JAMB UTME - Physics - 2025

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Thermal conductivity measures how readily a material passes heat on by conduction, that is by the transfer of energy from particle to particle without bulk movement of the material. Conduction depends on how closely and how strongly the particles are coupled, so it is best in solids (and outstanding in metals, where free electrons also carry energy), poorer in liquids, and worst in gases, whose molecules are far apart and rarely interact.

Air has by far the smallest value, so air is the poorest conductor of the four. This is exactly why insulating materials are designed to trap air rather than to be dense: cotton wool, fur, feathers, cavity walls and vacuum-flask jackets all work by holding air still. Ash insulates well for the same reason, but its own solid particles still conduct, so it cannot be a better insulator than the air within it.

A caution worth remembering: still air is a superb insulator, yet moving air carries heat away rapidly by convection. Conduction and convection are separate mechanisms, and a question about conductivity is asking only about the first. When the choices span different states of matter, rank them gas, liquid, non-metallic solid, metal in increasing order of conductivity and the answer usually follows at once.

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

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

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

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

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

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

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When two surfaces are in contact, the contact force between them can be resolved into two parts. The component perpendicular (normal) to the surface is the normal reaction, and the component along the surface, that is tangential to it, is friction. The definition given in the question specifies a force that is tangential to the surface and that opposes sliding, and that is precisely the frictional force.

Friction arises from the interlocking of microscopic irregularities and from attraction between the molecules of the two surfaces at the points where they genuinely touch. It always acts along the surface and always in the direction that opposes relative sliding, or the tendency to slide, which is why a stationary block on a rough incline does not slip and why a pushed box eventually stops. For a body on the point of sliding, \(F = \mu R\), where \(R\) is the normal reaction and \(\mu\) the coefficient of friction, a relation which itself shows that friction and the normal reaction are two distinct, mutually perpendicular quantities.

The normal force is the tempting alternative, but it acts at right angles to the surface and pushes the body away from the surface; it supports the body rather than resisting its sliding. Weight, \(W = mg\), is the gravitational pull of the Earth and acts vertically downwards regardless of any surface, so it is not tangential except in the special case of a vertical wall. Upthrust is the upward force a fluid exerts on a body immersed in it, again vertical and not a surface-contact tangential force. In the examination, use the direction words as your key: perpendicular to the surface means normal reaction, along the surface means friction.

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

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

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

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

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

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

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

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

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Light travels so much faster than sound that the flash of lightning reaches the observer effectively at the instant it is produced. The 4 s delay is therefore the whole time the sound took to cover the distance from the storm to the observer, and the distance follows from the definition of speed: \[d = v \times t.\]

Substituting the values given in the question: \[d = 60 \times 4 = 240\ \text{m}.\] The observer is 240 m from the centre of the storm.

Two points are worth fixing. First, use the speed value the question supplies, not the familiar figure for air at room temperature; this question deliberately sets the speed at \(60\ \text{m s}^{-1}\), so answering 340 m by using \(340\ \text{m s}^{-1}\) and a time of 1 s, or by multiplying the wrong pair of numbers, ignores the given data. Second, do not halve the time as you would in an echo calculation. An echo travels to a reflector and back, so there the distance is \(\frac{vt}{2}\); thunder makes a one-way trip, so the full time is used. Exam reminder: decide first whether the sound path is one-way or a there-and-back journey before applying the speed equation.

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

Work out each side separately. The heat required is

The power of the kettle is

Since power is energy per second, the time taken is

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

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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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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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The critical angle \(C\) is the angle of incidence inside the denser medium at which the refracted ray just grazes along the boundary, so the angle of refraction in air is \(90^\circ\). Applying Snell's law at the glass-air boundary, \[n_{g}\sin C = n_{a}\sin 90^\circ.\] Taking \(n_a = 1\) for air and \(\sin 90^\circ = 1\), this rearranges to the standard result \[n = \frac{1}{\sin C}.\]

Substituting \(C = 45^\circ\), for which \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\): \[n = \frac{1}{1/\sqrt{2}} = \sqrt{2} \approx 1.41.\] The refractive index of the glass is \(\sqrt{2}\).

The frequent error is to write \(n = \sin C\), giving \(0.71\), a value less than one that would describe a medium in which light travels faster than in air. A refractive index for a denser medium relative to air is always greater than \(1\), so \(n = 1/\sin C\) is the correct arrangement, and \(\sin C\) small means \(n\) large. Keep the physical consequence in mind too: at any angle of incidence greater than \(45^\circ\) inside this glass, no light escapes and total internal reflection occurs, which is the principle behind optical fibres, prism periscopes and the sparkle of cut gemstones.

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

Given:

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

Calculate the cubical expansivity:

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

Apply the volume expansion formula:

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

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

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

The new bore volume is approximately 3.56 cm³.

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

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The energy stored in a capacitor charged to a potential difference \(V\) is

For a fixed supply voltage the stored energy therefore depends only on the capacitance of the combination, so that is the quantity to examine.

For capacitors in series the effective capacitance obeys

which always gives a value smaller than the smallest individual capacitance. For two \(4\,\mu\mathrm{F}\) capacitors, for instance, the series value is \(2\,\mu\mathrm{F}\), so across a \(10\,\mathrm{V}\) supply the pair stores \(\tfrac{1}{2}(2\times10^{-6})(10)^{2} = 1.0\times10^{-4}\,\mathrm{J}\), whereas one of them alone across the same supply would store \(2.0\times10^{-4}\,\mathrm{J}\). The fall in stored energy therefore traces directly to the fall in overall capacitance produced by the series connection. Physically, the applied p.d. is shared among the capacitors, so no single capacitor receives the full \(V\), and each stores less than it would on its own.

The suggestion that unequal charges are deposited is the misconception worth clearing up: in a series chain the charge on every capacitor is the same, because the plates between neighbouring capacitors are isolated and can only separate charge, not create it. What differs between unequal capacitors in series is the voltage each carries, from \(V = Q/C\). Internal resistance of the source affects how quickly charging happens and causes heating in the wires, but it is not the reason the fully charged combination holds less energy. In the examination, tie any energy comparison for capacitors back to \(E = \tfrac{1}{2}CV^{2}\) and ask what has changed, \(C\) or \(V\).

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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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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.

Maelezo ya Majibu

T x 80 + 15 x 10 = W x 50

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

T + 15 = W  - - - - - - - - - - - (2)

80T + 150 = 50(T + 15)

80T + 150 = 50T + 750

30T = 750 - 150

30T = 600

T = 20N

The closest option is 19.2N

Maelezo ya Majibu

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

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

Substituting the masses and the known electrochemical equivalent of copper,

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

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

Maelezo ya Majibu

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.

Maelezo ya Majibu

Reverberation is the prolonging of a sound in an enclosed space caused by repeated reflections from the walls, floor and ceiling arriving at the listener slightly after the direct sound. In a large hall with hard, smooth surfaces the reflected sound persists for a long time, so syllables overlap and speech becomes blurred. Reducing reverberation means reducing the energy of those reflections.

Perforated materials, along with soft boards, curtains and padded seats, are good absorbers of sound. Sound waves entering the small holes are repeatedly reflected inside the pores and against the fibres, and the energy is gradually converted into heat by friction, so very little is reflected back into the hall. Lining the walls with such material therefore shortens the reverberation time and improves the clarity of speech and music.

The other effects listed are not what the lining changes. Diffraction is the spreading of a wave as it passes an obstacle or through a gap, and it depends on the wavelength compared with the size of the gap, not on absorption. Refraction is the change in direction of a wave when its speed changes on entering a different medium, which is not the phenomenon at work here. There is no recognised acoustic quantity called an auditorium pulse. Keep the distinction sharp in the examination: echoes and reverberation are reflection phenomena, so they are controlled by absorbers, whereas diffraction and refraction are controlled by geometry and by the medium.

Maelezo ya Majibu

The decisive word in this question is "erect". A concave mirror forms an upright (and therefore virtual) image in one situation only: when the object lies between the pole and the principal focus. Any object placed at or beyond the focus gives a real, inverted image. So even before calculating, the object distance must be smaller than the focal length of 20 cm.

The arithmetic confirms it. Magnification is \(m = \dfrac{v}{u}\) in size, and for an erect image from a concave mirror the image is virtual, so the image distance is negative: \(v = -4u\) when the image is four times the object. Substituting into the mirror formula \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\): \[\frac{1}{20} = \frac{1}{u} + \frac{1}{-4u} = \frac{4 - 1}{4u} = \frac{3}{4u}.\] Cross-multiplying gives \(4u = 60\), so \(u = 15\ \text{cm}\), which is indeed less than 20 cm. The image is then 60 cm behind the mirror, virtual, erect and magnified, which is how a shaving or make-up mirror works.

The usual error is to take \(v = +4u\), which gives \(\frac{1}{20} = \frac{5}{4u}\) and \(u = 25\ \text{cm}\), an object distance between the focus and the centre of curvature. That answer describes a real, inverted, magnified image and so contradicts the word "erect" in the question. Exam takeaway: read the image description first, use it to fix the sign of \(v\) before substituting, and remember that for a concave mirror upright means virtual and means the object is inside the focal length.

Maelezo ya Majibu

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.

Maelezo ya Majibu

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

Maelezo ya Majibu

For a spring obeying Hooke's law, the elastic potential energy stored is:

\[ E = \frac{1}{2}kx^2 \]

where \( k \) is the spring constant and \( x \) is the compression (or extension). Since the applied force \( F = kx \), we can write \( x = \frac{F}{k} \), and substituting:

\[ E = \frac{1}{2}k\left(\frac{F}{k}\right)^2 = \frac{F^2}{2k} \]

This shows that the energy stored is proportional to the square of the applied force: \( E \propto F^2 \).

For two different forces applied to the same spring:

\[ \frac{E_2}{E_1} = \left(\frac{F_2}{F_1}\right)^2 \]

Substituting the given values:

\[ \frac{E_2}{0.16} = \left(\frac{700}{200}\right)^2 = (3.5)^2 = 12.25 \]

\[ E_2 = 0.16 \times 12.25 = 1.96 \text{ J} \]

The spring stores 1.96 J of energy when compressed by 700 N.

The critical insight is that energy depends on the square of the force, not linearly. Tripling the force does not triple the energy - it increases it by a factor of nine.

Maelezo ya Majibu

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

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

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

Maelezo ya Majibu

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.

Maelezo ya Majibu

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

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

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

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

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

Maelezo ya Majibu

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.