JAMB UTME - Physics - 2025

For small oscillations a simple pendulum has period

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

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

Hence

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

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

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.

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.

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.

A hydraulic press works by Pascal's principle: pressure applied to an enclosed incompressible liquid is transmitted equally throughout, so the pressure under the small piston equals the pressure under the large piston. \[\frac{F_1}{A_1} = \frac{F_2}{A_2}.\] Since each piston is circular, \(A = \pi r^2\), and the \(\pi\) cancels: \[\frac{F_1}{r_1^{2}} = \frac{F_2}{r_2^{2}} \quad\Rightarrow\quad \frac{r_2^{2}}{r_1^{2}} = \frac{F_2}{F_1}.\]

Substituting the given forces, \[\frac{r_2^{2}}{r_1^{2}} = \frac{3200}{200} = 16 \quad\Rightarrow\quad \frac{r_2}{r_1} = \sqrt{16} = 4.\] So \(r_1 : r_2 = 1 : 4\).

The decisive step is the square root. Forces in a hydraulic press scale with area, and area scales with the square of the radius, so a force multiplication of \(16\) needs a radius ratio of only \(4\), not \(16\). Reading off \(1:16\) is the classic error, made by matching the force ratio straight to the radii; \(1:2\) comes from taking the square root twice.

Remember also that the press multiplies force but not energy: the small piston must travel \(16\) times as far as the large one, since the same volume of liquid is displaced, \(A_1 d_1 = A_2 d_2\). In an examination, decide first whether the ratio you are asked for is one of areas, radii or diameters, and insert or remove the square accordingly.

This is a direct application of the ideal gas equation in molar form:

The one conversion that must be made is the temperature, because \(T\) in this equation is the absolute temperature:

Now substitute, keeping the units consistent in the SI system (\(V\) in \(\text{m}^3\), \(R\) in \(\text{J K}^{-1}\text{mol}^{-1}\), giving \(P\) in pascals):

The pressure is about \(4445\ \text{Pa}\).

The commonest error is substituting \(17\) for the temperature, which gives roughly \(260\ \text{Pa}\), a value so small it should look wrong at once. A second slip is confusing the two very similar numbers in the data: \(8.3\) is the number of moles while \(8.31\ \text{J K}^{-1}\text{mol}^{-1}\) is the molar gas constant, and both appear in the numerator, so neither may be dropped. Before dividing, check that only the volume sits in the denominator, and always convert Celsius to kelvin as your first line of working in any gas calculation.

When you look down into water, light from the bottom bends away from the normal as it leaves the water, so the bottom appears to be nearer the surface than it really is. The depth you seem to see is the apparent depth; the depth actually there is the real depth. For an object viewed almost vertically, the refractive index of the liquid links the two: \[n = \frac{\text{real depth}}{\text{apparent depth}}.\] Because \(n\) for water is greater than 1, the real depth must always be the larger of the two numbers.

Rearranging and substituting the given values: \[\text{real depth} = n \times \text{apparent depth} = 1.33 \times 10 = 13.3\ \text{cm}.\] So the pool is 13.3 cm deep, and the water makes it look only 10 cm deep.

The tempting error is to divide instead of multiply, giving \(10 / 1.33 = 7.5\) cm. That answer would mean the water made the bottom look deeper than it is, which never happens for a denser medium viewed from air. Before you compute, decide which depth is missing: if you are told the apparent depth, multiply by \(n\); if you are told the real depth and want the apparent one, divide by \(n\). The apparent shift itself is real depth minus apparent depth, here 3.3 cm.

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.

T x 80 + 15 x 10 = W x 50

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

T + 15 = W  - - - - - - - - - - - (2)

80T + 150 = 50(T + 15)

80T + 150 = 50T + 750

30T = 750 - 150

30T = 600

T = 20N

The closest option is 19.2N

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

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

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

At constant pressure a fixed mass of gas obeys Charles' law: the volume is directly proportional to the absolute temperature, so

Converting the initial temperature to kelvin is the essential first step, because a ratio of Celsius temperatures is meaningless:

The gas shrinks to one-fifth of its original volume, so \(V_2 = \tfrac{1}{5}\times 2.00 = 0.40\,\mathrm{cm^{3}}\). Rearranging Charles' law:

Converting back to the Celsius scale asked for in the options:

Notice how the wording controls the arithmetic. Read as "the volume becomes one-fifth of the original", the volume ratio is \(1/5\) and the temperature falls by the same factor, giving \(-207\,^{\circ}\mathrm{C}\). Read instead as "the volume falls by one-fifth", the ratio would be \(4/5\) and the answer would be \(330\times0.8 = 264\,\mathrm{K} = -9\,^{\circ}\mathrm{C}\), which is not offered, so the first reading is the intended one.

The commonest error in this topic is to work in degrees Celsius, which here would give \(57/5 \approx 11\,^{\circ}\mathrm{C}\) and is completely wrong because the gas laws are proportionalities measured from absolute zero, not from the ice point. Always convert to kelvin before forming any ratio, and convert back only at the last line.

The circuit diagram shows a 16 V battery connected to a single 2 Ω resistor in a closed loop. To find the power dissipated in the resistor, use the formula:

\(P = \frac{V^2}{R}\)

Substituting the values:

\(P = \frac{(16)^2}{2} = \frac{256}{2} = 128 \text{ W}\)

The total power dissipated in the circuit is 128 W.

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.

In the kinetic theory, gas molecules move randomly and collide elastically with the container walls. Each collision reverses the component of a molecule's momentum normal to the wall, so one molecule of mass \(m\) striking a wall at speed \(u\) and rebounding changes its momentum by \(2mu\), and the wall receives that momentum. Newton's second law in its general form says force is the rate at which momentum is transferred: \[F = \frac{\Delta p}{\Delta t}.\] Pressure is force spread over area, \(P = F/A\), so combining the two gives \[P = \frac{1}{A}\,\frac{\Delta p}{\Delta t}.\] Pressure is therefore the rate of change of momentum imparted to the walls, per unit area, which is the description the question requires.

The time factor is the part most often dropped. Momentum imparted per unit area alone is an impulse per unit area, with units \(\text{N s m}^{-2}\), not \(\text{N m}^{-2}\); it would grow without limit the longer you waited, whereas the pressure of a gas in a sealed vessel is steady. Only by dividing the momentum transfer by the time over which it happens do you obtain a constant force and hence a constant pressure. Dividing momentum change by volume instead of area is wrong in the same way, and also produces the units of momentum density rather than pressure.

This reasoning is what leads to the kinetic-theory result \[P = \frac{1}{3}\rho \overline{c^{2}} = \frac{1}{3}\frac{Nm}{V}\overline{c^{2}},\] where \(\overline{c^{2}}\) is the mean square speed. When a question offers several verbal definitions, test each one by its units: the correct statement for pressure must reduce to \(\text{N m}^{-2}\), and only "momentum per second per unit area" does so.

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.

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.

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.

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

This is a vector-addition problem, so the two journeys must be resolved into perpendicular components before they are combined. The bearing notation \(N30^\circ E\) means the direction is measured \(30^\circ\) away from north, turning towards the east. For a displacement of \(10.0\,\text{m}\) in that direction, north is the adjacent side and east the opposite side of the \(30^\circ\) angle:

• eastward part \(= 10.0\sin 30^\circ = 5.0\,\text{m}\)
• northward part \(= 10.0\cos 30^\circ = 8.66\,\text{m}\)

The first leg is entirely eastward, so the totals are \[x = 6.0 + 5.0 = 11.0\,\text{m (east)},\qquad y = 0 + 8.66 = 8.66\,\text{m (north)}.\] These two totals are at right angles, so Pythagoras gives the straight-line distance from the start: \[r = \sqrt{11.0^2 + 8.66^2} = \sqrt{121 + 75.0} = \sqrt{196} = 14.0\,\text{m}.\] The man is \(14.0\,\text{m}\) from his starting point.

The usual error is to add the magnitudes, \(6.0 + 10.0 = 16.0\,\text{m}\), or to interchange the sine and cosine because the angle was assumed to be measured from the east line. In bearings written as \(N\theta E\) the angle is measured from north, so north takes the cosine. Sketching the two arrows head-to-tail, as above, shows at once which component belongs to which trigonometric ratio.

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

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

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

sinr = 0.3

Dimensions describe a quantity in terms of the base quantities mass \(M\), length \(L\) and time \(T\), independent of the units used. The key physical insight here is that tension in a string is simply the force the string exerts along its length. It is not a special new quantity, so it must have exactly the dimensions of force.

Get those dimensions from Newton's second law, \(F = ma\). Mass contributes \(M\). Acceleration is velocity change per unit time, that is \(\frac{L\,T^{-1}}{T} = L\,T^{-2}\). Multiplying: \[[F] = M \times L\,T^{-2} = M\,L\,T^{-2}.\] So the dimensional formula of tension is \(M\,L\,T^{-2}\), whose SI unit, the newton, is correspondingly \(\text{kg}\,\text{m}\,\text{s}^{-2}\).

Watch the sign of the time index. Writing \(M\,L\,T^{2}\) would mean force grows with the square of time, which is dimensionally the same as mass times length times time squared and matches no mechanical quantity here; the index is negative because time appears in the denominator of acceleration twice. An expression with no \(M\) at all, such as \(L\,T^{-2}\), is the dimension of acceleration alone, not of a force, and raising \(M\) to a power other than one has no justification since force is directly proportional to a single mass. Exam takeaway: whenever a question asks for the dimensions of tension, thrust, weight, upthrust or any pull or push, answer with the dimensions of force, \(M\,L\,T^{-2}\).

Energy stored in the bonds between atoms in a substance is called chemical energy. Food consists of carbohydrates, fats and proteins, which are large molecules whose covalent bonds hold energy. During respiration these molecules are broken down and reorganised into carbon dioxide and water, and because the products have lower bond energy than the reactants, energy is released for the body to use. The energy in food is therefore chemical energy.

The confusion in this question comes from the fact that chemical energy is a form of stored energy, so it feels reasonable to call it potential energy. In physics, however, potential energy at this level means energy due to position in a field or due to elastic deformation, such as gravitational potential energy \(E = mgh\) or the energy in a stretched spring. Food is not raised or stretched, so labelling it simply potential energy misses the actual store. Kinetic energy is the energy of a body in motion, \(E = \tfrac{1}{2}mv^2\), and a plate of food at rest has none. Mechanical energy is the sum of kinetic and gravitational potential energy of a body, which again is not what makes food nourishing.

A useful check in the examination is to ask what physical change would be needed to release the energy. If the store is released by a chemical reaction, as in food, fuels, and batteries, the energy is chemical. If it is released by letting an object fall or a spring relax, it is potential. If it is already present in motion, it is kinetic.

Clouds are collections of condensed water droplets or ice crystals suspended in the air, but high above the ground (typically hundreds of meters to kilometers up).

Fog is the same phenomenon (condensed suspended droplets, often with dust), but at ground level, reducing visibility.

The question specifies "in the air" near the surface (implied by dust particles and suspension context), so fog is correct, not cloud.

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

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

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

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

Given:

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

Calculate the cubical expansivity:

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

Apply the volume expansion formula:

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

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

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

The new bore volume is approximately 3.56 cm³.

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

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

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

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

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.

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

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.

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.

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\(^2\).

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

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

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

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.

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.

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.

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.

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

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

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

The key words here are very small. Measuring an ordinary resistance is one problem; measuring a resistance of a fraction of an ohm is a harder one, because the resistance of the connecting leads and of the sliding or soldered contacts is itself of that same order. Any method in which those stray resistances are counted along with the unknown will give a badly wrong result. The instrument that avoids this is the potentiometer.

In the potentiometer method the unknown low resistance \(R\) is joined in series with a known low standard resistance \(S\), so that exactly the same current \(I\) flows through both. The potential difference across each is then tapped off and balanced against a length of the potentiometer wire, giving balancing lengths \(l_1\) and \(l_2\). Since \(V = IR\) and the potentiometer reading is proportional to the potential difference,

Two features make this accurate for tiny resistances. At balance the galvanometer carries no current, so the potentiometer draws nothing from the circuit and does not disturb it, and the tappings are made directly across the resistance itself, so the lead and contact resistances lie outside the measured section and cancel out of the ratio.

A Wheatstone bridge, in the metre-bridge form, is the standard circuit for a moderate resistance of a few ohms upwards, and that familiarity is what makes it tempting here. It becomes unreliable at the extremes, however: for a very small unknown, the end corrections and the resistance of the jockey contact and connecting wires are comparable with the quantity being measured, so the balance point loses its meaning. A voltmeter is unsuitable because a real voltmeter draws some current from the circuit and the potential difference across a very small resistance is minute, so the reading would be dominated by instrument error. A rheostat is not a measuring instrument at all; it is a variable resistor used to control the current in a circuit.

The examination point to retain is that the range of the resistance decides the method: a bridge for middling values, and a potentiometer, whose null reading excludes lead and contact resistance, for very small ones.

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.

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.