JAMB UTME - Physics - 2025

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

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

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

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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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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 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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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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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.

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

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

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

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

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

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

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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.

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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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Each statement has to be tested separately against the physics of dispersion.

1. Colour depends on frequency. Correct. Frequency is the property of a light wave that does not change when the light passes from one medium into another; wavelength and speed both change since \(v = f\lambda\). Because the sensation of colour is unchanged when light travels from air into water or glass, colour must be tied to frequency rather than to wavelength in a particular medium.
2. Yellow deviated more than green by a prism. Incorrect. Deviation by a prism increases as the refractive index of the glass for that light increases, and the index rises as wavelength falls. Green (about \(510\,\text{nm}\)) has a shorter wavelength than yellow (about \(580\,\text{nm}\)), so green has the larger index and is deviated more, not less. The order of increasing deviation is red, orange, yellow, green, blue, indigo, violet, with violet bent most.
3. Rainbows form when rain falls heavily. Incorrect as stated. A rainbow requires sunlight entering water droplets, refracting, reflecting internally once at the back of the drop, and refracting again on leaving, with the observer standing with the Sun behind and the drops in front. Fine drizzle or spray from a waterfall or hose produces a rainbow perfectly well, while heavy rain with a fully overcast sky produces none because no direct sunlight reaches the drops. Heaviness of the rain is not the condition; sunlight plus suspended droplets is.

Only the claim about frequency survives, so the correct response is the one that accepts statement I alone.

The misconception worth correcting is the assumption that longer-wavelength light bends more, which reverses the whole dispersion sequence. Anchor it with one fact: red is deviated least, violet most, because \(n\) is largest for the shortest wavelength. That single rule settles most prism and dispersion questions in an examination.

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

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

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

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

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

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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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The visible spectrum is the narrow band of electromagnetic radiation the eye can detect, roughly from about \(400\,\text{nm}\) to \(700\,\text{nm}\). Within that band, colour is decided by wavelength, and the colours run in a fixed order of decreasing wavelength: red, orange, yellow, green, blue, indigo, violet. Red therefore sits at the long-wavelength (low-frequency) end and violet at the short-wavelength (high-frequency) end, so the colour with the longest wavelength here is red.

Approximate values make the ordering concrete: red is near \(700\,\text{nm}\), yellow near \(580\,\text{nm}\), blue near \(470\,\text{nm}\) and violet near \(400\,\text{nm}\). Because all colours travel at the same speed \(c\) in vacuum, wavelength and frequency are linked by \[c = f\lambda \quad\Rightarrow\quad f = \frac{c}{\lambda},\] so the longest wavelength automatically carries the lowest frequency and the smallest photon energy \(E = hf\). Violet is the exact opposite: shortest wavelength, highest frequency, most energetic photon.

A common slip is to assume that the brightest or most striking colour must have the longest wavelength, or to reverse the spectral order and choose violet. Fix the mnemonic ROYGBIV in memory and attach one fact to it: wavelength decreases from R to V while frequency and energy increase. In an examination this single ordering answers questions on longest or shortest wavelength, greatest or least deviation by a prism, and highest photon energy.

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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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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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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.

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

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

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

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

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

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

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

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

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

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

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

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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.

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The pressure law (Gay-Lussac's law) states that for a fixed mass of gas at constant volume the pressure is directly proportional to the absolute temperature: \[P \propto T \quad\Rightarrow\quad \frac{P}{T} = \text{constant}.\] A graph of \(P\) against \(T\) can only be a straight line through the origin if the temperature axis is zeroed at the point where the pressure itself would be zero, that is at absolute zero. The scale defined that way, independent of any particular substance, is the thermodynamic (absolute, kelvin) scale, so that is the scale the question is after.

Plotting the same experimental data on the Celsius scale gives a straight line of the same gradient, but its zero of temperature is displaced: the line cuts the temperature axis at \(-273\,^\circ\text{C}\) and cuts the pressure axis at a positive intercept, so it does not pass through the origin.

Note that Fahrenheit shares the Celsius problem in a worse form, since its zero lies at about \(-459\,^\circ\text{F}\) below the pressure-zero point. Rankine is genuinely an absolute scale as well (\(0\,^\circ\text{R}\) is absolute zero, with degrees the size of Fahrenheit degrees), so a Rankine plot would also pass through the origin; it is not, however, the scale physics defines the gas laws on, and "thermodynamic scale" is the standard name for the absolute scale used in \(P \propto T\). The examination point to carry away is that every gas-law calculation and graph requires temperature in kelvin: convert with \(T/\text{K} = \theta/^\circ\text{C} + 273\) before substituting.

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

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

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

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

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

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

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

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

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

For small oscillations a simple pendulum has period

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

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

Hence

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

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

Detalhes da Resposta

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of water:

Because it is a ratio of two densities, it is a pure number with no unit. Making the density of the liquid the subject gives

The density of the liquid is \(760\ \text{kg m}^{-3}\), and since this is less than \(1000\ \text{kg m}^{-3}\) the liquid would float on water, which is a sensible check on the result.

The other values are the sort produced by a misplaced decimal point, for example dividing by \(10\) or multiplying by \(10\,000\) instead of \(1000\). A quick way to guard against this is to reason with the definition rather than with the arithmetic: a specific gravity of \(0.76\) means the liquid is a little over three quarters as dense as water, so its density must be a little over three quarters of \(1000\ \text{kg m}^{-3}\). Remember also that if the density of water is quoted as \(1\ \text{g cm}^{-3}\) the same specific gravity gives \(0.76\ \text{g cm}^{-3}\), which is the identical physical density expressed in different units.

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

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.

Detalhes da Resposta

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

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

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

Detalhes da Resposta

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.

Detalhes da Resposta

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.