JAMB UTME - Physics - 2025

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

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

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

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

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

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

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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.

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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.

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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.

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

Substituting the given half-life,

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

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

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

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

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

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The power of a lens measures how strongly it converges or diverges light. A lens that bends rays sharply brings them to a focus close to the lens, so it has a short focal length; a weak lens focuses rays far away. Power is therefore defined as the reciprocal of the focal length, \[P = \frac{1}{f},\] with \(f\) in metres. The unit of \(P\) is the dioptre (\(\text{D}\)), which is simply \(\text{m}^{-1}\). So the power in dioptres is \(\frac{1}{f}\).

Two details make the definition work. First, \(f\) must be expressed in metres before taking the reciprocal: a lens of focal length \(20\,\text{cm} = 0.20\,\text{m}\) has \[P = \frac{1}{0.20} = +5.0\,\text{D}.\] Second, the sign of \(f\) carries through, so a converging (convex) lens has positive power and a diverging (concave) lens has negative power. Powers also add for thin lenses placed in contact, \(P = P_1 + P_2\), which is exactly why opticians quote lenses in dioptres rather than in centimetres.

Expressions such as \(f\), \(2f\) or \(3f\) cannot be correct because they grow as the focal length grows, which would say that a lens focusing light far away is the more powerful one. They also have the wrong unit: metres instead of \(\text{m}^{-1}\). A quick unit check on any formula offered in an optics question will usually eliminate the distractors immediately, and remember to convert centimetres to metres before computing a dioptre value.

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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.

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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.

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

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The 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.

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

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

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

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This question links the electrical energy supplied by the kettle to the heat energy gained by the water. Assuming no heat is lost, the electrical energy delivered in time \(t\) equals the heat needed to raise the water's temperature:

Work out each side separately. The heat required is

The power of the kettle is

Since power is energy per second, the time taken is

Two slips account for the other figures. Using the final temperature \(65\ ^\circ\text{C}\) instead of the temperature rise of \(35\ \text{K}\) inflates the energy badly, and halving or doubling the power (for instance by dividing by \(2400\) instead of \(1200\)) gives \(245\ \text{s}\), which is the trap set here. Also note that a temperature change of \(35\ ^\circ\text{C}\) is numerically identical to \(35\ \text{K}\), so the specific heat capacity in \(\text{J kg}^{-1}\text{K}^{-1}\) can be used directly without converting to kelvin. Always compute the temperature difference first and write it down before substituting.

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

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

The force is the weight of the block:

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

Therefore

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

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

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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.

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Thermal conductivity measures how readily a material passes heat on by conduction, that is by the transfer of energy from particle to particle without bulk movement of the material. Conduction depends on how closely and how strongly the particles are coupled, so it is best in solids (and outstanding in metals, where free electrons also carry energy), poorer in liquids, and worst in gases, whose molecules are far apart and rarely interact.

Air has by far the smallest value, so air is the poorest conductor of the four. This is exactly why insulating materials are designed to trap air rather than to be dense: cotton wool, fur, feathers, cavity walls and vacuum-flask jackets all work by holding air still. Ash insulates well for the same reason, but its own solid particles still conduct, so it cannot be a better insulator than the air within it.

A caution worth remembering: still air is a superb insulator, yet moving air carries heat away rapidly by convection. Conduction and convection are separate mechanisms, and a question about conductivity is asking only about the first. When the choices span different states of matter, rank them gas, liquid, non-metallic solid, metal in increasing order of conductivity and the answer usually follows at once.

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In a liquid or a gas the molecules are not held in fixed positions, so they move continuously in random directions and collide with one another and with anything suspended in the fluid. A small visible particle, such as a smoke particle in air or a pollen grain in water, is struck unequally from different sides at each instant, and so it jiggles along an irregular zig-zag path. This ceaseless random movement of particles in fluids is called Brownian motion, named after the botanist who first observed it, and it is the standard experimental evidence for the kinetic theory of matter.

The other terms describe something different. Translational motion is one particular type of molecular movement, namely motion of the whole molecule from place to place, and it is only part of the picture; it is not the name given to the observed random movement in fluids, and it says nothing about randomness. Vibrational motion is the to-and-fro oscillation of particles about fixed mean positions, which is characteristic of a solid, where the particles are too tightly packed to wander. An isobaric process is not a kind of motion at all: it is a thermodynamic change that takes place at constant pressure.

A helpful way to keep this straight is to link each state of matter to its dominant motion: solids vibrate about fixed points, while liquids and gases show free random movement, which is Brownian motion. Also note that Brownian motion becomes more vigorous when the temperature is raised or the suspended particle is smaller, because the average kinetic energy of the molecules increases and a lighter particle responds more to each uneven collision.

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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.

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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.

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Net force in the horizontal (x) direction:
\(F_x = 8 \, \text{N} - 4 \, \text{N} = 4 \, \text{N} \quad \text{(to the right)}\)

Net force in the vertical (y) direction:
\(F_y = 15 \, \text{N} - 12 \, \text{N} = 3 \, \text{N} \quad \text{(3 N upward)}\)

Magnitude of the resultant force: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}\)

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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.

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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.

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

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

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

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In a pinhole camera light travels in straight lines through the small hole, so the object, the pinhole and the image form two similar triangles with the pinhole at the common apex. Similar triangles give the magnification directly as a ratio of distances: \[m = \frac{\text{image height}}{\text{object height}} = \frac{\text{image distance }v}{\text{object distance }u},\] where the image distance is simply the length of the camera box, because the screen is the back of the box.

Here the box length gives \(v = 25\ \text{cm}\), and the image is one-fifth the size of the object, so \(m = \frac{1}{5}\). Substituting: \[\frac{1}{5} = \frac{25}{u} \quad \Rightarrow \quad u = 5 \times 25 = 125\ \text{cm} = 1.25\ \text{m}.\] The object stands 1.25 m in front of the pinhole. The result is sensible: an image smaller than the object means the object must be further from the pinhole than the screen is, and here it is five times as far.

The likeliest error is inverting the ratio and writing \(u = 25/5 = 5\ \text{cm}\), which would place the object nearer the pinhole than the screen and would make the image larger, not smaller. A second trap is the unit change: the options are in metres while the camera length is in centimetres, so the final conversion \(125\ \text{cm} = 1.25\ \text{m}\) must be made. Note also that no focal length or lens formula is involved, since a pinhole has no focal length; only the straight-line propagation of light and similar triangles are needed.

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

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

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

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The whole question turns on how cells combine, so begin with the two rules and the reasoning behind them. E.m.f. is energy supplied per unit charge, so when cells are joined in series a single charge is driven through all of them in turn and receives energy from each, giving

When identical cells are joined in parallel a charge passes through only one of them on its way round the circuit, so the total e.m.f. is that of a single cell:

What the parallel grouping reduces is the internal resistance, \(r_{\text{eff}} = r/n\), which is why it is used to obtain a larger current at the same voltage.

The condition stated in the question, that the three e.m.f.s are equal, is the decisive clue. In a series chain the e.m.f.s add whether they are equal or not, so no such assumption would be needed. The assumption is required only for a parallel grouping, because cells of unequal e.m.f. connected in parallel drive current through one another and the combination no longer has a single well-defined e.m.f. With the cells equal, the parallel arrangement has a total e.m.f. equal to that of any one cell, so the relation \(E = E_1 = E_2 = E_3\) is the one that describes it.

The two reciprocal expressions offered can be rejected on principle rather than by inspecting the wiring. Reciprocals of that kind belong to resistors in parallel and to capacitors in series; e.m.f.s never combine reciprocally, because e.m.f. adds as energy per unit charge along a path. In the examination, first decide from the diagram whether charge must pass through every cell (series, so add) or through only one cell (parallel, so take a single cell's value), and never transfer the reciprocal formula from resistance to e.m.f.

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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.

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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.

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The graph is a velocity-time graph with three stages of motion. From 0 to 10 seconds the body moves at a constant velocity of 12 m/s. From 10 to 14 seconds the velocity drops linearly from 12 m/s to 0 m/s, representing uniform deceleration. This final segment is the last stage of the motion.

The distance covered during any stage equals the area under the velocity-time curve for that interval. The last stage forms a right triangle with base \(\Delta t = 14 - 10 = 4\) s and height \(v = 12\) m/s.

Distance = \(\frac{1}{2} \times 4 \times 12 = 24\) m.

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

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

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

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

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

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

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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.

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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.

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For a spring obeying Hooke's law, the elastic potential energy stored is:

\[ E = \frac{1}{2}kx^2 \]

where \( k \) is the spring constant and \( x \) is the compression (or extension). Since the applied force \( F = kx \), we can write \( x = \frac{F}{k} \), and substituting:

\[ E = \frac{1}{2}k\left(\frac{F}{k}\right)^2 = \frac{F^2}{2k} \]

This shows that the energy stored is proportional to the square of the applied force: \( E \propto F^2 \).

For two different forces applied to the same spring:

\[ \frac{E_2}{E_1} = \left(\frac{F_2}{F_1}\right)^2 \]

Substituting the given values:

\[ \frac{E_2}{0.16} = \left(\frac{700}{200}\right)^2 = (3.5)^2 = 12.25 \]

\[ E_2 = 0.16 \times 12.25 = 1.96 \text{ J} \]

The spring stores 1.96 J of energy when compressed by 700 N.

The critical insight is that energy depends on the square of the force, not linearly. Tripling the force does not triple the energy - it increases it by a factor of nine.

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

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

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

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

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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.

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

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

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

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Specific heat capacity is the heat needed to raise the temperature of one kilogram of a substance by one kelvin. It comes from the heat equation \[Q = mc\Delta\theta,\] where \(Q\) is the heat supplied in joules, \(m\) the mass in kilograms and \(\Delta\theta\) the temperature rise. Because heat loss to the surroundings is stated to be negligible, all 1000 J supplied goes into the rod, so no correction is needed.

Making \(c\) the subject and substituting: \[c = \frac{Q}{m\Delta\theta} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\ \text{J kg}^{-1}\text{K}^{-1}.\] So the specific heat capacity is \(2666.7\ \text{J kg}^{-1}\text{K}^{-1}\).

Note that the temperature rise needs no conversion. A change of \(15\ ^\circ\text{C}\) is a change of 15 K because the two scales have the same size of degree, so adding 273 here is a wasted step that produces a badly wrong answer. The other frequent slip is working out the denominator carelessly: \(0.025 \times 15 = 0.375\), not 0.0375 or 3.75. Exam reminder: distinguish specific heat capacity \(c\), measured in \(\text{J kg}^{-1}\text{K}^{-1}\), from heat capacity \(C = mc\), measured in \(\text{J K}^{-1}\); the units in the options tell you which one is wanted.

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

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

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

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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.