JAMB UTME - Physics - 2025

Antwortdetails

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

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

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

Antwortdetails

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

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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

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

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

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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

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

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

Antwortdetails

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.

Antwortdetails

Efficiency measures how much of the energy put into a machine comes out as useful work, expressed as a percentage:

Rearranging for the output and substituting the given values,

The remaining \(200 - 160 = 40\ \text{J}\) is not destroyed; it is wasted mainly as heat and sound through friction in the moving parts, which is why no real machine reaches \(100\%\) efficiency.

The value \(250\ \text{J}\) comes from dividing by \(0.8\) instead of multiplying, and it should be rejected immediately on physical grounds: an output larger than the input would mean the machine creates energy, which violates the conservation of energy. Use that check every time. In any efficiency question the useful output must be smaller than the input, so the correct operation is always the one that reduces the number.

Antwortdetails

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

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

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

Antwortdetails

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

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

Substituting the masses and the known electrochemical equivalent of copper,

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

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

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

A thermometric property is any physical property that varies measurably, continuously and reproducibly with temperature, so that its value can be used as a scale of temperature. Different thermometers exploit different properties: a constant-volume gas thermometer uses pressure, a resistance thermometer uses electrical resistance, a thermocouple uses emf, and a liquid-in-glass thermometer uses the expansion of the liquid.

Mercury is used in liquid-in-glass thermometers, where the mercury is sealed in a bulb attached to a fine capillary tube. As the temperature rises the mercury expands, and because the bore is narrow a small increase in the volume of mercury produces a long, easily read movement of the thread. The property being used is therefore the change of volume with temperature, and mercury suits the job because it expands almost uniformly over a wide range (\(-39\,^\circ\text{C}\) to \(357\,^\circ\text{C}\)), is opaque and easily seen, is a good conductor of heat so it responds quickly, and does not wet glass.

Density does change with temperature, but only as a consequence of the volume change at fixed mass, and density is not what the instrument reads; the length of the mercury thread is a direct measure of volume. Pressure change belongs to gas thermometers, and resistance change belongs to platinum resistance thermometers, not to mercury in glass. When a question names a specific thermometric substance, identify the instrument it is used in first, because the instrument fixes which property is being measured.

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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 \(V\).

Antwortdetails

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

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

The moon's gravitational field strength is

so the weight on the moon is

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

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

Antwortdetails

The speed of sound in a gas is governed by how stiff the gas is compared with how heavy it is, expressed as \[v = \sqrt{\frac{\gamma P}{\rho}},\] where \(P\) is the pressure, \(\rho\) the density and \(\gamma\) a constant for the gas. It looks as though raising \(P\) should raise \(v\), and that is exactly the trap in this question.

Pressure and density are not independent. For a fixed mass of gas at constant temperature, Boyle's law gives \(PV = \text{constant}\), and since \(\rho = m/V\) the density rises in exact proportion to the pressure. So the ratio \(P/\rho\) stays the same when the pressure is doubled: the gas becomes stiffer, but it also becomes correspondingly heavier per unit volume, and the two effects cancel. The speed of sound is therefore unchanged when pressure increases at constant temperature.

The same equation shows what does change the speed. Writing \(P/\rho = RT/M\) for an ideal gas gives \(v = \sqrt{\gamma RT/M}\), so the speed depends on the absolute temperature and on the molar mass of the gas, and it is proportional to \(\sqrt{T}\). This is why sound travels faster on a hot day and faster in a light gas such as helium, but is not altered by simply pumping the air to a higher pressure at the same temperature. Exam reminder: whenever a question changes the pressure of a gas at constant temperature, check whether the density changes with it before concluding that a quantity depending on \(P/\rho\) has changed.

Antwortdetails

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.

Antwortdetails

Whether a body floats or sinks is decided by comparing its weight with the maximum upthrust available. By Archimedes' principle the upthrust equals the weight of fluid displaced. When a body is fully submerged it displaces its own volume \(V\) of water, so

The body sinks when \(W > U_{\max}\), that is when \(\rho_b V g > \rho_w V g\). The common volume \(V\) and \(g\) cancel, leaving the condition \(\rho_b > \rho_w\). An object sinks in water precisely because its density is greater than the density of water.

This is why a small steel nail sinks while a large wooden log floats: what matters is density, not size or weight on its own. A steel ship floats only because its hull encloses air, which lowers the average density of the whole ship below that of water.

The statement that upthrust equals weight describes a body in equilibrium, which is the condition for floating or for remaining suspended at rest in the fluid, not for sinking; a sinking body has an upthrust smaller than its weight and so has a net downward force. Saying the density is less than that of water gives the condition for floating, the exact opposite. Comparing water pressure with weight is meaningless because pressure and force are different quantities with different units, so they can never be equated. In the examination, reduce every flotation question to a comparison of two densities, and check the units of any quantities you are asked to compare.

Antwortdetails

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.

Antwortdetails

When a block slides on a horizontal surface and comes to rest, the only horizontal force acting on it is the kinetic friction force. This friction force produces a deceleration that brings the block to a stop.

The friction force on a horizontal surface is given by:

\[ f = \mu_k m g \]

where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity. By Newton's second law, the deceleration \( a \) equals \( \mu_k g \).

Using the kinematic equation for motion with constant deceleration:

\[ v^2 = u^2 - 2as \]

The block starts at \( u = 10 \) m/s and comes to rest (\( v = 0 \)) after travelling \( s = 25 \) m. Substituting:

\[ 0 = (10)^2 - 2 \times a \times 25 \]

\[ 0 = 100 - 50a \]

\[ a = \frac{100}{50} = 2 \text{ m/s}^2 \]

Since \( a = \mu_k g \):

\[ \mu_k = \frac{a}{g} = \frac{2}{9.8} \approx 0.204 \]

Rounding to one decimal place, the coefficient of kinetic friction is approximately 0.2.

A common mistake is to forget that the deceleration on a horizontal surface due to friction depends only on \( \mu_k \) and \( g \), not on the mass of the block (mass cancels out). This is why the question does not need to provide the mass.

Antwortdetails

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.

Antwortdetails

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\(^2\).

Antwortdetails

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

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

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

Antwortdetails

Faraday's first law of electrolysis states that the mass deposited at an electrode is proportional to the quantity of charge passed, \(m = ZQ = ZIt\), where \(Z\) is the electrochemical equivalent of the substance. The whole calculation therefore begins with the charge.

Convert the time to seconds first, since the ampere is a coulomb per second:

For silver, one mole of \(\text{Ag}^{+}\) ions carries one faraday of charge, so depositing \(108\ \text{g}\) requires \(96\,500\ \text{C}\). This gives

and hence

The mass of silver deposited is about \(1.34\ \text{g}\).

The most frequent error is leaving the time in minutes, which makes the charge \(20\ \text{C}\) and the mass a hundredth of the true value, landing near the small figures offered here. A second error is dividing by a valency of \(2\); silver is monovalent, unlike copper in \(\text{Cu}^{2+}\), so no factor of two appears. Remember the routine: seconds, then coulombs, then multiply by the electrochemical equivalent.

Antwortdetails

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.

Antwortdetails

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

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

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

Substituting,

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

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

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

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.

Antwortdetails

This is a method-of-mixtures problem, and the governing statement is the principle of conservation of energy: with the container ignored and no loss to the surroundings, \[\text{heat lost by the hot water} = \text{heat gained by the cold water}.\] Each term is calculated from \(Q = mc\,\Delta\theta\). Boiled water is at \(100\,^\circ\text{C}\), and the final mixture temperature is \(60\,^\circ\text{C}\), so the temperature changes are:

• hot water cools from \(100\,^\circ\text{C}\) to \(60\,^\circ\text{C}\): \(\Delta\theta = 40\,\text{K}\)
• cold water warms from \(25\,^\circ\text{C}\) to \(60\,^\circ\text{C}\): \(\Delta\theta = 35\,\text{K}\)

Both liquids are water, so the specific heat capacity \(c\) is the same on each side and cancels: \[m \times c \times 40 = 8 \times c \times 35\] \[40m = 280 \quad\Rightarrow\quad m = 7\,\text{kg}.\] Seven kilograms of boiled water is required.

Three points decide this question. First, "boiled water" fixes the hot temperature at \(100\,^\circ\text{C}\); it is data given in words rather than symbols. Second, the two temperature changes are different (\(40\,\text{K}\) against \(35\,\text{K}\)), so the masses cannot simply be equal, and the hot mass must be the smaller multiple: \(m/8 = 35/40\). Third, because both substances are water, \(c\) never needs a numerical value, so quoting \(4200\,\text{J kg}^{-1}\text{K}^{-1}\) adds arithmetic but no information. Also note that no latent heat appears here: nothing changes state, the steam having already condensed. In an examination, write out both \(\Delta\theta\) values explicitly before forming the equation, since reversing them is the commonest source of a wrong mass.

Antwortdetails

Every simple cell has three parts to identify separately: two electrodes, the electrolyte that conducts by ion movement between them, and often a depolariser that removes hydrogen gas from the positive electrode. The question asks only for the electrolyte, so the answer must be a substance that ionises in solution and carries charge inside the cell.

In the wet Leclanche cell the positive electrode is a carbon rod, the negative electrode is a zinc rod, and the electrolyte is a strong solution of ammonium chloride, \(\mathrm{NH_4Cl}\). It dissociates to give \(\mathrm{NH_4^+}\) and \(\mathrm{Cl^-}\) ions, which carry the current through the liquid while zinc dissolves at the negative electrode and hydrogen is released at the carbon rod. Manganese(IV) oxide, \(\mathrm{MnO_2}\), is packed round the carbon rod as the depolariser, oxidising the hydrogen to water and slowing down polarisation. The cell gives an e.m.f. of about 1.5 V but has a large internal resistance, so it suits work needing brief currents such as ringing a bell.

The other substances belong to different cells or to different parts of a cell. Carbon is the positive electrode of this same cell, which is why it is a tempting choice: an electrode is a conductor, not the ion-carrying solution. Lead(IV) oxide is the positive plate of the lead-acid accumulator, whose electrolyte is dilute sulphuric acid, and nickel hydroxide belongs to the alkaline nickel-cadmium or nickel-iron cell, whose electrolyte is potassium hydroxide. When revising cells, learn each one as a set of four labels: negative electrode, positive electrode, electrolyte, depolariser.

Antwortdetails

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.