (3.0\ \text{A}) and (2.0\ \text{A}) currents both feed into the junction (as drawn), so the current that must leave via the only other branch (through (P)) is their sum:

I_P = 3.0 + 2.0 = 5.0\ \text{A}.

2) Potential difference across (P)

Take the left-hand loop containing the (9.0\ \text{V}) battery, the (1.0\ \text{V}) cell and the (2.0\ \Omega) resistor, plus (P).

From the diagram, the (1.0\ \text{V}) cell is opposing the (9.0\ \text{V}) battery in that loop, so the net e.m.f. driving the loop is:

\mathcal{E}_\text{net} = 9.0 - 1.0 = 8.0\ \text{V}.

The p.d. across the (2.0\ \Omega) resistor (carrying (3.0\ \text{A})) is:

V_{2\Omega} = IR = 3.0 \times 2.0 = 6.0\ \text{V}.

The remaining p.d. in the loop must be across (P):

V_P = 8.0 - 6.0 = 2.0\ \text{V}.

3) Resistance of (P)

Finally,

R_P = \frac{V_P}{I_P} = \frac{2.0}{5.0} = 0.40\ \Omega.

So the correct option is A.

Key Takeaways

Use KCL first to find unknown branch currents.
Use KVL to relate sources and resistor drops to find an unknown p.d.
Opposing sources subtract: (\mathcal{E}_\text{net} = \mathcal{E}_1 - \mathcal{E}_2) when they act in opposite directions around the chosen loop.

Common Mistakes

Subtracting currents instead of adding them at the junction when both are directed into (or out of) the junction.
Adding e.m.f.s that oppose each other, instead of subtracting.
Using (V=IR) with the wrong current (e.g. using (5.0\ \text{A}) through the (2.0\ \Omega) resistor when it actually has (3.0\ \text{A})).

Things to Be Careful About

Keep track of the direction you traverse a loop: crossing a source from (-) to (+) is a rise, and from (+) to (-) is a drop.
Ensure the current used in (V=IR) is the current through that specific resistor.
Quote the final resistance to a sensible number of significant figures matching the data (here, (0.40\ \Omega)).

Techniques used

apply Kirchhoff’s first law at a junction to relate branch currentsapply Kirchhoff’s second law around a loop to find the potential difference across a componentuse Ohm’s law to determine resistance from potential difference and current

Q351MMedium-EasyD.C. Circuits

The diagram shows a circuit with a battery connected to a resistor $R$ . The battery has an internal resistance represented by resistor $r$ .

Fig. 16

A second resistor, identical to $R$ , is connected in parallel with $R$ .

Which row describes the changes to the potential difference (p.d.) across $r$ and the current shown on the ammeter when the second resistor is connected?

Options

	p.d. across $r$	current shown on ammeter
A	decreases	decreases
B	decreases	increases
C	increases	decreases
D	increases	increases

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Initially total resistance $= R + r$ so current

I_1 = \frac{\mathcal{E}}{R+r}

With a second identical resistor in parallel, external resistance

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

Total resistance $= r + \frac{R}{2}$ so

I_2 = \frac{\mathcal{E}}{r+\frac{R}{2}} > I_1

p.d. across internal resistance:

V_r = Ir

So as $I$ increases, $V_r$ increases.

Answer

D

Final answer

D

Detailed explanation

Background Concept

A real battery can be modelled as an ideal source of electromotive force (e.m.f.) $\mathcal{E}$ in series with an internal resistance $r$ . When a current $I$ flows, there is a “lost volts” p.d. across the internal resistance:

V_r = Ir

The current in the circuit is set by the total series resistance:

I = \frac{\mathcal{E}}{R_\text{total}}

When resistors are connected in parallel, the equivalent resistance decreases. For two identical resistors $R$ in parallel:

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

Understanding the Question

Initially, the battery (with internal resistance $r$ ) is in series with an external resistor $R$ and an ammeter. Then a second resistor, identical to $R$ , is added in parallel with the first $R$ .

The question asks how two quantities change:

the p.d. across the internal resistor $r$ (inside the battery model), and
the current reading on the ammeter (the total current supplied by the battery).

Approach

Work out how the external load resistance changes when the second $R$ is connected in parallel.
Use $I = \mathcal{E}/R_\text{total}$ to decide whether the total current increases or decreases.
Use $V_r = Ir$ to decide how the p.d. across $r$ changes when the current changes.

Step-by-Step Reasoning

1. Before adding the second resistor
External resistance is $R$ , so total resistance is:

R_{\text{total},1} = r + R

Current:

I_1 = \frac{\mathcal{E}}{r+R}

2. After adding the second identical resistor in parallel
Two identical resistors in parallel give:

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

So the new total resistance is:

R_{\text{total},2} = r + \frac{R}{2}

Since $r + \frac{R}{2} < r + R$ , the total resistance has decreased, so the total current increases:

I_2 = \frac{\mathcal{E}}{r+\frac{R}{2}} > I_1

Therefore the ammeter reading increases.

3. p.d. across the internal resistance $r$
The p.d. across $r$ is:

V_r = Ir

Because $r$ is constant and $I$ increases, $V_r$ increases.

So both quantities increase: option D.

Key Takeaways

Adding a parallel resistor reduces the external resistance, so the total circuit current increases.
The p.d. across the internal resistance is $V_r = Ir$ , so it increases when the current increases.

Common Mistakes

Thinking the p.d. across $r$ must decrease because the terminal p.d. across the external circuit decreases; in fact, the “lost volts” $Ir$ increases when current increases.
Treating the two resistors as series instead of parallel (giving $2R$ instead of $R/2$ ).

Things to Be Careful About

Distinguish between terminal p.d. across the external load and p.d. across internal resistance: they change in opposite directions when current changes.
The ammeter reads the total current supplied by the battery (the current through $r$ ), not the current in just one parallel branch.

Techniques used

combine identical resistors in parallel to find the new external resistanceapply Ohm's law to the whole circuit to compare the total currentuse the internal resistance p.d. relation $V_r = Ir$

Q361MMedium-EasyElectricityD.C. Circuits

Two identical wires $X$ and $Y$ , each of length $L$ and radius $r$ , are connected in parallel as shown.

Fig. 17

The total resistance of this combination is $R_1$ .

Wire $X$ is replaced with a wire of the same material with length $L$ and radius $2r$ .

The total resistance of the new combination is $R_2$ .

What is the ratio $\dfrac{R_1}{R_2}$ ?

Options

A $\dfrac{2}{5}$
B $\dfrac{2}{3}$
C $\dfrac{3}{2}$
D $\dfrac{5}{2}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For each original wire,

R = \frac{\rho L}{\pi r^2}

Two identical wires in parallel:

R_1 = \frac{R}{2}

Replacement wire has radius $2r$ so area is $4\pi r^2$ :

R_X' = \frac{\rho L}{\pi (2r)^2} = \frac{R}{4}

New parallel combination:

R_2 = \frac{R\left(\frac{R}{4}\right)}{R + \frac{R}{4}} = \frac{\frac{R^2}{4}}{\frac{5R}{4}} = \frac{R}{5}

Hence

\frac{R_1}{R_2} = \frac{R/2}{R/5} = \frac{5}{2}

Answer

D

Final answer

D

Detailed explanation

Background Concept

For a uniform wire of resistivity $\rho$ , length $L$ and cross-sectional area $A$ , the resistance is

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

For a circular wire, $A = \pi r^2$ , so for fixed material ( $\rho$ ) and fixed length ( $L$ ), the resistance varies as

R \propto \frac{1}{r^2}

When two resistors are connected in parallel, the potential difference across each is the same and the currents add. The equivalent resistance $R_{\text{eq}}$ satisfies

\frac{1}{R_{\text{eq}}} = \frac{1}{R_a} + \frac{1}{R_b}

For two equal resistors $R$ in parallel, this gives $R_{\text{eq}} = R/2$ .

Understanding the Question

Initially, wires $X$ and $Y$ are identical (same $L$ , same $r$ , same material), and connected in parallel, so the combination has total resistance $R_1$ .

Then wire $X$ is replaced by a wire of the same material and length, but with radius $2r$ . This changes only its cross-sectional area, hence its resistance. The new parallel combination has total resistance $R_2$ .

We are asked for the ratio $R_1/R_2$ and must pick the matching option.

Approach

Write the resistance of one original wire using $R = \rho L/(\pi r^2)$ .
Find $R_1$ for two identical resistors in parallel.
Find the new resistance of wire $X$ after doubling its radius (area increases by a factor of 4, so resistance decreases by a factor of 4).
Combine the new $X$ resistance in parallel with the unchanged $Y$ resistance to get $R_2$ .
Take the ratio $R_1/R_2$ and match to the options.

Step-by-Step Reasoning

Let the resistance of each original wire be $R$ .

Because the wires have the same material and length,

R = \frac{\rho L}{\pi r^2}

Initial combination (two identical wires in parallel):
For two equal resistances $R$ in parallel,

R_1 = \frac{R}{2}

After replacing wire $X$ (radius $2r$ ):
The new cross-sectional area of $X$ is

A' = \pi (2r)^2 = 4\pi r^2

So its resistance becomes

R_X' = \frac{\rho L}{A'} = \frac{\rho L}{4\pi r^2} = \frac{R}{4}

Wire $Y$ is unchanged, so it still has resistance $R$ .

New parallel combination:
Using the product-over-sum form for two resistors in parallel,

R_2 = \frac{R\left(\frac{R}{4}\right)}{R + \frac{R}{4}}

Compute:

R_2 = \frac{\frac{R^2}{4}}{\frac{5R}{4}} = \frac{R}{5}

Ratio:

\frac{R_1}{R_2} = \frac{R/2}{R/5} = \frac{5}{2}

So the correct option is D.

Key Takeaways

For fixed $\rho$ and $L$ , wire resistance varies as $1/r^2$ .
Doubling a wire’s radius quarters its resistance.
Two equal resistors in parallel give half the resistance.
Parallel combinations are often easiest using either reciprocal addition or product-over-sum.

Common Mistakes

Treating resistance as proportional to $1/r$ instead of $1/r^2$ (forgetting area is $\pi r^2$ ).
Combining parallel resistances by adding them directly (that is for series, not parallel).
Correctly finding $R_1$ and $R_2$ but inverting the ratio to get $R_2/R_1$ .

Things to Be Careful About

The resistivity $\rho$ and length $L$ do not change, so only the area change affects resistance.
Keep track of factors of 4 carefully when the radius doubles.
For a 1-mark MCQ, the method should be clean enough to avoid arithmetic slips: simplify symbolically (in terms of $R$ ) before calculating the ratio.

Techniques used

use the resistivity relation to compare resistances when cross-sectional area changescalculate equivalent resistance for two resistors in parallelform a ratio of two equivalent resistances

Q371MMedium-EasyD.C. Circuits

The diagram shows the currents in part of an electric circuit.

Fig. 18

The resistors are identical.

Which equation is not correct?

Options

A $I_1 = I_2 + I_4 + I_5$
B $I_2 = I_1 - I_3$
C $I_2 = I_3 + I_4 + I_5$
D $I_4 = I_3 - I_5$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

At the left junction,

I_1 = I_2 + I_3

At the lower-left junction,

I_3 = I_4 + I_5

So $I_1 = I_2 + I_4 + I_5$ (A correct), and $I_2 = I_1 - I_3$ (B correct), and $I_4 = I_3 - I_5$ (D correct).

Option C would give

I_2 = I_3 + I_4 + I_5 = I_3 + (I_4+I_5) = I_3 + I_3 = 2I_3

which is not generally true.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Kirchhoff’s first law (KCL) states that at any junction (node) in a circuit, the total current flowing into the junction equals the total current flowing out.

In equation form, if currents into the node are taken as positive and currents out as negative (or vice versa), then

\sum I = 0

This is a direct consequence of conservation of charge: charge cannot build up at a junction in a steady d.c. circuit.

Understanding the Question

You are given a circuit section with currents labelled $I_1, I_2, I_3, I_4, I_5$ . The question asks which proposed equation is not correct.

The key idea is: ignore the resistor values at first; to check these equations, you only need current conservation at the junctions where currents split or recombine.

Approach

Identify each junction where one current splits into two (or more) currents.
Apply KCL at each junction to write true relationships between the currents.
Compare each option with these relationships (and any simple substitutions between them) to see which one contradicts KCL.

Step-by-Step Reasoning

diagram-1

At the left junction, current $I_1$ arrives and splits into the top branch current $I_2$ and the lower branch current $I_3$ .

So by KCL:

I_1 = I_2 + I_3

Rearranging gives:

I_2 = I_1 - I_3

So option B is correct.

At the lower-left junction, current $I_3$ splits into the two parallel currents $I_4$ and $I_5$ .

So by KCL:

I_3 = I_4 + I_5

Rearranging gives:

I_4 = I_3 - I_5

So option D is correct.

Substitute $I_3 = I_4 + I_5$ into the left-junction equation:

I_1 = I_2 + I_3 = I_2 + (I_4 + I_5)

So:

I_1 = I_2 + I_4 + I_5

So option A is correct.

Now test option C:

I_2 = I_3 + I_4 + I_5

But since $I_3 = I_4 + I_5$ , then

I_3 + I_4 + I_5 = (I_4 + I_5) + I_4 + I_5 = 2(I_4 + I_5) = 2I_3

So option C claims $I_2 = 2I_3$ , which is not required by any junction rule and will not be true in general.

Therefore, C is the incorrect equation.

Key Takeaways

Use Kirchhoff’s first law at every junction: current in = current out.
You can often decide MCQs about currents without using resistor values at all.
When multiple junction equations exist, substitute between them to simplify and test statements.

Common Mistakes

Writing $I_1 = I_2 + I_4 + I_5$ directly without first justifying that $I_3 = I_4 + I_5$ .
Mixing up which currents meet at which junction (using KCL on the wrong node).
Assuming equal resistors means equal currents everywhere; equal resistors only imply equal currents if they have the same potential difference across them.

Things to Be Careful About

KCL must be applied at a specific junction; always identify the node first.
Be consistent with directions: here the equations use magnitudes with the directions already shown by the arrows.
An equation can look plausible but still be wrong if it effectively “double counts” a branch current (as in option C, which includes $I_3$ as well as its components $I_4$ and $I_5$ ).

Techniques used

apply Kirchhoff's first law at a junctionwrite current-conservation equations for multiple nodessubstitute one junction equation into anotheridentify the option that contradicts junction equations

Q381MMedium-EasyParticle Physics

The diagram shows a sequence of radioactive decays involving three $\alpha$ -particles and a $\beta^{-}$ particle.

Fig. 19

What is nuclide $T$ ?

Options

A $^{225}_{88}\text{Ra}$
B $^{231}_{88}\text{Ra}$
C $^{225}_{90}\text{Th}$
D $^{229}_{90}\text{Th}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For $\alpha$ decay: $A \to A-4$ , $Z \to Z-2$ .

For $\beta^{-}$ decay: $A$ unchanged, $Z \to Z+1$ .

Start: $^{237}_{93}\text{Np}$

After $\alpha$ : $A=237-4=233$ , $Z=93-2=91$ .

After $\beta^{-}$ : $A=233$ , $Z=91+1=92$ .

After $\alpha$ : $A=233-4=229$ , $Z=92-2=90$ .

After $\alpha$ : $A=229-4=225$ , $Z=90-2=88$ .

So $T = {}^{225}_{88}\text{Ra}$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

In nuclear equations, two numbers must balance:

Nucleon (mass) number $A$ : total number of protons + neutrons.
Proton (atomic) number $Z$ : number of protons (this identifies the element).

For common decays:

Alpha decay emits an $\alpha$ $α$ particle $\left( {}^{4}_{2}\text{He} \right)$ $(_{2}^{4} He)$ , so the nucleus loses 2 protons and 2 neutrons:
- $A \to A - 4$
- $Z \to Z - 2$
Beta-minus decay emits an electron ( $\beta^-$ $β^{-}$ ). A neutron turns into a proton (plus an electron and an antineutrino), so:
- $A$ unchanged
- $Z \to Z + 1$

Understanding the Question

You are given a decay sequence starting from $^{237}_{93}\text{Np}$ and proceeding through four steps:

$\alpha$ to $Q$
$\beta^{-}$ to $R$
$\alpha$ to $S$
$\alpha$ to $T$

The task is to find the nuclide $T$ (its $A$ , $Z$ , and hence its element symbol), then match it to the options.

Approach

Track $A$ and $Z$ through the chain step-by-step:

Subtract $(4,2)$ from $(A,Z)$ for each $\alpha$ .
Add $(0,1)$ to $(A,Z)$ for the $\beta^{-}$ .
Finally, convert the final $Z$ into the element name (here, $Z=88$ corresponds to radium, Ra).

Step-by-Step Reasoning

Start with neptunium:

{}^{237}_{93}\text{Np}

After first $\alpha$ decay:

A = 237 - 4 = 233, \qquad Z = 93 - 2 = 91

So $Q$ is $^{233}_{91}\text{Pa}$ (protactinium), though you don’t actually need the symbol.

After $\beta^{-}$ decay:

A = 233, \qquad Z = 91 + 1 = 92

So $R$ is $^{233}_{92}\text{U}$ .

After next $\alpha$ decay:

A = 233 - 4 = 229, \qquad Z = 92 - 2 = 90

So $S$ is $^{229}_{90}\text{Th}$ .

After final $\alpha$ decay:

A = 229 - 4 = 225, \qquad Z = 90 - 2 = 88

So

T = {}^{225}_{88}\text{Ra}

Comparing with the options, this is option A.

Key Takeaways

$\alpha$ decay changes $(A,Z)$ by $(-4,-2)$ .
$\beta^{-}$ decay changes $(A,Z)$ by $(0,+1)$ .
Following a decay chain is usually just careful bookkeeping of $A$ and $Z$ .

Common Mistakes

Treating $\beta^{-}$ decay as decreasing $Z$ (it increases $Z$ by 1).
Changing $A$ during $\beta^{-}$ decay (it must stay the same).
Forgetting that each $\alpha$ decay reduces $Z$ by 2 (not by 1).
Identifying the wrong element for a given $Z$ (e.g. confusing $Z=88$ Ra with $Z=90$ Th).

Things to Be Careful About

Apply the decay steps in the correct order; don’t combine them incorrectly.
Keep track of both $A$ and $Z$ at every step.
For MCQ, the final check is matching the final $(A,Z)$ pair to the listed nuclides exactly.

Techniques used

apply nucleon number and proton number changes for alpha decayapply nucleon number and proton number changes for beta-minus decaytrack nuclide changes through a decay chainidentify an element from its proton number

Q391MMedium-EasyParticle PhysicsElectricity

The number of electrons in a neutral atom of an isotope of plutonium, $^{239}_{94}\text{Pu}$ , is changed to produce a charged atom (ion) $W$ .

$W$ has an overall charge of $+1.6 \times 10^{-19}\text{ C}$ .

How many protons, neutrons and electrons are in $W$ ?

Options

	protons	neutrons	electrons
A	94	145	93
B	94	145	95
C	238	94	239
D	239	94	238

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For $^{239}_{94}\text{Pu}$ :

protons $= 94$
neutrons $= 239 - 94 = 145$

Charge on $W$ is $+1.6 \times 10^{-19}\ \text{C} = +e$ , so it has lost $1$ electron.

Electrons $= 94 - 1 = 93$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

In nuclide notation $^{A}_{Z}X$ :

$Z$ is the proton number (number of protons).
$A$ is the nucleon number (total number of protons + neutrons).
So the neutron number is

N = A - Z.

A neutral atom has equal numbers of protons and electrons, so its net charge is zero. If an atom becomes an ion with charge $+q$ , it must have fewer electrons than protons (it has lost electrons). The smallest possible charge magnitude is the elementary charge

e = 1.6 \times 10^{-19}\ \text{C}.

Understanding the Question

The atom is an isotope of plutonium $^{239}_{94}\text{Pu}$ . We change the number of electrons only (protons and neutrons in the nucleus stay the same) to form an ion $W$ with overall charge $+1.6 \times 10^{-19}\ \text{C}$ .

We must find how many protons, neutrons, and electrons the ion $W$ has, then match to the table.

Approach

Read $Z = 94$ directly as the number of protons.
Compute neutrons using $N = A - Z$ .
Use the given ion charge and the value of $e$ to decide how many electrons were removed from the neutral atom.

Step-by-Step Reasoning

From $^{239}_{94}\text{Pu}$ :
- Proton number $Z = 94$ , so protons $= 94$ .
- Nucleon number $A = 239$ .
Neutrons:

N = A - Z = 239 - 94 = 145.

Electrons:
- A neutral plutonium atom would have electrons $= 94$ (to balance the $94$ protons).
- The ion has charge $+1.6 \times 10^{-19}\ \text{C} = +e$ , meaning it is missing one electron compared to neutral.

So electrons in $W$ :

94 - 1 = 93.

This corresponds to option A: $94$ protons, $145$ neutrons, $93$ electrons.

Key Takeaways

In $^{A}_{Z}X$ , protons $= Z$ and neutrons $= A - Z$ .
Ionisation changes electrons only, not protons or neutrons.
A $+e$ ion has one fewer electron than the neutral atom.

Common Mistakes

Swapping $A$ and $Z$ and claiming protons $= 239$ .
Changing the number of protons to account for charge (charge changes here are due to electron loss/gain).
Forgetting that $+1.6 \times 10^{-19}\ \text{C}$ corresponds to exactly one elementary charge.

Things to Be Careful About

Always compute neutrons using $A - Z$ , not $Z - A$ .
A positive ion means electrons have been removed (electron count smaller than proton count).
The charge given is already in coulombs, so compare directly with $e = 1.6 \times 10^{-19}\ \text{C}$ to find the number of electrons lost.

Techniques used

interpret nuclide notation to identify proton number and nucleon numbercalculate neutron number using nucleon number minus proton numberuse the elementary charge to relate ion charge to the change in electron number

Q401MEasyParticle Physics

Which particle is a lepton?

Options

A meson
B positron
C proton
D quark

Pick an answer

DifficultyEasy

Worked solution

Mesons and protons are hadrons, and a quark is a constituent of hadrons.
A positron is the antiparticle of the electron, and electrons/positrons are leptons.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Particles are grouped into broad families.

Leptons are fundamental particles that do not experience the strong interaction. Examples: electron $e^-$ , muon $\mu^-$ , tau $\tau^-$ , and their neutrinos, plus their antiparticles (e.g. positron $e^+$ ).
Hadrons are particles that do experience the strong interaction and are made of quarks. There are two main types:
- Baryons (three quarks), e.g. proton.
- Mesons (a quark and an antiquark), e.g. pions/kaons.
Quarks are fundamental particles, but they are not leptons; they are a different family of fundamental particles.

Understanding the Question

We must choose which option is a lepton. The options include a hadron type (meson), a specific antiparticle (positron), a specific baryon (proton), and a fundamental particle type (quark). Only one belongs to the lepton family.

Approach

Classify each option:

decide whether it is a lepton, hadron, or quark,
use known examples (electron/positron are leptons; proton/meson are hadrons).

Step-by-Step Reasoning

A meson: a hadron made of a quark–antiquark pair, so not a lepton.
B positron: this is $e^+$ , the antiparticle of the electron. Electrons and positrons are leptons, so this is a lepton.
C proton: a baryon (hadron) made of three quarks ( $uud$ ), so not a lepton.
D quark: a fundamental particle, but it belongs to the quark family and participates in the strong interaction, so not a lepton.

Therefore the lepton is the positron.

Key Takeaways

Leptons include $e^-$ and $e^+$ (and neutrinos, muons, taus and their antiparticles).
Protons and mesons are hadrons (made of quarks).
Quarks are fundamental but are not leptons.

Common Mistakes

Thinking “fundamental particle” automatically means “lepton” (quarks are fundamental but not leptons).
Forgetting that antiparticles belong to the same family as their particles (positron is a lepton like the electron).
Confusing mesons with leptons because both can be short-lived (mesons are hadrons).

Things to Be Careful About

Particle family classification is about interactions/composition: hadrons are composite (quarks), leptons are not.
The proton is specifically a baryon, while “meson” is a category—both are hadrons and therefore not leptons.

Techniques used

recall the definition and examples of leptonsclassify each option as lepton, hadron, or constituent particleidentify an antiparticle as belonging to the same family as its particle

Physics 9702/13 — October/November 2025

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

1) Calculate XXX

2) Fractional uncertainty in XXX

3) Absolute uncertainty in XXX

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

1) Calculate $X$

2) Fractional uncertainty in $X$

3) Absolute uncertainty in $X$