Physics 9702/13 — May/June 2024

Background Concept

SI prefixes represent powers of 10. In particular,

$$\mu = 10^{-6}$$

So

$$1\ \mu\text{J} = 10^{-6}\ \text{J}$$

Converting a small decimal into standard form (scientific notation) makes it easy to identify which prefix applies.

Understanding the Question

We are asked to rewrite

$$0.000005\ \text{J}$$

using one of the prefixed units listed in the options (mJ, MJ, \mu J, nJ). So we need the corresponding power of 10.

Approach

1. Convert the given value in joules into standard form $a \times 10^n$.
2. Recognise which prefix corresponds to that power of 10.
3. Choose the option with the same numerical value.

Step-by-Step Reasoning

Rewrite the decimal:

$$0.000005 = 5 \times 10^{-6}$$

So the energy is:

$$0.000005\ \text{J} = 5 \times 10^{-6}\ \text{J}$$

Now use the definition $1\ \mu\text{J} = 10^{-6}\ \text{J}$:

$$5 \times 10^{-6}\ \text{J} = 5\ \mu\text{J}$$

Therefore the correct option is C.

Key Takeaways

• Convert decimals to standard form to spot the power of 10.
• Memorise common prefixes: $\text{m} = 10^{-3}$, $\mu = 10^{-6}$, $\text{n} = 10^{-9}$, $\text{M} = 10^{6}$.

Common Mistakes

• Confusing $\mu$ (micro, $10^{-6}$) with m (milli, $10^{-3}$).
• Treating $\text{MJ}$ as “microjoule” instead of megajoule (it is $10^{6}\ \text{J}$).
• Miscounting zeros when converting $0.000005$ to standard form.

Things to Be Careful About

• Case matters in prefixes: $\text{M}$ (mega) is not the same as $\text{m}$ (milli).
• Check the exponent sign: small numbers correspond to negative powers of 10.
• Ensure the unit stays as joules; only the prefix changes.

Background Concept

Measurements are affected by two main types of error:

• Random errors cause readings to vary unpredictably above and below the true value due to uncontrollable fluctuations (e.g. reaction time, small environmental changes, limited instrument resolution). They mainly affect precision (the spread of readings).
• Systematic errors shift all readings in the same direction away from the true value (e.g. zero error, miscalibration, parallax due to consistent viewing angle). They mainly affect accuracy (closeness to the true value).

A key idea is how to reduce each type:

• Random errors are reduced by repetition and averaging (the mean tends to be closer to the true value than a single reading).
• Systematic errors are reduced by correcting the cause (e.g. calibrating, zeroing, adjusting the apparatus/procedure).

Understanding the Question

You are given four statements about random and systematic errors. The task is to choose the one statement that correctly describes the nature of these errors or how to reduce them.

Approach

Use the definitions:

1. Check whether each statement matches random or systematic behaviour.
2. Check whether the proposed method (if any) is actually effective for that type of error.
3. Pick the only statement that is fully correct.

Step-by-Step Reasoning

• A: “A systematic error cannot be reduced by adjusting the apparatus.” This is false. Many systematic errors (e.g. zero error, calibration error) can be reduced/corrected by adjusting or calibrating the apparatus.

• B: “A systematic error results in a different reading each time…” This describes random error (scatter), not systematic error. So false.

• C: “Random errors are always caused by the person…” False. Random errors can come from many sources (e.g. electrical noise in sensors, fluctuations in temperature, limited scale resolution), not only the observer.

• D: “Random errors can be reduced by taking the average of several measurements.” True: averaging repeated readings reduces the effect of random fluctuations and improves precision.

Therefore, the correct option is D.

Key Takeaways

• Random errors → scatter in readings → reduced by repeats and mean.
• Systematic errors → consistent offset → reduced by calibration/zero correction/improving method.
• Precision relates to random error; accuracy relates to systematic error.

Common Mistakes

• Saying averaging reduces systematic error (it does not; the mean would still be shifted).
• Thinking systematic error causes readings to vary each time (that is random error).
• Assuming random errors must be due to “human error” only.

Things to Be Careful About

• A set of repeated readings with a consistent offset suggests a systematic error even if the readings are tightly clustered.
• When you take many readings, the spread decreases in importance for the mean, but any systematic offset remains unless corrected.
• In MCQs like this, look for statements that are true without any extra conditions (e.g. “always” is often a clue to a false statement unless it is genuinely universal).

Background Concept

When a quantity is calculated from measured values, its uncertainty depends on (i) how uncertain each measurement is and (ii) how strongly the result depends on that measurement.

For multiplication and division, fractional (or percentage) uncertainties add:

• If $Q = ab/c$, then

$$\frac{\Delta Q}{Q} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}.$$

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

• If $Q = a^n$, then

$$\frac{\Delta Q}{Q} \approx |n|\frac{\Delta a}{a}.$$

These are the standard A-Level “worst-case” (maximum) rules used for uncertainty propagation.

Understanding the Question

You are given

$$E = \frac{4FL}{\pi d^2 x}$$

and the measured values with their absolute uncertainties for $L$, $d$, $F$ and $x$.

The question asks: which one measurement contributes the most to the uncertainty in the calculated value of $E$? So we compare the size of each term in the fractional uncertainty of $E$.

Approach

1. Write $E$ as a proportionality to see the dependence:

$$E \propto \frac{FL}{d^2 x}.$$

1. Use the fractional-uncertainty addition rule for products/quotients.
2. Apply the power rule to the $d^2$ term (this doubles the fractional uncertainty from $d$).
3. Calculate the percentage uncertainty for each measurement and identify the largest.

Step-by-Step Reasoning

Start from

$$E \propto \frac{FL}{d^2 x}.$$

Hence the fractional uncertainty is approximately

$$\frac{\Delta E}{E} \approx \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta x}{x}.$$

Now compute each fractional uncertainty:

• For $L = 2.043 \pm 0.002\ \text{m}$:

$$\frac{\Delta L}{L} = \frac{0.002}{2.043} \approx 9.8 \times 10^{-4} \approx 0.10\%.$$

• For $F = 19.62 \pm 0.01\ \text{N}$:

$$\frac{\Delta F}{F} = \frac{0.01}{19.62} \approx 5.1 \times 10^{-4} \approx 0.051\%.$$

• For $d = 0.54 \pm 0.02\ \text{mm}$, first find $\Delta d/d$ then double it because of $d^2$:

$$\frac{\Delta d}{d} = \frac{0.02}{0.54} \approx 0.037 \approx 3.7\%,$$

so

$$2\frac{\Delta d}{d} \approx 7.4\%.$$

• For $x = 5.2 \pm 0.2\ \text{mm}$:

$$\frac{\Delta x}{x} = \frac{0.2}{5.2} \approx 0.0385 \approx 3.85\%.$$

Comparing contributions: $0.051\%$, $0.10\%$, $7.4\%$, $3.85\%$. The largest is clearly the diameter contribution, because it is relatively uncertain and also squared in the formula.

Therefore the measurement with the largest effect on the uncertainty in $E$ is $d$ (option B).

Key Takeaways

• Compare uncertainties using fractional (percentage) uncertainty, not absolute uncertainty.
• For a quantity in the denominator, its fractional uncertainty still adds.
• Powers matter: if $d$ is squared, its fractional uncertainty contribution doubles.
• The dominant uncertainty is often from the measurement with the largest percentage uncertainty and/or highest power.

Common Mistakes

• Forgetting the factor of $2$ for $d^2$.
• Comparing absolute uncertainties directly (e.g. $0.2$ vs $0.02$) even though the quantities have different sizes and units.
• Thinking quantities in the denominator “subtract” uncertainties (they do not for the standard worst-case method).

Things to Be Careful About

• Use consistent significant figures when comparing percentages; you only need enough precision to see which is largest.
• The units (m vs mm) do not matter for fractional uncertainty as long as numerator and denominator are in the same unit.
• The question asks for the largest effect on uncertainty, so you must compare the fractional contributions, not compute $E$ itself.

Background Concept

A scalar has magnitude only (e.g. mass, time, pressure, energy). A vector has both magnitude and direction (e.g. displacement, velocity, acceleration, force).

When multiplying quantities:

• scalar (\times) scalar (\to) scalar
• scalar (\times) vector (\to) vector
• vector (\times) vector is not a simple “multiplication” in physics unless a specific product is defined. The most common relevant one here is the dot product:

$$\vec{F} \cdot \vec{s} = Fs\cos\theta$$

This produces a scalar (work done / energy transferred).

Understanding the Question

We are given four products of two physical quantities and asked which product produces a scalar quantity.

So for each option we decide whether the result is scalar or vector, using the scalar/vector nature of each quantity and (where appropriate) known physics definitions.

Approach

Check each option:

1. Identify whether each factor is scalar or vector.
2. If a factor is a vector, decide whether the product is intended as a scalar multiple (giving a vector) or a dot-product-like combination (giving a scalar).
3. Pick the one that must be scalar.

Step-by-Step Reasoning

Option A: (force) (\times) (displacement in the direction of the force)

• Force is a vector (\vec{F}).
• “Displacement in the direction of the force” means the component of displacement along (\vec{F}), i.e. (s\cos\theta), which is a scalar.
• The product is

$$W = F(s\cos\theta)$$

This is work done, and work/energy is a scalar.
So A produces a scalar.

Option B: (mass) (\times) (acceleration)

• Mass (m) is scalar.
• Acceleration (\vec{a}) is vector.

$$\vec{F} = m\vec{a}$$

This gives force, a vector.

Option C: (pressure) (\times) (area)

• Pressure (p) is scalar.
• Area (A) is scalar.
• The product (pA) gives the magnitude of the force on an area in a uniform pressure, but the actual force has a direction (normal to the surface). In vector form you would write (\vec{F} = pA,\hat{n}).
  So the physical quantity “force” is vector; (pA) alone is not the full vector.

Option D: (velocity) (\times) (time)

• Velocity (\vec{v}) is vector.
• Time (t) is scalar.

$$\vec{s} = \vec{v}t$$

This gives displacement, a vector.

Therefore the only clearly scalar outcome is A.

Key Takeaways

• Work done is a scalar: (W = \vec{F} \cdot \vec{s} = Fs\cos\theta).
• Multiplying a vector by a scalar gives a vector (e.g. (\vec{s} = \vec{v}t), (\vec{F} = m\vec{a})).
• Always check whether the question implies a dot product (scalar) or just a scalar multiple (vector).

Common Mistakes

• Treating (m\times a) as a scalar because (m) is scalar, forgetting that acceleration is a vector so (m\vec{a}) is a vector.
• Thinking (v\times t) is a scalar “distance” rather than the vector displacement (unless speed is stated instead of velocity).
• For pressure (\times) area: giving “scalar” without noting that force is a vector; (pA) is only the magnitude unless a direction is specified.

Things to Be Careful About

• Wording like “in the direction of the force” is a strong clue that a component is being taken and the product is the dot product, giving a scalar.
• Distinguish velocity (vector) from speed (scalar). If the option had said speed (\times) time, that would give a scalar distance.
• In pressure questions, direction is typically perpendicular to the surface; without a direction unit vector (\hat{n}), you have only the magnitude (pA).

Background Concept

For motion in a straight line, a velocity–time graph shows how velocity $v$ changes with time $t$.

The key fact is:

• Displacement over a time interval equals the area under the $v$–$t$ graph for that interval.

This works because for a small time $\Delta t$, the displacement is approximately $v\,\Delta t$; adding (integrating) these contributions gives the total displacement.

Understanding the Question

The graph is a straight line from $(0,0)$ to $(4\ \text{s}, 12\ \text{m s}^{-1})$, so velocity increases uniformly.

“Last $2$ seconds of the motion” means the time interval from:

$$t = 2\ \text{s} \text{ to } t = 4\ \text{s}.$$

We must find the area under the graph between these times.

Approach

1. Read (or calculate) the velocities at $t=2\ \text{s}$ and $t=4\ \text{s}$.
2. The region under the straight-line section between $t=2$ and $t=4$ is a trapezium.
3. Use trapezium area = average of parallel sides $\times$ width in time.

Step-by-Step Reasoning

1. Read velocities from the graph:

   • At $t=2\ \text{s}$, the dashed line shows $v=6\ \text{m s}^{-1}$.
   • At $t=4\ \text{s}$, the end point is $v=12\ \text{m s}^{-1}$.

2. Displacement between $t=2$ and $t=4$ is the trapezium area:

$$\text{displacement} = \frac{1}{2}(v_1+v_2)\Delta t$$

Here $v_1 = 6\ \text{m s}^{-1}$, $v_2 = 12\ \text{m s}^{-1}$, and $\Delta t = 2\ \text{s}$:

$$\text{displacement} = \frac{1}{2}(6+12)\times 2 = 18\ \text{m}.$$

So the correct option is C.

Key Takeaways

• Displacement over an interval is the area under the velocity–time graph over that interval.
• For a straight-line segment, the area is often easiest as a trapezium.

Common Mistakes

• Finding the area from $t=0$ to $t=2$ (first 2 seconds) instead of the last 2 seconds.
• Using the gradient (acceleration) instead of area (displacement).
• Treating the answer as a velocity and giving units of $\text{m s}^{-1}$ instead of $\text{m}$.

Things to Be Careful About

• Make sure you use the correct time interval: here it is $2\to 4\ \text{s}$.
• Always include units: area under $v$–$t$ has units $(\text{m s}^{-1})\times(\text{s})=\text{m}$.
• When using trapezium area, use $\frac{1}{2}(v_1+v_2)\Delta t$ (not $\frac{1}{2}v_1v_2\Delta t$ or similar mix-ups).

Background Concept

Momentum $p$ is defined by $p = mv$ and is a vector. Newton's second law can be written in momentum form:

$$F_{\text{net}} = \frac{\mathrm{d}p}{\mathrm{d}t}$$

So, on a momentum–time graph, the gradient gives the resultant (net) force in the direction chosen as positive. A negative gradient means the net force is opposite to the positive direction.

Understanding the Question

A block moves to the right. There is an applied constant force $F$ to the right and a constant resistive force of $5.0\ \text{N}$ to the left.

The momentum decreases linearly from $2.7\ \text{kg m s}^{-1}$ at $t=0$ to $1.5\ \text{kg m s}^{-1}$ at $t=0.40\ \text{s}$. You are asked to find the magnitude of $F$.

Approach

1. Use the momentum–time graph to find $\Delta p/\Delta t$, which equals $F_{\text{net}}$.
2. Choose rightwards as positive and write the force balance horizontally: $F_{\text{net}} = F - 5.0$.
3. Solve for $F$.

Step-by-Step Reasoning

From the graph:

$$\Delta p = 1.5 - 2.7 = -1.2\ \text{kg m s}^{-1}$$ $$\Delta t = 0.40\ \text{s}$$

Hence the net force:

$$F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{-1.2}{0.40} = -3.0\ \text{N}$$

The negative sign shows the resultant force is to the left.

Now resolve forces horizontally (right positive):

$$F_{\text{net}} = F - 5.0$$

Substitute $F_{\text{net}}=-3.0\ \text{N}$:

$$-3.0 = F - 5.0 \;\Rightarrow\; F = 2.0\ \text{N}$$

So the correct option is A.

Key Takeaways

• The gradient of a $p$-vs-$t$ graph equals the net force.
• A decreasing momentum (negative gradient) means the net force is opposite the chosen positive direction.
• When multiple forces act, $F_{\text{net}}$ is the algebraic sum using a sign convention.

Common Mistakes

• Using $F = \Delta p \times \Delta t$ instead of $F = \Delta p/\Delta t$.
• Forgetting the sign of the gradient and treating the net force as $+3.0\ \text{N}$.
• Adding the resistive force instead of subtracting it (mixing up directions).

Things to Be Careful About

• Keep a clear positive direction (here, rightwards).
• Remember $1\ \text{N} = 1\ \text{kg m s}^{-2}$, so $\Delta p/\Delta t$ is already in newtons.
• The question asks for the magnitude of $F$ (a positive number), but the intermediate $F_{\text{net}}$ can be negative depending on direction.

Background Concept

Newton’s third law states: when two bodies interact, they exert forces on each other that are equal in magnitude and opposite in direction.

Key properties of a Newton’s third-law pair:

• equal magnitude
• opposite direction
• act on two different objects (one force on each object)
• same type of force (e.g. both contact forces, or both gravitational forces)

Equilibrium is different: an object is in equilibrium when the resultant force on that same object is zero (and, for full equilibrium, resultant moment is also zero). The forces that balance for equilibrium all act on the same object.

Understanding the Question

We are given four statements about the forces in a Newton’s third-law pair, and we must choose the one that is not correct.

So we compare each option against the defining properties above.

Approach

1. Write down the defining features of a Newton’s third-law pair.
2. Check each option:
   • if it matches the definition, it is correct;
   • if it confuses Newton’s third law with equilibrium (balanced forces on one object), it will be incorrect.

Step-by-Step Reasoning

• A: “They act in opposite directions.”
  This matches Newton’s third law (forces are opposite in direction). So A is correct.

• B: “They act on different objects.”
  This is essential: one force acts on one body, the other acts on the other body. So B is correct.

• C: “They are the same type of force.”
  Also correct: if one is a frictional/contact/normal/gravitational force, the partner is the corresponding frictional/contact/normal/gravitational force on the other object.

• D: “They cause an object to be in equilibrium.”
  This is not correct because the two forces in a Newton’s third-law pair do not act on the same object. Equilibrium requires forces on the same object to sum to zero. A third-law pair therefore cannot directly make one object have zero resultant force.

Hence the incorrect description is D.

Key Takeaways

• Newton’s third-law forces act on different objects; they do not cancel each other on a single object.
• Equilibrium requires balancing forces acting on the same object.
• Action–reaction pairs are equal and opposite and of the same type of force.

Common Mistakes

• Saying the third-law pair “cancels out”: they do not cancel because they act on different objects.
• Treating any equal-and-opposite forces as a third-law pair (they might instead be two different forces on the same object producing equilibrium).

Things to Be Careful About

• Always identify which object each force acts on before deciding if forces form a Newton’s third-law pair.
• For equilibrium questions, focus only on forces acting on the chosen object; do not include the reaction force acting on the other object.

Background Concept

On a speed–time graph, the acceleration $a$ at any instant is the gradient:

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

For a falling object in air, the forces are weight $mg$ downward and air resistance (drag) upward. The resultant force is

$$F_\text{resultant} = mg - F_\text{drag}$$

Newton’s second law links resultant force to acceleration:

$$F_\text{resultant} = ma$$

Terminal speed occurs when forces balance ($mg = F_\text{drag}$), so the resultant force is zero and hence $a = 0$.

Understanding the Question

The graph starts at zero speed, rises quickly, and then gradually becomes horizontal. You are asked which statement correctly describes what is happening as time increases. The key is that the curve becomes horizontal, meaning the speed becomes constant.

Approach

1. Use the speed–time graph to deduce how acceleration changes (look at the gradient).
2. Connect acceleration to resultant force using $F = ma$.
3. Use the idea of terminal speed: constant speed implies zero acceleration and zero resultant force.

Step-by-Step Reasoning

• At the start, the curve is steep, so the gradient is large: acceleration is large.
• As time goes on, the curve becomes less steep: the gradient decreases, so acceleration decreases.
• Eventually the curve is horizontal: gradient $= 0$, so acceleration $= 0$.
• Zero acceleration means resultant force is zero (from $F = ma$).
• Therefore the raindrop reaches a steady (terminal) speed because the acceleration decreases to zero.

So statement A is correct.

(Why the others are wrong, briefly: if acceleration were increasing (B), the curve would get steeper; drag does not decrease with speed (C), it increases; resultant force does not increase with speed (D), it decreases to zero as terminal speed is reached.)

Key Takeaways

• Gradient of a speed–time graph gives acceleration.
• Terminal speed corresponds to zero acceleration and hence zero resultant force.
• In air, increasing speed generally increases drag, reducing the resultant force over time.

Common Mistakes

• Thinking a high speed means a high acceleration: acceleration depends on how quickly speed changes, not the speed itself.
• Saying the resultant force is constant because weight is constant: drag changes with speed, so the resultant changes.
• Confusing “levels off” with “increases more slowly”: in fact it tends to a constant value, so acceleration tends to zero.

Things to Be Careful About

• The graph is speed–time, not distance–time: you must use gradient as acceleration, not speed.
• A horizontal section means $a = 0$, not that forces are absent; it means forces balance.
• For terminal speed, you should state both: constant speed and zero acceleration (and thus zero resultant force).

Background Concept

In collisions, two key ideas are often used:

1. Conservation of momentum (always true if there is no external resultant impulse):

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

1. Elasticity of the collision, which describes what happens to kinetic energy and to relative speeds.

A perfectly elastic collision is defined by:

• total kinetic energy conserved, and equivalently
• coefficient of restitution

$$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = 1$$

So, for a perfectly elastic collision:

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

Understanding the Question

You are asked which statement correctly describes a perfectly elastic collision between two objects. The options combine two possible properties:

• whether total kinetic energy is conserved
• whether relative speed of approach equals relative speed of separation

You must choose the one option that matches the definition of a perfectly elastic collision.

Approach

Recall the defining properties of a perfectly elastic collision:

1. Kinetic energy is conserved.
2. Coefficient of restitution $e=1$, meaning relative speed of separation equals relative speed of approach.

Then compare these two properties against the four options.

Step-by-Step Reasoning

• For a perfectly elastic collision, by definition, kinetic energy is conserved.
• Also, for a perfectly elastic collision, $e=1$:

$$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = 1$$

which implies:

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

• Option A states both of these facts, so it is correct.

Key Takeaways

• Perfectly elastic collision: total kinetic energy conserved.
• Perfectly elastic collision: relative speed of approach = relative speed of separation (because $e=1$).

Common Mistakes

• Thinking momentum conservation alone implies kinetic energy conservation (it does not).
• Mixing up elastic and inelastic:
  • inelastic: kinetic energy is not conserved and typically relative speeds are not equal.
  • perfectly inelastic: objects stick together; kinetic energy is definitely not conserved.

Things to Be Careful About

• The “relative speed” is the speed of one object as seen from the other, i.e. the difference in their velocities along the line of impact.
• Kinetic energy conservation refers to total kinetic energy of the system, not each object separately.

Background Concept

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

$$\tau = Fr_\perp$$

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

If two equal forces act in opposite directions at different points (a couple), they produce rotation without a resultant force. The turning effects (torques) of the two forces add, giving a non-zero net torque.

Understanding the Question

A straight tube of total length $0.15\ \text{m}$ is pivoted at its midpoint. Each end experiences a horizontal force of magnitude $3.0\ \text{N}$, and each force is at right angles to the tube.

We are asked for the magnitude of the net torque about the pivot due to both forces.

Key information:

• Pivot is at the middle, so each force acts at a distance $0.15/2$ from the pivot.
• Force is perpendicular to the tube, so the lever arm is simply that half-length.
• The two forces act in opposite directions at opposite ends, which makes a couple; their torques act in the same rotational sense and therefore add.

Approach

1. Find the distance from the pivot to one end of the tube.
2. Use $\tau = Fr$ for one jet (since force is perpendicular).
3. Double it because there are two ends producing the same sense of rotation.
4. Match the numerical value to the options.

Step-by-Step Reasoning

Distance from pivot to either end:

$$r = \frac{0.15\ \text{m}}{2} = 0.075\ \text{m}$$

Torque from one end (perpendicular force, so no sine factor needed):

$$\tau_1 = Fr = (3.0\ \text{N})(0.075\ \text{m}) = 0.225\ \text{N m}$$

Now consider both jets. Although the forces are opposite directions, they act on opposite sides of the pivot. This is exactly a couple: each force tends to rotate the tube in the same direction about the pivot, so the torques add:

$$\tau = 2\tau_1 = 2(0.225) = 0.45\ \text{N m}$$

So the correct option is C.

Key Takeaways

• Use $\tau = Fr_\perp$ (perpendicular distance to the line of action).
• For a pivot at the midpoint, the lever arm to an end is half the total length.
• Two equal and opposite forces at opposite ends form a couple with non-zero net torque: the moments add.

Common Mistakes

• Using the full length $0.15\ \text{m}$ as the lever arm instead of $0.075\ \text{m}$.
• Subtracting the two torques because the forces are opposite, forgetting that they act at different points and produce rotation in the same sense.
• Forgetting that the force is perpendicular and unnecessarily introducing a sine/cosine factor incorrectly.

Things to Be Careful About

• Torque depends on the perpendicular distance from pivot; here that is half the tube length.
• Ensure the final unit is $\text{N m}$.
• When forces form a couple, the resultant force is zero but the resultant torque is not.

Background Concept

For an object in equilibrium under forces in a plane,

$$\sum \vec F = \vec 0$$

This means the forces add as vectors to form a closed polygon (often a triangle for three forces). Equivalently, if you add any two forces to get a resultant $\vec R$, then the third force must be

$$\vec F_3 = -\vec R$$

i.e. same magnitude as $\vec R$ but opposite direction.

A reliable way to add forces is to resolve each force into perpendicular components (e.g. horizontal $x$ and vertical $y$), add the components, then reconstruct the resultant using Pythagoras and trigonometry.

Understanding the Question

Two forces are shown acting from the same point:

• $43\ \text{N}$ vertically upward.
• $17\ \text{N}$ downwards and to the left, making an angle of $125^\circ$ with the upward vertical.

A third force is missing. Since the object is in equilibrium, the third force must exactly cancel the resultant of the two shown forces. The options provide different magnitudes and directions; we must find which matches.

Approach

1. Choose axes: up as +$y$, right as +$x$.
2. Resolve the $17\ \text{N}$ force into $x$ and $y$ components using the given angle to the vertical.
3. Add components with the $43\ \text{N}$ force to get the resultant $\vec R$.
4. The third force is $-\vec R$ (same size, opposite direction). Match that magnitude/direction to the options.

Step-by-Step Reasoning

1. Resolve the $17\ \text{N}$ force.

The angle between the $17\ \text{N}$ force and the upward vertical is $125^\circ$. The component along the vertical (upwards positive) is

$$F_{2y} = 17\cos 125^\circ$$

Since $125^\circ$ is obtuse, $\cos 125^\circ$ is negative, so this correctly gives a downward vertical component.

Numerically,

$$F_{2y} = 17\cos 125^\circ = -9.75\ \text{N}$$

The horizontal component has magnitude $17\sin 125^\circ$. The force points to the left, so the $x$-component is negative:

$$F_{2x} = -17\sin 125^\circ = -13.9\ \text{N}$$

1. Add components to get the resultant of the two shown forces.

The $43\ \text{N}$ force has components $F_{1x}=0$ and $F_{1y}=43$.

So the resultant components are

$$R_y = 43 + (-9.75) = 33.25\ \text{N}$$ $$R_x = 0 + (-13.9) = -13.9\ \text{N}$$

This means the resultant points up-and-left.

1. Magnitude of resultant:

$$R = \sqrt{R_x^2+R_y^2} = \sqrt{(13.9)^2+(33.25)^2} \approx 36\ \text{N}$$

1. Direction of resultant relative to the vertical.

Using the right triangle with vertical side $R_y$ and horizontal side $|R_x|$,

$$\tan\theta = \frac{|R_x|}{R_y} = \frac{13.9}{33.25} \Rightarrow \theta \approx 23^\circ$$

So $\vec R$ is $23^\circ$ to the left of the upward vertical.

1. Use equilibrium.

The third force must be equal and opposite to $\vec R$, so it must be:

• magnitude $\approx 36\ \text{N}$
• direction down-and-right, at $23^\circ$ to the downward vertical.

Many options label the angle to the upward vertical instead; that would be

$$180^\circ - 23^\circ = 157^\circ$$

So the correct option is the $36\ \text{N}$ force directed down-right with an angle marked $157^\circ$: option B.

Key Takeaways

• Equilibrium in a plane means $\sum \vec F = 0$.
• For three forces, the “missing” force is the negative of the resultant of the other two.
• Resolving into components is often the quickest and least ambiguous method.

Common Mistakes

• Using $17\cos 125^\circ$ and then treating it as upward (forgetting it is negative for an obtuse angle).
• Taking the angle incorrectly (e.g. using $55^\circ$ instead of $125^\circ$ without thinking about the force direction).
• Finding the resultant correctly but forgetting to reverse it for the equilibrant (the third force).
• Mixing up which angle is being quoted in the options (angle to upward vertical vs angle to downward vertical).

Things to Be Careful About

• Always assign signs to components based on direction (left negative $x$, down negative $y$).
• Check your component results are physically sensible: the $17\ \text{N}$ force is down-left, so both components should be negative.
• When matching to options, convert between equivalent angle descriptions (e.g. $23^\circ$ from downward vertical is the same direction as $157^\circ$ from upward vertical).

Background Concept

An object is in equilibrium only if:

1. The resultant (net) force is zero: (\sum \vec{F} = 0).
2. The resultant moment (torque) about any point is zero: (\sum \tau = 0).

If exactly two forces act on a rigid object, these two conditions strongly restrict how those forces can be arranged:

• (\sum \vec{F} = 0) requires the forces to be equal in magnitude and opposite in direction.
• (\sum \tau = 0) requires the forces to have the same line of action; otherwise they produce a turning effect (a couple).

Understanding the Question

You are told there are only two forces on the object and the object is in equilibrium. You must decide which of the three statements must be true.

The key is that “equilibrium” is not just “no acceleration” (no net force), but also “no tendency to rotate” (no net moment).

Approach

Check each statement against the two equilibrium requirements:

• Use (\sum \vec{F} = 0) to test whether magnitudes/directions are constrained.
• Use (\sum \tau = 0) to test whether their lines of action are constrained.

Then match the true statements to the options A–D.

Step-by-Step Reasoning

1. Statement 1: the two forces must have equal magnitudes.

With only two forces, equilibrium means:

$$\vec{F}_1 + \vec{F}_2 = 0$$

So (\vec{F}_2 = -\vec{F}_1), which implies equal magnitudes. Statement 1 is correct.

1. Statement 2: the two forces must act in the same direction.

From (\vec{F}_2 = -\vec{F}_1), they must be in opposite directions, not the same direction. If they acted in the same direction, the resultant would be non-zero. Statement 2 is incorrect.

1. Statement 3: the two forces must act through the same point.

For a rigid object, it is not enough that forces cancel; they must also not create a couple. If two equal and opposite forces act along different parallel lines, they form a couple and cause rotation.

Therefore, for equilibrium with only two forces, they must act along the same line of action (collinear). Many mark schemes express this as “act through the same point/line” (i.e. no separation between their lines of action). Hence statement 3 is taken as correct in this context.

So the correct set is statements 1 and 3 only (\Rightarrow) option B.

Key Takeaways

• Equilibrium requires both (\sum \vec{F}=0) and (\sum \tau = 0).
• With only two forces, equilibrium (\Rightarrow) forces are equal, opposite, and collinear.
• Equal and opposite but not collinear (\Rightarrow) a couple (\Rightarrow) rotation, so not equilibrium.

Common Mistakes

• Thinking equilibrium means only “resultant force is zero” and forgetting the moment condition.
• Concluding “same direction” instead of “opposite direction” when (\sum \vec{F}=0).
• Believing “equal and opposite forces always give equilibrium” (they do not if they form a couple).

Things to Be Careful About

• For an extended object, “same point” is often shorthand for “same line of action” in MCQs; the physical requirement is no turning effect.
• If the question had said “particle” instead of “object”, moments are irrelevant (all forces act at one point), but here it says “object”, so rotation must be considered.

Background Concept

For a crystal lattice, we often model the solid as made of repeating "unit cells". The density is related to the mass and volume of one unit cell:

$$\rho = \frac{m_{\text{cell}}}{V_{\text{cell}}}$$

For a cubic unit cell with edge length $a$,

$$V_{\text{cell}} = a^3$$

A key idea is that atoms at the corners of a cube are shared between neighbouring cubes. In a simple cubic lattice, each corner atom belongs to 8 different cubes, so each cube effectively contains $\tfrac{1}{8}$ of each corner atom.

Understanding the Question

You are told:

• atoms are only at the corners of a cube (simple cubic),
• mass of each atom is $3.5 \times 10^{-25}\ \text{kg}$,
• density of the crystal is $9.2 \times 10^3\ \text{kg m}^{-3}$.

The question asks for the shortest distance between the centres of two adjacent atoms. In a simple cubic lattice, the nearest neighbours are along the cube edges, so the shortest centre-to-centre distance is just the cube edge length $a$.

Approach

1. Work out how many atoms belong to one unit cell (taking into account sharing of corner atoms).
2. Find the unit cell mass $m_{\text{cell}}$.
3. Use $\rho = m_{\text{cell}} / a^3$ to find $a^3$.
4. Take the cube root to get $a$, which is the nearest-neighbour distance.

Step-by-Step Reasoning

1) Atoms per unit cell
There are 8 corners. Each corner atom is shared by 8 cubes.

So the number of atoms in one unit cell is

$$N = 8 \times \frac{1}{8} = 1$$

2) Mass per unit cell

$$m_{\text{cell}} = N m_{\text{atom}} = 1 \times 3.5 \times 10^{-25} = 3.5 \times 10^{-25}\ \text{kg}$$

3) Use density to find $a^3$

$$\rho = \frac{m_{\text{cell}}}{a^3} \;\Rightarrow\; a^3 = \frac{m_{\text{cell}}}{\rho}$$ $$a^3 = \frac{3.5 \times 10^{-25}}{9.2 \times 10^{3}} \approx 3.8 \times 10^{-29}\ \text{m}^3$$

4) Cube root

$$a = \left(3.8 \times 10^{-29}\right)^{1/3} \approx 3.4 \times 10^{-10}\ \text{m}$$

This matches option C.

Key Takeaways

• In a simple cubic lattice, effective atoms per unit cell $= 1$ (because $8 \times \tfrac{1}{8}$).
• Nearest-neighbour distance in simple cubic is the cube edge length $a$.
• Use $\rho = m_{\text{cell}}/a^3$ to connect microscopic mass to lattice spacing.

Common Mistakes

• Forgetting that corner atoms are shared, and using 8 atoms per unit cell (gives a distance too large).
• Using $a^2$ instead of $a^3$ for the volume of a cube.
• Arithmetic slips when taking the cube root of a number in standard form.

Things to Be Careful About

• The "shortest distance between adjacent atoms" depends on the lattice type; here it is along an edge, not a face diagonal or body diagonal.
• Check units: $\rho$ in $\text{kg m}^{-3}$ and mass in $\text{kg}$ ensures $a^3$ comes out in $\text{m}^3$.
• Cube roots of powers of ten: $10^{-29/3} = 10^{-9.666...}$, so the result should be around $10^{-10}\ \text{m}$ (an atomic-scale spacing).

Background Concept

In a stationary liquid, pressure increases with depth because the liquid lower down must support the weight of the liquid above it. The hydrostatic pressure at depth $h$ below the free surface is

$$p = \rho g h$$

where:

• $p$ is the (gauge) pressure due to the liquid column,
• $\rho$ is the density of the liquid,
• $g$ is gravitational field strength,
• $h$ is the vertical depth below the liquid surface.

Key idea: for the same liquid and same $g$, pressure depends only on depth $h$, not on the container shape or total volume.

Understanding the Question

You are shown four measuring cylinders containing the same liquid, but with two different liquid heights ($30\ \text{cm}$ and $10\ \text{cm}$). Points $A$, $B$, $C$, and $D$ are at different positions in the liquids.

The question asks which point has the greatest pressure. Since all cylinders are open to the atmosphere, the one at greatest depth below its own liquid surface has greatest pressure.

Approach

1. Use $p = \rho g h$.
2. Since $\rho$ and $g$ are the same for all, compare the depth $h$ for each point.
3. Choose the point with the largest $h$.

Step-by-Step Reasoning

• Point $A$ is near the top surface of a $30\ \text{cm}$ column, so its depth $h$ is small.
• Point $C$ is at the surface of a $10\ \text{cm}$ column, so $h = 0$ (pressure due to liquid is zero there).
• Point $D$ is near the bottom of a $10\ \text{cm}$ column, so $h \approx 10\ \text{cm}$.
• Point $B$ is well below the surface in a $30\ \text{cm}$ column. From the diagram description, $B$ is about level with the surface of the $10\ \text{cm}$ column, i.e. about $10\ \text{cm}$ above the base, so its depth below the $30\ \text{cm}$ surface is roughly

$$h \approx 30 - 10 = 20\ \text{cm}$$

This is greater than the depth at $D$ (about $10\ \text{cm}$), so $B$ has the greatest pressure.

Key Takeaways

• In a liquid at rest, pressure increases with vertical depth: $p = \rho g h$.
• For the same liquid, comparing pressures is the same as comparing depths below the surface.
• Container shape does not affect hydrostatic pressure at a given depth.

Common Mistakes

• Choosing the cylinder with the greatest total height without checking the point’s depth below the surface.
• Thinking pressure depends on the amount/volume of water rather than depth.
• Comparing heights above the base instead of depths below the surface.

Things to Be Careful About

• $h$ is measured from the liquid surface down to the point, not from the base up.
• Points at the surface have $h=0$ so no hydrostatic (gauge) pressure.
• If the question expects absolute pressure, it would be $p = p_{\text{atm}} + \rho g h$; however, the ranking is unchanged because $p_{\text{atm}}$ is the same for all points.

Background Concept

Efficiency is the fraction of the input energy (or power) that is converted into the desired (useful) form.

$$\eta = \frac{\text{useful energy output}}{\text{total energy input}}$$

To express efficiency as a percentage:

$$\eta_{\%} = \eta \times 100\%$$

In a filament lamp, the useful output is light energy; thermal energy is wasted from the point of view of producing light.

Understanding the Question

For every $8\ \text{J}$ emitted as light, the lamp emits $92\ \text{J}$ as thermal energy. These two outputs together account for the electrical energy supplied (assuming other losses are negligible), so:

• Useful output: $E_{\text{light}} = 8\ \text{J}$
• Wasted output: $E_{\text{thermal}} = 92\ \text{J}$
• Total input: $E_{\text{in}} = 8 + 92\ \text{J}$

The question asks for the percentage of electrical energy converted to light.

Approach

1. Add the light and thermal energies to get the total electrical energy input.
2. Use $\eta = E_{\text{useful}}/E_{\text{in}}$.
3. Multiply by $100\%$ to get a percentage.
4. Match the result to the given options.

Step-by-Step Reasoning

Total energy supplied to the lamp is the sum of the energies it outputs as light and as heat:

$$E_{\text{in}} = 8\ \text{J} + 92\ \text{J} = 100\ \text{J}$$

Efficiency (useful fraction) is:

$$\eta = \frac{E_{\text{light}}}{E_{\text{in}}} = \frac{8}{100} = 0.08$$

Convert to a percentage:

$$\eta_{\%} = 0.08 \times 100\% = 8\%$$

So the correct option is A.

Key Takeaways

• Useful energy for a lamp (in this context) is the light energy, not the heat.
• Total input energy equals the sum of all output energy forms.
• Efficiency is a ratio; percentage efficiency is the ratio multiplied by $100\%$.

Common Mistakes

• Using $92\ \text{J}$ as the input energy instead of adding $8 + 92$.
• Calculating $92/100$ (wasted fraction) instead of $8/100$ (useful fraction).
• Forgetting to multiply by $100\%$ when asked for a percentage.

Things to Be Careful About

• The wording “for every $8\ \text{J}$ of light” indicates a ratio; you must form the corresponding total ($8 + 92$).
• Efficiency must always be based on useful output divided by total input (not total output of one type).

Background Concept

Power $P$ is the rate at which energy is transferred (or work is done). By definition,

$$P = \frac{\text{energy transferred}}{\text{time taken}} = \frac{W}{t}$$

So the SI unit of power must be “joules per second”, which is also called the watt ($\text{W}$).

Understanding the Question

You are given four possible compound units and asked which one is a unit of power. Since power is energy per unit time, you should look for a unit of the form $\text{J}$ divided by $\text{s}$.

Approach

1. Write down the defining equation for power: $P = W/t$.
2. Convert that directly into units by replacing $W$ with joules ($\text{J}$) and $t$ with seconds ($\text{s}$).
3. Choose the option that matches $\text{J s}^{-1}$.

Step-by-Step Reasoning

Start with the definition:

$$P = \frac{W}{t}$$

• $W$ (work done / energy transferred) has unit joule, $\text{J}$.
• $t$ has unit second, $\text{s}$.

Therefore the unit of $P$ is

$$\frac{\text{J}}{\text{s}} = \text{J s}^{-1}$$

This corresponds to option D.

Key Takeaways

• Power is a rate: energy per time.
• Unit of power: $\text{J s}^{-1}$, which is the watt ($\text{W}$).

Common Mistakes

• Confusing power with energy: choosing $\text{J}$ instead of $\text{J s}^{-1}$.
• Mixing up other derived quantities: e.g. $\text{J C}^{-1}$ is volt ($\text{V}$), not power.

Things to Be Careful About

• Look for “per second” ($\text{s}^{-1}$) when the quantity is a rate.
• Remember that many compound units have special names (e.g. $\text{J C}^{-1}$ is $\text{V}$), but you should always be able to get the unit from the defining equation.