JAMB UTME - Physics - 2025

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Faraday's first law of electrolysis states that the mass deposited at an electrode is proportional to the quantity of charge passed, \(m = ZQ = ZIt\), where \(Z\) is the electrochemical equivalent of the substance. The whole calculation therefore begins with the charge.

Convert the time to seconds first, since the ampere is a coulomb per second:

For silver, one mole of \(\text{Ag}^{+}\) ions carries one faraday of charge, so depositing \(108\ \text{g}\) requires \(96\,500\ \text{C}\). This gives

and hence

The mass of silver deposited is about \(1.34\ \text{g}\).

The most frequent error is leaving the time in minutes, which makes the charge \(20\ \text{C}\) and the mass a hundredth of the true value, landing near the small figures offered here. A second error is dividing by a valency of \(2\); silver is monovalent, unlike copper in \(\text{Cu}^{2+}\), so no factor of two appears. Remember the routine: seconds, then coulombs, then multiply by the electrochemical equivalent.

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This question tests magnetic permeability, which is a measure of how easily a material allows magnetic field lines to pass through it. Magnetic shielding does not work by blocking field lines, because magnetic field lines cannot simply be stopped. It works by offering the field lines a much easier path that carries them around the region you want to protect.

Soft iron has a very high relative permeability, several thousand times that of air. When an instrument is enclosed in a soft iron case, nearly all of the external field lines are pulled into the iron walls and guided around the cavity, leaving the space inside with an extremely weak field. Soft iron rather than steel is used because soft iron has low retentivity: it magnetises strongly while the external field is present, but loses almost all of that magnetism once the field is removed, so the screen itself does not become a permanent magnet that would disturb the instrument.

Aluminium, brass and copper are non-magnetic. Their relative permeability is essentially the same as that of air, so field lines pass straight through them and the enclosed region is not protected. Copper and aluminium do oppose a changing magnetic field through induced eddy currents, which is why they appear in electrical screening, but against a steady magnetic field they provide no shielding. A useful examination link is: magnetic screening requires high permeability with low retentivity, and that combination describes soft iron.

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Refractive index is always a ratio of two lengths measured in the same figure, and it is greater than one for light passing from air into glass. In the two standard constructions used with a glass block, the value is obtained as the larger measured length divided by the smaller:

• Snell construction: \(n = \dfrac{\sin i}{\sin r}\), where the two sines are represented by the perpendicular distances of two points, one on the incident ray and one on the refracted ray, from the normal at the point of incidence.
• Apparent-depth construction: \(n = \dfrac{\text{real depth}}{\text{apparent depth}}\), the real depth of a mark under the glass being the larger of the two.

With \(SQ = 10\,\mathrm{cm}\) and \(SR = 6\,\mathrm{cm}\), the ratio is

The value is dimensionless, which is why the centimetres cancel and no unit is quoted. It is also physically sensible: glass has a refractive index of about \(1.5\) to \(1.7\), and the corresponding speed of light in the glass would be \(v = c/n = 3.0\times10^{8}/1.67 = 1.8\times10^{8}\,\mathrm{m\,s^{-1}}\).

The most tempting wrong answer comes from inverting the ratio, \(6/10 = 0.60\). A refractive index less than one would mean light travels faster in the glass than in air, which cannot happen for light entering a denser medium; that value belongs to the reverse passage, glass to air, where \(n_{\text{glass}\to\text{air}} = 1/1.67 = 0.60\). Use this check every time: when light passes into the optically denser medium, divide so that the answer exceeds one, and remember that the ray bends towards the normal on entering the glass, so the angle in air is the larger one.

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A wavelength \(\lambda\) is defined as the distance between any two successive points on a wave that are in phase, that is, points that are at the same stage of the vibration and moving in the same direction. Two neighbouring troughs satisfy that definition exactly: each is a point of maximum downward displacement, so the separation between them is one complete wavelength. The same is true of two neighbouring crests.

The distance that equals half a wavelength is the separation between a crest and the trough next to it, because those two points are exactly out of phase, one at maximum positive displacement and the other at maximum negative displacement. Confusing these two measurements is the usual source of error, and it matters in calculations: in a resonance-tube or standing-wave experiment the distance between consecutive nodes is \(\lambda/2\), whereas the distance between consecutive troughs of a travelling wave is \(\lambda\).

A quick check with numbers makes this secure. If \(\lambda = 0.5\,\mathrm{m}\), successive troughs are \(0.5\,\mathrm{m}\) apart and each trough is \(0.25\,\mathrm{m}\) from the crest beside it. In the examination, decide first whether the two marked points are in phase or out of phase, and only then attach \(\lambda\) or \(\lambda/2\) to the distance.

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A bore is a cavity in the aluminium block, and it expands as though it were a solid piece of the same material. Since the bore has a volume (cm³), we use cubical (volume) expansivity, \( \gamma = 3\alpha \).

Given:

• Initial volume, \( V_0 = 3.48 \text{ cm}^3 \)
• Linear expansivity, \( \alpha = 24 \times 10^{-6} \text{ K}^{-1} \)
• Temperature change, \( \Delta T = 340 - 34 = 306 \text{ K} \)

Calculate the cubical expansivity:

\[ \gamma = 3\alpha = 3 \times 24 \times 10^{-6} = 72 \times 10^{-6} \text{ K}^{-1} \]

Apply the volume expansion formula:

\[ V = V_0(1 + \gamma \Delta T) = 3.48(1 + 72 \times 10^{-6} \times 306) \]

\[ V = 3.48(1 + 0.022032) = 3.48 \times 1.022032 \]

\[ V \approx 3.56 \text{ cm}^3 \]

The new bore volume is approximately 3.56 cm³.

Remember: a hole or bore in a material expands exactly as if it were filled with the same material. The linear expansivity given must be converted to cubical expansivity (\( \gamma = 3\alpha \)) whenever the quantity expanding is a volume.

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The key words here are very small. Measuring an ordinary resistance is one problem; measuring a resistance of a fraction of an ohm is a harder one, because the resistance of the connecting leads and of the sliding or soldered contacts is itself of that same order. Any method in which those stray resistances are counted along with the unknown will give a badly wrong result. The instrument that avoids this is the potentiometer.

In the potentiometer method the unknown low resistance \(R\) is joined in series with a known low standard resistance \(S\), so that exactly the same current \(I\) flows through both. The potential difference across each is then tapped off and balanced against a length of the potentiometer wire, giving balancing lengths \(l_1\) and \(l_2\). Since \(V = IR\) and the potentiometer reading is proportional to the potential difference,

Two features make this accurate for tiny resistances. At balance the galvanometer carries no current, so the potentiometer draws nothing from the circuit and does not disturb it, and the tappings are made directly across the resistance itself, so the lead and contact resistances lie outside the measured section and cancel out of the ratio.

A Wheatstone bridge, in the metre-bridge form, is the standard circuit for a moderate resistance of a few ohms upwards, and that familiarity is what makes it tempting here. It becomes unreliable at the extremes, however: for a very small unknown, the end corrections and the resistance of the jockey contact and connecting wires are comparable with the quantity being measured, so the balance point loses its meaning. A voltmeter is unsuitable because a real voltmeter draws some current from the circuit and the potential difference across a very small resistance is minute, so the reading would be dominated by instrument error. A rheostat is not a measuring instrument at all; it is a variable resistor used to control the current in a circuit.

The examination point to retain is that the range of the resistance decides the method: a bridge for middling values, and a potentiometer, whose null reading excludes lead and contact resistance, for very small ones.

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T x 80 + 15 x 10 = W x 50

80T + 150 = 50W - - -- - - - - -(1)

T + 15 = W  - - - - - - - - - - - (2)

80T + 150 = 50(T + 15)

80T + 150 = 50T + 750

30T = 750 - 150

30T = 600

T = 20N

The closest option is 19.2N

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This question tests Charles' law: for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute (kelvin) temperature, so \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\).

The temperatures must be converted to kelvin before they are substituted, because the proportionality only holds on a scale whose zero is absolute zero:

• \(T_1 = 0 + 273 = 273\ \text{K}\)
• \(T_2 = 273 + 273 = 546\ \text{K}\)

Substituting,

The absolute temperature doubles, so the volume doubles to \(400\ \text{m}^3\).

Two mistakes are common. The first is using the Celsius values directly, which produces the meaningless ratio \(273/0\) and tempts a student into a wrong figure. The second is assuming that because the mass is fixed the volume cannot change; a fixed mass only means no gas enters or leaves, and the gas is still free to expand. A volume of \(200\ \text{m}^3\) would require the temperature to be unchanged, and \(100\ \text{m}^3\) would require the absolute temperature to be halved, neither of which happens here. In every gas-law calculation, convert to kelvin as the very first step.

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\(_ag_g\) = \(\frac{\text{sini}}{\text{sinr}}\)

sinr = \(\frac{\text{sini}}{_ag_g}\)

sinr = \(\frac{0.5}{1.5}\)

sinr = 0.3

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Melting changes the arrangement of the molecules but not how many there are, so the mass is conserved while the volume changes because the density changes. The route through the problem is therefore: use \(V = \dfrac{m}{\rho}\) for the ice, use it again for the water formed, then subtract.

The change in volume is \[\Delta V = 100 - 90 = 10\,\text{cm}^3,\] and it is a decrease, because water is denser than ice. This is the well-known anomaly of water: the open hydrogen-bonded lattice of ice collapses on melting, so a given mass of ice shrinks when it turns to liquid. It is also why ice floats and why a full bottle of water bursts when it freezes.

Two traps are worth naming. Answering \(90\,\text{cm}^3\) means the volume of the water was quoted instead of the change in volume. Answering \(9\,\text{cm}^3\) comes from taking \(10\%\) of \(90\), which wrongly assumes the water volume is the starting figure; the \(10\%\) difference in density applies to the ice volume of \(100\,\text{cm}^3\). In any density question, work out each volume separately from \(m/\rho\) and only then subtract, and state clearly whether the change is an increase or a decrease.

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The temperature gradient of a lagged rod is the rate at which temperature falls along its length, defined as

Lagging matters because it stops heat escaping through the sides, so in the steady state the same heat flows through every cross-section and the temperature falls uniformly from one end to the other. That uniform fall is what makes a single gradient value meaningful.

Rearranging for the temperature difference across the whole rod:

A difference of \(83.2\ \text{K}\) is numerically the same as a difference of \(83.2\ ^\circ\text{C}\), because the kelvin and the Celsius degree are the same size; only the zeros of the two scales differ. Taking the given end as the cooler end, the other end is

so the temperature at the other end is about \(316\ ^\circ\text{C}\). The listed value of \(316.28\ ^\circ\text{C}\) is this result, the tiny difference in the final digit arising from rounding in the printed data.

Two points are worth noting. First, arithmetically the far end could also have been the cooler one, giving \(233-83.2 = 149.8\ ^\circ\text{C}\); that value is not among the choices, which fixes the given end as the cold end. Second, do not convert \(233\ ^\circ\text{C}\) to kelvin and then add the gradient result and forget to convert back, and do not multiply by the gradient without the length: \(83.2\) is a temperature difference, never a temperature. Always separate the difference calculation from the final scale reading.

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A floating body sinks until the upthrust equals its weight. By Archimedes' principle the upthrust equals the weight of liquid displaced, so for a block of volume \(V\) with a fraction \(f\) of that volume submerged in a liquid of density \(\rho_L\):

The fraction immersed is simply the ratio of the density of the body to the density of the liquid.

Relative density is a density compared with that of water, so a relative density of \(0.4\) means

Therefore

A quarter of the block's volume is below the liquid surface, and the other three quarters stay above it.

The mistake this question is designed to catch is using the relative density \(0.4\) directly as though it were the density in \(\text{kg m}^{-3}\), or dividing \(0.4\) by \(1600\), both of which give far too small a fraction. Relative density has no unit, so it must be multiplied by \(1000\ \text{kg m}^{-3}\) before it is compared with a liquid density given in \(\text{kg m}^{-3}\). Also remember the sanity check: since the block floats, the fraction immersed must lie between \(0\) and \(1\), and a denser liquid means less of the block is submerged.

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Whether a body floats or sinks is decided by comparing its weight with the maximum upthrust available. By Archimedes' principle the upthrust equals the weight of fluid displaced. When a body is fully submerged it displaces its own volume \(V\) of water, so

The body sinks when \(W > U_{\max}\), that is when \(\rho_b V g > \rho_w V g\). The common volume \(V\) and \(g\) cancel, leaving the condition \(\rho_b > \rho_w\). An object sinks in water precisely because its density is greater than the density of water.

This is why a small steel nail sinks while a large wooden log floats: what matters is density, not size or weight on its own. A steel ship floats only because its hull encloses air, which lowers the average density of the whole ship below that of water.

The statement that upthrust equals weight describes a body in equilibrium, which is the condition for floating or for remaining suspended at rest in the fluid, not for sinking; a sinking body has an upthrust smaller than its weight and so has a net downward force. Saying the density is less than that of water gives the condition for floating, the exact opposite. Comparing water pressure with weight is meaningless because pressure and force are different quantities with different units, so they can never be equated. In the examination, reduce every flotation question to a comparison of two densities, and check the units of any quantities you are asked to compare.

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Two plane mirrors inclined at an angle \(\theta\) produce multiple images by repeated reflection. For a single object the number of images is

The reason for subtracting one is that the \(360^{\circ}/\theta\) positions found by successive reflection include a pair that coincide on the line bisecting the angle behind the mirrors, so one of the counted images is not separate from another.

Here \(\theta = 30^{\circ}\), so

Each of the tinsels acts as an independent object, and each is imaged by the same mirror system, so the total number of images is

The frequent mistake is to use \(360^{\circ}/\theta\) itself, which gives \(12\) per object and \(120\) in total, or to forget to multiply by the number of objects and answer \(11\). Note also the rule for the other case: if \(360^{\circ}/\theta\) turns out to be an odd whole number, the object on the bisector still gives \(360^{\circ}/\theta - 1\) images, but an object placed off the bisector gives \(360^{\circ}/\theta\). In the examination, always evaluate \(360^{\circ}/\theta\) first, check whether it is even, apply the subtraction once, and only then multiply by the number of objects.

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This question is about streamlining. When a body moves through a fluid such as air or water, the fluid must be pushed aside and made to flow round the body. A blunt front forces the fluid to change direction abruptly, the flow behind it breaks up into swirling eddies, and the pressure in front becomes much higher than the pressure behind. That pressure difference, together with the rubbing of the fluid layers along the surface, makes up the resistive force called drag or fluid friction.

A pointed, tapered front lets the fluid part smoothly and rejoin gradually behind the body, so the flow stays streamlined instead of turbulent and the pressure difference between front and back is much smaller. The result is a reduction in the fluid friction acting on the boat or aircraft, which means less driving force is needed for a given speed, less fuel is used, and a higher top speed becomes possible for the same engine power. This is why fast-moving objects in nature and in engineering, from fish and birds to aircraft and racing hulls, all share the same tapered shape.

The suggestion that the shape increases fluid friction reverses the physics: increasing drag would waste energy, and shapes deliberately made blunt, such as a parachute canopy, are used precisely when large drag is wanted. Stopping depends on reverse thrust, brakes or drag devices, not on the shape of the nose, and appearance is not a physical explanation. In the examination, treat any question about the shape of a moving vehicle as a question about minimising drag, and be ready to name the mechanism as smooth, streamlined flow replacing turbulent flow.

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A hydraulic press works by Pascal's principle: pressure applied to an enclosed incompressible liquid is transmitted equally throughout, so the pressure under the small piston equals the pressure under the large piston. \[\frac{F_1}{A_1} = \frac{F_2}{A_2}.\] Since each piston is circular, \(A = \pi r^2\), and the \(\pi\) cancels: \[\frac{F_1}{r_1^{2}} = \frac{F_2}{r_2^{2}} \quad\Rightarrow\quad \frac{r_2^{2}}{r_1^{2}} = \frac{F_2}{F_1}.\]

Substituting the given forces, \[\frac{r_2^{2}}{r_1^{2}} = \frac{3200}{200} = 16 \quad\Rightarrow\quad \frac{r_2}{r_1} = \sqrt{16} = 4.\] So \(r_1 : r_2 = 1 : 4\).

The decisive step is the square root. Forces in a hydraulic press scale with area, and area scales with the square of the radius, so a force multiplication of \(16\) needs a radius ratio of only \(4\), not \(16\). Reading off \(1:16\) is the classic error, made by matching the force ratio straight to the radii; \(1:2\) comes from taking the square root twice.

Remember also that the press multiplies force but not energy: the small piston must travel \(16\) times as far as the large one, since the same volume of liquid is displaced, \(A_1 d_1 = A_2 d_2\). In an examination, decide first whether the ratio you are asked for is one of areas, radii or diameters, and insert or remove the square accordingly.

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Myopia (short-sightedness) is a defect of vision in which distant objects cannot be seen clearly because the eye focuses light in front of the retina. The far point (the farthest distance at which objects are seen clearly) is closer than infinity - in this case, 95 cm.

To correct myopia, a diverging (concave) lens is placed before the eye. The lens diverges incoming parallel rays from distant objects so that they appear to come from the person's far point, which the eye can then focus on the retina.

The required focal length of the correcting lens equals the far point distance. Since the lens must produce a virtual image at 95 cm for an object at infinity:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-95} - \frac{1}{\infty} = -\frac{1}{95} \]

So \( f = -95 \text{ cm} \) (negative sign confirms a diverging lens).

The correction is a diverging lens of focal length 95 cm. A converging lens would worsen myopia, and a mirror is not used to correct refractive eye defects.

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The steam gives out energy in two distinct stages, and both must be included:

1. Condensation at \(100\,^\circ\text{C}\). Changing state releases latent heat at constant temperature: \(Q_1 = mL\). The supplied value \(2.26\times10^{6}\,\text{J kg}^{-1}\) is the specific latent heat associated with the liquid-vapour change for water, which is the change happening here; note that the subscript printed with it conventionally denotes fusion, whose value for water is \(3.34\times10^{5}\,\text{J kg}^{-1}\), so use the number given and read it as the latent heat of vaporisation, since no melting or freezing occurs in this process.
2. Cooling of the condensed water from \(100\,^\circ\text{C}\) to \(15\,^\circ\text{C}\): \(Q_2 = mc\,\Delta\theta\) with \(\Delta\theta = 85\,\text{K}\).

The total energy released is therefore \[Q = m\left(L + c\,\Delta\theta\right).\] Evaluating the bracket first: \[L + c\Delta\theta = 2.26\times10^{6} + 4200 \times 85 = 2.26\times10^{6} + 3.57\times10^{5} = 2.617\times10^{6}\,\text{J kg}^{-1}.\] With \(Q = 5400\,\text{kJ} = 5.4\times10^{6}\,\text{J}\), \[m = \frac{5.4\times10^{6}}{2.617\times10^{6}} = 2.06\,\text{kg}.\] About \(2.06\,\text{kg}\) of water is collected.

Two errors account for the other figures. Using the latent heat alone gives \(5.4\times10^{6}/2.26\times10^{6} = 2.39\,\text{kg}\), because it ignores the cooling from \(100\,^\circ\text{C}\) to \(15\,^\circ\text{C}\); using the cooling term alone gives \(5.4\times10^{6}/(4200\times85) = 15.1\,\text{kg}\), because it ignores the far larger latent heat. Notice the scale of the two contributions: condensing \(1\,\text{kg}\) of steam releases roughly six times as much energy as cooling that same kilogram of boiling water down to room temperature, which is why steam scalds so severely. Always convert kilojoules to joules before dividing, and check that a latent-heat stage has no temperature change attached to it.

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A neutral point is a point in a magnetic field where the resultant magnetic flux density is zero, because two fields of equal magnitude act there in exactly opposite directions and cancel. A compass needle placed at a neutral point feels no turning force, and when iron filings are sprinkled around a magnet they leave such a spot bare, since no field line passes through it.

In the standard experiment a bar magnet is laid on paper in the Earth's field. The field of the magnet and the horizontal component of the Earth's field are equal and opposite at certain positions: with the north pole of the magnet pointing north, the neutral points lie on either side of the magnet near its middle, and with the north pole pointing south they lie on the axis beyond each pole. Because the cancellation is between two specific fields, the position of a neutral point depends on the strength of the magnet, which is why it can be used to compare magnetic field strengths.

The other terms belong to different topics. Near point and far point are optical terms describing the closest and furthest distances at which the eye can focus clearly, and they have nothing to do with magnetism. There is no defined quantity called an infinite point; in any case the field of a magnet only approaches zero at very large distances rather than vanishing at an identifiable point, and that gradual weakening is not what the term neutral point means. The key idea to carry into the examination is that a neutral point is created by cancellation of two fields, not by simple distance from the magnet.

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Current carrying straight conductor (Concentric circles typical of straight wire magnetic field.

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Given: Inductance: \( L = 0.5 \, \text{H} \), Resistance: \( R = 45 \, \Omega \), Frequency: \( f = 50 \, \text{Hz} \)

inductive reactance \( X_L \): \(X_L = 2 \pi f L = 2 \times \frac{22}{7} \times 50 \times 0.5 = \frac{1100}{7} \approx 157.14 \, \Omega\)
The impedance \( Z \):

\(Z = \sqrt{R^2 + X_L^2} = \sqrt{(45)^2 + (157.14)^2} = \sqrt{2025 + 24642.82} \approx \sqrt{26667.82} \approx 163.5 \, \Omega\).

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An annular eclipse is a particular kind of solar eclipse. Like every solar eclipse it happens only when the sun, the moon and the earth lie on the same straight line with the moon in the middle, so that the moon's shadow falls on the earth. What makes it annular rather than total is the moon's distance: because the moon's orbit is elliptical, its angular size varies. When it is near its farthest point it appears slightly smaller than the sun, so the dark umbra does not quite reach the earth's surface and a bright ring (Latin annulus) of the sun's disc remains visible round the black moon.

Among the statements offered, the one that correctly describes the condition for this event is that the sun, moon and earth come into a straight line. That alignment, called syzygy, is the necessary geometry for both the total and the annular solar eclipse; the difference between them is only the apparent size of the moon at the time.

The statement that the earth comes between the moon and the sun describes a lunar eclipse, in which the earth's shadow falls on the moon; that is the commonest confusion in this topic, so fix the order clearly: in a solar eclipse the moon blocks the sun from the earth, in a lunar eclipse the earth blocks the sun from the moon. A gathering of stars is a cluster or constellation and has nothing to do with eclipses, and simple invisibility of one body is not a definition of an eclipse, since the moon is invisible at new moon in every month without any eclipse occurring. In the examination, first identify which body is being shadowed, then decide whether the shadow is total, partial or annular.

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A charge moving through a magnetic field feels a force that depends on how much charge is moving, how fast it moves, how strong the field is, and crucially the angle between the velocity and the field: \[F = qvB\sin\theta.\] The \(\sin\theta\) factor is the part most often dropped. It is largest when the charge cuts straight across the field lines (\(\theta = 90^\circ\)) and zero when the charge moves along the field lines.

Make the flux density the subject and substitute, converting the charge from microcoulombs to coulombs first (\(5\ \mu\text{C} = 5 \times 10^{-6}\ \text{C}\)): \[B = \frac{F}{qv\sin\theta} = \frac{6}{(5 \times 10^{-6})(3 \times 10^{4})\sin 45^\circ}.\] The product \(qv = (5 \times 10^{-6})(3 \times 10^{4}) = 0.15\), and \(\sin 45^\circ = 0.7071\), so the denominator is \(0.15 \times 0.7071 = 0.1061\). Hence \[B = \frac{6}{0.1061} = 56.57\ \text{T},\] so the flux density is about 56.6 T.

The most likely wrong route is to ignore the angle altogether and use \(B = F/qv = 6/0.15 = 40\) T, which is too small; forgetting \(\sin\theta\) always understates \(B\) because \(\sin\theta < 1\) for any angle other than a right angle. A second common slip is leaving the charge in microcoulombs, which shifts the answer by a factor of a million. Exam reminder: in \(F = qvB\sin\theta\), \(\theta\) is measured between the velocity and the field direction, not between the velocity and the force.

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Doping means adding a controlled trace of impurity to a pure semiconductor such as silicon or germanium to increase the number of mobile charge carriers. Silicon has four valence electrons and forms four covalent bonds.

• Adding a pentavalent impurity (phosphorus, arsenic, antimony) supplies five valence electrons per atom. Four complete the bonds and the fifth is left loosely bound, becoming a free electron. The material is n-type and electrons are its majority carriers.
• Adding a trivalent impurity (boron, aluminium, gallium) supplies only three valence electrons, so one bond is left short of an electron. That vacancy is a hole, and it moves through the lattice as neighbouring electrons hop into it, behaving like a mobile positive charge of magnitude \(+e\). The material is p-type and holes are its majority carriers.

Both kinds of carrier are present in any doped sample, one as the majority and the other as the minority produced by thermal generation, so the charge carriers in doped semiconductors are electrons and holes.

The distractors rest on real misconceptions. Protons and neutrons are locked in the nuclei of the fixed lattice atoms and cannot migrate, so they never carry current in a solid. Anions and cations do carry charge, but that is electrolytic conduction in a solution or molten salt, where whole ions drift; a semiconductor crystal keeps its atoms in place and moves only electrons and the holes they leave behind. Remember for the examination that conventional current in a p-type region is described as a flow of holes in the direction of the field, while the electrons that actually move travel the opposite way.

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A current-carrying inductor stores energy in the magnetic field of its coil. The energy stored is

where \(L\) is the inductance in henries and \(I\) the steady current. This is the magnetic counterpart of the energy \(\tfrac{1}{2}CV^2\) stored in a capacitor's electric field, and like it the energy depends on the square of the current.

Rearrange for the current before substituting:

So a steady current of about \(1.29\ \text{A}\) stores \(2.5\ \text{J}\) in a \(3\ \text{H}\) coil.

The trap is forgetting the square root and dividing instead, for example \(2E/L = 1.67\) or \(E/L\) style combinations, or forgetting the factor \(\tfrac{1}{2}\), which would give \(\sqrt{2.5/3}=0.91\ \text{A}\). Because the relationship is quadratic, doubling the current stores four times the energy, and that squared dependence is exactly what the examiner is checking. Write the formula down, make the unknown the subject, then substitute.

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This is a vector-addition problem, so the two journeys must be resolved into perpendicular components before they are combined. The bearing notation \(N30^\circ E\) means the direction is measured \(30^\circ\) away from north, turning towards the east. For a displacement of \(10.0\,\text{m}\) in that direction, north is the adjacent side and east the opposite side of the \(30^\circ\) angle:

• eastward part \(= 10.0\sin 30^\circ = 5.0\,\text{m}\)
• northward part \(= 10.0\cos 30^\circ = 8.66\,\text{m}\)

The first leg is entirely eastward, so the totals are \[x = 6.0 + 5.0 = 11.0\,\text{m (east)},\qquad y = 0 + 8.66 = 8.66\,\text{m (north)}.\] These two totals are at right angles, so Pythagoras gives the straight-line distance from the start: \[r = \sqrt{11.0^2 + 8.66^2} = \sqrt{121 + 75.0} = \sqrt{196} = 14.0\,\text{m}.\] The man is \(14.0\,\text{m}\) from his starting point.

The usual error is to add the magnitudes, \(6.0 + 10.0 = 16.0\,\text{m}\), or to interchange the sine and cosine because the angle was assumed to be measured from the east line. In bearings written as \(N\theta E\) the angle is measured from north, so north takes the cosine. Sketching the two arrows head-to-tail, as above, shows at once which component belongs to which trigonometric ratio.

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Demagnetization is the process of removing or reducing the magnetism of a magnet. The standard methods include:

• Heating the magnet to a high temperature (above its Curie point) and allowing it to cool while oriented in the east-west direction, so it is perpendicular to the Earth's magnetic field.
• Hammering or dropping the magnet repeatedly while it is oriented in the east-west direction.
• Passing a decreasing alternating current through a solenoid containing the magnet.

The key requirement is that the magnet must be oriented in the east-west direction during demagnetization. This ensures the Earth's magnetic field does not re-magnetize the bar as its domains are disrupted.

Heating a magnetic bar red hot and allowing it to cool in the east-west direction is a valid demagnetization method. Heating disrupts the alignment of magnetic domains, and cooling in the E-W orientation prevents re-alignment along the Earth's field.

Placing the bar in a solenoid alone does not demagnetize it - it would magnetize it. Stroking or hammering in the north-south direction would tend to magnetize the bar rather than demagnetize it, because the N-S orientation aligns with the Earth's magnetic field.

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Refraction happens at a boundary between media of different refractive index, and the larger the difference in index and the more curved the surface, the greater the bending. Light entering the eye meets its largest index change at the front surface of the cornea, where it passes from air (\(n \approx 1.00\)) into corneal tissue (\(n \approx 1.38\)) across a strongly curved surface. That single boundary provides roughly two thirds of the eye's total converging power. The crystalline lens (\(n \approx 1.41\)) supplies the remaining power, and it is the only part whose power can be varied: the ciliary muscles change its curvature so that objects at different distances are focused on the retina, a process called accommodation. The features that chiefly refract the light are therefore the cornea and the lens.

The aqueous humour behind the cornea and the vitreous humour in front of the retina are watery fluids of index about \(1.34\). Their indices are so close to those of the cornea and the lens that the boundaries with them cause very little further bending; their jobs are to keep the eyeball firm, maintain its shape and nourish the tissues, not to focus light. That is why pairings built around a humour are weaker answers.

A useful examination check: whenever a question asks which structure refracts, look for the surface with the biggest refractive-index step. In the eye that step is at air-to-cornea, which also explains why vision is blurred under water, since water and cornea have nearly the same index and the cornea then loses most of its power.

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The electric field intensity at a distance \(r\) from a point charge obeys an inverse-square law: \[E = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q}{r^{2}} = \frac{kQ}{r^{2}},\] with \(k = 9.0 \times 10^{9}\ \text{N m}^{2}\text{C}^{-2}\). Since the distance is wanted, make \(r\) the subject: \[r = \sqrt{\frac{kQ}{E}}.\]

Work out the numerator first. \[kQ = (9.0 \times 10^{9})(1.2 \times 10^{-7}) = 1.08 \times 10^{3}\ \text{N m}^{2}\text{C}^{-1}.\] Dividing by the field strength gives \[r^{2} = \frac{1.08 \times 10^{3}}{4.8 \times 10^{-4}} = 2.25 \times 10^{6}\ \text{m}^{2},\] so \[r = \sqrt{2.25 \times 10^{6}} = 1.5 \times 10^{3}\ \text{m} = 1.5\ \text{km}.\] The field intensity falls to \(4.8 \times 10^{-4}\ \text{N C}^{-1}\) at 1.5 km from the charge.

The commonest error is forgetting the square root and quoting \(2.25 \times 10^{6}\), or taking the root of only part of the expression. Handle the powers of ten deliberately: to take the square root of a number in standard form, first arrange the index to be even, as with \(2.25 \times 10^{6}\), so that halving it gives \(10^{3}\) exactly. Because the relationship is inverse-square, notice also that reducing the field to a quarter of a value doubles the distance, and the final answer had to be converted from metres to kilometres to match the way the alternatives are written.

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Two ideas must be combined: Newton's second law for the horizontal motion, and the definition of the coefficient of friction. On a horizontal floor the applied pull \(F\) is opposed by friction \(F_f\), and the leftover force produces the acceleration:

1. Net force: \(F - F_f = ma\), so \[F_f = F - ma = 18 - (40)(0.3) = 18 - 12 = 6\,\text{N}.\]
2. Normal reaction on a horizontal surface: \[N = mg = 40 \times 10 = 400\,\text{N}.\]
3. Coefficient of friction: \[\mu = \frac{F_f}{N} = \frac{6}{400} = 0.015.\]

So the coefficient of friction is \(0.015\), a dimensionless number, since it is a ratio of two forces.

The step most often skipped is subtracting \(ma\) from the applied force. Using the full \(18\,\text{N}\) as the frictional force gives \(18/400 = 0.045\), which ignores the fact that the mass is accelerating; friction can only equal the applied force when the body moves at constant velocity. The other frequent slip is dividing by the mass instead of the weight, which yields \(6/40 = 0.15\) and confuses a mass in kilograms with a normal reaction in newtons.

In an examination, set out the horizontal equation and the vertical equation separately: \(F - \mu mg = ma\) horizontally and \(N = mg\) vertically. Rearranged in one line this reads \[\mu = \frac{F - ma}{mg},\] and substituting the given values reproduces \(0.015\) directly. Always check that \(\mu\) comes out with no unit.

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The SI system is built on seven base (fundamental) units which are defined independently of one another: the metre, kilogram, second, ampere, kelvin, mole and candela. Every other unit is a derived unit, meaning it can be written as a combination of these base units. So the task here is simply to test each unit for whether it can be broken down further.

The ampere is the base unit of electric current, so it cannot be expressed in terms of anything more fundamental. The other three all reduce to combinations of base units:

• joule: work \(=\) force \(\times\) distance, so \(1\,\mathrm{J} = 1\,\mathrm{N\,m} = 1\,\mathrm{kg\,m^{2}\,s^{-2}}\);
• volt: \(V = W/Q\), so \(1\,\mathrm{V} = 1\,\mathrm{J\,C^{-1}} = 1\,\mathrm{kg\,m^{2}\,s^{-3}\,A^{-1}}\);
• ohm: \(R = V/I\), so \(1\,\Omega = 1\,\mathrm{kg\,m^{2}\,s^{-3}\,A^{-2}}\).

A common misconception is that any unit with its own special name, such as the joule or the volt, must be fundamental. The special name is only a convenience; what matters is whether the unit can be written in terms of others. Notice too that the coulomb is not a base unit even though charge feels more basic than current: the SI system defines the ampere first and then treats \(1\,\mathrm{C} = 1\,\mathrm{A\,s}\). Memorise the seven base units and their quantities, then any question of this type becomes a single-step elimination.

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Stimulated emission is the process in which an incoming photon of a specific energy causes an excited atom to release a second photon that is identical in energy, phase, direction, and polarisation. This mechanism is the fundamental operating principle of a laser.

The word "laser" is itself an acronym: Light Amplification by Stimulated Emission of Radiation. The entire device is designed around achieving and sustaining stimulated emission through population inversion and an optical cavity.

The other light sources listed operate on different principles:

• An incandescent bulb produces light by heating a filament until it glows - this is thermal radiation (black-body emission).
• A fluorescent lamp works by passing an electric current through mercury vapour, which emits ultraviolet radiation. A phosphor coating on the tube absorbs the UV and re-emits it as visible light through spontaneous emission (fluorescence).
• An LED (light-emitting diode) produces light through electroluminescence, where electrons recombine with holes across a semiconductor junction and release photons through spontaneous emission.

Only the laser relies on stimulated emission as its primary mechanism of light production.

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For a parallel-plate capacitor with air (or vacuum) between the plates, the capacitance depends only on the geometry: \[C = \frac{\varepsilon_0 A}{d},\] where \(A\) is the area of overlap of one plate and \(d\) the separation. Wider plates store more charge for the same voltage, and closer plates do too, which is why \(A\) is on top and \(d\) underneath. Because \(\varepsilon_0\) is quoted in SI units, both the area and the separation must be converted to metres before substituting.

The plates are circular, so the area is \[A = \pi r^{2} = 3.142 \times (0.06)^{2} = 3.142 \times 3.6 \times 10^{-3} = 1.131 \times 10^{-2}\ \text{m}^{2},\] using \(r = 6\ \text{cm} = 0.06\ \text{m}\). The separation is \(d = 0.12\ \text{cm} = 1.2 \times 10^{-3}\ \text{m}\). Substituting: \[C = \frac{(8.85 \times 10^{-12})(1.131 \times 10^{-2})}{1.2 \times 10^{-3}} = (8.85 \times 10^{-12}) \times 9.426 = 8.34 \times 10^{-11}\ \text{F}.\] So the capacitance is about \(8.3 \times 10^{-11}\ \text{F}\), which is 83 pF.

Two traps sit in this question. The first is using the diameter as the radius or forgetting to square the radius, which changes the area by a factor of four. The second is leaving centimetres in place: since \(1\ \text{cm}^{2} = 10^{-4}\ \text{m}^{2}\) and \(1\ \text{cm} = 10^{-2}\ \text{m}\), a mixed substitution shifts the power of ten. Note as well that any physically real capacitance of a small air capacitor must come out as a tiny fraction of a farad, so a positive index such as \(10^{11}\ \text{F}\) can be rejected on sight.

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This question uses de Broglie's idea that a moving particle has a wavelength linked to its momentum: \[\lambda = \frac{h}{p} = \frac{h}{mv},\] so that \[m = \frac{h}{\lambda v}.\] Everything therefore depends on getting the wavelength into metres, because \(h\) is in \(\text{J s}\) and the speed in \(\text{m s}^{-1}\).

The wavelength is given in millimetres, so convert first: \[\lambda = 4.0 \times 10^{-7}\ \text{mm} = 4.0 \times 10^{-7} \times 10^{-3}\ \text{m} = 4.0 \times 10^{-10}\ \text{m}.\] Now substitute: \[m = \frac{6.63 \times 10^{-34}}{(4.0 \times 10^{-10})(2.7 \times 10^{8})} = \frac{6.63 \times 10^{-34}}{1.08 \times 10^{-1}} = 6.1 \times 10^{-33}\ \text{kg}.\] The significant figures come out as 6.1, so the intended choice is the value quoted with those figures; its power of ten appears to be misprinted, since the correct working gives \(6.1 \times 10^{-33}\ \text{kg}\) rather than \(10^{-31}\). Quote \(6.1 \times 10^{-33}\ \text{kg}\) as your worked answer and select the value beginning 6.1.

The step that costs most marks is the millimetre-to-metre conversion. Skipping it, and using \(4.0 \times 10^{-7}\ \text{m}\), gives \(6.1 \times 10^{-36}\ \text{kg}\), a thousand times too small. A second slip is inverting the relation and multiplying by \(\lambda v\) instead of dividing. As a check on the physics, remember the inverse proportionality: a shorter wavelength means a larger momentum, so a heavier or faster particle always has the smaller de Broglie wavelength, which is why wave behaviour is only observed for very light particles such as electrons.

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The magnetic effect of an electric current is the fact that a current-carrying conductor produces a magnetic field around itself. When the conductor is wound into a coil around a soft-iron core, the arrangement becomes an electromagnet, which attracts iron only while current flows. Any device that works by this principle must contain a coil whose magnetism does mechanical work.

The electric bell is exactly that device. Current from the supply passes through the coils of an electromagnet, which attracts a soft-iron armature carrying a hammer, and the hammer strikes the gong. The movement of the armature breaks the circuit at a contact screw, the electromagnet loses its magnetism, a spring pulls the armature back and remakes the contact, and the cycle repeats rapidly to give continuous ringing. Remove the magnetic effect of the current and nothing at all happens, so the bell is the device that operates on it.

The alternatives depend on different effects. A rheostat is simply a variable resistor used to control current by changing resistance, and it uses the heating effect at most, not magnetism. A thermostat relies on the differential thermal expansion of a bimetallic strip, which bends with temperature and opens or closes a circuit. A carbon microphone works because the resistance of loosely packed carbon granules changes as sound waves compress them, so it converts sound into a varying current through a resistance change, not through magnetism. When a question asks which device uses a named effect, look for the component that produces it: a soft-iron core with a winding signals the magnetic effect, a bimetallic strip signals expansion, and a resistance wire signals the heating effect.

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Heat capacity (thermal capacity) is defined as the heat energy needed to raise the temperature of a whole body by one kelvin:

Its unit, \(\text{J K}^{-1}\), is itself a reminder of the definition: joules per kelvin of temperature change.

Convert the energy to joules before dividing, since the answer is wanted in \(\text{J K}^{-1}\):

A temperature difference of \(53\ ^\circ\text{C}\) is the same size as a difference of \(53\ \text{K}\), because one Celsius degree and one kelvin represent the same interval; only the zero points of the two scales differ. There is therefore no need to add \(273\) to a temperature change. Substituting,

Two errors are worth guarding against. Adding \(273\) to the \(53\) gives \(326\ \text{K}\) and a much smaller capacity, and leaving the energy as \(48\ \text{kJ}\) gives \(0.906\), which is in \(\text{kJ K}^{-1}\) rather than the requested unit. Note also that this quantity applies to this particular body only; to obtain the specific heat capacity of the material you would additionally divide by the mass, using \(c = C/m\).

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In a simple pendulum, kinetic energy is maximum at the lowest point (equilibrium position) where potential energy is minimum and speed is maximum.Here, Q is the control point (mean/equilibrium position), so maximum kinetic energy occurs at Q. At extremes P and S, kinetic energy is zero (velocity = 0). At R, it is between the extreme and the equilibrium.

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An alternating current has no single fixed value: it grows to a maximum in one direction, falls to zero, grows to a maximum in the opposite direction, and repeats. To describe such a current with one useful number we quote its root-mean-square (r.m.s.) value, which is the steady direct current that would produce the same average heating effect in the same resistor. For a sinusoidal current the r.m.s. value is tied to the peak (maximum) value \(I_0\) by \[I_{r.m.s} = \frac{I_0}{\sqrt{2}} = 0.707\,I_0.\]

The single current value quoted in the question, 7 A, has to be read as the greatest value the current reaches, because an r.m.s. value can only be obtained from the peak. Substituting: \[I_{r.m.s} = \frac{7}{\sqrt{2}} = \frac{7}{1.414} = 4.95\ \text{A} \approx 5\ \text{A}.\] So the r.m.s. current is about 5 A.

Two slips account for most wrong answers here. Dividing by 2 instead of \(\sqrt{2}\) gives 3.5 A, and multiplying by \(\sqrt{2}\) gives 9.9 A, which is the route from r.m.s. back to peak rather than peak to r.m.s. A quick safety check in the exam: for a sinusoidal current the r.m.s. value is always about 70% of the peak, so it must come out smaller than the peak, never equal to it or larger.

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Parallax is the apparent shift in the relative positions of two things at different distances from the eye when the eye is moved sideways. The nearer of the two appears to move more, so the two seem to separate. This is the basis of the no-parallax method used in optics to locate an image: a search pin is moved until it and the image appear to stay locked together as the head moves from side to side, and at that setting the pin is exactly where the image is.

If the object and its image move together in the same direction, at the same apparent rate, then there is no relative displacement between them as the eye moves. They lie at the same distance from the observer, which is precisely the condition described as no parallax error, and it is the signal that the image position has been found correctly.

The tempting choice is parallax error, on the grounds that something appears to be moving. What matters is not that the pair appears to move as the eye moves, but whether they move relative to each other. Movement in the same direction together means zero relative shift. A related use of the same idea in measurement is reading a scale: to avoid parallax error on a metre rule or an ammeter, look along a line perpendicular to the scale so that the pointer and its position on the scale coincide. In the examination, remember that parallax is judged by relative displacement, never by absolute apparent motion.

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The eye must form a sharp image on the retina whether the object is close or far away. Since the distance from lens to retina is fixed, the only way to keep the image in focus is to change the focal length of the lens itself. This adjustment is called accommodation, and it is carried out by the ciliary muscles, the ring of muscle attached to the lens through the suspensory ligaments.

The mechanism works as follows. When the ciliary muscles contract, the ring they form becomes smaller, the tension in the suspensory ligaments falls, and the elastic lens is allowed to bulge. A fatter lens is more strongly converging, so its focal length shortens and its power \(P = 1/f\) rises, which is what is needed for a near object. When the muscles relax, the ligaments pull the lens flatter, the focal length lengthens, and distant objects come into focus. In the thin-lens relation \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), the image distance \(v\) is fixed by the eyeball, so a change in \(u\) must be answered by a change in \(f\).

The other structures play different roles. The vitreous humour is the transparent jelly filling the eyeball behind the lens; it helps maintain the shape of the eye and refracts light slightly, but its shape is not adjustable. The aqueous and vitreous fluids have fixed refractive indices, so they cannot vary the focal length. The retina and its nerves detect the image and transmit signals to the brain; they take no part in focusing. When a question mentions a variable focal length in the eye, the required answer is always the ciliary muscle changing the curvature of the lens.