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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Power is the rate of doing work, \(P = \dfrac{W}{t}\). When a load is lifted steadily, the work done against gravity in a time \(t\) is \(W = mgh\), and since \(h = vt\) for constant speed, \[P = \frac{mg(vt)}{t} = mgv.\] This is the useful shortcut: the power needed to raise a load at constant velocity is the weight of the load multiplied by the speed of lifting.

Substituting the data, with \(m = 20\,\text{kg}\), \(g = 10\,\text{m s}^{-2}\) and \(v = 5\,\text{m s}^{-1}\): \[P = 20 \times 10 \times 5 = 1000\,\text{W} = 1\,\text{kW}.\]

The phrase "single string pulley system" matters less than it appears. A pulley system can reduce the force the boy must apply, but it cannot reduce the work or the power required, because whatever force is saved is paid for in extra distance pulled. For an ideal system the power developed is still \(mgv\); a real system with friction would need more input power, never less. Note also that \(100\,\text{W}\) comes from omitting \(g\) and computing \(mv\), while \(250\,\text{W}\) comes from using the kinetic-energy expression \(\tfrac{1}{2}mv^{2}\), which is an energy in joules, not a power.

The examination takeaway is to check the unit of the expression you build: \(mgv\) has units \(\text{kg}\cdot\text{m s}^{-2}\cdot\text{m s}^{-1} = \text{J s}^{-1} = \text{W}\), which confirms it is a power before any numbers are substituted.

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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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Efficiency measures how much of the energy put into a machine comes out as useful work, expressed as a percentage:

Rearranging for the output and substituting the given values,

The remaining \(200 - 160 = 40\ \text{J}\) is not destroyed; it is wasted mainly as heat and sound through friction in the moving parts, which is why no real machine reaches \(100\%\) efficiency.

The value \(250\ \text{J}\) comes from dividing by \(0.8\) instead of multiplying, and it should be rejected immediately on physical grounds: an output larger than the input would mean the machine creates energy, which violates the conservation of energy. Use that check every time. In any efficiency question the useful output must be smaller than the input, so the correct operation is always the one that reduces the number.

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When a solid melts at its melting point, the heat supplied is used to break down the rigid arrangement of the particles rather than to raise the temperature, so a thermometer in the mixture stays at \(0\ ^\circ\text{C}\) throughout. Heat that produces a change of state at constant temperature is called latent heat, the word latent meaning hidden, because it produces no temperature reading.

The distinction the question turns on is between a total quantity and a per-kilogram quantity. The specific latent heat of fusion \(l\) is the heat needed to melt one kilogram of the solid at its melting point, with the unit \(\text{J kg}^{-1}\). The latent heat of fusion is the heat needed to melt the whole given mass, so

with the unit joule. Because the question fixes a definite mass of \(5\ \text{kg}\), the quantity described is the latent heat of fusion of that ice, not the specific latent heat. Naming it as the specific quantity would be wrong by a factor of \(5\).

The heat-capacity terms do not apply at all, because both describe heat that causes a temperature change: heat capacity is \(Q/\Delta\theta\) in \(\text{J K}^{-1}\) and specific heat capacity is \(Q/(m\Delta\theta)\) in \(\text{J kg}^{-1}\text{K}^{-1}\). Here the temperature does not change, so any formula containing \(\Delta\theta\) is ruled out immediately.

Carry two habits into the examination. First, the word specific always means per unit mass, so it can only be used when no particular mass is mentioned. Second, decide whether the heat causes a temperature change or a change of state: use \(Q = mc\Delta\theta\) for the first and \(Q = ml\) for the second.

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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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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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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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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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The circuit diagram shows a 16 V battery connected to a single 2 Ω resistor in a closed loop. To find the power dissipated in the resistor, use the formula:

\(P = \frac{V^2}{R}\)

Substituting the values:

\(P = \frac{(16)^2}{2} = \frac{256}{2} = 128 \text{ W}\)

The total power dissipated in the circuit is 128 W.

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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 thermometric property is any physical property that varies measurably, continuously and reproducibly with temperature, so that its value can be used as a scale of temperature. Different thermometers exploit different properties: a constant-volume gas thermometer uses pressure, a resistance thermometer uses electrical resistance, a thermocouple uses emf, and a liquid-in-glass thermometer uses the expansion of the liquid.

Mercury is used in liquid-in-glass thermometers, where the mercury is sealed in a bulb attached to a fine capillary tube. As the temperature rises the mercury expands, and because the bore is narrow a small increase in the volume of mercury produces a long, easily read movement of the thread. The property being used is therefore the change of volume with temperature, and mercury suits the job because it expands almost uniformly over a wide range (\(-39\,^\circ\text{C}\) to \(357\,^\circ\text{C}\)), is opaque and easily seen, is a good conductor of heat so it responds quickly, and does not wet glass.

Density does change with temperature, but only as a consequence of the volume change at fixed mass, and density is not what the instrument reads; the length of the mercury thread is a direct measure of volume. Pressure change belongs to gas thermometers, and resistance change belongs to platinum resistance thermometers, not to mercury in glass. When a question names a specific thermometric substance, identify the instrument it is used in first, because the instrument fixes which property is being measured.

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Energy stored in the bonds between atoms in a substance is called chemical energy. Food consists of carbohydrates, fats and proteins, which are large molecules whose covalent bonds hold energy. During respiration these molecules are broken down and reorganised into carbon dioxide and water, and because the products have lower bond energy than the reactants, energy is released for the body to use. The energy in food is therefore chemical energy.

The confusion in this question comes from the fact that chemical energy is a form of stored energy, so it feels reasonable to call it potential energy. In physics, however, potential energy at this level means energy due to position in a field or due to elastic deformation, such as gravitational potential energy \(E = mgh\) or the energy in a stretched spring. Food is not raised or stretched, so labelling it simply potential energy misses the actual store. Kinetic energy is the energy of a body in motion, \(E = \tfrac{1}{2}mv^2\), and a plate of food at rest has none. Mechanical energy is the sum of kinetic and gravitational potential energy of a body, which again is not what makes food nourishing.

A useful check in the examination is to ask what physical change would be needed to release the energy. If the store is released by a chemical reaction, as in food, fuels, and batteries, the energy is chemical. If it is released by letting an object fall or a spring relax, it is potential. If it is already present in motion, it is kinetic.

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When you look down into water, light from the bottom bends away from the normal as it leaves the water, so the bottom appears to be nearer the surface than it really is. The depth you seem to see is the apparent depth; the depth actually there is the real depth. For an object viewed almost vertically, the refractive index of the liquid links the two: \[n = \frac{\text{real depth}}{\text{apparent depth}}.\] Because \(n\) for water is greater than 1, the real depth must always be the larger of the two numbers.

Rearranging and substituting the given values: \[\text{real depth} = n \times \text{apparent depth} = 1.33 \times 10 = 13.3\ \text{cm}.\] So the pool is 13.3 cm deep, and the water makes it look only 10 cm deep.

The tempting error is to divide instead of multiply, giving \(10 / 1.33 = 7.5\) cm. That answer would mean the water made the bottom look deeper than it is, which never happens for a denser medium viewed from air. Before you compute, decide which depth is missing: if you are told the apparent depth, multiply by \(n\); if you are told the real depth and want the apparent one, divide by \(n\). The apparent shift itself is real depth minus apparent depth, here 3.3 cm.

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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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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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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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In a liquid or a gas the molecules are not held in fixed positions, so they move continuously in random directions and collide with one another and with anything suspended in the fluid. A small visible particle, such as a smoke particle in air or a pollen grain in water, is struck unequally from different sides at each instant, and so it jiggles along an irregular zig-zag path. This ceaseless random movement of particles in fluids is called Brownian motion, named after the botanist who first observed it, and it is the standard experimental evidence for the kinetic theory of matter.

The other terms describe something different. Translational motion is one particular type of molecular movement, namely motion of the whole molecule from place to place, and it is only part of the picture; it is not the name given to the observed random movement in fluids, and it says nothing about randomness. Vibrational motion is the to-and-fro oscillation of particles about fixed mean positions, which is characteristic of a solid, where the particles are too tightly packed to wander. An isobaric process is not a kind of motion at all: it is a thermodynamic change that takes place at constant pressure.

A helpful way to keep this straight is to link each state of matter to its dominant motion: solids vibrate about fixed points, while liquids and gases show free random movement, which is Brownian motion. Also note that Brownian motion becomes more vigorous when the temperature is raised or the suspended particle is smaller, because the average kinetic energy of the molecules increases and a lighter particle responds more to each uneven collision.

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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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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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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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Newton's law of universal gravitation states that the attractive force between two point masses is

where \(G\) is the universal gravitational constant and \(r\) is the distance between their centres. The force is therefore directly proportional to the product of the two masses, and this question asks only how that product changes.

Doubling each mass replaces \(m_1 m_2\) by \((2m_1)(2m_2) = 4m_1 m_2\), while \(r\) is unchanged. Writing the new force as \(F_2\) and dividing one expression by the other lets \(G\) and \(r\) cancel:

The mistake to guard against is doubling the force to \(20\,\mathrm{N}\), which comes from doubling only one mass, or from treating the force as proportional to the sum of the masses rather than their product. A second useful habit for this formula is to keep the two dependences separate: the force scales with each mass to the first power but with distance to the power \(-2\). So if the masses were doubled and the separation also doubled, the factor would be \(4 \times \tfrac{1}{4} = 1\) and the force would stay at \(10\,\mathrm{N}\). Setting up the ratio \(F_2/F_1\) rather than trying to find \(G\) or the actual masses is always the fastest and safest method in these proportionality questions.

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All electromagnetic waves travel at the same speed \(c = 3.0\times 10^{8}\ \text{m s}^{-1}\) in a vacuum, and they satisfy \(c = f\lambda\). Since \(c\) is fixed, wavelength and frequency are inversely related: the shortest wavelength belongs to the highest frequency, and therefore to the most energetic radiation, because \(E = hf\).

Ordering the members of the spectrum given here from long wavelength to short: infrared, then visible light, then ultraviolet, then X-rays. Of these, X-rays have the shortest wavelength, of the order of \(10^{-10}\ \text{m}\), compared with about \(10^{-8}\ \text{m}\) for ultraviolet, \(4\times 10^{-7}\) to \(7\times 10^{-7}\ \text{m}\) for visible light and around \(10^{-5}\ \text{m}\) for infrared. The table below sets out the comparison.

Ultraviolet is the tempting alternative because it is the one most students associate with harmful, penetrating radiation from the Sun, but it sits between visible light and X-rays. The very short wavelength of X-rays is precisely why they penetrate soft tissue and are diffracted by the regular spacing of atoms in crystals, an effect that only works when the wavelength is comparable with atomic spacing. A reliable method in the examination is to recite the spectrum in a fixed order, from radio waves through microwaves, infrared, visible light, ultraviolet and X-rays to gamma rays, remembering that wavelength decreases and frequency increases along that sequence, then read off whichever end the question asks for.

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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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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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Electrical power delivered to a device is the product of the potential difference across it and the current through it: \[P = IV.\] The rating on an appliance states the power it consumes at its working voltage, so the current follows by rearranging: \[I = \frac{P}{V} = \frac{500}{220} = 2.27\,\text{A}\ (3\ \text{s.f.}).\] The oven therefore draws about \(2.27\,\text{A}\).

It is worth seeing where the other numbers could come from, because each represents a specific error. Dividing the voltage by the power, \(220/500\), gives \(0.44\), while using a mains value of \(110\,\text{V}\) instead of \(220\,\text{V}\) would double the answer to \(4.55\,\text{A}\). Only the direct substitution into \(I = P/V\) with the values actually given is defensible.

Two related results are often needed in the same question and follow from the same data: the resistance of the heating element at working temperature is \[R = \frac{V^2}{P} = \frac{220^2}{500} = 96.8\,\Omega,\] and the energy consumed in, for example, half an hour is \[E = Pt = 500 \times 1800 = 9.0 \times 10^{5}\,\text{J} = 0.25\,\text{kWh}.\] Keep the three forms \(P = IV = I^2R = \dfrac{V^2}{R}\) at hand, and choose the one whose quantities are actually given rather than working through an intermediate you do not need.

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Dimensions describe a quantity in terms of the base quantities mass \(M\), length \(L\) and time \(T\), independent of the units used. The key physical insight here is that tension in a string is simply the force the string exerts along its length. It is not a special new quantity, so it must have exactly the dimensions of force.

Get those dimensions from Newton's second law, \(F = ma\). Mass contributes \(M\). Acceleration is velocity change per unit time, that is \(\frac{L\,T^{-1}}{T} = L\,T^{-2}\). Multiplying: \[[F] = M \times L\,T^{-2} = M\,L\,T^{-2}.\] So the dimensional formula of tension is \(M\,L\,T^{-2}\), whose SI unit, the newton, is correspondingly \(\text{kg}\,\text{m}\,\text{s}^{-2}\).

Watch the sign of the time index. Writing \(M\,L\,T^{2}\) would mean force grows with the square of time, which is dimensionally the same as mass times length times time squared and matches no mechanical quantity here; the index is negative because time appears in the denominator of acceleration twice. An expression with no \(M\) at all, such as \(L\,T^{-2}\), is the dimension of acceleration alone, not of a force, and raising \(M\) to a power other than one has no justification since force is directly proportional to a single mass. Exam takeaway: whenever a question asks for the dimensions of tension, thrust, weight, upthrust or any pull or push, answer with the dimensions of force, \(M\,L\,T^{-2}\).

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When a block slides on a horizontal surface and comes to rest, the only horizontal force acting on it is the kinetic friction force. This friction force produces a deceleration that brings the block to a stop.

The friction force on a horizontal surface is given by:

\[ f = \mu_k m g \]

where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity. By Newton's second law, the deceleration \( a \) equals \( \mu_k g \).

Using the kinematic equation for motion with constant deceleration:

\[ v^2 = u^2 - 2as \]

The block starts at \( u = 10 \) m/s and comes to rest (\( v = 0 \)) after travelling \( s = 25 \) m. Substituting:

\[ 0 = (10)^2 - 2 \times a \times 25 \]

\[ 0 = 100 - 50a \]

\[ a = \frac{100}{50} = 2 \text{ m/s}^2 \]

Since \( a = \mu_k g \):

\[ \mu_k = \frac{a}{g} = \frac{2}{9.8} \approx 0.204 \]

Rounding to one decimal place, the coefficient of kinetic friction is approximately 0.2.

A common mistake is to forget that the deceleration on a horizontal surface due to friction depends only on \( \mu_k \) and \( g \), not on the mass of the block (mass cancels out). This is why the question does not need to provide the mass.

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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 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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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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Stress is the force acting per unit cross-sectional area of the wire:

measured in \(\text{N m}^{-2}\) (pascals). Two quantities must be prepared before substituting: the stretching force and the area of the circular cross-section.

The force is the weight of the block:

The radius must be converted from centimetres to metres, since the answer is required in \(\text{N m}^{-2}\):

Therefore

Note that \(53\times 10^{4}\) and \(5.3\times 10^{5}\) are the same number written differently, so compare powers of ten carefully rather than glancing only at the digits.

Three traps are set here. Using the diameter in place of the radius quarters the stress. Forgetting to square the \(10^{-2}\) when converting the radius, so that the area comes out a hundred times too large, produces a figure a hundred times too small. And a negative power of ten in the answer should be rejected on sight: a force of \(15\ \text{N}\) spread over an area far smaller than \(1\ \text{m}^2\) must give a stress much larger than \(15\ \text{N m}^{-2}\), not a tiny fraction of it. Always convert lengths to metres before squaring.

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

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This is a direct application of the ideal gas equation in molar form:

The one conversion that must be made is the temperature, because \(T\) in this equation is the absolute temperature:

Now substitute, keeping the units consistent in the SI system (\(V\) in \(\text{m}^3\), \(R\) in \(\text{J K}^{-1}\text{mol}^{-1}\), giving \(P\) in pascals):

The pressure is about \(4445\ \text{Pa}\).

The commonest error is substituting \(17\) for the temperature, which gives roughly \(260\ \text{Pa}\), a value so small it should look wrong at once. A second slip is confusing the two very similar numbers in the data: \(8.3\) is the number of moles while \(8.31\ \text{J K}^{-1}\text{mol}^{-1}\) is the molar gas constant, and both appear in the numerator, so neither may be dropped. Before dividing, check that only the volume sits in the denominator, and always convert Celsius to kelvin as your first line of working in any gas calculation.

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Clouds are collections of condensed water droplets or ice crystals suspended in the air, but high above the ground (typically hundreds of meters to kilometers up).

Fog is the same phenomenon (condensed suspended droplets, often with dust), but at ground level, reducing visibility.

The question specifies "in the air" near the surface (implied by dust particles and suspension context), so fog is correct, not cloud.

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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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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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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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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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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.