JAMB UTME - Physics - 2025

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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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Every simple cell has three parts to identify separately: two electrodes, the electrolyte that conducts by ion movement between them, and often a depolariser that removes hydrogen gas from the positive electrode. The question asks only for the electrolyte, so the answer must be a substance that ionises in solution and carries charge inside the cell.

In the wet Leclanche cell the positive electrode is a carbon rod, the negative electrode is a zinc rod, and the electrolyte is a strong solution of ammonium chloride, \(\mathrm{NH_4Cl}\). It dissociates to give \(\mathrm{NH_4^+}\) and \(\mathrm{Cl^-}\) ions, which carry the current through the liquid while zinc dissolves at the negative electrode and hydrogen is released at the carbon rod. Manganese(IV) oxide, \(\mathrm{MnO_2}\), is packed round the carbon rod as the depolariser, oxidising the hydrogen to water and slowing down polarisation. The cell gives an e.m.f. of about 1.5 V but has a large internal resistance, so it suits work needing brief currents such as ringing a bell.

The other substances belong to different cells or to different parts of a cell. Carbon is the positive electrode of this same cell, which is why it is a tempting choice: an electrode is a conductor, not the ion-carrying solution. Lead(IV) oxide is the positive plate of the lead-acid accumulator, whose electrolyte is dilute sulphuric acid, and nickel hydroxide belongs to the alkaline nickel-cadmium or nickel-iron cell, whose electrolyte is potassium hydroxide. When revising cells, learn each one as a set of four labels: negative electrode, positive electrode, electrolyte, depolariser.

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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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A radiation detector is any device that responds to the ionisation, excitation or chemical change produced when nuclear radiation passes through matter. To answer an "odd one out" question like this, check each device against that definition rather than against how familiar the name sounds.

An electrophorus is an electrostatic instrument. It is a flat insulating disc (or slab) with a metal plate and an insulating handle, used to produce charge repeatedly by friction and then by induction: the slab is charged by rubbing, the metal plate is placed on it and earthed briefly, and the plate carries away a charge of opposite sign. It measures nothing and detects nothing about radioactivity, so it is the device that does not belong in this list.

The other three are genuine detectors. A Geiger-Muller counter uses a gas-filled tube at high voltage in which an entering particle ionises the gas and triggers a pulse of current that is counted electronically. A scintillation counter or chamber uses a phosphor that emits a tiny flash of light when radiation strikes it; a photomultiplier converts each flash into an electrical pulse. A film badge contains photographic film that darkens in proportion to the dose received, so it records the total exposure of a worker over time. In the examination, sort nuclear-physics apparatus by the effect it exploits: ionisation of a gas, light emission, or blackening of photographic emulsion. Anything based only on charging by friction or induction belongs to electrostatics.

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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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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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Waves divide into two families. Mechanical waves, such as water waves, sound and waves on a string, are oscillations of the particles of a material medium, so they cannot exist without that medium. Electromagnetic waves, such as light, radio waves and X-rays, are oscillations of electric and magnetic fields, which need no particles at all and therefore travel through a vacuum at \(3.0\times10^{8}\,\mathrm{m\,s^{-1}}\). This is the general difference between water waves and light waves: the medium of propagation each requires.

The evidence for it is everyday. Sunlight reaches the earth across the emptiness of space, whereas a water wave dies out the moment the water ends at a shoreline, and a ripple tank produces no waves when the water is drained. This is also why light from distant stars reaches us but their sound never does.

The other suggested differences do not hold. Both kinds of wave can be reflected, water waves from a barrier in a ripple tank and light from a mirror, and both can be diffracted, water waves spreading through a narrow gap between barriers and light spreading at the edge of an obstacle or through a fine slit. The direction of vibration is not a general point of difference either, because water surface waves and light waves are both transverse: the displacement is perpendicular to the direction of travel in each case. In the examination, when asked to distinguish two waves, first classify each as mechanical or electromagnetic, since that single classification decides the need for a medium, the possible speeds, and whether the wave can be polarised.

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Newton's law of universal gravitation states that the force between two masses obeys an inverse-square law: \[F = \frac{Gm_1m_2}{r^{2}}.\] The masses and \(G\) are unchanged, so only the separation matters, and \(F \propto \dfrac{1}{r^{2}}\). Doubling \(r\) multiplies \(r^{2}\) by \(4\), so the force falls to a quarter of its former value.

Working with a ratio avoids needing any of the constants: \[\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^{2} = \left(\frac{r}{2r}\right)^{2} = \frac{1}{4},\] so \[F_2 = \frac{20}{4} = 5\,\text{N}.\]

The frequent error is halving the force to \(10\,\text{N}\), which treats the relationship as \(F \propto 1/r\) and forgets the square. Test any inverse-square question with the same ratio method: at three times the separation the force becomes \(1/9\) of the original, and at half the separation it becomes four times as large. The identical reasoning applies to the electrostatic force between point charges and to the intensity of light or sound from a point source, so the technique is worth making automatic.

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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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Radioactive decay is random, so the number of undecayed nuclei falls exponentially: \(N = N_0 e^{-\lambda t}\), where \(\lambda\) is the decay constant, the probability per second that a given nucleus decays. The half-life \(t_{1/2}\) is the time for \(N\) to fall to \(N_0/2\). Putting \(N = N_0/2\) and \(t = t_{1/2}\) into the exponential law gives

Substituting the given half-life,

which to two significant figures is \(5.0\times 10^{-3}\ \text{s}^{-1}\). Note that the decay constant has the unit \(\text{s}^{-1}\), the reciprocal of time, because it is a rate per nucleus rather than a time.

The neighbouring values here are all within a couple of per cent of one another, so they are testing whether the constant \(0.693\) is used rather than a rounded \(0.7\) (which would give \(5.05\times 10^{-3}\)) or an inverted formula such as \(t_{1/2}/\ln 2\). Keep \(\ln 2 = 0.693\) and remember that a short half-life means a large decay constant, since the two are inversely proportional.

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The two voltameters are in the same circuit in series, so the same current flows through both for the same length of time. That means the quantity of charge \(Q = It\) passed through each is identical, and this shared value of \(Q\) is the bridge between the two metals.

By Faraday's first law, \(m = ZQ\), so for each metal \(Q = m/Z\). Equating the charges,

Substituting the masses and the known electrochemical equivalent of copper,

Notice that neither the current nor the time was needed, and neither was given: because the charge is common to both cells, it cancels out of the ratio. Recognising that cancellation is the real skill being tested here.

The likely error is inverting the mass ratio, using \(0.2/1.2\), which would give a value smaller than the copper figure. Check the sense of your answer physically: silver has a much larger mass deposited for the same charge, so its electrochemical equivalent, the mass per coulomb, must be the larger of the two. That is consistent with the chemistry, since each silver ion \(\text{Ag}^{+}\) carries only one elementary charge while each \(\text{Cu}^{2+}\) ion carries two.

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The diagram shows a gold-leaf electroscope that is already positively charged, as indicated by the diverged leaves marked with positive (+) signs. When a positively charged rod is brought near the cap, electrostatic induction occurs.

Since the electroscope already carries a net positive charge, the approaching positive rod repels additional positive charges from the cap region down through the stem and onto the leaves. This increases the concentration of positive charge on both leaves, causing the electrostatic repulsion between them to grow stronger.

As a result, the leaves spread further apart and the divergence increases. This is a standard demonstration of charge interaction: like charges repel, and adding more of the same sign of charge to the leaves amplifies their mutual repulsion.

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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 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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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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Specific heat capacity is the heat needed to raise the temperature of one kilogram of a substance by one kelvin. It comes from the heat equation \[Q = mc\Delta\theta,\] where \(Q\) is the heat supplied in joules, \(m\) the mass in kilograms and \(\Delta\theta\) the temperature rise. Because heat loss to the surroundings is stated to be negligible, all 1000 J supplied goes into the rod, so no correction is needed.

Making \(c\) the subject and substituting: \[c = \frac{Q}{m\Delta\theta} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\ \text{J kg}^{-1}\text{K}^{-1}.\] So the specific heat capacity is \(2666.7\ \text{J kg}^{-1}\text{K}^{-1}\).

Note that the temperature rise needs no conversion. A change of \(15\ ^\circ\text{C}\) is a change of 15 K because the two scales have the same size of degree, so adding 273 here is a wasted step that produces a badly wrong answer. The other frequent slip is working out the denominator carelessly: \(0.025 \times 15 = 0.375\), not 0.0375 or 3.75. Exam reminder: distinguish specific heat capacity \(c\), measured in \(\text{J kg}^{-1}\text{K}^{-1}\), from heat capacity \(C = mc\), measured in \(\text{J K}^{-1}\); the units in the options tell you which one is wanted.

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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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Impulse is defined as the change in momentum of a body. Mathematically, impulse \( J = \Delta p = m(v - u) \), where \( m \) is mass, \( v \) is final velocity, and \( u \) is initial velocity.

First, find the final velocity using the equation of motion:

\[ v = u + at = 4 + (5 \times 2) = 4 + 10 = 14 \text{ m/s} \]

Now substitute into the impulse equation:

\[ J = m(v - u) \]

\[ 60 = m(14 - 4) \]

\[ 60 = 10m \]

\[ m = \frac{60}{10} = 6 \text{ kg} \]

The mass of the body is 6 kg.

A common error is to confuse impulse with final momentum (\( mv \)) rather than the change in momentum (\( m \Delta v \)). Impulse equals \( F \times t \) or equivalently \( m(v - u) \); both routes give the same result here since \( F \times t = ma \times t = m \times 5 \times 2 = 10m \).

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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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The thermal capacity (heat capacity) of a body is the quantity of heat needed to raise the temperature of the whole body by one kelvin, measured in \(\text{J K}^{-1}\). It is related to the specific heat capacity \(c\) by

Reading that equation tells you exactly what \(C\) depends on. It depends on the mass \(m\) of the body, and on \(c\), which is fixed by the substance the body is made of, that is by its nature or material. So thermal capacity depends on the mass and the nature of the body, and on nothing else.

The quantity of heat supplied is not a factor, because \(C\) is a ratio, \(C = Q/\Delta\theta\); supplying twice the heat produces twice the temperature rise and leaves \(C\) unchanged. Temperature is not a factor either: \(C\) tells you how much heat is needed per kelvin, whichever kelvin you start from, so a body at \(20\ ^\circ\text{C}\) and the same body at \(80\ ^\circ\text{C}\) have essentially the same thermal capacity. Volume is not an independent factor because, for a given material, volume is only another way of stating mass through the density, \(m = \rho V\); once the mass and the material are named, the volume adds nothing.

A concrete check makes this memorable. Two blocks of the same mass, one aluminium and one lead, need very different amounts of heat for the same rise, which shows the nature matters, and two aluminium blocks of different masses also need different amounts, which shows the mass matters. Distinguish carefully in the examination: specific heat capacity \(c\), in \(\text{J kg}^{-1}\text{K}^{-1}\), depends only on the nature of the substance, while thermal capacity \(C\), in \(\text{J K}^{-1}\), depends on the nature and on how much of it there is.

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The speed of sound in a gas is governed by how stiff the gas is compared with how heavy it is, expressed as \[v = \sqrt{\frac{\gamma P}{\rho}},\] where \(P\) is the pressure, \(\rho\) the density and \(\gamma\) a constant for the gas. It looks as though raising \(P\) should raise \(v\), and that is exactly the trap in this question.

Pressure and density are not independent. For a fixed mass of gas at constant temperature, Boyle's law gives \(PV = \text{constant}\), and since \(\rho = m/V\) the density rises in exact proportion to the pressure. So the ratio \(P/\rho\) stays the same when the pressure is doubled: the gas becomes stiffer, but it also becomes correspondingly heavier per unit volume, and the two effects cancel. The speed of sound is therefore unchanged when pressure increases at constant temperature.

The same equation shows what does change the speed. Writing \(P/\rho = RT/M\) for an ideal gas gives \(v = \sqrt{\gamma RT/M}\), so the speed depends on the absolute temperature and on the molar mass of the gas, and it is proportional to \(\sqrt{T}\). This is why sound travels faster on a hot day and faster in a light gas such as helium, but is not altered by simply pumping the air to a higher pressure at the same temperature. Exam reminder: whenever a question changes the pressure of a gas at constant temperature, check whether the density changes with it before concluding that a quantity depending on \(P/\rho\) has changed.

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The power of a lens measures how strongly it converges or diverges light. A lens that bends rays sharply brings them to a focus close to the lens, so it has a short focal length; a weak lens focuses rays far away. Power is therefore defined as the reciprocal of the focal length, \[P = \frac{1}{f},\] with \(f\) in metres. The unit of \(P\) is the dioptre (\(\text{D}\)), which is simply \(\text{m}^{-1}\). So the power in dioptres is \(\frac{1}{f}\).

Two details make the definition work. First, \(f\) must be expressed in metres before taking the reciprocal: a lens of focal length \(20\,\text{cm} = 0.20\,\text{m}\) has \[P = \frac{1}{0.20} = +5.0\,\text{D}.\] Second, the sign of \(f\) carries through, so a converging (convex) lens has positive power and a diverging (concave) lens has negative power. Powers also add for thin lenses placed in contact, \(P = P_1 + P_2\), which is exactly why opticians quote lenses in dioptres rather than in centimetres.

Expressions such as \(f\), \(2f\) or \(3f\) cannot be correct because they grow as the focal length grows, which would say that a lens focusing light far away is the more powerful one. They also have the wrong unit: metres instead of \(\text{m}^{-1}\). A quick unit check on any formula offered in an optics question will usually eliminate the distractors immediately, and remember to convert centimetres to metres before computing a dioptre value.

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For small oscillations a simple pendulum has period

and frequency is the reciprocal of period, \(f = 1/T\). Two preparation steps decide whether the arithmetic will be right: the length must be converted to metres, and the frequency must be taken at the end rather than confused with the period.

With \(L = 120\,\mathrm{cm} = 1.20\,\mathrm{m}\), \(g = 10\,\mathrm{m\,s^{-2}}\) and \(\pi = \frac{22}{7}\):

Hence

A useful sense check is that a pendulum about a metre long swings roughly once every two seconds, so its frequency must be about half a hertz. Any answer of a few hertz would mean several complete swings each second, which is physically impossible for a pendulum this long.

Two errors produce the other figures. Leaving the length as \(120\) instead of \(1.20\) inflates \(\sqrt{L/g}\) by a factor of about ten and drives the frequency badly wrong, and stopping at \(T\) and quoting \(2.2\) as though it were the frequency confuses seconds with hertz. Note also that the mass of the bob and the amplitude do not appear in the formula, so they never affect the answer for small swings.

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At constant pressure a fixed mass of gas obeys Charles' law: the volume is directly proportional to the absolute temperature, so

Converting the initial temperature to kelvin is the essential first step, because a ratio of Celsius temperatures is meaningless:

The gas shrinks to one-fifth of its original volume, so \(V_2 = \tfrac{1}{5}\times 2.00 = 0.40\,\mathrm{cm^{3}}\). Rearranging Charles' law:

Converting back to the Celsius scale asked for in the options:

Notice how the wording controls the arithmetic. Read as "the volume becomes one-fifth of the original", the volume ratio is \(1/5\) and the temperature falls by the same factor, giving \(-207\,^{\circ}\mathrm{C}\). Read instead as "the volume falls by one-fifth", the ratio would be \(4/5\) and the answer would be \(330\times0.8 = 264\,\mathrm{K} = -9\,^{\circ}\mathrm{C}\), which is not offered, so the first reading is the intended one.

The commonest error in this topic is to work in degrees Celsius, which here would give \(57/5 \approx 11\,^{\circ}\mathrm{C}\) and is completely wrong because the gas laws are proportionalities measured from absolute zero, not from the ice point. Always convert to kelvin before forming any ratio, and convert back only at the last line.

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This is a method-of-mixtures problem, and the governing statement is the principle of conservation of energy: with the container ignored and no loss to the surroundings, \[\text{heat lost by the hot water} = \text{heat gained by the cold water}.\] Each term is calculated from \(Q = mc\,\Delta\theta\). Boiled water is at \(100\,^\circ\text{C}\), and the final mixture temperature is \(60\,^\circ\text{C}\), so the temperature changes are:

• hot water cools from \(100\,^\circ\text{C}\) to \(60\,^\circ\text{C}\): \(\Delta\theta = 40\,\text{K}\)
• cold water warms from \(25\,^\circ\text{C}\) to \(60\,^\circ\text{C}\): \(\Delta\theta = 35\,\text{K}\)

Both liquids are water, so the specific heat capacity \(c\) is the same on each side and cancels: \[m \times c \times 40 = 8 \times c \times 35\] \[40m = 280 \quad\Rightarrow\quad m = 7\,\text{kg}.\] Seven kilograms of boiled water is required.

Three points decide this question. First, "boiled water" fixes the hot temperature at \(100\,^\circ\text{C}\); it is data given in words rather than symbols. Second, the two temperature changes are different (\(40\,\text{K}\) against \(35\,\text{K}\)), so the masses cannot simply be equal, and the hot mass must be the smaller multiple: \(m/8 = 35/40\). Third, because both substances are water, \(c\) never needs a numerical value, so quoting \(4200\,\text{J kg}^{-1}\text{K}^{-1}\) adds arithmetic but no information. Also note that no latent heat appears here: nothing changes state, the steam having already condensed. In an examination, write out both \(\Delta\theta\) values explicitly before forming the equation, since reversing them is the commonest source of a wrong mass.

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Resistance thermometers respond faster because they have small sensor mass and use direct electrical detection. Liquid-in-glass and gas thermometers are slower due to thermal expansion and larger thermal inertia, often taking minutes to equilibrate.

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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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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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Waves are classified in two independent ways. By the medium they need, a wave is either mechanical (it requires matter to travel through) or electromagnetic (it does not). By the direction of vibration relative to the direction of travel, a wave is either transverse (particles vibrate at right angles to the direction of energy flow) or longitudinal (particles vibrate along the direction of energy flow).

A plucked string carries a wave along the length of the string, while each element of the string moves up and down, perpendicular to that length. The vibration is therefore at right angles to the propagation, which makes the wave transverse, and since it travels through the material of the string it is also mechanical. Being transverse, it has the humps and hollows that we call crests and troughs. The one statement that does not fit is the claim that the wave is longitudinal, so that is the untrue statement.

The usual confusion is to assume that because a plucked string produces sound, and sound in air is longitudinal, the wave on the string must be longitudinal too. They are two different waves: the transverse wave on the string sets the surrounding air into longitudinal compressions and rarefactions. Keep the classifications separate in an examination, and remember that only transverse waves can be polarised, which is another quick way to test a claim about wave type.

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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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A standing (stationary) wave is not a wave that travels; it is the pattern formed when two identical progressive waves of the same frequency and amplitude travel through the same region in opposite directions and superpose. In practice the second wave is supplied by reflection: a wave sent along a stretched string or down a pipe bounces back from the fixed end or the closed end and overlaps the incoming wave. So a standing wave is produced when a wave reflects off a boundary and interferes with itself.

Where the two waves always arrive in step, constructive interference gives points of maximum displacement called antinodes; where they always arrive exactly out of step, destructive interference gives points of permanently zero displacement called nodes. Because the nodes and antinodes stay in fixed positions, no energy is carried along the medium, which is exactly what distinguishes a standing wave from a progressive one. This is why a guitar string, an organ pipe and a microwave oven cavity all show fixed loud and quiet or bright and dark positions.

The alternatives describe different physics. A wave vibrating in a vertical plane is simply a plane-polarised transverse wave, and the word "standing" in the term refers to the pattern not moving along the medium, not to the direction of vibration. Motion of the source towards or away from the observer changes the observed frequency and is the Doppler effect, which involves a single travelling wave and no superposition at all. Exam reminder: link standing waves to the two conditions of reflection and superposition, and to the presence of fixed nodes and antinodes.

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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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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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Net force in the horizontal (x) direction:
\(F_x = 8 \, \text{N} - 4 \, \text{N} = 4 \, \text{N} \quad \text{(to the right)}\)

Net force in the vertical (y) direction:
\(F_y = 15 \, \text{N} - 12 \, \text{N} = 3 \, \text{N} \quad \text{(3 N upward)}\)

Magnitude of the resultant force: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}\)

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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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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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Electromagnetic induction is described by Faraday's law: whenever the magnetic flux linking a conductor changes, a voltage is set up across the conductor. That voltage is called an induced e.m.f. (electromotive force), and its size is given by \[\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\] where \(N\) is the number of turns and \(\Delta\Phi/\Delta t\) is the rate of change of magnetic flux. The minus sign is Lenz's law: the induced e.m.f. acts in the direction that opposes the change producing it.

The key distinction the question is testing is that induction produces a voltage, not a current, as its primary effect. A current only flows if that e.m.f. is connected to a complete circuit. This is why the e.m.f. still exists across the ends of a rod moved through a field even when the ends are not joined, and it is why a generator is rated by its e.m.f.

An eddy current is a circulating current, not a voltage; it is one of the consequences of induction inside a solid block of metal, and it is measured in amperes. Watts measure power, so that quantity cannot be a voltage at all. "Inductor voltage" is not a standard term in this topic; the recognised name for the quantity produced by a changing flux is the induced e.m.f. When a question names a unit or a quantity, check the dimensions first: only a quantity measured in volts can answer "a voltage called ...".

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The diagram shows a body of mass 10 kg submerged in a liquid. Three forces act on it:

• Weight (downward) = \(mg = 10 \times 10 = 100\) N
• Upthrust (upward) = 10 N
• Viscous drag (upward, opposing motion) = 15 N

The net downward force is:

\(F_{net} = W - U - F_d = 100 - 10 - 15 = 75\) N

Applying Newton's second law:

\(a = \frac{F_{net}}{m} = \frac{75}{10} = 7.5\) m/s\(^2\)

The body accelerates downward at 7.5 m/s\(^2\).

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This question links the electrical energy supplied by the kettle to the heat energy gained by the water. Assuming no heat is lost, the electrical energy delivered in time \(t\) equals the heat needed to raise the water's temperature:

Work out each side separately. The heat required is

The power of the kettle is

Since power is energy per second, the time taken is

Two slips account for the other figures. Using the final temperature \(65\ ^\circ\text{C}\) instead of the temperature rise of \(35\ \text{K}\) inflates the energy badly, and halving or doubling the power (for instance by dividing by \(2400\) instead of \(1200\)) gives \(245\ \text{s}\), which is the trap set here. Also note that a temperature change of \(35\ ^\circ\text{C}\) is numerically identical to \(35\ \text{K}\), so the specific heat capacity in \(\text{J kg}^{-1}\text{K}^{-1}\) can be used directly without converting to kelvin. Always compute the temperature difference first and write it down before substituting.