JAMB UTME - Physics - 2025

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

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

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

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

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

The net downward force is:

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

Applying Newton's second law:

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

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

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All the relationships follow from the ideal gas equation \[PV = nRT.\] To decide whether two quantities are directly or inversely proportional, hold the other two constant and see what the equation demands.

Only pressure and volume sit on the same side of the equation as a product, and a product held constant is the definition of inverse proportionality. So the inversely proportional pair is pressure and volume: squeeze a fixed mass of gas at constant temperature into half the space and the pressure doubles, because the molecules strike the walls twice as often.

A practical way to confirm the type of proportionality is the shape of the graph. Pressure against volume gives a curve (a hyperbola), while pressure against \(1/V\) gives a straight line through the origin. Pressure against absolute temperature and volume against absolute temperature both give straight lines through the origin directly. In an examination, always state which quantities are being held constant before quoting a gas law, since the same two variables can behave differently if a third is allowed to vary.

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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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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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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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Demagnetization is the process of removing or reducing the magnetism of a magnet. The standard methods include:

• Heating the magnet to a high temperature (above its Curie point) and allowing it to cool while oriented in the east-west direction, so it is perpendicular to the Earth's magnetic field.
• Hammering or dropping the magnet repeatedly while it is oriented in the east-west direction.
• Passing a decreasing alternating current through a solenoid containing the magnet.

The key requirement is that the magnet must be oriented in the east-west direction during demagnetization. This ensures the Earth's magnetic field does not re-magnetize the bar as its domains are disrupted.

Heating a magnetic bar red hot and allowing it to cool in the east-west direction is a valid demagnetization method. Heating disrupts the alignment of magnetic domains, and cooling in the E-W orientation prevents re-alignment along the Earth's field.

Placing the bar in a solenoid alone does not demagnetize it - it would magnetize it. Stroking or hammering in the north-south direction would tend to magnetize the bar rather than demagnetize it, because the N-S orientation aligns with the Earth's magnetic field.

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

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

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

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

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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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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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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 temperature gradient of a lagged rod is the rate at which temperature falls along its length, defined as

Lagging matters because it stops heat escaping through the sides, so in the steady state the same heat flows through every cross-section and the temperature falls uniformly from one end to the other. That uniform fall is what makes a single gradient value meaningful.

Rearranging for the temperature difference across the whole rod:

A difference of \(83.2\ \text{K}\) is numerically the same as a difference of \(83.2\ ^\circ\text{C}\), because the kelvin and the Celsius degree are the same size; only the zeros of the two scales differ. Taking the given end as the cooler end, the other end is

so the temperature at the other end is about \(316\ ^\circ\text{C}\). The listed value of \(316.28\ ^\circ\text{C}\) is this result, the tiny difference in the final digit arising from rounding in the printed data.

Two points are worth noting. First, arithmetically the far end could also have been the cooler one, giving \(233-83.2 = 149.8\ ^\circ\text{C}\); that value is not among the choices, which fixes the given end as the cold end. Second, do not convert \(233\ ^\circ\text{C}\) to kelvin and then add the gradient result and forget to convert back, and do not multiply by the gradient without the length: \(83.2\) is a temperature difference, never a temperature. Always separate the difference calculation from the final scale reading.

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

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

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

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

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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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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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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.

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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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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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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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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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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.

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

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

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

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An annular eclipse is a particular kind of solar eclipse. Like every solar eclipse it happens only when the sun, the moon and the earth lie on the same straight line with the moon in the middle, so that the moon's shadow falls on the earth. What makes it annular rather than total is the moon's distance: because the moon's orbit is elliptical, its angular size varies. When it is near its farthest point it appears slightly smaller than the sun, so the dark umbra does not quite reach the earth's surface and a bright ring (Latin annulus) of the sun's disc remains visible round the black moon.

Among the statements offered, the one that correctly describes the condition for this event is that the sun, moon and earth come into a straight line. That alignment, called syzygy, is the necessary geometry for both the total and the annular solar eclipse; the difference between them is only the apparent size of the moon at the time.

The statement that the earth comes between the moon and the sun describes a lunar eclipse, in which the earth's shadow falls on the moon; that is the commonest confusion in this topic, so fix the order clearly: in a solar eclipse the moon blocks the sun from the earth, in a lunar eclipse the earth blocks the sun from the moon. A gathering of stars is a cluster or constellation and has nothing to do with eclipses, and simple invisibility of one body is not a definition of an eclipse, since the moon is invisible at new moon in every month without any eclipse occurring. In the examination, first identify which body is being shadowed, then decide whether the shadow is total, partial or annular.

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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.

Akọwa Nkọwa

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

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

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

Akọwa Nkọwa

An alternating current has no single fixed value: it grows to a maximum in one direction, falls to zero, grows to a maximum in the opposite direction, and repeats. To describe such a current with one useful number we quote its root-mean-square (r.m.s.) value, which is the steady direct current that would produce the same average heating effect in the same resistor. For a sinusoidal current the r.m.s. value is tied to the peak (maximum) value \(I_0\) by \[I_{r.m.s} = \frac{I_0}{\sqrt{2}} = 0.707\,I_0.\]

The single current value quoted in the question, 7 A, has to be read as the greatest value the current reaches, because an r.m.s. value can only be obtained from the peak. Substituting: \[I_{r.m.s} = \frac{7}{\sqrt{2}} = \frac{7}{1.414} = 4.95\ \text{A} \approx 5\ \text{A}.\] So the r.m.s. current is about 5 A.

Two slips account for most wrong answers here. Dividing by 2 instead of \(\sqrt{2}\) gives 3.5 A, and multiplying by \(\sqrt{2}\) gives 9.9 A, which is the route from r.m.s. back to peak rather than peak to r.m.s. A quick safety check in the exam: for a sinusoidal current the r.m.s. value is always about 70% of the peak, so it must come out smaller than the peak, never equal to it or larger.

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

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

The force is the weight of the block:

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

Therefore

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

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

Akọwa Nkọwa

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

Akọwa Nkọwa

In a simple pendulum, kinetic energy is maximum at the lowest point (equilibrium position) where potential energy is minimum and speed is maximum.Here, Q is the control point (mean/equilibrium position), so maximum kinetic energy occurs at Q. At extremes P and S, kinetic energy is zero (velocity = 0). At R, it is between the extreme and the equilibrium.

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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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Stimulated emission is the process in which an incoming photon of a specific energy causes an excited atom to release a second photon that is identical in energy, phase, direction, and polarisation. This mechanism is the fundamental operating principle of a laser.

The word "laser" is itself an acronym: Light Amplification by Stimulated Emission of Radiation. The entire device is designed around achieving and sustaining stimulated emission through population inversion and an optical cavity.

The other light sources listed operate on different principles:

• An incandescent bulb produces light by heating a filament until it glows - this is thermal radiation (black-body emission).
• A fluorescent lamp works by passing an electric current through mercury vapour, which emits ultraviolet radiation. A phosphor coating on the tube absorbs the UV and re-emits it as visible light through spontaneous emission (fluorescence).
• An LED (light-emitting diode) produces light through electroluminescence, where electrons recombine with holes across a semiconductor junction and release photons through spontaneous emission.

Only the laser relies on stimulated emission as its primary mechanism of light production.

Akọwa Nkọwa

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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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.

Akọwa Nkọwa

\(_ag_g\) = \(\frac{\text{sini}}{\text{sinr}}\)

sinr = \(\frac{\text{sini}}{_ag_g}\)

sinr = \(\frac{0.5}{1.5}\)

sinr = 0.3

Akọwa Nkọwa

The steam gives out energy in two distinct stages, and both must be included:

1. Condensation at \(100\,^\circ\text{C}\). Changing state releases latent heat at constant temperature: \(Q_1 = mL\). The supplied value \(2.26\times10^{6}\,\text{J kg}^{-1}\) is the specific latent heat associated with the liquid-vapour change for water, which is the change happening here; note that the subscript printed with it conventionally denotes fusion, whose value for water is \(3.34\times10^{5}\,\text{J kg}^{-1}\), so use the number given and read it as the latent heat of vaporisation, since no melting or freezing occurs in this process.
2. Cooling of the condensed water from \(100\,^\circ\text{C}\) to \(15\,^\circ\text{C}\): \(Q_2 = mc\,\Delta\theta\) with \(\Delta\theta = 85\,\text{K}\).

The total energy released is therefore \[Q = m\left(L + c\,\Delta\theta\right).\] Evaluating the bracket first: \[L + c\Delta\theta = 2.26\times10^{6} + 4200 \times 85 = 2.26\times10^{6} + 3.57\times10^{5} = 2.617\times10^{6}\,\text{J kg}^{-1}.\] With \(Q = 5400\,\text{kJ} = 5.4\times10^{6}\,\text{J}\), \[m = \frac{5.4\times10^{6}}{2.617\times10^{6}} = 2.06\,\text{kg}.\] About \(2.06\,\text{kg}\) of water is collected.

Two errors account for the other figures. Using the latent heat alone gives \(5.4\times10^{6}/2.26\times10^{6} = 2.39\,\text{kg}\), because it ignores the cooling from \(100\,^\circ\text{C}\) to \(15\,^\circ\text{C}\); using the cooling term alone gives \(5.4\times10^{6}/(4200\times85) = 15.1\,\text{kg}\), because it ignores the far larger latent heat. Notice the scale of the two contributions: condensing \(1\,\text{kg}\) of steam releases roughly six times as much energy as cooling that same kilogram of boiling water down to room temperature, which is why steam scalds so severely. Always convert kilojoules to joules before dividing, and check that a latent-heat stage has no temperature change attached to it.

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This question uses de Broglie's idea that a moving particle has a wavelength linked to its momentum: \[\lambda = \frac{h}{p} = \frac{h}{mv},\] so that \[m = \frac{h}{\lambda v}.\] Everything therefore depends on getting the wavelength into metres, because \(h\) is in \(\text{J s}\) and the speed in \(\text{m s}^{-1}\).

The wavelength is given in millimetres, so convert first: \[\lambda = 4.0 \times 10^{-7}\ \text{mm} = 4.0 \times 10^{-7} \times 10^{-3}\ \text{m} = 4.0 \times 10^{-10}\ \text{m}.\] Now substitute: \[m = \frac{6.63 \times 10^{-34}}{(4.0 \times 10^{-10})(2.7 \times 10^{8})} = \frac{6.63 \times 10^{-34}}{1.08 \times 10^{-1}} = 6.1 \times 10^{-33}\ \text{kg}.\] The significant figures come out as 6.1, so the intended choice is the value quoted with those figures; its power of ten appears to be misprinted, since the correct working gives \(6.1 \times 10^{-33}\ \text{kg}\) rather than \(10^{-31}\). Quote \(6.1 \times 10^{-33}\ \text{kg}\) as your worked answer and select the value beginning 6.1.

The step that costs most marks is the millimetre-to-metre conversion. Skipping it, and using \(4.0 \times 10^{-7}\ \text{m}\), gives \(6.1 \times 10^{-36}\ \text{kg}\), a thousand times too small. A second slip is inverting the relation and multiplying by \(\lambda v\) instead of dividing. As a check on the physics, remember the inverse proportionality: a shorter wavelength means a larger momentum, so a heavier or faster particle always has the smaller de Broglie wavelength, which is why wave behaviour is only observed for very light particles such as electrons.

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

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

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