Physical Quantities and Measurement Techniques Worked ExamplesIGCSE Physics 0625

(a) Work out the value of one small division first: there are $5$ divisions between $40$ and $50$, so each division is $2\text{ cm}^3$. Read at the bottom of the meniscus (the curved surface). Before: three divisions above $40$.

After: four divisions above $60$.

(b) The stone pushes aside (displaces) its own volume of water, so the rise in level is the volume of the stone.

(c) Marking points: the eye is level with the bottom of the meniscus (1), looking at the scale at right angles, i.e. horizontally, to avoid parallax error (1).

(a) The row does not start at zero, so read both ends and subtract. Left end: $2.0\text{ cm}$; right end: $9.2\text{ cm}$.

(b) Divide the length of the row by the number of beads.

Convert to millimetres ($1\text{ cm} = 10\text{ mm}$).

(c) A ruler can only be read to about $1\text{ mm}$. That uncertainty is almost as big as one bead ($3.6\text{ mm}$), but is small compared with the whole row ($72\text{ mm}$). Dividing by $20$ shares the reading uncertainty between $20$ beads, so the average diameter is much more accurate.

Recall the syllabus lists. Scalars (magnitude only): distance, speed, time, mass, energy, temperature. Vectors (magnitude and direction): force, weight, velocity, acceleration, momentum, electric field strength, gravitational field strength.

Check each option for a scalar. B contains mass. C contains distance and speed. D contains energy and temperature.

Only A has no scalar in it: acceleration, momentum and weight all have a direction. Weight is a vector because it is a force — it always acts towards the centre of the Earth.

(a) A scalar has magnitude (size) only (1). A vector has magnitude and direction (1).

(b) Scalars: time, energy, temperature. Vectors: gravitational field strength, momentum, velocity. All six must be correct for the mark (1).

(c) The report gives a magnitude ($12\text{ m/s}$) and a direction (from the west, i.e. moving towards the east), so it is a velocity (1). "$12\text{ m/s}$" on its own would be a speed.

A fails because one pellet has a volume of only about $0.05\text{ cm}^3$ — far too small to see as a rise on a $100\text{ cm}^3$ scale.

B is tempting because it uses multiples, but a pile of pellets has air gaps between them, so the level of the pile gives a volume that is too large.

D fails twice: a $4.5\text{ mm}$ width cannot be measured precisely with a ruler, and a pellet is roughly a sphere, not a cube.

C combines displacement (water fills the gaps, so the rise is exactly the pellets' volume) with multiples (a rise large enough to read) and a narrow cylinder whose scale is finer. For example, a rise from $12.0$ to $14.5\text{ cm}^3$ gives $2.5 \div 50 = 0.050\text{ cm}^3$ per pellet.

(a) First find the average time for $20$ oscillations.

The period is the time for one complete oscillation, so divide by $20$.

(b) Starting and stopping a stopwatch by hand adds a reaction-time error of roughly $0.1$ to $0.3\text{ s}$. Over one swing (about $1.5\text{ s}$) that is a large fraction; over $20$ swings (about $29\text{ s}$) it is a small fraction, so the period is more accurate.

(c) Repeating and averaging reduces the effect of random errors (such as a slightly late start one time) and shows up any anomalous reading.

(d) Any one: start and stop timing as the bob passes a fixed marker (fiducial mark) at the centre of the swing, where it moves fastest and the moment is easiest to judge; count "zero" as the watch starts; keep the swings small.

(a) Any two of these marking points (1 each): the metre rule is vertical (check with a set square on the bench) and close to the string and bob; use a set square against the rule to line up the bottom of the bob with the scale; view each mark with the eye level with it, at right angles to the scale, to avoid parallax error.

(b) The distance from the point of suspension to the bottom of the bob is the difference of the two readings.

The centre of the bob is one radius above its bottom, so subtract half the diameter.

(c) The ruler cannot touch the bob along its widest part, so it is hard to judge where the diameter is (1), and the curved surface is some distance from the scale, so the reading depends on where the eye is — parallax error (1). Placing the bob between two blocks and measuring the gap between them avoids both problems.

The forces are at right angles, so the resultant is the hypotenuse of a right-angled triangle whose other sides are the two forces. Use Pythagoras.

Take the square root, and give the unit.

Find the direction as an angle measured from the $9.0\text{ N}$ (east) force, using $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$.

Choose a scale that makes the diagram large: here $1$ square $= 5\text{ m/s}$ (or $1\text{ cm} = 5\text{ m/s}$). Convert each velocity to a length.

Draw the $65\text{ m/s}$ arrow north. From its tip, draw the $18\text{ m/s}$ arrow east (tip to tail). The resultant goes from the start of the first arrow to the tip of the second.

Measure the resultant: about $13.5$ squares. Convert back using the scale.

Measure $\theta$ at the start point with a protractor: about $15^{\circ}$ east of north. Check with trigonometry.

Check the calculation first.

Each measurement has only $2$ significant figures, because the ruler reads to the nearest millimetre. A result cannot be more precise than the data it comes from.

Marking points: Student B / $28\text{ cm}^3$ (1), because the measurements are given to $2$ significant figures, which is the precision of the ruler (1). The extra digits in $27.968$ are not meaningful.

(a) The two velocities are at right angles, so combine them with Pythagoras.

(b) The angle from straight across uses the current as the opposite side.

(c) The key idea: only the $3.2\text{ m/s}$ part of the motion carries the boat across. So the crossing time is the width divided by $3.2\text{ m/s}$ — the current does not change it.

During those $15\text{ s}$ the current carries the boat downstream.

Check: the path length is $\sqrt{48^2 + 21^2} = 52.4\text{ m}$, and $3.49\text{ m/s} \times 15\text{ s} = 52.4\text{ m}$ — the two agree.

(a) Convert the stack height to metres first ($1\text{ cm} = 0.01\text{ m}$), then divide by the number of sheets.

Divide by $40$.

(b) Convert the other two lengths to metres before multiplying — this avoids converting $\text{cm}^3$ to $\text{m}^3$, where the factor is $10^6$, not $100$.

Multiply length × width × thickness.

Round to $2$ significant figures — the stack height ($1.8\text{ cm}$) has only $2$.