9702/14 Oct/Nov 2025 — Physics MCQ Answers

Q11MEasyPhysical Quantities and Units

What is essential to accurately represent all physical quantities?

Options

A a base unit and a number
B a unit and a number expressed in standard form (scientific notation)
C a unit and a numerical magnitude
D an SI unit and a numerical magnitude

Pick an answer

DifficultyEasy

Worked solution

A physical quantity must be stated as a numerical magnitude together with a unit (it does not have to be a base unit, SI unit, or in standard form).

Answer

C

Final answer

C

Detailed explanation

Background Concept

A physical quantity is something that can be measured and expressed numerically. To communicate it unambiguously, you must include:

a numerical value (the magnitude), and
a unit (what that number is measured in).

For example, saying “length = 2.0” is incomplete because it could mean $2.0\ \text{m}$ , $2.0\ \text{cm}$ , etc. But saying “length = $2.0\ \text{m}$ ” is a complete representation.

Understanding the Question

The question asks what is essential for accurately representing all physical quantities. That means we choose the option that always must be present, with no extra unnecessary conditions.

Approach

Start from the definition: every physical quantity needs “number + unit”. Then check each option to see whether it matches that requirement without adding restrictions that are not always necessary (like “must be SI” or “must be standard form”).

Step-by-Step Reasoning

Option A: “a base unit and a number” is not essential because many quantities are commonly expressed with derived units (e.g. $\text{N}$ , $\text{J}$ ) rather than base units explicitly.
Option B: “a unit and a number expressed in standard form” is not essential because numbers do not have to be written in standard form (e.g. $0.25\ \text{s}$ is fine).
Option C: “a unit and a numerical magnitude” is exactly the definition: magnitude + unit.
Option D: “an SI unit and a numerical magnitude” is not essential because quantities can be expressed in non-SI units where appropriate (e.g. time in minutes, pressure in mmHg in some contexts), and the key requirement is still just unit + magnitude.

Therefore, the correct choice is C.

Key Takeaways

A complete statement of a physical quantity is always numerical magnitude + unit.
Extra conditions like “SI”, “base units”, or “standard form” are often helpful but are not strictly essential in all cases.

Common Mistakes

Thinking the unit must always be SI (SI is preferred in exams, but not part of the definition of “physical quantity”).
Confusing “base units” with the fact that derived units are also valid units.

Things to Be Careful About

In exams, you should normally use SI units unless told otherwise, but the question asks what is essential, not what is preferred.
“Standard form” is only a formatting choice for convenience with very large/small numbers; it is not required for correctness.

Techniques used

use the definition of a physical quantity as magnitude with uniteliminate options that impose unnecessary restrictions

Q21MMedium-EasyPhysical Quantities and Units

A steel rule can be read to the nearest millimetre. It is used to measure the length of a bar whose true length is $895 \text{ mm}$ . Repeated measurements give the following readings.

length / mm
892, 891, 892, 891, 891, 892

Are the readings accurate and precise to within $1 \text{ mm}$ ?

Options

	results are accurate to within $1 \text{ mm}$	results are precise to within $1 \text{ mm}$
A	no	no
B	no	yes
C	yes	no
D	yes	yes

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Mean reading:

\bar{L} = \frac{892+891+892+891+891+892}{6} = 891.5\ \text{mm}

Accuracy: difference from true length

|895 - 891.5| = 3.5\ \text{mm} \; (> 1\ \text{mm}) \Rightarrow \text{not accurate to within } 1\ \text{mm}

Precision: readings lie between $891\ \text{mm}$ and $892\ \text{mm}$ (range $= 1\ \text{mm}$ ), so repeated readings agree to within $1\ \text{mm}$ .

Answer

B

Final answer

B

Detailed explanation

Background Concept

Accuracy describes how close a measurement (or the mean of many measurements) is to the true value.

Precision describes how close repeated measurements are to each other (small scatter). Precision is limited by random errors and by the instrument resolution.

To test “within $1\ \text{mm}$ ”:

for accuracy, compare the measured value (often the mean) with the true value and see if the difference is $\le 1\ \text{mm}$ .
for precision, look at the spread of repeated readings (e.g. range, or half-range) and see if the scatter is $\le 1\ \text{mm}$ .

Understanding the Question

The steel rule reads to the nearest millimetre, and six repeated readings of a bar length are given: $891\ \text{mm}$ and $892\ \text{mm}$ .

The true length is stated to be $895\ \text{mm}$ . We must decide whether the set of readings is:

accurate to within $1\ \text{mm}$ , and
precise to within $1\ \text{mm}$ ,
then choose the correct option A–D.

Approach

Find a representative measured value from repeats (the mean) and compare it with $895\ \text{mm}$ to judge accuracy.
Examine the scatter of the six readings (max–min, or how tightly clustered they are) to judge precision.
Convert each judgement into yes/no for the table.

Step-by-Step Reasoning

1) Accuracy

Compute the mean:

\bar{L} = \frac{892+891+892+891+891+892}{6} = 891.5\ \text{mm}

Compare with the true length:

\text{error} = 895 - 891.5 = 3.5\ \text{mm}

The magnitude of the error is $3.5\ \text{mm}$ , which is greater than $1\ \text{mm}$ , so the readings are not accurate to within $1\ \text{mm}$ .

This large offset of all readings below $895\ \text{mm}$ suggests a systematic error (e.g. zero error, worn rule end, misaligned reading).

2) Precision

Look at the spread:

minimum reading $= 891\ \text{mm}$
maximum reading $= 892\ \text{mm}$

So the range is

\text{range} = 892 - 891 = 1\ \text{mm}

All readings are within $1\ \text{mm}$ of each other (they only differ by $1\ \text{mm}$ at most), so they are precise to within $1\ \text{mm}$ .

3) Choose option

Accuracy: no. Precision: yes. Therefore the correct option is B.

Key Takeaways

Use the mean vs true value to judge accuracy.
Use the scatter/range of repeats to judge precision.
A set of measurements can be very precise but still inaccurate if there is a systematic offset.

Common Mistakes

Judging accuracy from a single reading instead of comparing the mean (or representative value) with the true value.
Thinking “precise” means “close to the true value” (that is accuracy, not precision).
Using the instrument resolution alone to claim precision without looking at the actual spread in the repeated readings.

Things to Be Careful About

The true length is $895\ \text{mm}$ , while readings cluster around $891$ – $892\ \text{mm}$ ; this is a clear sign of inaccuracy even though the readings are consistent.
When a question specifies “to within $1\ \text{mm}$ ”, explicitly compare the relevant differences (error for accuracy, spread for precision) to $1\ \text{mm}$ .
Because readings are to the nearest millimetre, small variation (only $1\ \text{mm}$ range) is consistent with good precision.

Techniques used

calculate the mean of repeated measurementscompare the mean value with the true value to assess accuracyuse the spread of repeated readings to assess precisioncompare deviations with the stated tolerance

Q31MMedium-EasyPhysical Quantities and Units

A stone is released from rest and falls vertically to the ground.

The time taken to fall to the ground and the distance travelled are measured. The measurements are used to determine the acceleration of free fall.

The percentage uncertainty in the measured time is $0.05\%$ . The percentage uncertainty in the measured distance fallen is $0.6\%$ .

What is the percentage uncertainty in the calculated value of the acceleration of free fall?

Options

A $0.5\%$
B $0.7\%$
C $1.1\%$
D $1.3\%$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For release from rest:

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

For $g \propto s\,t^{-2}$ , percentage uncertainty:

\frac{\Delta g}{g}(\%) = \frac{\Delta s}{s}(\%) + 2\frac{\Delta t}{t}(\%)

= 0.6\% + 2(0.05\%) = 0.70\%

Answer

B

Final answer

B

Detailed explanation

Background Concept

When a calculated quantity depends on measured quantities, its uncertainty comes from how measurement uncertainties propagate through the mathematical relationship.

For products and quotients, fractional (or percentage) uncertainties add. If a measured quantity is raised to a power, its fractional (or percentage) uncertainty is multiplied by the magnitude of that power:

If $Q = AB$ , then $\frac{\Delta Q}{Q} \approx \frac{\Delta A}{A} + \frac{\Delta B}{B}$ .
If $Q = A^n$ , then $\frac{\Delta Q}{Q} \approx |n|\frac{\Delta A}{A}$ .

These are the standard A-Level rules for uncertainty propagation in derived quantities.

Understanding the Question

A stone falls from rest through a measured distance $s$ in a measured time $t$ . From $s$ and $t$ , we calculate $g$ . You are given:

percentage uncertainty in time: $0.05\%$
percentage uncertainty in distance: $0.6\%$

You must find the percentage uncertainty in the calculated $g$ and then choose the correct option.

Approach

Use the constant-acceleration equation for an object released from rest to relate $s$ , $t$ , and $g$ .
Rearrange to make $g$ the subject.
Identify how $g$ depends on $s$ and $t$ (especially the power of $t$ ).
Add percentage uncertainties, multiplying the time percentage uncertainty by $2$ because of the square.

Step-by-Step Reasoning

From rest, $u=0$ , so the displacement is

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

Rearrange for $g$ :

g = \frac{2s}{t^2}

The constant $2$ has no uncertainty, so ignore it. The dependence is

g \propto s\,t^{-2}

So the percentage uncertainty in $g$ is the percentage uncertainty in $s$ plus $2$ times the percentage uncertainty in $t$ :

\frac{\Delta g}{g}(\%) = \frac{\Delta s}{s}(\%) + 2\frac{\Delta t}{t}(\%)

Substitute values:

\frac{\Delta g}{g}(\%) = 0.6\% + 2(0.05\%) = 0.6\% + 0.10\% = 0.70\%

This matches option B.

Key Takeaways

Rearrange the physics equation first so you can see the algebraic dependence of the derived quantity.
For $t^2$ in the denominator, the time percentage uncertainty contributes twice.
Add percentage uncertainties for multiplication/division relationships.

Common Mistakes

Forgetting to multiply the time uncertainty by $2$ (because of $t^2$ ).
Subtracting uncertainties because $t^2$ is in the denominator (you still add fractional uncertainties).
Using absolute uncertainties instead of percentage/fractional uncertainties in this type of question.

Things to Be Careful About

The power rule uses the magnitude of the power: $t^{-2}$ still gives a factor of $2$ .
Do not include constants (like $2$ or $\frac{1}{2}$ ) in uncertainty propagation.
Ensure you keep the answer as a percentage and match it to the closest option.

Techniques used

select the relevant kinematics equation for motion from restrearrange to express the required quantity in terms of measured variablesapply percentage uncertainty rules for products and powerscombine percentage uncertainties for multiplication and division

Q41MMedium-EasyDynamics

An object falls from rest towards the ground.

Air resistance is negligible.

Which graph shows the variation of the momentum $p$ of the object with time $t$ until it hits the ground?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For free fall with negligible air resistance, $a = g$ (constant) and starting from rest:

v = gt

Momentum:

p = mv = mgt

So $p$ is directly proportional to $t$ (straight line through the origin).

Answer

C

Final answer

C

Detailed explanation

Background Concept

Momentum is defined by

p = mv

For an object falling with negligible air resistance, the only significant force is its weight $mg$ , so the acceleration is constant and equal to $g$ .

Constant acceleration means velocity changes linearly with time. Starting from rest ( $u=0$ ), the kinematics result is

v = u + at = gt

Combining these, momentum in free fall is proportional to time:

p = m(gt) = mgt

So a momentum–time graph must be a straight line through the origin with constant positive gradient.

Understanding the Question

You are told:

the object falls from rest (so initial momentum is zero),
air resistance is negligible (so acceleration is constant at $g$ ),
you must choose which of four graphs correctly shows how momentum $p$ varies with time $t$ until impact.

So the key is: what is the shape of $p(t)$ for constant acceleration starting from rest?

Approach

Use free-fall physics to write velocity as a function of time ( $v = gt$ ).
Convert velocity to momentum using $p = mv$ .
Decide whether this relationship is constant, linear, or curved, then match to the option.

Step-by-Step Reasoning

Negligible air resistance means the net force is constant:

F = mg

So acceleration is constant:

a = \frac{F}{m} = g

Starting from rest ( $u=0$ ), constant acceleration gives

v = u + at = 0 + gt = gt

Momentum is

p = mv = m(gt) = mgt

Since $m$ and $g$ are constants, $p \propto t$ . That is a straight line through the origin with constant gradient $mg$ .
The only option showing a straight line through the origin is Option C.

Key Takeaways

With no air resistance, free fall has constant acceleration $g$ .
For constant acceleration from rest, $v$ increases linearly with $t$ .
Since $p = mv$ , momentum also increases linearly with $t$ .

Common Mistakes

Choosing a curved graph: curvature would imply changing acceleration (e.g. due to air resistance approaching terminal velocity).
Choosing a non-zero intercept: the object starts from rest, so initial momentum must be zero.
Confusing momentum–time with velocity–time: they have the same shape here because mass is constant, but the gradient differs by a factor of $m$ .

Things to Be Careful About

The phrase “air resistance is negligible” is the crucial clue that acceleration is constant.
Ensure the graph starts at the origin: at $t=0$ , $v=0$ so $p=0$ .
A straight line means constant gradient, which corresponds to constant resultant force because

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

Techniques used

use constant acceleration in free fall to relate velocity to timeuse the definition of momentum to relate momentum to velocitymatch the mathematical relationship to the correct graph shape

Q51MMedium-EasyKinematics

A projectile is fired at an angle of $45^\circ$ upwards from horizontal ground. Air resistance is negligible.

Which row describes the horizontal motion and the vertical motion of the projectile after it is fired until immediately before it reaches the ground again?

Options

	horizontal motion	vertical motion
A	constant velocity	constant acceleration
B	constant velocity	varying acceleration
C	varying velocity	constant acceleration
D	varying velocity	varying acceleration

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

With negligible air resistance, the only force is weight $mg$ vertically downward.

Horizontal: no force $\Rightarrow a_x = 0 \Rightarrow$ horizontal velocity is constant.

Vertical: acceleration is due to gravity only $\Rightarrow a_y = -g$ (constant).

Answer

A

Final answer

A

Detailed explanation

Background Concept

Projectile motion (with negligible air resistance) is motion under a single constant force: the weight $mg$ acting vertically downward. This means:

There is no horizontal force, so the horizontal acceleration is $a_x = 0$ .
There is a constant vertical force $mg$ , so the vertical acceleration is constant and equal to $a_y = -g$ (taking upward as positive).

A key idea is that horizontal and vertical motions are independent: you analyse them separately using the acceleration in each direction.

Understanding the Question

You are told:

The projectile is launched at $45^\circ$ above the horizontal.
Air resistance is negligible.
You must choose the row that correctly describes the projectile’s horizontal motion and vertical motion from just after launch until just before it returns to the ground.

So we only need to decide whether velocity/acceleration are constant or varying in each direction.

Approach

Identify forces acting after the projectile is fired.
Deduce accelerations in horizontal and vertical directions from the forces.
Convert accelerations into descriptions of motion:
- If $a = 0$ , velocity is constant.
- If $a$ is constant and non-zero, velocity varies linearly with time.

Step-by-Step Reasoning

After firing (in flight), with negligible air resistance, the only force is the weight $mg$ downward.
Horizontal direction:
- There is no horizontal force.
- Therefore $a_x = 0$ .
- Hence horizontal velocity stays constant throughout the flight.
Vertical direction:
- The vertical force is $mg$ downward (constant near Earth’s surface).
- Therefore $a_y = -g$ , a constant.
- Hence the vertical velocity changes continuously (decreases on the way up, becomes zero at the top, then increases downward), but the acceleration remains constant.
Matching to the options:

Horizontal motion: constant velocity.
Vertical motion: constant acceleration.
So the correct row is A.

Key Takeaways

With no air resistance, projectile motion has:
- constant horizontal velocity ( $a_x = 0$ )
- constant vertical acceleration ( $a_y = -g$ )
Velocity can vary even when acceleration is constant (as in vertical motion here).

Common Mistakes

Thinking the vertical acceleration becomes zero at the top of the trajectory (it does not; it remains $-g$ ).
Confusing “vertical velocity changes” with “vertical acceleration changes”. The acceleration stays constant.
Assuming the projectile slows down horizontally (that would require air resistance).

Things to Be Careful About

The question explicitly says air resistance is negligible; that is the crucial condition that makes horizontal velocity constant.
“Immediately before it reaches the ground again” is still during free flight, so the acceleration assumptions are unchanged (no contact forces yet).
Sign convention: $a_y$ is downward; whether you write $-g$ or $g$ downward, it is still constant.

Techniques used

resolve the motion into horizontal and vertical componentsidentify the acceleration components when air resistance is negligibleapply constant-acceleration kinematics to the vertical component

Q61MMedium-EasyKinematics

An aircraft on a runway accelerates uniformly from rest to its take-off speed of $58 \text{ m s}^{-1}$ .

The acceleration of the aircraft is $4.2 \text{ m s}^{-2}$ , and the aircraft uses $74\%$ of the length of the runway to reach its take-off speed.

What is the length of the runway?

Options

A $300 \text{ m}$
B $540 \text{ m}$
C $590 \text{ m}$
D $800 \text{ m}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

From rest, $u=0$ , $v=58\ \text{m s}^{-1}$ , $a=4.2\ \text{m s}^{-2}$ .

v^2 = u^2 + 2as

s = \frac{v^2}{2a} = \frac{58^2}{2\times 4.2} = \frac{3364}{8.4} \approx 4.00\times 10^2\ \text{m}

This is $74\%$ of runway length $L$ :

0.74L = 4.00\times 10^2

L = \frac{4.00\times 10^2}{0.74} \approx 5.41\times 10^2\ \text{m}

Answer

B

Final answer

B

Detailed explanation

Background Concept

For motion with uniform (constant) acceleration, the kinematics (SUVAT) equations relate displacement $s$ , initial speed $u$ , final speed $v$ , acceleration $a$ , and time $t$ . A key one that eliminates time is

v^2 = u^2 + 2as

This applies when acceleration is constant along a straight line.

Understanding the Question

The aircraft starts from rest ( $u=0$ ) and accelerates uniformly to a take-off speed $v=58\ \text{m s}^{-1}$ with acceleration $a=4.2\ \text{m s}^{-2}$ . It reaches this speed after travelling a distance that is $74\%$ of the total runway length. We must find the total runway length.

So we first find the distance needed to reach $58\ \text{m s}^{-1}$ , then scale it up because that distance corresponds to only $0.74$ of the runway.

Approach

Use $v^2 = u^2 + 2as$ with $u=0$ to find the distance $s$ travelled to reach take-off speed.
Interpret “uses $74\%$ of the runway length” as $s = 0.74L$ .
Rearrange for $L$ .

Step-by-Step Reasoning

Substitute $u=0$ into the time-free SUVAT equation:

v^2 = 2as

Solve for $s$ :

s = \frac{v^2}{2a}

Substitute values:

s = \frac{58^2}{2\times 4.2} = \frac{3364}{8.4} \approx 400\ \text{m}

This $400\ \text{m}$ is the distance covered while accelerating to take-off speed.

This distance is $74\%$ of the full runway length $L$ :

0.74L = 400

Hence

L = \frac{400}{0.74} \approx 541\ \text{m}

The closest option is $540\ \text{m}$ (option B).

Key Takeaways

For constant acceleration, choose an SUVAT equation that contains the unknown you want and avoids variables you don’t need (here, eliminate time).
“Uses $74\%$ of the runway” translates directly to $s = 0.74L$ .
After calculating, select the nearest option in an MCQ.

Common Mistakes

Using the wrong equation (e.g. involving time) and getting stuck because $t$ is not given.
Forgetting that $400\ \text{m}$ is not the runway length but only $74\%$ of it.
Treating $74\%$ as $74$ instead of $0.74$ .

Things to Be Careful About

Square the speed correctly: $58^2 = 3364$ .
Keep units consistent (they already are in SI here).
Don’t round too early: rounding $s$ too aggressively can shift the final answer near an option boundary; here it still clearly gives option B.

Techniques used

apply a constant-acceleration kinematics equationrearrange to find displacement from speed and accelerationuse a fraction of total length to scale to full runway length

Q71MEasyKinematics

How can the acceleration of an object be determined?

Options

A from the area under a displacement–time graph
B from the area under a velocity–time graph
C from the gradient of a displacement–time graph
D from the gradient of a velocity–time graph

Pick an answer

DifficultyEasy

Worked solution

Working

Acceleration is the rate of change of velocity, so it is given by the gradient of a velocity–time graph.

Answer

D

Final answer

D

Detailed explanation

Background Concept

For straight-line motion, the key definitions are:

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

So acceleration is the rate of change of velocity with time.

On common motion graphs:

The gradient (slope) of a graph shows “rate of change” of the vertical quantity with respect to the horizontal quantity.
The area under a graph represents an integral (accumulated change), which depends on what quantities are on the axes.

Understanding the Question

You are asked which graph feature (area or gradient) lets you determine an object’s acceleration.

The options involve displacement–time and velocity–time graphs. You must recall what the gradient/area represent for each.

Approach

Use the definitions linking each graph to physical quantities:

On a displacement–time graph: gradient relates displacement change to time.
On a velocity–time graph: gradient relates velocity change to time.

Then identify which one equals acceleration.

Step-by-Step Reasoning

Consider a velocity–time graph: the gradient is

\text{gradient} = \frac{\Delta v}{\Delta t}

By definition, acceleration is

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

Therefore acceleration is obtained from the gradient of a velocity–time graph.

So the correct option is D.

(For completeness: the area under a velocity–time graph gives displacement, and the gradient of a displacement–time graph gives velocity.)

Key Takeaways

Acceleration is found from the gradient of a $v$ – $t$ graph.
Area under a $v$ – $t$ graph gives displacement.
Gradient of an $s$ – $t$ graph gives velocity.

Common Mistakes

Choosing the area under a $v$ – $t$ graph for acceleration (that area gives displacement).
Choosing the gradient of an $s$ – $t$ graph for acceleration (that gradient gives velocity, not acceleration).

Things to Be Careful About

Always check what each axis represents before deciding whether gradient or area is needed.
Use the “rate of change” idea: acceleration is the rate of change of velocity, so it must come from a graph involving velocity on the vertical axis.

Techniques used

identify which kinematic quantity corresponds to a graph gradientidentify which kinematic quantity corresponds to a graph areamatch acceleration to the appropriate motion graph

Q81MEasyDynamics

The acceleration of free fall on the Earth is different to the acceleration of free fall on the Moon.

How do the mass and weight of an object on the Earth compare to its mass and weight on the Moon?

Options

	mass	weight
A	different	different
B	different	same
C	same	different
D	same	same

Pick an answer

DifficultyEasy

Worked solution

Working

Mass is an intrinsic property of the object, so it is the same on Earth and on the Moon.

Weight $W$ is the gravitational force:

W = mg

Since $g$ is smaller on the Moon than on Earth, the weight is different.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Mass $m$ measures an object's inertia (resistance to acceleration). For the same object, its mass does not change just because it is moved to a different location.

Weight $W$ is the gravitational force acting on a mass in a gravitational field. It depends on the local gravitational field strength $g$ :

W = mg

So if $g$ changes (Earth vs Moon), the weight changes even though the mass stays the same.

Understanding the Question

The question tells you that the acceleration of free fall is different on Earth and on the Moon. In A Level terms, that means $g$ is different in the two places.

You must compare, for the same object:

its mass on Earth vs mass on the Moon,
its weight on Earth vs weight on the Moon.

Approach

Decide whether mass changes with location (it does not).
Use $W = mg$ to see how weight depends on $g$ .
Since $g_\text{Earth} \neq g_\text{Moon}$ , conclude the weights are different.

Step-by-Step Reasoning

Mass: the object is the same object, so its mass $m$ is unchanged when moved from Earth to Moon.
Weight:

W_\text{Earth} = m g_\text{Earth}, \qquad W_\text{Moon} = m g_\text{Moon}

Given $g_\text{Earth} \neq g_\text{Moon}$ , it follows that $W_\text{Earth} \neq W_\text{Moon}$ , so the weight is different.

Therefore the correct row is: mass same, weight different (\rightarrow) option C.

Key Takeaways

Mass is an intrinsic property of an object and does not depend on location.
Weight is a force and depends on the local gravitational field strength: $W = mg$ .
Different $g$ values imply different weights for the same mass.

Common Mistakes

Saying mass is different because the weight is different.
Confusing mass (kg) with weight (N).
Thinking that “acceleration of free fall” affects mass; it affects weight through $g$ .

Things to Be Careful About

Use the correct units: mass in $\text{kg}$ , weight in $\text{N}$ .
The question is about comparing Earth vs Moon, so you only need to know that $g$ differs, not the actual numerical values.
Weight is the gravitational force, not “how heavy it feels” (though it relates to that).

Techniques used

recall that mass is an intrinsic property of an objectuse the relation between weight, mass and gravitational field strengthcompare weights by comparing gravitational field strengths

Q91MMedium-EasyForces, Density and Pressure

A sphere moves vertically downwards at terminal (constant) velocity in a liquid.

Which statement about the magnitude of the upthrust acting on the sphere is correct?

Options

A It is proportional to the acceleration of free fall.
B It is equal to the weight of the sphere.
C It is proportional to the density of the sphere.
D It is proportional to the square of the radius of the sphere.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Upthrust equals weight of displaced liquid:

U = \rho_{\text{liquid}} g V

So $U \propto g$ .

At terminal velocity, $W = U + D$ so $U \neq W$ .

Answer

A

Final answer

A

Detailed explanation

Background Concept

Upthrust (buoyant force) on an object in a fluid arises because fluid pressure increases with depth, giving a larger upward force on the bottom than the downward force on the top. Archimedes’ principle states:

The upthrust on a body immersed in a fluid is equal to the weight of the fluid displaced.

Mathematically,

U = \rho_{\text{fluid}} g V_{\text{displaced}}

where $\rho_{\text{fluid}}$ is the density of the fluid, $g$ is the acceleration of free fall (gravitational field strength), and $V_{\text{displaced}}$ is the volume of fluid displaced (for a fully submerged sphere, this is the sphere’s volume).

Terminal velocity means the object moves at constant velocity, so acceleration is zero and the resultant force is zero.

Understanding the Question

A sphere is moving vertically downward through a liquid at terminal (constant) velocity. The question asks which statement about the magnitude of the upthrust is correct.

You are given four possible proportionalities/equalities. You must decide which matches the physics of buoyancy.

Approach

Write the expression for upthrust using Archimedes’ principle.
From that expression, identify what upthrust is proportional to.
Use the terminal-velocity force balance to check whether upthrust equals weight (a common distractor).
Compare the result with the options.

Step-by-Step Reasoning

By Archimedes’ principle,

U = \rho_{\text{liquid}} g V

So the upthrust is directly proportional to $g$ :

U \propto g

This matches option A.

At terminal velocity, acceleration $a=0$ , so resultant force is zero. The forces on the falling sphere are:

weight $W$ downward,
upthrust $U$ upward,
drag/resistive force $D$ upward (because motion is downward).

Force balance:

W = U + D

So $U$ is not equal to $W$ unless $D=0$ (which is not true when moving through a liquid at terminal speed). Therefore option B is incorrect.

Option C says $U$ is proportional to density of the sphere. But from $U = \rho_{\text{liquid}} g V$ , buoyant force depends on the liquid density, not the object density (provided it is immersed and displaces the fluid). So C is incorrect.
Option D says proportional to $r^2$ . For a sphere,

V = \frac{4}{3}\pi r^3

So

U \propto V \propto r^3

not $r^2$ . So D is incorrect.

Therefore the correct option is A.

Key Takeaways

Upthrust is given by $U = \rho_{\text{fluid}} g V_{\text{displaced}}$ .
Terminal velocity implies zero acceleration and hence zero resultant force, but it does not mean upthrust equals weight.
For a sphere, buoyant force scales with volume ( $\propto r^3$ ), not surface area.

Common Mistakes

Assuming that at terminal velocity, upthrust must equal weight (forgetting drag also acts upward).
Using the object’s density instead of the fluid’s density in Archimedes’ principle.
Thinking upthrust depends on cross-sectional area ( $r^2$ ) rather than displaced volume ( $r^3$ ).

Things to Be Careful About

Terminal velocity means $a=0$ , so sum of forces is zero, not that any particular pair of forces are equal.
Upthrust depends on the fluid and displaced volume; it is independent of the object’s speed (unlike drag).
Check proportionalities carefully: for spheres, many quantities scale as different powers of $r$ (area $\propto r^2$ , volume $\propto r^3$ ).

Techniques used

apply Archimedes' principle to express upthrust in terms of fluid density, gravitational field strength and displaced volumeuse the condition for terminal velocity to relate forces without changing the expression for upthrusteliminate incorrect options by comparing proportionalities

Q101MMedium-EasyDynamicsWork, Energy and Power

An object of mass $2.0 \text{ kg}$ is travelling at a speed of $3.0 \text{ m s}^{-1}$ on a horizontal frictionless surface. This object collides head-on with a stationary object of mass $1.0 \text{ kg}$ . The two objects stick together on impact.

Fig. 1

How much kinetic energy is lost on impact?

Options

A zero
B $2.0 \text{ J}$
C $2.4 \text{ J}$
D $3.0 \text{ J}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Conservation of momentum:

(2.0)(3.0) + (1.0)(0) = (2.0+1.0)v

v = \frac{6.0}{3.0} = 2.0\ \text{m s}^{-1}

Initial kinetic energy:

E_{\text{i}} = \frac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}

Final kinetic energy:

E_{\text{f}} = \frac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}

Kinetic energy lost:

\Delta E = 9.0 - 6.0 = 3.0\ \text{J}

Answer

D

Final answer

D

Detailed explanation

Background Concept

In a collision on a frictionless surface, there is (ideally) no external horizontal force on the two-object system, so total momentum is conserved:

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

If two objects stick together, the collision is perfectly inelastic. Momentum is still conserved, but kinetic energy is not: some kinetic energy is converted into internal energy (deformation, heat, sound).

Kinetic energy of an object of mass $m$ moving at speed $v$ is

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

Understanding the Question

Object 1: $m_1 = 2.0\ \text{kg}$ , speed $u_1 = 3.0\ \text{m s}^{-1}$ .

Object 2: $m_2 = 1.0\ \text{kg}$ , initially at rest so $u_2 = 0$ .

They collide head-on and stick together, so after the collision they move as one combined mass $(m_1+m_2)$ with a common speed $v$ .

The question asks for the kinetic energy lost:

\Delta E = E_{\text{i}} - E_{\text{f}}.

Approach

Use conservation of momentum to find the shared speed $v$ after impact.
Calculate the initial kinetic energy using $m_1, u_1$ (the second object contributes zero initially).
Calculate the final kinetic energy using the combined mass $(m_1+m_2)$ moving at speed $v$ .
Subtract to get kinetic energy lost, then match to the options.

Step-by-Step Reasoning

1) Momentum conservation
Before collision:

p_{\text{before}} = m_1u_1 + m_2u_2 = (2.0)(3.0) + (1.0)(0) = 6.0\ \text{kg m s}^{-1}

After collision, they stick so total mass is $3.0\ \text{kg}$ :

p_{\text{after}} = (m_1+m_2)v = (3.0)v

Equate:

6.0 = 3.0v \Rightarrow v = 2.0\ \text{m s}^{-1}.

2) Initial kinetic energy
Only the $2.0\ \text{kg}$ mass is moving:

E_{\text{i}} = \frac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}.

3) Final kinetic energy
Combined mass $3.0\ \text{kg}$ moving at $2.0\ \text{m s}^{-1}$ :

E_{\text{f}} = \frac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}.

4) Energy lost

\Delta E = 9.0 - 6.0 = 3.0\ \text{J}.

So the correct option is D.

Key Takeaways

In the absence of external forces, momentum is conserved in collisions.
If objects stick together, the collision is inelastic, so kinetic energy decreases.
Energy lost is found from $E_{\text{i}} - E_{\text{f}}$ , not from momentum.

Common Mistakes

Assuming kinetic energy is conserved because momentum is conserved.
Forgetting the two masses stick and incorrectly using separate final velocities.
Calculating final kinetic energy using only one mass instead of the combined mass.
Using $p=mv^2$ (incorrect) instead of $p=mv$ .

Things to Be Careful About

Use the correct final mass: $(m_1+m_2)$ .
Keep track of units: momentum in $\text{kg m s}^{-1}$ , kinetic energy in $\text{J}$ .
In MCQs, compute the numerical value and then match it to the option exactly (here $3.0\ \text{J}$ ).

Techniques used

apply conservation of momentum for a perfectly inelastic collisioncompute kinetic energy before and after the collisionfind energy lost as the difference between initial and final kinetic energies

Q111MMedium-EasyDynamicsWork, Energy and Power

A moving object X collides with a stationary object Y.

The objects separate after the collision.

The collision is perfectly elastic and there are no external forces acting.

Which word equation is not correct?

Options

A during the collision, (force acting on object X) + (force acting on object Y) = zero
B (relative speed of approach of X and Y) + (relative speed of separation of X and Y) = zero
C (total kinetic energy before collision) = (total kinetic energy after collision)
D (total momentum before collision) = (total momentum after collision)

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a perfectly elastic collision:

total momentum is conserved (no external forces).
total kinetic energy is conserved.
coefficient of restitution $e = 1$ , so

\text{relative speed of separation} = \text{relative speed of approach}

Option B states the sum of these (speeds) is zero, which is not correct for speeds (magnitudes).

Answer

B

Final answer

B

Detailed explanation

Background Concept

In a collision with no external forces, the resultant external force on the system is zero, so the system’s total momentum is conserved:

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

A collision is perfectly elastic if total kinetic energy is also conserved:

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

Another way to express “how elastic” a collision is uses the coefficient of restitution $e$ :

e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}

For a perfectly elastic collision, $e = 1$ , hence:

\text{relative speed of separation} = \text{relative speed of approach}

Also, during the collision the two objects exert forces on each other which form a Newton’s third law pair:

\vec{F}_{X\text{ on }Y} = -\vec{F}_{Y\text{ on }X}

So their vector sum is zero.

Understanding the Question

Object $X$ is moving and collides with stationary object $Y$ . After collision they separate. The collision is perfectly elastic and there are no external forces.

You are given four word-equations (A–D) and must identify which one is not correct under these conditions.

Approach

Check each option against the standard collision results:

Newton’s third law for interaction forces (option A).
Momentum conservation when no external forces (option D).
Kinetic energy conservation for perfectly elastic collisions (option C).
Relative speed relation for $e=1$ (option B), paying attention to whether it uses speeds (magnitudes) or velocities (signed quantities).

Step-by-Step Reasoning

Option A: “during the collision, (force on X) + (force on Y) = zero”

The forces between the objects are equal in magnitude and opposite in direction (Newton’s third law pair).
So as vectors, they sum to zero. This is correct.

Option D: “(total momentum before) = (total momentum after)”

With no external forces, momentum of the system is conserved. Correct.

Option C: “(total kinetic energy before) = (total kinetic energy after)”

Perfectly elastic means kinetic energy is conserved. Correct.

Option B: “(relative speed of approach) + (relative speed of separation) = zero”

For a perfectly elastic collision, the correct relationship using speeds is:

\text{relative speed of separation} = \text{relative speed of approach}

Since “speed” means a magnitude (non-negative), adding them cannot give zero unless both are zero, which is not the collision described.
Therefore B is the incorrect word-equation.

Key Takeaways

No external forces on a system (\Rightarrow) total momentum conserved.
Perfectly elastic (\Rightarrow) total kinetic energy conserved and (e=1).
Newton’s third law gives equal and opposite interaction forces during the collision.
For elastic collisions, it is equality of relative speeds (not a zero sum).

Common Mistakes

Treating “relative speed” as if it were a signed “relative velocity” and accepting the sum-zero form.
Thinking kinetic energy is always conserved in collisions (it is only conserved in perfectly elastic collisions).
Confusing “no external forces” with “no forces at all” (there are large internal forces during impact).

Things to Be Careful About

The word speed implies a positive magnitude; if the question had said “relative velocity” (with sign), a sum-to-zero statement could be made by choosing a sign convention. Here it says relative speed, so the mark-scheme intent is the standard statement: separation speed equals approach speed.
When checking option A, interpret the forces as vectors (equal and opposite), not magnitudes.

Techniques used

apply Newton's third law to the interaction forces during a collisionuse conservation of momentum when there is no external resultant forceuse the criterion for a perfectly elastic collision in terms of kinetic energyuse the coefficient of restitution relation between relative speeds

Q121MMedium-EasyForces, Density and Pressure

The diagram shows a stationary sphere that is just fully submerged in a liquid. The radius of the sphere is $r$ and the density of the liquid is $\rho$ . The acceleration of free fall is $g$ .

The air exerts pressure $P_A$ on the surface of the liquid.

Fig. 1

What is the pressure at the lowest point of the sphere?

Options

A $P_A + \frac{4}{3}\pi r^3 \rho g$
B $P_A - \frac{4}{3}\pi r^3 \rho g$
C $2r\rho g + P_A$
D $2r\rho g - P_A$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Hydrostatic pressure at depth $h$ is

P = P_A + \rho g h

Lowest point is at depth $h = 2r$ (sphere just fully submerged).

P = P_A + \rho g (2r) = P_A + 2r\rho g

Answer

C

Final answer

C

Detailed explanation

Background Concept

In a liquid at rest, pressure increases with depth because the liquid above a point has weight. For a point a vertical depth $h$ below the free surface,

P = P_\text{surface} + \rho g h

where:

$P$ is the pressure at the point,
$P_\text{surface}$ is the pressure at the free surface (here the air pressure $P_A$ ),
$\rho$ is the liquid density,
$g$ is gravitational field strength,
$h$ is the vertical depth below the surface.

Only the depth matters; the shape/volume of the object is irrelevant for pressure at a point.

Understanding the Question

The sphere is stationary and just fully submerged, meaning its topmost point touches the liquid surface. The question asks for the pressure at the lowest point of the sphere. From the diagram, the lowest point is a vertical distance $2r$ below the surface.

diagram-1

So we need the pressure at depth $h = 2r$ given the surface pressure is $P_A$ .

Approach

Read the depth $h$ of the lowest point below the free surface.
Substitute this depth into the hydrostatic relation $P = P_A + \rho g h$ .
Match the resulting expression to the options.

Step-by-Step Reasoning

Since the sphere’s radius is $r$ and its top touches the surface, the centre is at depth $r$ and the bottom is a further $r$ below the centre. Hence bottom depth:

h = r + r = 2r

Use hydrostatic pressure increase:

P = P_A + \rho g h

Substitute $h = 2r$ :

P = P_A + \rho g (2r) = P_A + 2r\rho g

This matches option C.

Key Takeaways

Pressure in a static liquid depends only on vertical depth: $P = P_0 + \rho g h$ .
Use the surface pressure as the reference pressure ( $P_0 = P_A$ here).
Do not use volume-dependent expressions (those relate to upthrust/weight, not pressure at a point).

Common Mistakes

Using $\frac{4}{3}\pi r^3\rho g$ (involves volume) which is associated with upthrust magnitude, not pressure at a point.
Forgetting to add the surface pressure $P_A$ and giving only $\rho g h$ .
Taking the depth as $r$ instead of $2r$ .
Using $P = P_A - \rho g h$ (pressure decreases upward, not downward).

Things to Be Careful About

$h$ must be the vertical depth below the free surface.
Check the sign: pressure increases with depth, so the $\rho g h$ term must be added.
The sphere being “just fully submerged” is the key clue that the bottom is at depth $2r$ .

Techniques used

identify the vertical depth below a liquid surfaceapply the hydrostatic pressure relation p = p0 + rho g hselect the option matching the derived expression

Q131MMediumForces, Density and Pressure

A uniform rod is attached by a hinge at one end to a wall. The other end of the rod is supported by a wire so that the rod is horizontal and in equilibrium.

Fig. 1

Which arrow shows the direction of the force on the rod from the hinge?

Options

Pick an answer

DifficultyMedium

Worked solution

Working

Take moments about the hinge.

Vertical component of tension at the free end is $T_y$ .

T_y L = W \left(\frac{L}{2}\right)

T_y = \frac{W}{2}

Vertical equilibrium:

R_y + T_y = W \;\Rightarrow\; R_y = W - \frac{W}{2} = \frac{W}{2}

So the hinge force has an upward component.

Horizontal equilibrium: tension acts towards the wall (left), so the hinge force must have a rightward component.

Hence the hinge force is up and to the right.

Answer

D

Final answer

D

Detailed explanation

Background Concept

For a rigid body in equilibrium:

The resultant force must be zero: $\sum F_x = 0$ and $\sum F_y = 0$ .
The resultant moment about any point must be zero: $\sum \tau = 0$ .

A hinge (pin joint) can exert a force on the rod with both horizontal and vertical components (but it does not exert a moment about the hinge, because the rod can rotate freely there).

Understanding the Question

A uniform horizontal rod is hinged to a wall at the left end. The right end is held up by a wire going up to the wall, so the wire tension on the rod is along the wire (upwards and towards the wall).

We are asked for the direction of the force that the hinge exerts on the rod (the hinge reaction). The options are arrows showing possible directions.

Approach

Draw the free-body diagram of the rod.
Use moments about the hinge: this eliminates the unknown hinge force immediately.
The moment equation determines the vertical component of the wire tension.
Then use vertical and horizontal force balance to infer the signs of the hinge force components, hence its direction.

Step-by-Step Reasoning

diagram-1

Moments about the hinge

Weight $W$ acts at the centre of the rod (because the rod is uniform), i.e. at distance $L/2$ from the hinge, producing a clockwise moment $W(L/2)$ .
The wire tension $T$ at the far end has a vertical component $T_y$ producing an anticlockwise moment $T_y L$ .

Equilibrium of moments:

T_y L = W\left(\frac{L}{2}\right)

so

T_y = \frac{W}{2}.

Vertical force balance

Upward forces: $T_y$ and the hinge vertical component $R_y$ .
Downward force: $W$ .

R_y + T_y = W

R_y = W - \frac{W}{2} = \frac{W}{2}

So the hinge force must have an upward component.

Horizontal force balance

The tension in the wire pulls along the wire toward the wall, so it has a leftward horizontal component $T_x$ on the rod.
To make $\sum F_x = 0$ , the hinge must provide an equal rightward component.

Therefore the hinge force on the rod is upwards and to the right, which corresponds to option D.

Key Takeaways

Use moments about the hinge to avoid dealing with the unknown hinge reaction components.
For a uniform rod, weight acts at the midpoint.
A hinge reaction generally has both horizontal and vertical components; its direction is found from force balance.

Common Mistakes

Taking moments using the full tension $T$ instead of only its perpendicular (vertical) component $T_y$ .
Assuming the hinge force must point along the wire (it does not).
Forgetting that the weight acts at the centre of the uniform rod.
Guessing the hinge force direction without checking both horizontal and vertical equilibrium.

Things to Be Careful About

When taking moments, use the component of a force perpendicular to the rod (or equivalently use $\tau = Fr\sin\theta$ ).
Decide the direction of the wire tension correctly: it always pulls along the wire toward the support.
Ensure the sign of $R_y$ (up or down) is determined from the moment result; it is not always upward in different geometries.

Techniques used

draw a free-body diagram for a hinged rod in equilibriumtake moments about the hinge to relate tension component to weightresolve the tension into horizontal and vertical componentsapply net force equals zero in horizontal and vertical directions

Q141MMedium-EasyForces, Density and Pressure

A couple is applied to a tap, as shown.

Fig. 1

What is the torque of the couple?

Options

A $\frac{Fd}{2}$
B $Fd$
C $2Fd$
D $4Fd$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a couple, torque $\tau = F \times$ (perpendicular distance between the forces).

Here the forces act at distances $d$ on either side of the pivot, so separation $= 2d$ .

\tau = F(2d)=2Fd

Answer

C

Final answer

C

Detailed explanation

Background Concept

A couple is a pair of equal and opposite forces whose lines of action are parallel but not the same. A couple produces a turning effect (torque) but no resultant force (so it tends to rotate without translating).

The torque of a couple is defined as

\tau = F \times s

where $F$ is the magnitude of one of the forces and $s$ is the perpendicular separation between the two lines of action of the forces.

Understanding the Question

The tap handle has two equal forces $F$ acting in opposite directions at opposite ends. Each line of action is at perpendicular distance $d$ from the central pivot, so the two lines of action are separated by $2d$ .

The question asks for the torque produced by this couple in terms of $F$ and $d$ .

Approach

Use the definition of the moment/torque of a couple: multiply one force by the perpendicular separation of the forces.
From the diagram, determine the separation: since each is $d$ from the centre on opposite sides, separation is $2d$ .
Compute $\tau$ and match it to the given options.

Step-by-Step Reasoning

diagram-1

From the diagram:

left force: magnitude $F$ , line of action at distance $d$ from pivot.
right force: magnitude $F$ in opposite direction, line of action at distance $d$ from pivot.

Therefore the perpendicular separation between the forces is

s = d + d = 2d

Torque of a couple:

\tau = F \times s = F(2d) = 2Fd

This corresponds to option C.

Key Takeaways

Torque of a couple is $F$ multiplied by the separation of the lines of action, not the distance from the pivot to one force.
In many symmetric diagrams, the separation is twice the distance from the centre to one force.

Common Mistakes

Using $\tau = Fd$ by taking $d$ as the separation (it is only the distance from the pivot to one line of action).
Adding moments about the pivot incorrectly as $Fd + Fd$ but then forgetting that this is equivalent to $F(2d)$ (some then halve it by mistake).
Confusing the torque of a couple with the moment of a single force.

Things to Be Careful About

The separation must be perpendicular to the forces; check the distance shown is perpendicular to each line of action.
For a couple, the torque is independent of the choice of pivot: using the definition $\tau = F \times s$ avoids pivot confusion.

Techniques used

use the definition of torque of a coupleidentify the perpendicular separation of the lines of actioncalculate torque as force multiplied by separation

Q151MMedium-EasyForces, Density and Pressure

A uniform beam rests on two supports, X and Y, as shown.

Fig. 1

The beam has length $20.0 \text{ m}$ and weight $200 \text{ N}$ .

A man of weight $700 \text{ N}$ stands on the beam at a distance of $6.0 \text{ m}$ from support X.

The beam is in equilibrium.

What is the contact force of support X on the beam?

Options

A $310 \text{ N}$
B $450 \text{ N}$
C $590 \text{ N}$
D $900 \text{ N}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Let the contact forces (reactions) at supports be $R_X$ and $R_Y$ .

Vertical equilibrium:

R_X + R_Y = 700 + 200 = 900\ \text{N}

Take moments about $X$ :

R_Y(20.0) = 700(6.0) + 200(10.0)

R_Y = \frac{4200 + 2000}{20.0} = 310\ \text{N}

R_X = 900 - 310 = 590\ \text{N}

Answer

C

Final answer

C

Detailed explanation

Background Concept

For a body in static equilibrium:

Resultant force in any direction is zero (so for vertical forces, total up = total down).
Resultant moment about any point is zero (clockwise moments = anticlockwise moments).

A moment about a point is

\text{moment} = F \times d

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

Understanding the Question

A uniform horizontal beam of length $20.0\ \text{m}$ rests on two supports at its ends: support $X$ on the left and support $Y$ on the right.

Downward forces:

the man: $700\ \text{N}$ acting $6.0\ \text{m}$ from $X$ ,
the beam’s weight: $200\ \text{N}$ acting at the centre, i.e. $10.0\ \text{m}$ from $X$ .

Upward forces:

contact force from support $X$ on the beam: $R_X$ ,
contact force from support $Y$ on the beam: $R_Y$ .

We are asked to find $R_X$ .

diagram-15-1

Approach

Use the two equilibrium conditions:

Vertical force balance gives one equation linking $R_X$ and $R_Y$ .
Take moments about one support (choose $X$ ) so that $R_X$ produces zero moment and is eliminated, allowing $R_Y$ to be found directly.

Then substitute back to get $R_X$ .

Step-by-Step Reasoning

Vertical force equilibrium:

Total downward force $= 700 + 200 = 900\ \text{N}$ .

So total upward force must be $900\ \text{N}$ :

R_X + R_Y = 900

Moments about $X$ (anticlockwise = clockwise). Using distances from $X$ :

$R_Y$ acts at $20.0\ \text{m}$ and gives an anticlockwise moment $R_Y(20.0)$ .
The man gives a clockwise moment $700(6.0)$ .
The beam’s weight gives a clockwise moment $200(10.0)$ .

So

R_Y(20.0) = 700(6.0) + 200(10.0)

R_Y(20.0) = 4200 + 2000 = 6200

R_Y = \frac{6200}{20.0} = 310\ \text{N}

Substitute into the force balance:

R_X = 900 - 310 = 590\ \text{N}

This corresponds to option C.

Key Takeaways

In equilibrium: $\sum F = 0$ and $\sum \tau = 0$ .
Taking moments about a support is a powerful way to eliminate the reaction at that support.
For a uniform beam, its weight acts at the midpoint.

Common Mistakes

Using the wrong distance for the beam’s weight (it must be $10.0\ \text{m}$ from $X$ ).
Forgetting that $R_X$ has zero moment about $X$ .
Writing the moment equation with clockwise/anticlockwise swapped but not being consistent.
Assuming reactions are equal (they are only equal if loads are symmetrical).

Things to Be Careful About

Always use perpendicular distances to the line of action of each force.
Keep units consistent (here all distances are in $\text{m}$ , so moments are in $\text{N m}$ ).
After finding one reaction from moments, always check with $R_X + R_Y = \text{total weight}$ for consistency.

Techniques used

apply vertical force equilibriumtake moments about a support to eliminate an unknown reactionsolve simultaneous linear equilibrium equations

Q161MMedium-EasyWork, Energy and PowerDynamics

The diagram shows a block P of mass $M$ connected by a string over a frictionless pulley to a block Q of mass $m$ .

Fig. 1

Block P moves up the slope with a constant velocity $v$ . The slope is at angle $\theta$ to the horizontal.

The acceleration of free fall is $g$ .

The resistive forces on the blocks are negligible.

Which expressions give the energy transferred per unit time to block P?

1 $mgv$
2 $Mgv \sin \theta$
3 $(M + m)gv$

Options

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

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Constant velocity $\Rightarrow a=0$ .

For block Q: $mg - T = 0 \Rightarrow T = mg$ .

For block P along slope: $T - Mg\sin\theta = 0 \Rightarrow T = Mg\sin\theta$ .

Power transferred to P by the tension:

P = Tv = mgv = Mgv\sin\theta

Answer

A

Final answer

A

Detailed explanation

Background Concept

Energy transferred per unit time is power:

P = \frac{\text{energy transferred}}{t}

When a force acts in the direction of motion at constant speed, the rate of doing work is

P = Fv

Also, if an object gains gravitational potential energy $E_p = mgh$ , then the rate of increase is

P = \frac{\Delta E_p}{t} = mg\frac{\Delta h}{t} = mgv_\text{vertical}

Understanding the Question

Block P (mass $M$ ) is pulled up an incline at speed $v$ and block Q (mass $m$ ) moves down at the same speed because they are connected by a light string over a frictionless pulley. Resistive forces are negligible and the speed is constant, so there is no change in kinetic energy. The question asks which listed expressions equal the power transferred to block P.

Approach

Use “constant velocity” to set the acceleration to zero, so forces on each block balance.
Find the tension $T$ using Newton’s second law on each block.
Power delivered to block P is the work done per second by the tension: $P = Tv$ .
Compare the result to expressions 1–3.

Step-by-Step Reasoning

Since the blocks move with constant velocity, $a=0$ .

For block Q (hanging): forces are weight $mg$ downward and tension $T$ upward.

mg - T = 0 \Rightarrow T = mg

For block P (on the slope): along the slope, the tension pulls up the slope and the component of weight down the slope is $Mg\sin\theta$ .

T - Mg\sin\theta = 0 \Rightarrow T = Mg\sin\theta

So, in this steady motion, $mg = Mg\sin\theta$ .

Power transferred to P by the string is the rate at which the tension does work on P:

P = Tv

Substitute either expression for $T$ :

P = mgv

and also

P = Mgv\sin\theta

Hence expressions 1 and 2 are correct. Expression 3, $(M+m)gv$ , does not match $Tv$ and also has no reason to include $M+m$ in a simple power balance here.

Therefore the correct option is A (1 and 2).

Key Takeaways

Constant velocity implies $a=0$ , so net force is zero for each block.
Power delivered by a force in the direction of motion is $P = Fv$ .
With no losses and no change in kinetic energy, power lost by one part of a system equals power gained by another.

Common Mistakes

Using $Mg\cos\theta$ instead of $Mg\sin\theta$ for the component of weight along the slope.
Forgetting that constant velocity means acceleration is zero (so forces balance).
Treating $(M+m)gv$ as the power because two masses are involved; power depends on the force doing work on P (the tension) and the speed.

Things to Be Careful About

The power “to block P” is from the tension acting on P, not from adding powers of both blocks.
The vertical speed of P is $v\sin\theta$ , so the GPE gain rate of P is $Mg(v\sin\theta)$ , consistent with $Tv$ .
The equality $mgv = Mgv\sin\theta$ only holds here because there are no resistive forces and the motion is at constant speed (so tension values match both blocks’ equilibrium conditions).

Techniques used

apply Newton's second law with zero acceleration to each blockuse power as rate of work done (P = Fv)relate power to rate of change of gravitational potential energy

Q171MMedium-EasyWork, Energy and Power

A toy car travels around a vertical loop track.

Fig. 1

The toy car is released from rest at a height of $58 \text{ cm}$ above the bottom of the vertical loop.

The car is at a height of $32 \text{ cm}$ when it is at the top of the vertical loop.

Assume that no resistive forces act on the car.

What is the speed of the car at the top of the vertical loop?

Options

A $0.87 \text{ m s}^{-1}$
B $2.3 \text{ m s}^{-1}$
C $2.5 \text{ m s}^{-1}$
D $3.4 \text{ m s}^{-1}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Loss of GPE from release to top of loop:

\Delta h = 0.58 - 0.32 = 0.26\ \text{m}

Conservation of energy:

mg\Delta h = \frac{1}{2}mv^2

v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.81 \times 0.26} = 2.26\ \text{m s}^{-1} \approx 2.3\ \text{m s}^{-1}

Answer

B

Final answer

B

Detailed explanation

Background Concept

With no resistive forces, the toy car’s mechanical energy is conserved. That means the sum of gravitational potential energy (GPE) and kinetic energy (KE) stays constant.

Gravitational potential energy relative to some reference level is

E_p = mgh

where $m$ is mass, $g$ is gravitational field strength, and $h$ is vertical height.

Kinetic energy is

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

where $v$ is speed.

If the car starts from rest, its initial KE is zero, so any loss of GPE becomes gain in KE.

Understanding the Question

The car is released from rest at height $58\ \text{cm}$ above the bottom of the loop and later is at the top of the loop at height $32\ \text{cm}$ above the bottom.

We are asked for the car’s speed at the top. Since resistive forces are neglected, we use energy conservation between these two positions.

The key quantity is the vertical drop in height from start to the top of the loop:

\Delta h = 58\ \text{cm} - 32\ \text{cm}

Approach

Convert cm to m so SI units match $g$ .
Find the change in height $\Delta h$ from release point to the top.
Use conservation of energy: $mg\Delta h = \tfrac{1}{2}mv^2$ .
Solve for $v$ and match to the nearest option.

Step-by-Step Reasoning

Heights in metres:

0.58\ \text{m},\quad 0.32\ \text{m}

So the drop in height is

\Delta h = 0.58 - 0.32 = 0.26\ \text{m}

Initial KE is zero (released from rest). Loss of GPE equals gain in KE:

mg\Delta h = \frac{1}{2}mv^2

Mass cancels (so we never need $m$ ):

g\Delta h = \frac{1}{2}v^2

Hence

v = \sqrt{2g\Delta h}

Substitute $g = 9.81\ \text{m s}^{-2}$ and $\Delta h = 0.26\ \text{m}$ :

v = \sqrt{2 \times 9.81 \times 0.26} = \sqrt{5.1012} = 2.26\ \text{m s}^{-1}

Rounding to two significant figures gives

v \approx 2.3\ \text{m s}^{-1}

This corresponds to option B.

Key Takeaways

With no resistive forces, use conservation of mechanical energy.
Use height difference (not the absolute height) to find the KE gained.
In $mg\Delta h = \tfrac{1}{2}mv^2$ , the mass cancels, simplifying the calculation.

Common Mistakes

Using $0.58\ \text{m}$ or $0.32\ \text{m}$ as $\Delta h$ instead of the difference $0.26\ \text{m}$ .
Forgetting to convert cm to m (e.g. using $26$ instead of $0.26$ ).
Writing $mg\Delta h = mv^2$ (missing the $\tfrac{1}{2}$ factor), which would overestimate $v$ .

Things to Be Careful About

Heights are measured from the bottom of the loop in both cases, so subtraction is valid.
Keep units consistent: $g$ in $\text{m s}^{-2}$ requires $h$ in $\text{m}$ .
The question asks for speed at the top, not whether the car maintains contact; only energy is needed here.

Techniques used

apply conservation of mechanical energyconvert given heights into SI unitsequate loss of gravitational potential energy to gain in kinetic energysolve for speed from kinetic energy expression

Q181MEasyWork, Energy and Power

The total energy supplied to an electric motor is $E$ . Energy $Q$ is wasted and the remaining energy does useful work.

What is the efficiency of the motor?

Options

A $\frac{Q}{E}$
B $\left( \frac{Q}{E} \right) - 1$
C $1 - \left( \frac{Q}{E} \right)$
D $\frac{(1 - Q)}{E}$

Pick an answer

DifficultyEasy

Worked solution

Working

Total input energy $= E$ .

Wasted energy $= Q$ so useful energy $= E - Q$ .

\text{efficiency} = \frac{\text{useful}}{\text{input}} = \frac{E-Q}{E} = 1 - \frac{Q}{E}

Answer

C

Final answer

C

Detailed explanation

Background Concept

Efficiency describes how much of the input energy (or power) is converted into useful output.

\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}

It is often expressed as a percentage by multiplying by $100\%$ . The wasted energy is the part of the input that is not converted into the desired form.

Understanding the Question

The motor receives total energy $E$ . Of this, an amount $Q$ is wasted (e.g. as heat and sound). The rest is useful work done by the motor.

So we must find an expression for efficiency in terms of $E$ and $Q$ , then match it to the options.

Approach

Write the useful energy as: input minus wasted.
Substitute into the definition of efficiency.
Simplify and compare with the choices.

Step-by-Step Reasoning

Useful energy:

E_{\text{useful}} = E - Q

Efficiency:

\text{efficiency} = \frac{E_{\text{useful}}}{E} = \frac{E-Q}{E}

Split the fraction:

\frac{E-Q}{E} = \frac{E}{E} - \frac{Q}{E} = 1 - \frac{Q}{E}

This corresponds to option C.

Key Takeaways

Always start from $\text{efficiency} = \frac{\text{useful}}{\text{input}}$ .
If wasted energy is given, useful energy is $\text{input} - \text{wasted}$ .

Common Mistakes

Using $\frac{Q}{E}$ (this is the wasted fraction, not the efficiency).
Writing $\frac{1-Q}{E}$ (incorrect because $Q$ is an energy, not a dimensionless number; it must be subtracted from $E$ , not from $1$ ).
Forgetting that efficiency must be dimensionless.

Things to Be Careful About

Check dimensional consistency: $Q/E$ is dimensionless, so $1 - Q/E$ is valid.
Ensure you subtract energies before dividing: $\frac{E-Q}{E}$ , not $\frac{E}{E-Q}$ .
If asked for a percentage, multiply the final fraction by $100\%$ (not required here).

Techniques used

identify useful output energy from total input minus wasted energyapply efficiency = useful output / total inputsimplify an algebraic expression to match an option

Q191MMedium-EasyWork, Energy and Power

An object travels between two points. The change in gravitational potential energy $\Delta E_p$ of the object is given by

\Delta E_p = mg\Delta h

where $m$ is the mass of the object, $\Delta h$ is its change in height and $g$ is the acceleration of free fall.

What is a necessary condition in order for the above equation to be valid?

Options

A The object must have a constant acceleration of $9.81 \text{ m s}^{-2}$ .
B The object must be travelling in a uniform gravitational field.
C The object must be travelling only in a vertical direction.
D The resultant force on the object must be equal to its weight.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Using

\Delta E_p = mg\Delta h

requires $g$ to be constant over the height change, i.e. a uniform gravitational field.

Answer

B

Final answer

B

Detailed explanation

Background Concept

For small height changes near Earth’s surface, the gravitational field strength $g$ is approximately constant. In that situation, the increase in gravitational potential energy depends only on the vertical height change:

\Delta E_p = mg\Delta h

This comes from the work done against weight. If the gravitational field is uniform, the weight $mg$ is constant, so the work done in raising the object by $\Delta h$ is simply force $\times$ vertical distance.

Understanding the Question

You are given the formula

\Delta E_p = mg\Delta h

and asked for a necessary condition for it to be valid. The options mention acceleration, uniform field, vertical motion, and resultant force. The key issue is: when is it legitimate to treat $g$ (and hence weight $mg$ ) as constant?

Approach

Link $\Delta E_p$ to work done against gravity. The expression $mg\Delta h$ assumes the gravitational force is constant over the displacement, which is true in a uniform gravitational field (constant $g$ ). Then check each option against what is actually required.

Step-by-Step Reasoning

In a uniform gravitational field, the gravitational force on mass $m$ is constant:

W = mg

The work done against this force to raise the object through vertical height change $\Delta h$ is

\Delta E_p = W\,\Delta h = mg\Delta h

Therefore the required condition is that $g$ is constant over the motion, i.e. the object is travelling in a uniform gravitational field.
Check the distractors:

A is not required: the object does not need acceleration $9.81\ \text{m s}^{-2}$ (it could be lifted at constant speed, for example).
C is not required: the path can be non-vertical; only the change in vertical height matters.
D is not required: the resultant force need not equal weight (there may be tension, drag, thrust, etc.).

So the correct option is B.

Key Takeaways

$\Delta E_p = mg\Delta h$ assumes $g$ is constant (uniform field).
Gravitational potential energy change depends only on vertical height change, not on the path taken.
Force/acceleration conditions (like “resultant force equals weight”) are not conditions for defining $\Delta E_p$ .

Common Mistakes

Thinking the object must be in free fall (acceleration $= g$ ) for gravitational potential energy changes to apply.
Assuming the motion must be vertical; in fact only $\Delta h$ matters.
Confusing “weight is a force” with “resultant force”; the resultant can be different from $mg$ .

Things to Be Careful About

For large altitude changes, $g$ is not constant and the simple $mg\Delta h$ approximation becomes less valid.
Ensure you interpret “necessary condition” as something required by the assumptions behind the formula, not by a particular type of motion.

Techniques used

recall the expression for change in gravitational potential energy near Earthidentify the physical assumptions behind using constant geliminate options that confuse force/acceleration conditions with energy change

Q201MMedium-EasyDeformation of Solids

A cylindrical steel rod with negligible weight has a diameter of $1.5 \text{ cm}$ and a length of $5.2 \text{ cm}$ . The rod is firmly attached at the top end.

The rod is firmly attached to a machine at the other end that applies a constant force of $363 \text{ N}$ to the rod.

This causes the rod to extend to a length of $7.4 \text{ cm}$ and to have a minimum diameter of $0.60 \text{ cm}$ in the position shown.

Fig. 1

What is the maximum stress acting on the steel rod when the length is $7.4 \text{ cm}$ ?

Options

A $2.1 \times 10^6 \text{ Pa}$
B $3.2 \times 10^6 \text{ Pa}$
C $5.7 \times 10^6 \text{ Pa}$
D $1.3 \times 10^7 \text{ Pa}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Maximum stress occurs at the minimum diameter $d = 0.60\ \text{cm} = 6.0 \times 10^{-3}\ \text{m}$ .

A = \pi\left(\frac{d}{2}\right)^2 = \pi\left(3.0 \times 10^{-3}\right)^2 = 2.83 \times 10^{-5}\ \text{m}^2

\sigma = \frac{F}{A} = \frac{363}{2.83 \times 10^{-5}} = 1.28 \times 10^{7}\ \text{Pa} \approx 1.3 \times 10^{7}\ \text{Pa}

Answer

D

Final answer

D

Detailed explanation

Background Concept

Stress describes how concentrated a force is within a material. For a rod in tension,

\sigma = \frac{F}{A}

where $\sigma$ is the tensile stress (in Pa), $F$ is the tensile force (in N), and $A$ is the cross-sectional area perpendicular to the force (in $\text{m}^2$ ).

If the rod is not the same thickness everywhere, then the stress is not the same everywhere either. For the same applied force, a smaller area gives a larger stress. Therefore the maximum stress occurs where the rod has its minimum cross-sectional area.

Understanding the Question

A constant force of $363\ \text{N}$ stretches the steel rod so that it narrows in the middle to a minimum diameter of $0.60\ \text{cm}$ . The question asks for the maximum stress when the rod has stretched to $7.4\ \text{cm}$ . Since the force is constant, the length is not directly needed for stress; the key is the smallest diameter (smallest area).

Approach

Identify where stress is greatest: at the narrowest point (minimum diameter).
Convert the minimum diameter from cm to m.
Use the diameter to find area $A = \pi (d/2)^2$ .
Compute $\sigma = F/A$ and choose the matching option.

Step-by-Step Reasoning

Minimum diameter:

d = 0.60\ \text{cm} = 0.60 \times 10^{-2}\ \text{m} = 6.0 \times 10^{-3}\ \text{m}

Radius:

r = \frac{d}{2} = 3.0 \times 10^{-3}\ \text{m}

Cross-sectional area at the narrowest point:

A = \pi r^2 = \pi (3.0 \times 10^{-3})^2 = \pi (9.0 \times 10^{-6}) = 2.83 \times 10^{-5}\ \text{m}^2

Stress:

\sigma = \frac{F}{A} = \frac{363}{2.83 \times 10^{-5}} = 1.28 \times 10^{7}\ \text{Pa}

To 2 significant figures (matching the options):

\sigma \approx 1.3 \times 10^{7}\ \text{Pa}

This corresponds to option D.

Key Takeaways

Tensile stress is $\sigma = F/A$ .
With a varying diameter, the maximum stress occurs at the minimum area.
Always convert to SI units before substituting.

Common Mistakes

Using the original diameter $1.5\ \text{cm}$ instead of the minimum diameter $0.60\ \text{cm}$ (this would give a much smaller stress).
Forgetting to convert $\text{cm}$ to $\text{m}$ , leading to an answer wrong by factors of $10^4$ .
Using $A = \pi d^2$ instead of $A = \pi (d/2)^2$ .

Things to Be Careful About

The length change is a distraction for this particular question: stress here depends on force and cross-sectional area only.
The unit of stress is $\text{Pa} = \text{N m}^{-2}$ , so the area must be in $\text{m}^2$ .
Maximum stress means you must choose the smallest cross-section shown.

Techniques used

identify the location of maximum stress from the minimum cross-sectional areaconvert centimetres to metrescalculate cross-sectional area from diameterapply the definition of stress as force divided by areamatch the calculated value to the closest option

Q211MMedium-EasyDeformation of Solids

The graph shows the variation in extension with force for a sample of rubber.

Fig. 1

The top line shows the variation in extension as a force is applied.

The bottom line shows the variation in extension as the force is removed.

What is represented by the area between the two lines?

Options

A the work done as the force is applied
B the work done as the force is applied minus the work done as the force is removed
C the work done as the force is removed
D the work done as the force is removed plus the work done as the force is applied

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Work done when force is applied is the area under the loading curve.

Work done by the rubber when force is removed is the area under the unloading curve.

So, area between the curves = (area under loading curve) $-$ (area under unloading curve).

Answer

B

Final answer

B

Detailed explanation

Background Concept

For a force $F$ that causes an extension $x$ , the work done is

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

On a force–extension graph (force on the vertical axis, extension on the horizontal axis), the area under the curve between two extensions represents the work done (energy transferred).

For materials like rubber, the loading (increasing force) and unloading (decreasing force) curves are not the same. This is called hysteresis and indicates that not all the energy put in during stretching is recovered when the force is removed; some is dissipated (usually as internal heating).

Understanding the Question

You are shown two curves on the same $F$ – $x$ graph:

the upper curve for when the force is applied (loading),
the lower curve for when the force is removed (unloading).

The question asks what physical quantity is represented by the area between these two curves.

Approach

Use the fact that area under an $F$ – $x$ curve is work done.
Identify what the two separate areas (under loading and under unloading curves) represent.
Subtract to get what the area between the curves represents.

Step-by-Step Reasoning

When the force is applied and the rubber is stretched from $x=0$ to some extension $x=x_\text{max}$ , the work done on the rubber is

W_\text{apply} = \int_0^{x_\text{max}} F_\text{loading}\,\mathrm{d}x

This is the area under the top curve.

When the force is removed, the rubber contracts back. The work done by the rubber (energy returned) corresponds to the area under the bottom curve:

W_\text{remove} = \int_0^{x_\text{max}} F_\text{unloading}\,\mathrm{d}x

Because the unloading curve lies below the loading curve, $W_\text{remove} < W_\text{apply}$ . The difference

W_\text{apply} - W_\text{remove}

is the energy not recovered: it is dissipated (lost) in the rubber.

Geometrically, this difference is exactly the area between the two curves (the hysteresis loop).

Therefore the correct option is: the work done as the force is applied minus the work done as the force is removed.

Key Takeaways

Area under an $F$ – $x$ graph gives work done: $W=\int F\,\mathrm{d}x$ .
Different loading and unloading curves indicate hysteresis.
The area between the curves represents energy dissipated (work input minus work returned).

Common Mistakes

Saying it is just “the work done as the force is applied” (that is the area under the loading curve, not the area between curves).
Adding the two works rather than taking the difference.
Confusing the axes: it must be force vs extension (not force vs time) for area to represent work.

Things to Be Careful About

The unloading area corresponds to energy returned; the hysteresis area corresponds to energy lost per loading–unloading cycle.
The question wording “force is removed” refers to the unloading curve; the sign convention for work can be confusing, so focus on the geometric interpretation: difference in areas = energy dissipated.

Techniques used

interpret the area under a force–extension graph as work donecompare loading and unloading curves to identify energy dissipatedrelate the hysteresis loop area to energy loss per cycle

Q221MMedium-EasyDeformation of Solids

Two identical springs have the same spring constant $k$ . The springs are connected in parallel.

The length of the unstretched springs is $x$ .

A force $F$ is applied to the spring combination. The length of the springs is now $y$ .

Fig. 1

Both springs are deformed within their limits of proportionality.

Which expression gives the elastic potential energy stored in one of the springs?

Options

A $\frac{1}{4}F(y - x)$
B $\frac{1}{2}F(y - x)$
C $\frac{1}{4}k(y - x)^2$
D $k(y - x)^2$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Extension of each spring:

e = y - x

In parallel, the total force $F$ is shared equally, so force in one spring:

F_1 = \frac{F}{2}

Elastic potential energy in one spring:

E_1 = \frac{1}{2}F_1 e = \frac{1}{2}\left(\frac{F}{2}\right)(y-x)=\frac{1}{4}F(y-x)

Answer

A

Final answer

A

Detailed explanation

Background Concept

For a spring obeying Hooke’s law (within the limit of proportionality), the force and extension are related by

F = k e

where $k$ is the spring constant and $e$ is the extension.

The elastic potential energy stored is the work done in stretching it from $0$ to $e$ . Because the force increases linearly from $0$ to $F$ , this energy is the area under the force–extension graph:

E = \text{area} = \frac{1}{2}Fe = \frac{1}{2}ke^2

Understanding the Question

Two identical springs (each spring constant $k$ ) are connected in parallel and pulled so that each spring changes length from $x$ (unstretched) to $y$ .

So the extension of each spring is $e = y-x$ . A total downward force $F$ is applied to the combination. The question asks for the elastic potential energy stored in one spring, written in terms of the given symbols.

diagram-1

Approach

In a parallel arrangement, both springs have the same extension (they are attached to the same top support and the same bottom bar).
The applied force $F$ is the sum of the forces from the two springs, so identical springs share the force equally.
Use $E = \tfrac{1}{2}Fe$ for a single spring, with the single-spring force and the common extension.

Step-by-Step Reasoning

The extension of each spring is

e = y-x.

Because the springs are identical and stretched by the same amount, each exerts the same restoring force, so

F = F_1 + F_2 = 2F_1 \;\;\Rightarrow\;\; F_1 = \frac{F}{2}.

Elastic potential energy in one spring is

E_1 = \frac{1}{2}F_1 e.

Substitute $F_1 = \tfrac{F}{2}$ and $e = y-x$ :

E_1 = \frac{1}{2}\left(\frac{F}{2}\right)(y-x) = \frac{1}{4}F(y-x).

This matches option A.

(If you instead use $E_1 = \tfrac{1}{2}ke^2$ , you would get $E_1 = \tfrac{1}{2}k(y-x)^2$ , but that form is not offered; the options are expressed using $F$ .)

Key Takeaways

Springs in parallel have the same extension.
The total force is shared: for identical springs, each carries $F/2$ .
Elastic potential energy for a Hooke’s-law spring is $E = \tfrac{1}{2}Fe$ .

Common Mistakes

Using the total force $F$ as the force in one spring (forgetting the force is shared), leading to $E = \tfrac{1}{2}F(y-x)$ (option B).
Treating parallel springs like series springs (in series, the force is the same in each spring but extensions add).
Using $E = Fe$ instead of $E = \tfrac{1}{2}Fe$ (forgetting that force increases from $0$ to $F$ ).

Things to Be Careful About

Identify correctly whether the arrangement is series or parallel: here the bottom bar ensures equal extension.
Use $e = y-x$ (extension), not $y$ or $x$ alone.
Make sure the energy asked is for one spring, not the total for both springs.

Techniques used

recognise that identical springs in parallel share the loadapply Hooke's law to relate force and extensionuse elastic potential energy as the area under a force–extension graph

Q231MEasyDeformation of Solids

Which expression gives the formula for the spring constant?

Options

A $Fx$
B $2Fx^2$
C $\frac{F}{x}$
D $\frac{F}{x^2}$

Pick an answer

Q241MMediumWaves

A transverse progressive wave travels along a string.

The graph shows the variation with distance of the displacement of the string at time $t = 0$ .

Fig. 1

Which graph represents the variation with time of the velocity of point P on the string?

Options

Pick an answer

DifficultyMedium

Worked solution

Working

For a wave travelling to the right,

y(x,t)=f(x-vt)

So the transverse velocity of a point on the string is

v_y=\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}.

At $P$ (at $t=0$ ) the displacement is zero and the wave profile is rising with $x$ so $\partial y/\partial x>0$ .
Hence $v_y<0$ (maximum magnitude at a zero crossing for a sinusoid).

Answer

D

Final answer

D

Detailed explanation

Background Concept

A transverse progressive wave on a string can be written (for motion to the right) as

y(x,t)=f(x-vt),

where $y$ is the transverse displacement, $x$ is position along the string, and $v$ is the wave speed.

The transverse particle velocity of a particular point on the string is the time rate of change of its displacement:

v_y=\frac{\partial y}{\partial t}.

Differentiating $y(x,t)=f(x-vt)$ gives

\frac{\partial y}{\partial t}=-v\,f'(x-vt)

and since

\frac{\partial y}{\partial x}=f'(x-vt),

we get the important link (for a right-moving wave):

v_y=-v\frac{\partial y}{\partial x}.

So at any instant, the sign of the particle velocity is the opposite of the sign of the slope of the displacement–distance graph.

Understanding the Question

You are given the wave shape $y$ against distance $x$ at $t=0$ and told the wave travels to the right.

Point $P$ is located at a zero crossing where the curve is rising as $x$ increases (positive slope).

The question asks which of the four graphs shows how the velocity of point P varies with time.

Approach

Use the fact the wave travels to the right to choose $y(x,t)=f(x-vt)$ .
Use $v_y=\partial y/\partial t=-v\,\partial y/\partial x$ to find the sign of the velocity at $t=0$ from the slope at $P$ .
Use the fact that in a sinusoidal progressive wave, at a zero displacement the particle speed is maximum (so the velocity is at an extreme, not zero).
Pick the option whose $v_y(t)$ is sinusoidal with the correct initial value/sign.

Step-by-Step Reasoning

diagram-1

The displacement–distance graph at $t=0$ shows at point $P$ :
- $y=0$ (it is on the axis),
- the curve is rising with distance, so $\partial y/\partial x>0$ .
Because the wave travels to the right,

v_y=-v\frac{\partial y}{\partial x}.

So with $\partial y/\partial x>0$ we must have $v_y<0$ at $t=0$ .

Also, $P$ is at a zero crossing of a sinusoidal wave shape. For a sinusoidal motion of a particle, when displacement is zero, the speed is maximum. Therefore the velocity should be at its most negative value at that instant (an extreme), and then later will become less negative, pass through zero, become positive, etc. So the time graph must be sinusoidal and start negative.
Among the options, the only sinusoidal $v$ – $t$ curve that starts with a negative value at $t=0$ is D.

Key Takeaways

For a right-travelling wave, particle velocity and wave-profile slope are related by:

v_y=-v\frac{\partial y}{\partial x}.

- At a zero crossing of a sinusoidal wave, the particle speed is maximum. - The direction of travel matters: if the wave travelled left, the sign would reverse. ## Common Mistakes - Assuming velocity is zero whenever displacement is zero (it is actually maximum for SHM). - Ignoring the wave direction arrow: using $y=f(x+vt)$ instead of $y=f(x-vt)$ changes the sign. - Confusing wave speed (along the string) with particle velocity (transverse motion of a point on the string). ## Things to Be Careful About - Always distinguish: slope of $y$ vs $x$ (a spatial gradient) from the change of $y$ with $t$ (particle velocity). - At an extreme (maximum/minimum) of a sinusoid, velocity is zero; at a zero crossing, velocity magnitude is maximum. - Ensure you use the correct sign convention consistently (right-moving: $x-vt$).

Techniques used

use the right-travelling wave form y(x,t) = f(x - vt)differentiate a wave function to relate particle velocity to wave profile slopededuce the sign and phase of the particle velocity from the displacement–distance graphidentify the matching velocity–time graph from phase and sign at t = 0

Q251MMedium-EasyWaves

Which statement about electromagnetic waves is correct?

Options

A A wave of wavelength $5.0 \times 10^{-6} \text{ m}$ is invisible to the human eye.
B They can all travel at different speeds in free space.
C They cannot be polarised.
D They consist of vibrating atoms.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For $\lambda = 5.0 \times 10^{-6}\,\text{m}$ , this is infrared radiation, which is not visible to the human eye, so A is correct.

B is false because all electromagnetic waves travel at $c$ in free space. C is false because electromagnetic waves are transverse and can be polarised. D is false because electromagnetic waves are oscillating electric and magnetic fields, not vibrating atoms.

Answer

A

Final answer

A

Detailed explanation

Background Concept

Electromagnetic (EM) waves are transverse waves made of oscillating electric field ( $\vec{E}$ ) and magnetic field ( $\vec{B}$ ) components, perpendicular to each other and to the direction of travel. In free space (vacuum), all EM waves travel at the same speed:

c = 3.0 \times 10^8\,\text{m s}^{-1}

The electromagnetic spectrum is classified by wavelength (or frequency). Visible light is roughly $4 \times 10^{-7}\,\text{m}$ to $7 \times 10^{-7}\,\text{m}$ . Wavelengths longer than visible (e.g. micrometres) are infrared.

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

Understanding the Question

You are given four statements about EM waves and must choose the single correct one.

Option A gives a wavelength and asks whether it is visible.
Options B–D test key properties: speed in vacuum, polarisation, and what “is vibrating” in the wave.

Approach

Convert/interpret the wavelength in A and locate it on the EM spectrum relative to visible light.
Use core facts: in free space all EM waves have the same speed; EM waves are transverse so can be polarised; EM waves are fields rather than vibrations of atoms.
Pick the only statement that remains true.

Step-by-Step Reasoning

Option A:
Given $\lambda = 5.0 \times 10^{-6}\,\text{m} = 5\,\mu\text{m}$ .
Visible light is around $0.4$ to $0.7\,\mu\text{m}$ . Since $5\,\mu\text{m}$ is much longer, it lies in the infrared region, which is invisible to the human eye. So A is true.

Option B:
“All travel at different speeds in free space” is false: all EM waves travel at the same speed $c$ in vacuum.

Option C:
“They cannot be polarised” is false: EM waves are transverse, so they can be polarised.

Option D:
“They consist of vibrating atoms” describes mechanical waves (like sound). EM waves are oscillations of electric and magnetic fields, not atoms, so D is false.

Therefore, only A is correct.

Key Takeaways

Use wavelength ranges to identify parts of the EM spectrum (visible vs infrared).
In vacuum, all EM waves have the same speed $c$ .
EM waves are transverse and can be polarised.
EM waves are oscillating fields, not vibrations of matter.

Common Mistakes

Thinking different EM waves travel at different speeds in vacuum (confusing vacuum with materials where speed depends on refractive index).
Saying EM waves cannot be polarised (mixing up transverse and longitudinal waves).
Treating EM waves as mechanical waves needing a medium (vibrating atoms).
Misremembering visible wavelengths by orders of magnitude (e.g. confusing $10^{-6}$ with $10^{-9}$ ).

Things to Be Careful About

Compare wavelengths using powers of ten: $10^{-6}\,\text{m}$ (micrometre, IR) vs $10^{-7}\,\text{m}$ (visible).
The phrase “free space” means vacuum; the constant speed statement applies there.
Polarisation is a strong clue that the wave is transverse.

Techniques used

identify the region of the electromagnetic spectrum from wavelengthrecall that all electromagnetic waves have the same speed in free spaceuse the transverse nature of electromagnetic waves to assess polarisation statementsdistinguish electromagnetic waves from mechanical vibrations of matter

Q261MEasySuperposition

For a progressive transverse wave, what describes the term diffraction?

Options

A the change in observed frequency of a wave due to a movement of the wave source
B the spreading of a wave as it passes through a gap or around an obstacle
C when oscillations of a wave are confined to one plane
D the resultant displacement of two waves is equal to the sum of the displacements of the individual waves when the waves meet

Pick an answer

DifficultyEasy

Worked solution

Working

Diffraction is the spreading of a wave when it passes through a gap or around an obstacle.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Diffraction is a wave effect where a wave spreads out after passing through a narrow opening (gap) or around the edge of an obstacle. The effect is most noticeable when the size of the gap/obstacle is comparable to the wavelength.

Understanding the Question

You are asked which option correctly describes the term diffraction for a progressive transverse wave. The options include definitions of other wave phenomena (Doppler effect, polarisation, superposition/interference), so the task is to identify the correct definition.

Approach

Recall the definition of diffraction (spreading around gaps/edges) and compare it directly with each option.

Step-by-Step Reasoning

Option A describes a change in observed frequency due to motion of the source: this is the Doppler effect, not diffraction.
Option B states “the spreading of a wave as it passes through a gap or around an obstacle”: this is precisely diffraction.
Option C describes oscillations confined to one plane: this is polarisation.
Option D describes the displacements adding when waves meet: this is the principle of superposition.
Therefore, the correct option is B.

Key Takeaways

Diffraction means spreading of waves at apertures and edges.
It is distinct from Doppler effect (frequency shift), polarisation (plane of oscillation), and superposition (addition of displacements).

Common Mistakes

Confusing diffraction with interference/superposition (both involve waves, but diffraction is specifically spreading at gaps/edges).
Choosing polarisation (transverse-only property) because the question mentions “transverse wave”.

Things to Be Careful About

The word “spreading” is a strong clue for diffraction.
Diffraction is not defined by a frequency change (that is Doppler effect), nor by adding displacements (that is superposition).

Techniques used

recall the definition of diffractionmatch a physical definition to the correct multiple-choice option

Q271MMedium-EasyWaves

An aircraft flies at a velocity $v_s$ directly away from a stationary observer.

The aircraft emits a sound of constant frequency.

The speed of sound in air is $v$ .

The frequency of the sound heard by the observer on the ground is $500 \text{ Hz}$ .

The speed of the aircraft is increased so that it flies away from the observer at a greater velocity.

The observer now hears a sound of frequency $250 \text{ Hz}$ .

Which expression gives the new velocity of the aircraft?

Options

A $2(v + v_s)$
B $\frac{v + v_s}{2}$
C $2v_s + v$
D $2v_s - v$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a source moving away from a stationary observer,

f = f_0\frac{v}{v+v_s}

Initially:

500 = f_0\frac{v}{v+v_s}

After increasing the aircraft speed to $v_2$ :

250 = f_0\frac{v}{v+v_2}

Divide the equations:

\frac{250}{500} = \frac{v+v_s}{v+v_2}

So

\frac{1}{2} = \frac{v+v_s}{v+v_2} \Rightarrow v+v_2 = 2(v+v_s)

v_2 = 2(v+v_s) - v = v + 2v_s

Answer

C

Final answer

C

Detailed explanation

Background Concept

The Doppler effect for sound is the change in the observed frequency when there is relative motion between the source and the observer.

For a stationary observer and a moving source:

if the source moves towards the observer, wavefronts are compressed and the observed frequency increases,
if the source moves away, wavefronts are stretched and the observed frequency decreases.

For a source moving away at speed $v_s$ in a medium where the sound speed is $v$ , the observed frequency $f$ is

f = f_0\frac{v}{v+v_s}

where $f_0$ is the emitted (true) frequency of the source.

Understanding the Question

An aircraft (the sound source) moves directly away from a stationary observer.

When its speed is $v_s$ , the observer hears $500\ \text{Hz}$ .
The aircraft speeds up to a new speed (call it $v_2$ ), and the observer now hears $250\ \text{Hz}$ .

The emitted frequency stays constant, so we can use the two Doppler expressions to eliminate $f_0$ and solve for $v_2$ in terms of $v$ and $v_s$ .

Approach

Write the Doppler formula for a receding source for the first situation and the second situation.
Divide one equation by the other to cancel the unknown emitted frequency $f_0$ .
Rearrange to obtain $v_2$ , then match it to one of the options.

Step-by-Step Reasoning

For a receding source,

f = f_0\frac{v}{v+v_s}

First speed $v_s$ gives observed frequency $500\ \text{Hz}$ :

500 = f_0\frac{v}{v+v_s}

New speed $v_2$ gives observed frequency $250\ \text{Hz}$ :

250 = f_0\frac{v}{v+v_2}

Divide the second by the first to eliminate $f_0$ and $v$ :

\frac{250}{500} = \frac{v+v_s}{v+v_2}

So

\frac{1}{2} = \frac{v+v_s}{v+v_2}

Cross-multiply:

v+v_2 = 2(v+v_s)

Then

v_2 = 2(v+v_s)-v = v+2v_s

This corresponds to option C: $2v_s+v$ .

Key Takeaways

For a moving source and stationary observer, use $f = f_0\dfrac{v}{v\pm v_s}$ .
Moving away makes the denominator $v+v_s$ (frequency decreases).
When the emitted frequency is unknown but constant, forming a ratio is the fastest way to remove it.

Common Mistakes

Using the formula for a moving observer instead of a moving source.
Putting the wrong sign in the denominator (using $v-v_s$ when the source is moving away).
Assuming frequency halves implies speed doubles directly (it does not; the relationship is through $v+v_s$ ).

Things to Be Careful About

The speed of sound $v$ is relative to the air; the Doppler formula is written in terms of $v$ and the source speed relative to the air.
“Directly away” means the motion is along the line joining source and observer, so no component resolution is needed.
Check plausibility: halving the observed frequency requires a substantial increase in the denominator $v+v_s$ , so $v_2$ must be larger than $v_s$ .

Techniques used

apply the Doppler shift formula for a source moving away from a stationary observereliminate the emitted frequency by forming a ratio of two observationsrearrange an algebraic expression to solve for the new speed

Q281MEasyWaves

What may be observed with light waves but not with sound waves?

Options

A diffraction
B interference
C polarisation
D reflection

Pick an answer

DifficultyEasy

Worked solution

Working

Polarisation can occur only for transverse waves. Light is transverse but sound in air is longitudinal.

Answer

C

Final answer

C

Detailed explanation

Background Concept

Wave phenomena such as diffraction, interference and reflection can occur for many types of waves (both transverse and longitudinal), provided the physical conditions are right.

Polarisation is different: it is the restriction of the direction of oscillation of a transverse wave to one plane/direction. A longitudinal wave oscillates parallel to the direction of travel, so there is no “sideways” oscillation direction to restrict, and so longitudinal waves cannot be polarised.

Light (electromagnetic waves) is transverse. Sound in air is longitudinal.

Understanding the Question

We are asked to choose an effect that can be observed with light waves but cannot be observed with sound waves.

So we look for an effect that requires light’s transverse nature, and that sound (as a longitudinal wave in air) cannot show.

Approach

Check each option against what is true for both light and sound:

If an effect happens for both, it is not the answer.
If an effect requires a transverse wave, it will work for light but not for sound.

Step-by-Step Reasoning

Diffraction (A): both light and sound diffract when they pass through gaps/around edges of comparable size to their wavelength. So A is not unique to light.
Interference (B): both light and sound can interfere (superposition of waves) if coherent sources/conditions are used. So B is not unique to light.
Polarisation (C): requires transverse oscillations. Light is transverse so it can be polarised; sound in air is longitudinal so it cannot. Therefore C is the required observation.
Reflection (D): both light and sound reflect from surfaces. So D is not unique to light.

Hence the correct option is C.

Key Takeaways

Polarisation is a clear discriminator: transverse waves can be polarised; longitudinal waves cannot.
Diffraction, interference and reflection are general wave behaviours and are not restricted to light.

Common Mistakes

Choosing interference because it is often taught with light: sound also shows interference (e.g. beats, two-speaker patterns).
Choosing diffraction: sound diffracts very noticeably due to its relatively long wavelength.
Forgetting that typical sound in air is longitudinal (so polarisation does not apply).

Things to Be Careful About

The question says “sound waves” without qualification; for A-Level this normally means sound in air (longitudinal).
Some other media can support transverse mechanical waves, but ordinary airborne sound does not; the exam expects the standard classification.

Techniques used

identify whether a wave is transverse or longitudinalrecall which wave phenomena require a transverse waveeliminate options that apply to both types of wave

Q291MMedium-EasySuperposition

A wire is fixed at both ends and is vibrated by a source of frequency $f$ . A stationary wave is formed on the wire with a total of two antinodes.

The frequency of the source is increased to $3f$ and a new stationary wave is formed.

What is the total number of antinodes on the new wave?

Options

A 4
B 5
C 6
D 9

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For a string fixed at both ends, the $n$ th harmonic has $n$ antinodes and frequency $f_n \propto n$ .

Two antinodes means $n = 2$ , so initial $f = f_2 = 2f_1$ .

New frequency $= 3f = 3(2f_1) = 6f_1 \Rightarrow n = 6$ antinodes.

Answer

C

Final answer

C

Detailed explanation

Background Concept

A stationary wave on a string fixed at both ends must have nodes at both ends. Only certain wavelengths (\lambda) “fit” the length (L) of the string:

L = \frac{n\lambda}{2} \quad (n = 1,2,3,\dots)

Each allowed pattern is a harmonic (or normal mode). For the (n)th harmonic:

there are (n) antinodes (and (n+1) nodes including the ends),
the wavelength is (\lambda_n = \frac{2L}{n}),
the frequency is

f_n = \frac{v}{\lambda_n} = \frac{nv}{2L}

So, for the same string (same (L) and wave speed (v)), the harmonic frequency is directly proportional to (n): (f_n \propto n).

Understanding the Question

You are told that at driving frequency (f), the stationary wave pattern has a total of two antinodes. Then the source frequency is increased to (3f) and you must determine how many antinodes the new stationary wave will have.

Key clue: “fixed at both ends” means the allowed modes are harmonics with (f_n \propto n), and “number of antinodes” tells you which harmonic you are on.

Approach

Convert “two antinodes” into the harmonic number (n).
Use (f_n = nf_1) (or (f_n \propto n)) to relate the given (f) to the fundamental frequency (f_1).
Multiply the frequency by 3 and convert that new frequency back into a new harmonic number (n), which equals the number of antinodes.

Step-by-Step Reasoning

Two antinodes means the pattern is the 2nd harmonic:

\text{two antinodes} \Rightarrow n=2

So the given frequency (f) is:

f = f_2 = 2f_1

Now increase the source frequency to (3f):

3f = 3(2f_1) = 6f_1

That corresponds to the 6th harmonic (since (f_n = nf_1)):

3f = f_6 \Rightarrow n = 6

The total number of antinodes in the 6th harmonic is (6).

Therefore the correct option is C (6).

Key Takeaways

For a string fixed at both ends, the (n)th harmonic has (n) antinodes.
Harmonic frequencies scale as (f_n \propto n) for the same string.
Multiplying the driving frequency multiplies the harmonic number by the same factor (once you know which harmonic you started from).

Common Mistakes

Assuming “two antinodes” means the fundamental (it does not; the fundamental has 1 antinode).
Thinking the number of antinodes triples directly from 2 to 6 without justifying it via (f_n \propto n). (It happens to be true here, but the proportionality is the reason.)
Confusing antinodes with nodes (ends are always nodes for a string fixed at both ends).

Things to Be Careful About

The given (f) is not necessarily the fundamental frequency (f_1); it is just the source frequency that produced a particular mode.
The direct proportionality (f_n = nf_1) only holds when the wave speed (v) and the length (L) are unchanged (same wire under the same tension and linear density).

Techniques used

identify the harmonic number from the number of antinodesuse the proportionality of harmonic frequency to harmonic number for a fixed-length stringscale the harmonic number when the driving frequency is multiplied

Q301MMedium-EasySuperposition

A parallel beam of red light of wavelength $700 \text{ nm}$ is incident normally on a diffraction grating that has $400$ lines per millimetre.

What is the total number of intensity maxima from the grating?

Options

A 6
B 7
C 8
D 9

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Lines per metre $= 400\ \text{mm}^{-1} = 4.00 \times 10^{5}\ \text{m}^{-1}$

d = \frac{1}{4.00 \times 10^{5}} = 2.50 \times 10^{-6}\ \text{m}

For maxima, $d\sin\theta = n\lambda$ and $\sin\theta \le 1$ so

n_{\max} = \left\lfloor \frac{d}{\lambda} \right\rfloor = \left\lfloor \frac{2.50 \times 10^{-6}}{7.00 \times 10^{-7}} \right\rfloor = \left\lfloor 3.57 \right\rfloor = 3

Total maxima $= 2n_{\max}+1 = 2(3)+1 = 7$ .

Answer

B

Final answer

B

Detailed explanation

Background Concept

A diffraction grating has many equally spaced slits. Bright (principal) maxima occur when waves from adjacent slits arrive in phase.

The condition for a principal maximum is

d\sin\theta = n\lambda

where:

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

Because $\sin\theta \le 1$ , there is a highest possible order:

\sin\theta = \frac{n\lambda}{d} \le 1 \quad \Rightarrow \quad n \le \frac{d}{\lambda}

Understanding the Question

You are given:

red light wavelength $\lambda = 700\ \text{nm} = 7.00 \times 10^{-7}\ \text{m}$ ,
a grating with 400 lines per millimetre.

The question asks for the total number of intensity maxima. That means count the central maximum ( $n=0$ ) plus maxima on both sides ( $+n$ and $-n$ ), up to the maximum possible order.

Approach

Convert the line density (lines per mm) into grating spacing $d$ in metres.
Use $n \le d/\lambda$ (from $\sin\theta \le 1$ ) to find the largest integer order $n_{\max}$ .
Total number of maxima is

1 + 2n_{\max}

(the “1” is the central maximum, and the factor 2 is for left and right).

Step-by-Step Reasoning

1) Find the grating spacing $d$ .
400 lines per mm means 400 lines in $1.0 \times 10^{-3}\ \text{m}$ .

So lines per metre:

400\ \text{mm}^{-1} = 400 \times 10^{3}\ \text{m}^{-1} = 4.00 \times 10^{5}\ \text{m}^{-1}

Grating spacing is the inverse:

d = \frac{1}{4.00 \times 10^{5}} = 2.50 \times 10^{-6}\ \text{m}

2) Find the maximum order.
From $d\sin\theta = n\lambda$ and $\sin\theta \le 1$ :

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

Compute:

\frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{7.00 \times 10^{-7}} = 3.57

So the largest allowed integer order is

n_{\max} = 3

(orders $n=4$ and above would require $\sin\theta > 1$ , which is impossible).

3) Count the total maxima.
Orders present: $n=0, \pm 1, \pm 2, \pm 3$ .

Total number:

1 + 2n_{\max} = 1 + 2(3) = 7

So the correct option is B (7).

Key Takeaways

Convert “lines per unit length” into spacing using $d = 1/(\text{lines per metre})$ .
The highest observable grating order is limited by $\sin\theta \le 1$ , giving $n_{\max} = \left\lfloor d/\lambda \right\rfloor$ .
Total number of principal maxima is $2n_{\max} + 1$ .

Common Mistakes

Forgetting to convert $\text{mm}^{-1}$ to $\text{m}^{-1}$ (missing the factor $10^{3}$ ).
Using $n_{\max} = d/\lambda$ without taking the integer part (you must take the largest whole number).
Giving the number of orders on one side ( $n_{\max}$ ) instead of the total maxima ( $2n_{\max}+1$ ).

Things to Be Careful About

Wavelength conversion: $700\ \text{nm} = 700 \times 10^{-9}\ \text{m} = 7.00 \times 10^{-7}\ \text{m}$ .
The central maximum ( $n=0$ ) always exists and must be included in the total.
The negative orders are physically the maxima on the opposite side of the central maximum, so they double the count (except the central one).

Techniques used

convert line density to grating spacingapply the diffraction grating equationuse the condition sin(\theta) \le 1 to find the maximum ordercount the total number of principal maxima including \pm orders and the central maximum

Q311MEasySuperposition

Two light sources are used to produce an interference pattern.

Interference fringes appear when the two sources emit waves that are coherent.

What is meant by coherent waves?

Options

A The two waves are emitted with the same frequency.
B The two waves are emitted with constant phase difference.
C The two waves are emitted with the same intensity.
D The two waves are emitted with zero phase difference.

Pick an answer

DifficultyEasy

Worked solution

For stable interference fringes, the phase relationship between the two waves must not change with time, i.e. the waves have a constant phase difference.

Answer

B

Final answer

B

Detailed explanation

Background Concept

Coherent waves are waves that maintain a fixed (constant) phase relationship. In practice for two-source interference, this means:

same frequency (so the phase difference does not drift), and
constant phase difference (a fixed phase relation).

This is what produces a stable interference pattern (fringes that do not wash out over time).

Understanding the Question

The question asks for the meaning of “coherent waves” in the context of producing an interference pattern from two light sources. You must choose the statement that best matches the definition needed for sustained fringes.

Approach

Compare each option to the definition of coherence used for interference:

Check whether it guarantees a fixed phase relationship.
Reject statements that are only partially correct (necessary but not sufficient) or too restrictive.

Step-by-Step Reasoning

Option A (same frequency): having the same frequency is important, but by itself does not guarantee a fixed phase relationship. Two sources can emit at the same frequency but with a phase difference that fluctuates randomly.
Option B (constant phase difference): this is the defining condition for coherence and directly ensures stable fringes.
Option C (same intensity): intensity affects fringe visibility (contrast) but not whether the sources are coherent.
Option D (zero phase difference): zero phase difference is one special case of constant phase difference; coherence does not require it to be zero, only constant.

Therefore the correct choice is B.

Key Takeaways

Coherence means a constant phase difference (fixed phase relationship).
Same frequency alone is not the full definition for coherence in interference questions.

Common Mistakes

Choosing A because “same frequency” is remembered, but forgetting the need for constant phase difference.
Choosing D by thinking coherence means “in phase”; coherence allows any fixed phase difference.

Things to Be Careful About

In MCQs, pick the most precise definition: “constant phase difference” is the key phrase.
Intensity is not a criterion for coherence; it only changes the sharpness/visibility of fringes.

Techniques used

recall the definition of coherence for interferenceidentify the condition required for stable fringeseliminate options that are necessary but not sufficient

Q321MMedium-EasyElectricity

Which component has the $I$ – $V$ graph shown?

Fig. 1

Options

A filament lamp
B metallic conductor at constant temperature
C resistor of fixed resistance
D semiconductor diode

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

The curve passes through the origin but its gradient decreases as $V$ increases, so $R = V/I$ increases with current (non-ohmic, heating effect).

A diode would have very small current until a threshold then increasing gradient; a fixed resistor / metallic conductor at constant temperature would give a straight line.

Answer

A

Final answer

A

Detailed explanation

Background Concept

An $I$ – $V$ characteristic shows how current $I$ depends on potential difference $V$ for a component.

For an ohmic conductor at constant temperature (or an ideal fixed resistor), $V \propto I$ and the graph is a straight line through the origin.
The gradient of an $I$ – $V$ graph is

\text{gradient} = \frac{\Delta I}{\Delta V} \approx \frac{1}{R}

so a decreasing gradient means increasing resistance.

A filament lamp is non-ohmic because as current increases, the filament temperature rises; the resistivity (and hence resistance) increases.

A semiconductor diode (forward bias) typically has negligible current for small $V$ and then a rapid rise in $I$ after a threshold voltage, i.e. increasing gradient.

Understanding the Question

You are given an $I$ – $V$ curve that:

passes through the origin,
rises steeply at first,
then becomes flatter as $V$ increases (decreasing gradient),
is shown only in the first quadrant (positive $V$ and $I$ ).

You must choose which listed component has this type of $I$ – $V$ characteristic.

Approach

Decide whether the graph is linear (ohmic) or non-linear (non-ohmic).
Use how the gradient changes to determine whether resistance is increasing or decreasing with current.
Match that behaviour to the standard $I$ – $V$ shapes for: fixed resistor/metal at constant temperature, filament lamp, diode.

Step-by-Step Reasoning

The curve goes through the origin, so at $V = 0$ we have $I = 0$ . This is consistent with several components (resistor, filament lamp, diode in forward bias when only the forward region is considered).
The key feature is that the curve starts steep and then flattens. That means as $V$ increases, $I$ still increases but by a smaller amount per volt:

\frac{\Delta I}{\Delta V} \text{ decreases as } V \text{ increases.}

So $1/R$ decreases, therefore $R$ increases as current increases.

Compare with options:

Metallic conductor at constant temperature: constant $R$ , so straight line through origin (not correct).
Resistor of fixed resistance: also straight line through origin (not correct).
Semiconductor diode: forward characteristic has very small current initially then becomes very steep (gradient increases), opposite of the given curve (not correct).
Filament lamp: filament heats up as $I$ increases, causing resistance to increase, producing a curve that flattens (correct).

Therefore the component is a filament lamp.

Key Takeaways

A straight-line $I$ – $V$ graph through the origin indicates constant resistance (ohmic behaviour).
The gradient of an $I$ – $V$ graph is related to $1/R$ .
A filament lamp shows decreasing gradient because heating increases resistance.
A diode shows increasing gradient in forward bias after a threshold.

Common Mistakes

Choosing a diode because the graph is only in the first quadrant (the quadrant alone is not the deciding feature; the shape is).
Thinking “curved graph = diode” without checking whether the gradient increases (diode) or decreases (filament lamp).
Mixing up $R = V/I$ with the gradient: the gradient is $\Delta I/\Delta V$ , not $V/I$ .

Things to Be Careful About

Some exam graphs show only the positive- $V$ region even for components that would be symmetric for negative $V$ (e.g. filament lamp). Do not rely on quadrant symmetry unless it is explicitly shown.
Use the change in gradient (steeper vs flatter) as the main diagnostic:
- flattening curve $\Rightarrow R$ increasing (filament lamp),
- steepening curve $\Rightarrow R$ decreasing / diode turn-on.

Techniques used

interpret the shape of an I–V characteristicuse gradient of an I–V graph to infer resistance changesidentify non-ohmic behaviour and match to a component model

Q331MMedium-EasyD.C. Circuits

The diagram shows an electrical circuit containing a light-dependent resistor (LDR).

Fig. 1

What is the circuit diagram for this circuit?

Options

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

The apparatus shows a potentiometer connected across the power supply, with the sliding contact feeding a series circuit containing an ammeter and an LDR.

So the correct diagram must:

use the LDR symbol (resistor with arrows towards it), and
place the ammeter in series in the branch with the LDR (not in the main supply line).

Only option A matches this.

Answer

A

Final answer

A

Detailed explanation

Background Concept

A circuit diagram is a symbolic representation of the same electrical connections (nodes and branches) as a real circuit.

Key symbol facts used here:

Ammeter measures current and must be connected in series with the component whose current is being measured.
Potentiometer has three terminals: two ends of a resistive track and a sliding contact (wiper). If the supply is connected across the two ends, the slider provides a variable output potential difference (a potential divider arrangement).
LDR symbol is a resistor with two arrows pointing towards it (light incident), not a variable resistor (which is a resistor with a diagonal arrow through it).

Understanding the Question

You are given a photographed-style layout: a power supply, a potentiometer, an ammeter, and an LDR connected with wires.

You must choose which of the four circuit diagrams A–D represents the same connections.

From the description of the layout:

The power supply is connected across the potentiometer track (across the two end terminals).
The sliding contact is connected into a loop that contains the ammeter and the LDR in series.

So the correct option must show:

potentiometer ends across the supply, with a slider connection,
an LDR (not a variable resistor) in the lower branch,
an ammeter in series with the LDR branch, not measuring the total supply current.

Approach

Identify which options even contain an LDR symbol.
Check whether the ammeter is placed in the correct branch: it must be in series with the LDR path from the slider to the return node.
Eliminate diagrams where the ammeter is on the main line (it would then measure current in multiple branches, not just through the LDR).

Step-by-Step Reasoning

Options B and C use a variable resistor symbol (resistor with diagonal arrow), so they cannot represent an LDR. Eliminate B and C.
Options A and D contain the LDR symbol (resistor with arrows pointing towards it), so either could be correct.
Now check the ammeter position:
- In A, the ammeter is in the same lower branch as the LDR, so it is in series with the LDR, matching the apparatus.
- In D, the ammeter is placed above the junction (in the main branch), so it would measure the current supplied to more than just the LDR branch (depending on how the potentiometer is connected). This does not match “ammeter connected in series with an LDR”.

Therefore the only diagram consistent with the given circuit is A.

Key Takeaways

Always match connections (nodes), not just component order in the picture.
An ammeter must be in series in the branch of interest.
Recognise component symbols: LDR (arrows towards resistor) vs variable resistor (diagonal arrow through resistor).

Common Mistakes

Choosing a diagram with a variable resistor symbol instead of the LDR symbol.
Placing the ammeter in the main supply line, which makes it measure total current, not the current through the LDR.
Assuming the potentiometer is just a series variable resistor; here it is connected across the supply and used via its slider.

Things to Be Careful About

The crucial feature is the slider (wiper) connection on the potentiometer: the correct diagram must show the slider feeding the external branch.
For MCQs like this, first eliminate options by incorrect symbols, then by incorrect series/parallel placement.
Ensure the ammeter is not accidentally placed where it sits before a junction, because then it measures the combined current of multiple branches.

Techniques used

translate an apparatus layout into standard circuit symbolsidentify series connections and junctions in a circuitdistinguish an LDR symbol from a variable resistor symbolcheck that the ammeter is placed in series with the intended component

Q341MMedium-EasyElectricity

A copper wire of length $2 \text{ m}$ has a circular cross-section. There is a constant current in the wire when it is connected to a mobile phone in order to charge the battery.

The wire has a fault. A $1 \text{ m}$ length of the wire has half the diameter of the other $1 \text{ m}$ length.

The power dissipated in the thicker length of wire is $P$ .

What is the power dissipated in the thinner length of wire?

Options

A $\frac{1}{4}P$
B $\frac{1}{2}P$
C $2P$
D $4P$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For each 1 m length,

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

Thinner section has diameter halved, so

A \propto d^2 \Rightarrow A_{\text{thin}} = \frac{1}{4}A_{\text{thick}}

Hence

R_{\text{thin}} = 4R_{\text{thick}}

With constant current $I$ ,

P = I^2R \Rightarrow P_{\text{thin}} = I^2R_{\text{thin}} = 4I^2R_{\text{thick}} = 4P

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}.

So, for the same material and the same length, resistance is inversely proportional to cross-sectional area: smaller area means larger resistance.

Electrical power converted to thermal energy in a resistor can be written in several equivalent forms:

P = IV = I^2R = \frac{V^2}{R}.

Which form you use depends on what is kept constant. Here the question states the current is constant, so $P = I^2R$ is the most direct.

Understanding the Question

The 2 m copper wire is faulty: it consists of two 1 m sections in series, one thicker and one thinner. The thinner section has half the diameter of the thicker one.

You are told the power dissipated in the thicker 1 m section is $P$ . You must find the power dissipated in the thinner 1 m section.

Because the wire is charging a phone, the two sections carry the same current (they are in series), and the question explicitly says the current is constant.

Approach

Compare the cross-sectional areas of the thick and thin sections using the fact that the wire is circular.
Use $R = \rho L/A$ to get the ratio of resistances for the two 1 m lengths.
Use $P = I^2R$ (valid because the current is the same through both sections) to convert the resistance ratio into a power ratio.

Step-by-Step Reasoning

For a circular cross-section,

A = \pi\left(\frac{d}{2}\right)^2 \propto d^2.

If the diameter halves, $d_{\text{thin}} = \frac{1}{2}d_{\text{thick}}$ , then

A_{\text{thin}} = \left(\frac{1}{2}\right)^2 A_{\text{thick}} = \frac{1}{4}A_{\text{thick}}.

Now use resistance:

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

Both sections have the same material ( $\rho$ same) and the same length ( $L = 1\ \text{m}$ ), so

\frac{R_{\text{thin}}}{R_{\text{thick}}} = \frac{A_{\text{thick}}}{A_{\text{thin}}} = \frac{A_{\text{thick}}}{\frac{1}{4}A_{\text{thick}}} = 4.

So $R_{\text{thin}} = 4R_{\text{thick}}$ .

The current is the same in both sections (series circuit) and constant, so power in each section is

P = I^2R.

Therefore,

\frac{P_{\text{thin}}}{P_{\text{thick}}} = \frac{I^2R_{\text{thin}}}{I^2R_{\text{thick}}} = \frac{R_{\text{thin}}}{R_{\text{thick}}} = 4.

Given $P_{\text{thick}} = P$ , we get

P_{\text{thin}} = 4P.

This corresponds to option D.

Key Takeaways

For a circular wire, $A \propto d^2$ so halving diameter quarters area.
With the same material and length, $R \propto 1/A$ .
In series, the current is the same in each component, so at constant current $P \propto R$ via $P = I^2R$ .

Common Mistakes

Using $P = V^2/R$ (this would be relevant if the potential difference across each section were the same, which it is not here).
Thinking halving diameter halves area (it quarters it, because of the square).
Assuming power is the same in both sections because they are in the same wire.

Things to Be Careful About

Always decide whether $I$ or $V$ is constant before choosing between $I^2R$ and $V^2/R$ .
Remember the area of a circle depends on radius (or diameter) squared, not linearly.
The two 1 m lengths have equal $L$ , so the entire difference comes only from $A$ (and hence from diameter).

Techniques used

use resistance proportionality with cross-sectional arearelate diameter change to area change for a circular cross-sectionapply the power relation P = I^2R for a constant currentcompare power dissipation using ratios

Q351MMedium-EasyElectricity

A wire carries a current of $5.6 \text{ A}$ .

What is the number of conduction electrons that pass a point on the wire in a time of $20 \text{ s}$ ?

Options

A $1.8 \times 10^{18}$
B $2.2 \times 10^{19}$
C $3.5 \times 10^{19}$
D $7.0 \times 10^{20}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Q = It = 5.6 \times 20 = 112\ \text{C}

N = \frac{Q}{e} = \frac{112}{1.60 \times 10^{-19}} = 7.0 \times 10^{20}

Answer

D

Final answer

D

Detailed explanation

Background Concept

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

I = \frac{Q}{t}

so the charge that passes a point in time $t$ is

Q = It.

If the charge carriers are electrons, each electron has charge magnitude (elementary charge) $e = 1.60 \times 10^{-19}\ \text{C}$ . If a total charge $Q$ flows past a point, the number of electrons is

N = \frac{Q}{e}.

Understanding the Question

A steady current of $5.6\ \text{A}$ flows in a wire. You are asked how many conduction electrons pass a fixed point in the wire in $20\ \text{s}$ . This is a charge-flow counting problem: find total charge moved in 20 s, then divide by the charge per electron.

Approach

Use $Q = It$ to find the total charge that passes the point in $20\ \text{s}$ .
Convert charge to number of electrons using $N = Q/e$ .
Compare the calculated $N$ with the options.

Step-by-Step Reasoning

Calculate charge transferred:

Q = It = 5.6\ \text{A} \times 20\ \text{s} = 112\ \text{C}.

(Recall $1\ \text{A} = 1\ \text{C s}^{-1}$ , so the units are consistent.)

Convert charge to number of electrons:

N = \frac{Q}{e} = \frac{112}{1.60 \times 10^{-19}}.

Compute:

N = \left(\frac{112}{1.60}\right) \times 10^{19} = 70 \times 10^{19} = 7.0 \times 10^{20}.

This matches option D.

Key Takeaways

Use $Q = It$ to find how much charge flows in a given time.
The number of electrons is $N = Q/e$ .
Keep track of powers of ten when dividing by $10^{-19}$ .

Common Mistakes

Using $I = Q/t$ but rearranging incorrectly (e.g. $Q = I/t$ ).
Forgetting that electrons have charge $-e$ ; for a number of electrons you use the magnitude $e$ .
Power-of-ten slip: dividing by $10^{-19}$ increases the answer by a factor of $10^{19}$ .

Things to Be Careful About

Use $e = 1.60 \times 10^{-19}\ \text{C}$ (not $10^{19}$ ).
Make sure the final result is a pure number (no units) because it is a count of electrons.
Sensible check: $112\ \text{C}$ is large compared to $10^{-19}\ \text{C}$ , so the electron count should be around $10^{20}$ , not $10^{18}$ or $10^{22}$ .

Techniques used

use the definition of electric current to calculate charge transferreduse the elementary charge to convert total charge into number of electronsmatch the calculated value to the closest option

Q361MMedium-EasyD.C. Circuits

Four resistors of resistance $R$ , $2R$ , $3R$ and $4R$ are connected to form a network.

A battery of negligible internal resistance and a voltmeter are connected to the resistor network as shown.

Fig. 1

The voltmeter reading is $2 \text{ V}$ .

What is the electromotive force (e.m.f.) of the battery?

Options

A $2 \text{ V}$
B $4 \text{ V}$
C $6 \text{ V}$
D $10 \text{ V}$

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

The resistors $3R$ , $2R$ , and $R$ are in series, so their total is $6R$ .

For this series branch,

V_{2R} = \frac{2R}{6R}V_{\text{supply}} = \frac{1}{3}V_{\text{supply}}

Given $V_{2R} = 2\ \text{V}$ ,

2 = \frac{1}{3}V_{\text{supply}} \Rightarrow V_{\text{supply}} = 6\ \text{V}

With negligible internal resistance, e.m.f. $= V_{\text{supply}}$ .

Answer

C

Final answer

C

Detailed explanation

Background Concept

In resistor networks, two key ideas are used:

Series resistors carry the same current. The potential difference (p.d.) across each is

V = IR

so the p.d. divides in proportion to resistance (this is the potential divider idea).

Parallel branches share the same p.d. between the two common nodes, even though the current splits between branches.

A voltmeter connected across a component reads the p.d. across that component (it is assumed to have very large resistance so it does not significantly change the circuit).

Understanding the Question

The circuit has four resistors: $3R$ on the top branch, $R$ on the bottom branch, $2R$ on the right vertical side (with the voltmeter across it), and $4R$ on the left vertical side. The battery is connected across the left-hand top and bottom nodes of the network, so it provides the same p.d. across every branch connected between those two nodes.

We are told the voltmeter across the $2R$ resistor reads $2\ \text{V}$ , and we must find the battery e.m.f.

diagram-1

Approach

Identify a simple series chain that includes the $2R$ resistor, so we can use voltage division.
Relate the voltage across $2R$ to the total voltage across that series chain.
Note that this total voltage is the same as the battery terminal p.d. (and equals the e.m.f. because internal resistance is negligible).

Step-by-Step Reasoning

Consider the path from the top-left node to the bottom-left node going around the right-hand side:
- $3R$ (top) then $2R$ (right) then $R$ (bottom).
These three are in series, so the total resistance of this branch is

R_{\text{series}} = 3R + 2R + R = 6R

Let the battery (supply) p.d. across the left nodes be $V_{\text{supply}}$ . This same $V_{\text{supply}}$ appears across the entire series branch (because the series branch is connected between the same two supply nodes).
In a series chain, the p.d. across one resistor is the fraction of the total resistance it represents:

V_{2R} = \frac{2R}{6R}V_{\text{supply}} = \frac{1}{3}V_{\text{supply}}

Substitute the voltmeter reading $V_{2R} = 2\ \text{V}$ :

2 = \frac{1}{3}V_{\text{supply}} \Rightarrow V_{\text{supply}} = 6\ \text{V}

Since the battery has negligible internal resistance, its e.m.f. equals its terminal p.d., which is $V_{\text{supply}}$ .

So the e.m.f. is $6\ \text{V}$ (option C).

Key Takeaways

Identify series groups to apply voltage division: $V \propto R$ in series.
Parallel branches do not change the p.d. across each branch; they only affect currents.
With negligible internal resistance, battery e.m.f. equals the p.d. across the external network.

Common Mistakes

Treating the $4R$ resistor as if it were in series with the $2R$ resistor (it is in a different parallel branch).
Adding all four resistors as though they are all in one series loop.
Forgetting that the p.d. across the series branch is the same as the battery p.d. across the network.

Things to Be Careful About

Ensure you are taking the ratio using resistances in the same series chain as the measured resistor.
The voltmeter measures only across the $2R$ resistor, not across the entire right-hand branch.
“Negligible internal resistance” is the clue that e.m.f. equals the external terminal p.d.

Techniques used

identify series and parallel combinations in the networkuse voltage division in a series chainuse the fact that parallel branches share the same potential difference

Q371MMedium-EasyD.C. Circuits

The circuit diagram shows a battery with internal resistance $r$ , two resistors, a switch and a voltmeter. The switch is open.

Fig. 1

The switch is now closed.

What happens to the current in the battery, and what happens to the reading on the voltmeter, when the switch is closed?

Options

	current in battery	reading on voltmeter
A	increases	decreases
B	increases	remains the same
C	remains the same	decreases
D	remains the same	remains the same

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Closing the switch adds the lower resistor in parallel with the existing resistor, so $R_{\text{ext}}$ decreases.

I = \frac{\varepsilon}{r + R_{\text{ext}}}

So $I$ increases.

The voltmeter reads the terminal p.d.

V = \varepsilon - Ir

As $I$ increases, $Ir$ increases, so $V$ decreases.

Answer

A

Final answer

A

Detailed explanation

Background Concept

A real battery can be modelled as an ideal source of e.m.f. $\varepsilon$ in series with an internal resistance $r$ . When a current $I$ flows, some of the e.m.f. is “lost” inside the battery as a voltage drop $Ir$ .

The terminal potential difference (what a voltmeter reads across the battery terminals) is

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

Also, the current supplied by the battery depends on the total series resistance seen by the source:

I = \frac{\varepsilon}{r + R_{\text{ext}}}

where $R_{\text{ext}}$ is the equivalent resistance of everything outside the battery.

Understanding the Question

With the switch open, only the middle resistor is connected across the rails, so the battery supplies current through just that one resistor. The voltmeter is connected across the rails, so it measures the terminal p.d. across the external circuit (and hence across the battery terminals).

When the switch is closed, the lower branch (switch + resistor) becomes a second conducting branch across the same two rails. The question asks how this affects:

the current in the battery (total current supplied), and
the voltmeter reading (terminal p.d. across the rails).

Approach

Decide how closing the switch changes the external equivalent resistance $R_{\text{ext}}$ (series vs parallel reasoning).
Use $I = \varepsilon/(r+R_{\text{ext}})$ to determine whether the battery current increases or decreases.
Use $V_{\text{terminal}} = \varepsilon - Ir$ to determine how the voltmeter reading changes when $I$ changes.

Step-by-Step Reasoning

When the switch is open, the external resistance is just the single middle resistor, call it $R_1$ :

R_{\text{ext, open}} = R_1

When the switch is closed, the lower resistor (call it $R_2$ ) is connected across the same two rails, so $R_1$ and $R_2$ are in parallel:

R_{\text{ext, closed}} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1}

A parallel combination is always smaller than either resistor alone, so

R_{\text{ext, closed}} < R_{\text{ext, open}}

Now compare the battery currents:

I_{\text{open}} = \frac{\varepsilon}{r + R_{\text{ext, open}}}, \qquad I_{\text{closed}} = \frac{\varepsilon}{r + R_{\text{ext, closed}}}

Since $R_{\text{ext, closed}}$ is smaller, the denominator $r+R_{\text{ext}}$ is smaller, so $I_{\text{closed}}$ is larger. Therefore the current in the battery increases.

For the voltmeter reading (terminal p.d.):

V = \varepsilon - Ir

When the switch is closed, $I$ increases, so the internal drop $Ir$ increases, so $V$ must decrease. Therefore the voltmeter reading decreases.

So the correct option is: current increases, voltmeter reading decreases (A).

Key Takeaways

Closing a switch that adds an extra branch in parallel reduces the external equivalent resistance.
With internal resistance present, increasing current reduces terminal p.d. because $V = \varepsilon - Ir$ .
The voltmeter across the rails measures terminal p.d., not the e.m.f.

Common Mistakes

Assuming the voltmeter reading stays at $\varepsilon$ regardless of load (only true if $r=0$ or if $I=0$ ).
Thinking “more resistors means more resistance” without recognising that parallel resistors reduce the equivalent resistance.
Confusing the current in one branch with the total current supplied by the battery.

Things to Be Careful About

The switch closure changes the circuit topology: it creates a second parallel path, so you must use parallel-resistance logic.
The voltmeter is across the external circuit, so it reads $V_{\text{terminal}}$ , which changes with current if $r \neq 0$ .
If $r$ were negligibly small, the voltmeter would be almost constant; but with internal resistance included, the mark-scheme logic is that $V$ decreases when $I$ increases.

Techniques used

identify how closing a switch changes the external resistanceuse parallel-resistance reasoning to compare total resistanceapply $I = \varepsilon /(R_{\text{ext}}+r)$ to compare currentsuse $V_{\text{terminal}} = \varepsilon - Ir$ to compare voltmeter readings

Q381MMedium-EasyParticle Physics

What is the correct equation for $\beta^{+}$ decay?

Options

A neutron $\rightarrow$ proton + electron + electron antineutrino
B neutron $\rightarrow$ proton + electron + electron neutrino
C proton $\rightarrow$ neutron + positron + electron antineutrino
D proton $\rightarrow$ neutron + positron + electron neutrino

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In $\beta^+$ decay, a proton changes to a neutron and emits a positron and an electron neutrino.

Charge check: $+1 = 0 + (+1) + 0$ .

Answer

D

Final answer

D

Detailed explanation

Background Concept

In beta decay, a nucleon changes type via the weak interaction.

In $\beta^-$ decay: a neutron becomes a proton, emitting an electron ( $e^-$ ) and an electron antineutrino ( $\bar{\nu}_e$ ).
In $\beta^+$ decay: a proton becomes a neutron, emitting a positron ( $e^+$ ) and an electron neutrino ( $\nu_e$ ).

These decay equations must satisfy conservation laws:

Charge conservation (total electric charge before = after)
Lepton number conservation (electron family lepton number: $e^-$ and $\nu_e$ are $+1$ ; $e^+$ and $\bar{\nu}_e$ are $-1$ ).

(At quark level for $\beta^+$ : an up quark changes to a down quark, so $uud \to udd$ .)

Understanding the Question

You are asked to choose the correct nuclear equation for $\beta^+$ decay from four options that involve neutrons/protons and leptons (electron/positron and neutrino/antineutrino).

So we must identify which option represents:

\text{proton} \to \text{neutron} + e^+ + \nu_e

Approach

Recall what particle is emitted in $\beta^+$ decay (a positron, not an electron).
Decide whether the nucleon changes from proton to neutron or the reverse.
Use charge and lepton number conservation to decide between neutrino vs antineutrino.

Step-by-Step Reasoning

In $\beta^+$ decay, the nucleus emits a positron ( $e^+$ ), so any option with an electron is not $\beta^+$ (eliminates A and B).
Because a positron has charge $+1$ , to conserve charge the nucleon must change:
- starting with a proton ( $+1$ )
- ending with a neutron ( $0$ ) plus a positron ( $+1$ ) plus a neutral neutrino/antineutrino ( $0$ ).
So we need:
$p \to n + e^+ + (\text{neutrino/antineutrino})$
This matches options C and D.
Now use lepton number:
- Initially: no leptons, so total lepton number $=0$ .
- If we emit $e^+$ , lepton number contribution is $-1$ .
- To keep total at $0$ , we must also emit a particle with lepton number $+1$ , which is the electron neutrino $\nu_e$ .

Therefore the correct equation is:

p \to n + e^+ + \nu_e

which is option D.

Key Takeaways

$\beta^+$ decay: $p \to n + e^+ + \nu_e$ .
Distinguish neutrino ( $\nu_e$ ) from antineutrino ( $\bar{\nu}_e$ ) using lepton number conservation.
Always check charge conservation to confirm the nucleon change direction.

Common Mistakes

Swapping $\beta^-$ and $\beta^+$ : writing an electron instead of a positron.
Choosing antineutrino in $\beta^+$ decay (violates lepton number conservation).
Thinking a neutron decays to a proton in $\beta^+$ decay (that is $\beta^-$ decay).

Things to Be Careful About

Notation: positron is $e^+$ ; electron is $e^-$ .
Neutrino type matters: $\nu_e$ vs $\bar{\nu}_e$ is determined by lepton number.
In MCQs, quickly eliminate options by spotting the wrong emitted charged lepton (electron vs positron).

Techniques used

recall the particle emission in beta-plus decayapply conservation of chargeapply conservation of lepton number

Q391MMedium-EasyParticle Physics

A nucleus of polonium, $^{214}_{84}\text{Po}$ , decays by emitting an $\alpha$ -particle to become a nucleus of lead.

The nucleus of lead is also unstable and decays by emitting a $\beta^{-}$ particle to form a nucleus of bismuth which then decays by emitting a $\beta^{-}$ particle to produce a nucleus X.

What is the number of neutrons in nucleus X?

Options

A 126
B 128
C 130
D 210

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

$^{214}_{84}\text{Po} \rightarrow {}^{210}_{82}\text{Pb} + {}^{4}_{2}\alpha$

$^{210}_{82}\text{Pb} \rightarrow {}^{210}_{83}\text{Bi} + \beta^{-}$

$^{210}_{83}\text{Bi} \rightarrow {}^{210}_{84}X + \beta^{-}$

So $A_X = 210$ and $Z_X = 84$ .

Number of neutrons in $X$ :

N = A - Z = 210 - 84 = 126

Answer

A

Final answer

A

Detailed explanation

Background Concept

In nuclear decay, two conserved quantities are used to identify the daughter nucleus:

The nucleon (mass) number $A$ = total number of protons + neutrons.
The proton (atomic) number $Z$ = number of protons.

Key changes:

$\alpha$ -decay emits $^{4}_{2}\alpha$ (2 protons, 2 neutrons), so
- $A$ decreases by 4
- $Z$ decreases by 2
$\beta^{-}$ -decay converts a neutron into a proton (plus an electron and an antineutrino). So
- $A$ stays the same
- $Z$ increases by 1

Finally, the number of neutrons in a nucleus is

N = A - Z.

Understanding the Question

You are told:

$^{214}_{84}\text{Po}$ undergoes $\alpha$ -decay to become lead.
That lead nucleus then undergoes two successive $\beta^{-}$ decays (lead $\to$ bismuth $\to$ nucleus $X$ ).
The question asks for the number of neutrons in nucleus $X$ .

So we must track how $A$ and $Z$ change through the three decays, then compute $N$ .

Approach

Apply the $\alpha$ -decay changes to get the lead nucleus.
Apply two $\beta^{-}$ -decay changes to reach nucleus $X$ .
Use $N = A - Z$ for nucleus $X$ .

Step-by-Step Reasoning

Start with polonium:

^{214}_{84}\text{Po}

After $\alpha$ -decay:

$A: 214 \to 214 - 4 = 210$
$Z: 84 \to 84 - 2 = 82$
So the daughter is

^{210}_{82}\text{Pb}.

First $\beta^{-}$ -decay (lead to bismuth):

$A$ unchanged: $210$
$Z$ increases by 1: $82 \to 83$
So

^{210}_{82}\text{Pb} \to {}^{210}_{83}\text{Bi}.

Second $\beta^{-}$ -decay (bismuth to $X$ ):

$A$ unchanged: $210$
$Z$ increases by 1: $83 \to 84$
So

X = {}^{210}_{84}X.

Now find the neutrons in $X$ :

N = A - Z = 210 - 84 = 126.

This matches option A.

Key Takeaways

$\alpha$ -decay: $A - 4$ , $Z - 2$ .
$\beta^{-}$ -decay: $A$ unchanged, $Z + 1$ .
Neutron number is found from $N = A - Z$ .

Common Mistakes

Treating $\beta^{-}$ -decay as decreasing $Z$ (that is $\beta^{+}$ , not $\beta^{-}$ ).
Changing $A$ during $\beta^{-}$ -decay (it must stay the same).
Forgetting that neutron number is not $A/2$ or anything similar; it is specifically $A-Z$ .

Things to Be Careful About

Keep track of each decay step separately; two $\beta^{-}$ decays mean $Z$ increases by 2 in total.
Do not confuse the emitted particle’s numbers: an $\alpha$ particle is $^{4}_{2}\alpha$ .
Ensure you compute neutrons for the final nucleus $X$ , not for the intermediate lead or bismuth.

Techniques used

apply nucleon and proton number conservation in alpha decayapply nucleon and proton number conservation in beta-minus decaycalculate neutron number from nucleon and proton numbers

Q401MMedium-EasyParticle Physics

Which combination of three quarks has no overall charge?

Options

A down, strange, strange
B up, charm, strange
C up, charm, top
D up, down, strange

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Charges: $u,c,t = +\frac{2}{3}e$ and $d,s = -\frac{1}{3}e$ .

For option D:

Q = \left(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3}\right)e = 0

Answer

D

Final answer

D

Detailed explanation

Background Concept

Quarks have fractional electric charge (in units of the elementary charge $e$ ). For the quark flavours needed here:

Up-type quarks ( $u$ , $c$ , $t$ ) have charge $+\frac{2}{3}e$ .
Down-type quarks ( $d$ , $s$ , $b$ ) have charge $-\frac{1}{3}e$ .

A baryon is made of three quarks, and its overall charge is the algebraic sum of the three quark charges.

Understanding the Question

You are given four different combinations of three quarks. The task is to find which combination has total charge zero (i.e. the charges add to $0e$ ).

So you must (1) assign each quark its charge and (2) add the three values.

Approach

Write down the charge for each quark flavour that appears in the options.
For each option, add the three charges.
Select the option where the sum equals zero.

Step-by-Step Reasoning

Recall:

$u, c, t$ each contribute $+\frac{2}{3}e$ .
$d, s$ each contribute $-\frac{1}{3}e$ .

Now test the options:

A: $dss$ gives

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

Not zero.

B: $ucs$ gives

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

Not zero.

C: $uct$ gives

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

Not zero.

D: $uds$ gives

Q = \left(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3}\right)e = 0

So option D has no overall charge.

Key Takeaways

Up-type quarks ( $u,c,t$ ) have charge $+\frac{2}{3}e$ .
Down-type quarks ( $d,s,b$ ) have charge $-\frac{1}{3}e$ .
Net charge of a 3-quark combination is found by simple algebraic addition.

Common Mistakes

Mixing up which quarks are up-type and down-type (e.g. thinking $s$ is $+\frac{2}{3}e$ ).
Adding fractions incorrectly (a frequent error is treating $\frac{2}{3}-\frac{1}{3}-\frac{1}{3}$ as $\frac{0}{3}$ without showing the arithmetic).
Forgetting that the question asks for zero charge, not $+1e$ or $-1e$ .

Things to Be Careful About

Keep the sign on each contribution: $-\frac{1}{3}e$ must be negative.
It can help to add the two down-type charges first: in option D, $(-\frac{1}{3}-\frac{1}{3})e=-\frac{2}{3}e$ , which cancels the $+\frac{2}{3}e$ from the up quark.

Techniques used

recall the electric charges of quark flavourssum the charges of three quarks to find the net chargeidentify the combination with zero total charge

Physics 9702/14 — October/November 2025

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Understanding the Question

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Understanding the Question

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Understanding the Question

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Understanding the Question

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Step-by-Step Reasoning