JAMB UTME - Physics - 2025

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

Volume expansion is governed by the cubic expansivity \(\gamma\): \[\Delta V = V_1 \gamma\, \Delta\theta.\] For a solid the three expansivities are related by \(\gamma = 3\alpha\) and \(\beta = 2\alpha\), because a solid expands by the same fractional amount in each of its three perpendicular directions. Here the linear expansivity is given, so convert first: \[\gamma = 3\alpha = 3 \times 2.0\times10^{-5} = 6.0\times10^{-5}\,\text{K}^{-1}.\]

Now substitute the data, with \(V_1 = 1\,\text{cm}^3\) and \(\Delta V = 0.0018\,\text{cm}^3\): \[0.0018 = 1 \times 6.0\times10^{-5} \times \theta \quad\Rightarrow\quad \theta = \frac{0.0018}{6.0\times10^{-5}} = 30.\] The temperature rise is \(30\) kelvin, which is a rise of \(30\,^\circ\text{C}\). A change of temperature has the same numerical value on both scales because the degree sizes are identical, which is why an expansivity quoted in \(\text{K}^{-1}\) may be used directly with a Celsius temperature change.

The mistake that produces \(90\) is using \(\alpha\) itself in the volume formula, and the mistake that produces \(45\) is using \(\beta = 2\alpha\), the area expansivity. Match the expansivity to the dimension being measured: length with \(\alpha\), area with \(2\alpha\), volume with \(3\alpha\). Note also that only the ratio \(\Delta V / V_1\) matters, so the units of volume cancel and no conversion of cubic centimetres is needed.

Antwoorddetails

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.

Antwoorddetails

The magnetic effect of an electric current is the fact that a current-carrying conductor produces a magnetic field around itself. When the conductor is wound into a coil around a soft-iron core, the arrangement becomes an electromagnet, which attracts iron only while current flows. Any device that works by this principle must contain a coil whose magnetism does mechanical work.

The electric bell is exactly that device. Current from the supply passes through the coils of an electromagnet, which attracts a soft-iron armature carrying a hammer, and the hammer strikes the gong. The movement of the armature breaks the circuit at a contact screw, the electromagnet loses its magnetism, a spring pulls the armature back and remakes the contact, and the cycle repeats rapidly to give continuous ringing. Remove the magnetic effect of the current and nothing at all happens, so the bell is the device that operates on it.

The alternatives depend on different effects. A rheostat is simply a variable resistor used to control current by changing resistance, and it uses the heating effect at most, not magnetism. A thermostat relies on the differential thermal expansion of a bimetallic strip, which bends with temperature and opens or closes a circuit. A carbon microphone works because the resistance of loosely packed carbon granules changes as sound waves compress them, so it converts sound into a varying current through a resistance change, not through magnetism. When a question asks which device uses a named effect, look for the component that produces it: a soft-iron core with a winding signals the magnetic effect, a bimetallic strip signals expansion, and a resistance wire signals the heating effect.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

Electromagnetic induction is described by Faraday's law: whenever the magnetic flux linking a conductor changes, a voltage is set up across the conductor. That voltage is called an induced e.m.f. (electromotive force), and its size is given by \[\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\] where \(N\) is the number of turns and \(\Delta\Phi/\Delta t\) is the rate of change of magnetic flux. The minus sign is Lenz's law: the induced e.m.f. acts in the direction that opposes the change producing it.

The key distinction the question is testing is that induction produces a voltage, not a current, as its primary effect. A current only flows if that e.m.f. is connected to a complete circuit. This is why the e.m.f. still exists across the ends of a rod moved through a field even when the ends are not joined, and it is why a generator is rated by its e.m.f.

An eddy current is a circulating current, not a voltage; it is one of the consequences of induction inside a solid block of metal, and it is measured in amperes. Watts measure power, so that quantity cannot be a voltage at all. "Inductor voltage" is not a standard term in this topic; the recognised name for the quantity produced by a changing flux is the induced e.m.f. When a question names a unit or a quantity, check the dimensions first: only a quantity measured in volts can answer "a voltage called ...".

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

For a parallel-plate capacitor with air (or vacuum) between the plates, the capacitance depends only on the geometry: \[C = \frac{\varepsilon_0 A}{d},\] where \(A\) is the area of overlap of one plate and \(d\) the separation. Wider plates store more charge for the same voltage, and closer plates do too, which is why \(A\) is on top and \(d\) underneath. Because \(\varepsilon_0\) is quoted in SI units, both the area and the separation must be converted to metres before substituting.

The plates are circular, so the area is \[A = \pi r^{2} = 3.142 \times (0.06)^{2} = 3.142 \times 3.6 \times 10^{-3} = 1.131 \times 10^{-2}\ \text{m}^{2},\] using \(r = 6\ \text{cm} = 0.06\ \text{m}\). The separation is \(d = 0.12\ \text{cm} = 1.2 \times 10^{-3}\ \text{m}\). Substituting: \[C = \frac{(8.85 \times 10^{-12})(1.131 \times 10^{-2})}{1.2 \times 10^{-3}} = (8.85 \times 10^{-12}) \times 9.426 = 8.34 \times 10^{-11}\ \text{F}.\] So the capacitance is about \(8.3 \times 10^{-11}\ \text{F}\), which is 83 pF.

Two traps sit in this question. The first is using the diameter as the radius or forgetting to square the radius, which changes the area by a factor of four. The second is leaving centimetres in place: since \(1\ \text{cm}^{2} = 10^{-4}\ \text{m}^{2}\) and \(1\ \text{cm} = 10^{-2}\ \text{m}\), a mixed substitution shifts the power of ten. Note as well that any physically real capacitance of a small air capacitor must come out as a tiny fraction of a farad, so a positive index such as \(10^{11}\ \text{F}\) can be rejected on sight.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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

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

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

Antwoorddetails

Resistance thermometers respond faster because they have small sensor mass and use direct electrical detection. Liquid-in-glass and gas thermometers are slower due to thermal expansion and larger thermal inertia, often taking minutes to equilibrate.

Antwoorddetails

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 \(V\).

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

This question is about streamlining. When a body moves through a fluid such as air or water, the fluid must be pushed aside and made to flow round the body. A blunt front forces the fluid to change direction abruptly, the flow behind it breaks up into swirling eddies, and the pressure in front becomes much higher than the pressure behind. That pressure difference, together with the rubbing of the fluid layers along the surface, makes up the resistive force called drag or fluid friction.

A pointed, tapered front lets the fluid part smoothly and rejoin gradually behind the body, so the flow stays streamlined instead of turbulent and the pressure difference between front and back is much smaller. The result is a reduction in the fluid friction acting on the boat or aircraft, which means less driving force is needed for a given speed, less fuel is used, and a higher top speed becomes possible for the same engine power. This is why fast-moving objects in nature and in engineering, from fish and birds to aircraft and racing hulls, all share the same tapered shape.

The suggestion that the shape increases fluid friction reverses the physics: increasing drag would waste energy, and shapes deliberately made blunt, such as a parachute canopy, are used precisely when large drag is wanted. Stopping depends on reverse thrust, brakes or drag devices, not on the shape of the nose, and appearance is not a physical explanation. In the examination, treat any question about the shape of a moving vehicle as a question about minimising drag, and be ready to name the mechanism as smooth, streamlined flow replacing turbulent flow.

Antwoorddetails

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.

Antwoorddetails

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

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

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

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

A current-carrying inductor stores energy in the magnetic field of its coil. The energy stored is

where \(L\) is the inductance in henries and \(I\) the steady current. This is the magnetic counterpart of the energy \(\tfrac{1}{2}CV^2\) stored in a capacitor's electric field, and like it the energy depends on the square of the current.

Rearrange for the current before substituting:

So a steady current of about \(1.29\ \text{A}\) stores \(2.5\ \text{J}\) in a \(3\ \text{H}\) coil.

The trap is forgetting the square root and dividing instead, for example \(2E/L = 1.67\) or \(E/L\) style combinations, or forgetting the factor \(\tfrac{1}{2}\), which would give \(\sqrt{2.5/3}=0.91\ \text{A}\). Because the relationship is quadratic, doubling the current stores four times the energy, and that squared dependence is exactly what the examiner is checking. Write the formula down, make the unknown the subject, then substitute.

Antwoorddetails

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.

Antwoorddetails

Radioactive decay is random, so the number of undecayed nuclei falls exponentially: \(N = N_0 e^{-\lambda t}\), where \(\lambda\) is the decay constant, the probability per second that a given nucleus decays. The half-life \(t_{1/2}\) is the time for \(N\) to fall to \(N_0/2\). Putting \(N = N_0/2\) and \(t = t_{1/2}\) into the exponential law gives

Substituting the given half-life,

which to two significant figures is \(5.0\times 10^{-3}\ \text{s}^{-1}\). Note that the decay constant has the unit \(\text{s}^{-1}\), the reciprocal of time, because it is a rate per nucleus rather than a time.

The neighbouring values here are all within a couple of per cent of one another, so they are testing whether the constant \(0.693\) is used rather than a rounded \(0.7\) (which would give \(5.05\times 10^{-3}\)) or an inverted formula such as \(t_{1/2}/\ln 2\). Keep \(\ln 2 = 0.693\) and remember that a short half-life means a large decay constant, since the two are inversely proportional.

Antwoorddetails

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

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

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

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

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

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.

Antwoorddetails

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.