JAMB UTME - Physics - 2025

In the kinetic theory, gas molecules move randomly and collide elastically with the container walls. Each collision reverses the component of a molecule's momentum normal to the wall, so one molecule of mass \(m\) striking a wall at speed \(u\) and rebounding changes its momentum by \(2mu\), and the wall receives that momentum. Newton's second law in its general form says force is the rate at which momentum is transferred: \[F = \frac{\Delta p}{\Delta t}.\] Pressure is force spread over area, \(P = F/A\), so combining the two gives \[P = \frac{1}{A}\,\frac{\Delta p}{\Delta t}.\] Pressure is therefore the rate of change of momentum imparted to the walls, per unit area, which is the description the question requires.

The time factor is the part most often dropped. Momentum imparted per unit area alone is an impulse per unit area, with units \(\text{N s m}^{-2}\), not \(\text{N m}^{-2}\); it would grow without limit the longer you waited, whereas the pressure of a gas in a sealed vessel is steady. Only by dividing the momentum transfer by the time over which it happens do you obtain a constant force and hence a constant pressure. Dividing momentum change by volume instead of area is wrong in the same way, and also produces the units of momentum density rather than pressure.

This reasoning is what leads to the kinetic-theory result \[P = \frac{1}{3}\rho \overline{c^{2}} = \frac{1}{3}\frac{Nm}{V}\overline{c^{2}},\] where \(\overline{c^{2}}\) is the mean square speed. When a question offers several verbal definitions, test each one by its units: the correct statement for pressure must reduce to \(\text{N m}^{-2}\), and only "momentum per second per unit area" does so.

The whole question turns on how cells combine, so begin with the two rules and the reasoning behind them. E.m.f. is energy supplied per unit charge, so when cells are joined in series a single charge is driven through all of them in turn and receives energy from each, giving

When identical cells are joined in parallel a charge passes through only one of them on its way round the circuit, so the total e.m.f. is that of a single cell:

What the parallel grouping reduces is the internal resistance, \(r_{\text{eff}} = r/n\), which is why it is used to obtain a larger current at the same voltage.

The condition stated in the question, that the three e.m.f.s are equal, is the decisive clue. In a series chain the e.m.f.s add whether they are equal or not, so no such assumption would be needed. The assumption is required only for a parallel grouping, because cells of unequal e.m.f. connected in parallel drive current through one another and the combination no longer has a single well-defined e.m.f. With the cells equal, the parallel arrangement has a total e.m.f. equal to that of any one cell, so the relation \(E = E_1 = E_2 = E_3\) is the one that describes it.

The two reciprocal expressions offered can be rejected on principle rather than by inspecting the wiring. Reciprocals of that kind belong to resistors in parallel and to capacitors in series; e.m.f.s never combine reciprocally, because e.m.f. adds as energy per unit charge along a path. In the examination, first decide from the diagram whether charge must pass through every cell (series, so add) or through only one cell (parallel, so take a single cell's value), and never transfer the reciprocal formula from resistance to e.m.f.

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.

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.

Pressure measures how a force is spread over the surface it acts on: \[P = \frac{F}{A}.\] The water's weight, 5000 N, is the downward force pressing on the base of the reservoir, and the base is the rectangle on which that weight is distributed. So the whole problem is finding the base area and dividing.

The base area is \[A = \text{length} \times \text{breadth} = 10 \times 5 = 50\ \text{m}^{2},\] therefore \[P = \frac{5000}{50} = 100\ \text{N m}^{-2} = 100\ \text{Pa}.\] The pressure at the bottom of the reservoir is 100 Pa.

Notice that the depth of the water is never needed. Students often reach for \(P = \rho g h\) and stall because no depth or density is given, but that formula and \(P = F/A\) are the same statement: \(\rho g h\) is just the weight of the water column divided by the base area. When the weight of the liquid and the base dimensions are supplied, use \(F/A\) directly. Also check the units of the area: a pressure in pascals requires the area in square metres, so lengths given in centimetres must be converted before dividing.

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.

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

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

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

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

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.

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.

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.

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.

A photovoltaic cell (solar cell) turns light directly into electricity, and it can only do so because it is built from semiconductor material, typically silicon doped to form a p-n junction. In a semiconductor the valence and conduction bands are separated by a small energy gap, of the order of 1 eV. A photon of visible light carries just enough energy to lift an electron across that gap, creating a free electron and leaving a positive hole behind. The built-in electric field at the junction then sweeps the electron one way and the hole the other, and this separation of charge is what produces the cell's e.m.f. and drives current through an external circuit.

That mechanism explains why the other materials cannot do the job. In a metallic conductor the conduction band is already full of free electrons and there is no energy gap and no internal junction field, so light-generated charge carriers recombine at once and no sustained potential difference builds up. In an insulator the gap is far too wide, so ordinary light photons lack the energy to release any electrons at all. A chemical action is the basis of a primary or secondary cell, where energy comes from a redox reaction rather than from incident light, so it is not the operating principle of a photovoltaic cell.

Keep the distinction sharp between the two light-and-electron effects on the syllabus: in photoemission (the photoelectric effect) electrons are ejected completely from a metal surface into a vacuum, whereas in the photovoltaic effect electrons stay inside the semiconductor and are simply moved across a junction to create a voltage.

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.

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.

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 \( F \times t = ma \times t = m \times 5 \times 2 = 10m \).

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

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

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

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

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

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

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

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.

The graph is a velocity-time graph with three stages of motion. From 0 to 10 seconds the body moves at a constant velocity of 12 m/s. From 10 to 14 seconds the velocity drops linearly from 12 m/s to 0 m/s, representing uniform deceleration. This final segment is the last stage of the motion.

The distance covered during any stage equals the area under the velocity-time curve for that interval. The last stage forms a right triangle with base \(\Delta t = 14 - 10 = 4\) s and height \(v = 12\) m/s.

Distance = \(\frac{1}{2} \times 4 \times 12 = 24\) m.

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.

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.

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.

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.

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.

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.

For equilibrium, clockwise moment = anticlockwise moment about the pivot.

clockwise distance from pivot: 50cm - 10cm = 40cm

anticlockwise distance from pivot: 85cm - 10cm = 75cm

Applying the principle of moments

W x distance(w) = T x distance(T)

100 x 40 = T x 75

T = \(\frac{ 4000}{75}\) ? 53.33N

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

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

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

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

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

Electrical energy consumed is calculated using the formula:

\[ E = P \times t \]

where \( P \) is power in watts and \( t \) is time in hours (when the result is needed in watt-hours).

Given:

• Power, \( P = 150 \text{ W} \)
• Time, \( t = 336 \text{ hours} \)

\[ E = 150 \times 336 = 50{,}400 \text{ Wh} \]

Convert to kilowatt-hours by dividing by 1000:

\[ E = \frac{50{,}400}{1000} = 50.40 \text{ kWh} \]

The energy consumed is 50.40 kWh.

When calculating energy in kWh, ensure power is converted from watts to kilowatts (divide by 1000) either before or after multiplication. Using \( P \) in kW from the start: \( 0.15 \times 336 = 50.40 \text{ kWh} \), which confirms the answer.

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 ...".

In a pinhole camera light travels in straight lines through the small hole, so the object, the pinhole and the image form two similar triangles with the pinhole at the common apex. Similar triangles give the magnification directly as a ratio of distances: \[m = \frac{\text{image height}}{\text{object height}} = \frac{\text{image distance }v}{\text{object distance }u},\] where the image distance is simply the length of the camera box, because the screen is the back of the box.

Here the box length gives \(v = 25\ \text{cm}\), and the image is one-fifth the size of the object, so \(m = \frac{1}{5}\). Substituting: \[\frac{1}{5} = \frac{25}{u} \quad \Rightarrow \quad u = 5 \times 25 = 125\ \text{cm} = 1.25\ \text{m}.\] The object stands 1.25 m in front of the pinhole. The result is sensible: an image smaller than the object means the object must be further from the pinhole than the screen is, and here it is five times as far.

The likeliest error is inverting the ratio and writing \(u = 25/5 = 5\ \text{cm}\), which would place the object nearer the pinhole than the screen and would make the image larger, not smaller. A second trap is the unit change: the options are in metres while the camera length is in centimetres, so the final conversion \(125\ \text{cm} = 1.25\ \text{m}\) must be made. Note also that no focal length or lens formula is involved, since a pinhole has no focal length; only the straight-line propagation of light and similar triangles are needed.

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.

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.

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.

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

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

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

Substituting,

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

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

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.

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

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

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

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

Two identical cells joined in parallel behave as a single cell whose e.m.f. is the same as one of them, because their terminals are tied together so neither can raise the terminal voltage above its own e.m.f. What the parallel arrangement does change is the internal resistance: the two internal resistances are in parallel, so

Applying the circuit equation \(E = I(R + r_{\text{eff}})\) with the external resistor \(R = 2\,\Omega\):

This \(0.6\,\mathrm{A}\) is the total current delivered to the resistor; each cell supplies half of it, \(0.3\,\mathrm{A}\), which is why parallel grouping is used when a circuit needs a larger current than one cell can comfortably provide at the same voltage.

The trap in this question is to treat the cells as though they were in series. That would give \(E = 3.0\,\mathrm{V}\), \(r = 2\,\Omega\) and \(I = 3.0/4 = 0.75\,\mathrm{A}\), which rounds close to one of the other figures offered. A second common slip is to use \(r = 1\,\Omega\) unchanged and obtain \(1.5/3 = 0.5\,\mathrm{A}\). Fix the rule firmly: cells in series add their e.m.f.s and their internal resistances; identical cells in parallel keep the single-cell e.m.f. and divide the internal resistance by the number of cells.

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.

Weight is the force of gravity on a body, \(W = mg\). The mass is a fixed property of the body and does not change when the body is moved to the moon; only \(g\) changes. Since the moon's gravitational field strength is one-sixth of the earth's, the weight there must also be one-sixth of the weight on earth.

Working through the two steps explicitly, first find the mass on earth:

The moon's gravitational field strength is

so the weight on the moon is

The same result comes directly from \(W_{m} = \tfrac{1}{6}W_{e} = 2500/6 = 416.67\,\mathrm{N}\), which is the quicker route once you have noticed that mass cancels.

The misconception this question targets is the belief that mass itself becomes smaller on the moon. It does not: a \(250\,\mathrm{kg}\) body remains \(250\,\mathrm{kg}\) anywhere, and a beam balance comparing masses would read the same on the moon, while a spring balance measuring force would read one-sixth as much. In the examination, write down which quantity is being asked for, mass in kilograms or weight in newtons, before you start dividing by six.

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.