JAMB Physics

C: v/f − 1

For a real image, using positive object and image distances, 1/f = 1/u + 1/v. Multiplying by v gives v/f = v/u + 1. The magnitude of linear magnification is v/u = v/f − 1; the image itself is inverted.

D: Frequency

For a pure tone, pitch is primarily determined by frequency: higher frequency is heard as higher pitch. Timbre or quality distinguishes sounds with different waveform and harmonic content, rather than defining the frequency of a pure tone.

B: Be parallel

Two nonzero vectors with an angle of 0° point in the same direction and are parallel. Their cross product is zero and their dot product equals the product of their magnitudes; perpendicular vectors instead have a 90° angle.

A: I, II, III and IV

All four involve equal and opposite forces on different bodies. Engines push exhaust backwards and receive forward thrust; a gun pushes a bullet forward and recoils. When walking, the foot pushes the ground backwards and the ground exerts forward friction on the foot.

B: 4 m²

The transmitted pressure is F₁/A₁ = 40/0.4 = 100 Pa. The load piston must have area A₂ = F₂/P = 400/100 = 4 m². This gives the same pressure at both pistons in the ideal hydraulic press.

D: 13 N rightwards

Take rightwards as positive. The horizontal components add to −4 + 10 + 8 cos 60° + 6 cos 60° = −4 + 10 + 4 + 3 = 13 N. The angled forces both contribute rightwards, although their vertical components oppose each other.

C: X_L = X_C

At series resonance the inductive and capacitive reactances cancel. Since the net reactance is X_L − X_C, it vanishes when X_L = X_C. The ideal series impedance is then purely resistive.

A: λ = h/p

The de Broglie relation is λ = h/p, where p is particle momentum and h is Planck’s constant. It associates a wavelength with matter. The formulas c/f and hc/E apply directly to photons in vacuum, and 2d sin θ describes first-order Bragg diffraction.

D: Their resistance decreases with temperature

A negative temperature coefficient means resistance falls as temperature rises over the stated operating range. In many semiconductors, heating creates additional mobile carriers, which can increase conductivity and reduce resistance.

B: Excitation of the gas followed by fluorescence of the phosphor coating

An electric discharge excites mercury atoms in the tube. They emit ultraviolet radiation, which the phosphor coating absorbs and re-emits as visible light. The process uses excitation and fluorescence rather than simply heating a filament until it glows.

C: Minority carriers

Before reverse breakdown, the small reverse current in a junction diode is associated with minority carriers: electrons on the p side and holes on the n side. The reverse field sweeps these carriers across the junction.

D: 9.0 × 10⁻² J

The magnetic energy stored in a linear inductor is U = ½LI². With L = 5 × 10⁻³ H and I = 6 A, U = ½ × 0.005 × 36 = 0.090 J, or 9.0 × 10⁻² J.

A: Have wavelengths comparable to atomic-plane spacings

X-ray wavelengths can be comparable to the spacing between atomic planes in a crystal. The resulting diffraction pattern depends on that spacing and arrangement, allowing crystal structure to be studied. Speed and invisibility alone do not provide this spatial resolution.

C: 1.2 A

The net reactance is 16 − 10 = 6 Ω. Thus Z = √(8² + 6²) = 10 Ω, and the rms current is I = V/Z = 12/10 = 1.2 A. Reactances subtract because their voltage contributions have opposite phases.

B: Polarization

Polarization selects a direction of transverse oscillation. A purely longitudinal wave oscillates along its direction of propagation and cannot be polarized in this way. Diffraction, reflection and refraction can occur for both types of wave.

D: Increase in the volume of water

Water expands when it freezes into ordinary ice, whose open molecular arrangement gives it a lower density. In a full, rigid bottle there is insufficient space for the increased volume, so the resulting pressure can crack the bottle.

A: 0.05 m³

The ice volume is 450/900 = 0.50 m³. The same mass of water occupies 450/1000 = 0.45 m³. The volume therefore decreases by 0.50 − 0.45 = 0.05 m³; mass is conserved during melting.

C: 250 N

The first extension is 20.01 − 20.00 = 0.01 m. The required extension is 0.05 m, five times larger. Within the Hooke’s-law range, force is proportional to extension, so the required force is 5 × 50 = 250 N.

B: Radiation

A shiny silver surface has low thermal emissivity, so it emits less thermal radiation than a dull surface at the same temperature. The coating does not remove heat transfer by air convection or conduction through the pot and its supports.

A: 500 J kg⁻¹ K⁻¹

In 10 minutes the heater supplies Q = Pt = 50 × 600 = 30 000 J. Using Q = mcΔT gives c = 30 000/(5 × 12) = 500 J kg⁻¹ K⁻¹. A rise of 12°C equals a rise of 12 K.

B: Electrophorus

An electrophorus uses a charged insulating base and a removable conducting plate. Bringing the plate near the base, grounding it, disconnecting the ground and lifting it produces a charged plate. The sequence can be repeated to obtain charges of the same sign.

A: Characteristic X-rays arise from electronic transitions; nuclear gamma rays arise from nuclear transitions

Characteristic X-rays arise from transitions involving atomic electrons. Nuclear gamma rays arise from transitions between nuclear energy levels. Both are uncharged electromagnetic radiation, and their energy ranges can overlap, so origin is the useful distinction here.

B: 1 : 2

Both parallel resistors have the same 12 V across them. Their powers are P₆ = 12²/6 = 24 W and P₃ = 12²/3 = 48 W. Thus P₆ : P₃ = 24 : 48 = 1 : 2.

A: Grape juice

Grape juice contains mobile ions from dissolved acids and mineral salts, so it conducts electricity as an electrolytic solution. Pure sugar solution is molecular rather than ionic, while pure alcohol and paraffin do not supply comparable mobile ions.

C: 0.1 μF

Connecting all three capacitors in series gives the smallest equivalent capacitance. In μF units, 1/C = 1/0.3 + 1/0.5 + 1/0.2 = 31/3. Thus C = 3/31 ≈ 0.0968 μF, which rounds to 0.1 μF to one decimal place.

C: Surface tension

Water can form curved films across the small spaces in a fabric. Surface tension helps those films resist passage of water; touching the inside can disturb the film and promote wetting through the fabric.

A: 20 N

The fulcrum is 45 cm from either end. The load arm is 45 −15 =30 cm, while the effort arm is 45 cm. Balancing moments gives F ×45 =30 ×30, so F =20 N.

B: 5.1 ×10³ m/s

For longitudinal waves in a slender elastic bar, v =√(Y/ρ). Here v =√(7.0 ×10¹⁰/2.7 ×10³) ≈5092 m/s, or 5.1 ×10³ m/s to two significant figures.

C: The temperatures of both objects are equal

Objects in thermal equilibrium have equal temperatures and no net heat transfer between them. Their heat capacities, masses and internal energies can still differ.

B: 0.25

For a point source radiating uniformly without absorption, intensity is power divided by 4πr². Doubling distance makes the area four times larger, so intensity becomes one quarter of its original value.

D: Infinitely many

In the ideal model, two parallel facing mirrors reflect the object and each other’s images repeatedly. There is no final reflection order, so the construction gives infinitely many images.

D: 4.0 ×10⁴ J/K

The heater supplies Q =Pt =2000 ×600 =1.2 ×10⁶ J. Heat capacity of the whole object is C =Q/ΔT =1.2 ×10⁶/30 =4.0 ×10⁴ J/K. Mass is needed for specific heat capacity, not total heat capacity.

A: 2.5

The magnitude of linear magnification is image distance divided by object distance. Thus |m| =25/10 =2.5;the real image is inverted but 2.5 times as tall.

B: 1.0 ×10² m²

Fourier’s conduction law gives heat rate kA times the temperature-gradient magnitude. Therefore A =(1.2 ×10⁶/1)/(400 ×30) =100 m².

A: 57%

Relative humidity is actual vapour pressure divided by saturation vapour pressure at the same temperature, then multiplied by 100. Thus RH =100 ×10/17.50 ≈57.14%, or 57% to the nearest whole percent.

D: At the principal focus

In the paraxial model, a point at the principal focus of a concave mirror reflects into parallel rays. The image distance is therefore infinite.

D: v/(4 l)

The fundamental mode of a pipe closed at one end has a displacement node at the closed end and an antinode at the open end. Its length is one quarter wavelength, so f =v/(4 l).

B: 27 W

The single series 2 Ω resistor carries the full currentI, while each parallel 2 Ω branch carriesI/2. Its 18 W limit givesI² ×2 =18, orI =3 A. The branches each dissipate 4.5 W, totaling 27 W.

D: Blue and red

Red and blue occupy well-separated regions of the visible spectrum. Each other pair listed consists of neighbouring colours, so blue and red span the widest spectral interval among these choices.

D: Leclanché cell

The traditional Leclanché cell was widely used for intermittent electric-bell circuits. Rest periods allow it to recover from polarization, making brief intermittent operation suitable.

A: Narrows

Forward bias opposes the junction’s built-in electric field and lowers its potential barrier. The depletion region narrows, allowing greater carrier injection across the junction.

D: Binding energy

A bound nucleus has less rest mass than its separated nucleons. The mass defect multiplied byc² is its binding energy, the energy required to separate it completely into those nucleons.

C: 5.28 ×10⁻¹⁹ J

An emitted photon has energy E =hf. With h =6.6 ×10⁻³⁴ J·s and f =8.0 ×10¹⁴ Hz, E =5.28 ×10⁻¹⁹ J. This equals the electron’s energy-level decrease.

D: 9.0 ×10⁹ J

A mass loss of 0.01% of 1.0 g is 10⁻⁴ ×10⁻³ =10⁻⁷ kg. Hence E =Δmc² =10⁻⁷(3.0 ×10⁸)² =9.0 ×10⁹ J.

C: 600 W

For sinusoidal current of peak 10 A, the mean square current is 10²/2 =50 A². Average resistor power is I_RMS²R =50 ×12 =600 W.

A: Metal resistance increases with temperature, while intrinsic semiconductor resistance decreases

For ordinary metals near room temperature, heating increases lattice scattering and resistance. In an intrinsic semiconductor, heating creates more charge carriers, so resistance generally decreases over that range.

A: 0.05 A

Faraday’s law gives Q = m/z = 0.990/(3.3 × 10⁻⁴) = 3000 C. The deposition time is 2400 s, so the actual current is 1.25 A. Add 1.25 − 1.20 = 0.05 A to the meter reading.

C: 0.34 V

The stopping potential satisfies eV_s = K_max. An electron with kinetic energy 0.34 eV is stopped by a potential difference of 0.34 V. The work function helps determine the incident photon energy, but is not added to the stopping potential.

D: 30°

The magnetic force magnitude is F = |q|vB sin θ. Half the maximum force requires sin θ = 1/2. Of the listed angles, 30° satisfies this condition; 150° would also do so if it were offered.

A: 50 counts/min

Sixteen days contains 16/4 = 4 half-lives. The count rate due to the sample therefore falls to 800 × (1/2)⁴ = 50 counts per minute, with the same detector arrangement.

A: 1.5 m

For the stated hydrostatic model, ΔP =ρgh. Thus h =(1.65 ×10⁴)/(1.1 ×10³ ×10) =1.5 m.

A: Be constant

The net force equals the time rate of change of momentum: F_net =dp/dt. If the net force remains zero, momentum has zero rate of change and stays constant.

B: 7.5 ×10⁻³ J

Between extensions 0.05 m and 0.10 m, force rises linearly from 0.10 N to 0.20 N. The added elastic energy is the trapezium area:½(0.10 +0.20)(0.10 −0.05) =0.0075 J.

A: One ninth

Gravitational attraction varies inversely with separation squared. Replacing r by 3 r gives F_new/F_old =r²/(3 r)² =1/9. The masses remain unchanged.

C: Fx/2 J

Work is the area under the force-distance graph. A force rising linearly from zero toF over distancex forms a triangle, so W =½ ×x ×F =Fx/2.

C: 10√2 N,45° east of north

North and east components are perpendicular and both 10 N. The magnitude is√(10² +10²) =10√2 N, and equal components give 45° east of north.

D: Astigmatism

Regular astigmatism gives different focusing powers in different directions across the eye. A cylindrical lens supplies direction-dependent power to compensate for that difference.

D: Energy

A travelling wave transfers energy from one region to another. Amplitude, wavelength and frequency describe the wave, while energy is what the propagating disturbance transports.

A: 264 ms-1

The fundamental mode of a string fixed at both ends has half a wavelength along its length. Thus λ =2 L =1.2 m and v =fλ =220 ×1.2 =264 m/s.

D: Convection

Air flowing over radiator fins carries heat away by forced convection. Heat first reaches the fin surfaces through the metal, but its transfer from the surface into moving air is convective.

A: Reducing vapour partial pressure in the air just above the liquid

Moving relatively dry air sweeps vapour away from just above the liquid. This lowers the local vapour partial pressure and reduces condensation back onto the liquid, increasing net evaporation.

C: 7.47 ×10⁶ Pa

Using 27°C ≈300 K, the ideal-gas law gives P =nRT/V =(3 ×8.3 ×300)/10⁻³ =7.47 ×10⁶ Pa. Temperature must be in kelvin.

A: Between the principal focus and the pole

With an object inside the focal length of a concave mirror, reflected rays diverge and their backward extensions form an upright virtual image. Its magnification is greater than one.

C: Interference

Light reflected from the front and back surfaces of the thin soap film takes different optical paths. Their interference reinforces some wavelengths and suppresses others, producing colours that vary with film thickness.

A: Reverberation

Reverberation is the fading persistence of sound caused by many overlapping reflections after the source stops. A distinct delayed repetition is called an echo.

B: Difference in temperature between the body and its surrounding

Newton’s law of cooling models heat-loss rate as H =k(T_body −T_surroundings) when the heat-transfer coefficient is effectively constant. It is the temperature difference that drives the heat flow.

D: 52.9 Ohms

For a resistive element, P =V²/R. Therefore R =230²/1000 =52.9 Ω at its rated operating condition.

A: Both X and Y are positive

Electric field lines point away from positive charges and toward negative charges. Since the arrows emerge from bothX andY, both charges are positive.

A: Iris

Muscles in the iris change the pupil’s diameter, controlling how much light enters the eye and reaches the retina. The iris therefore acts as the adjustable aperture.

A: 5 ×10⁻⁴ J

Capacitor energy is U =Q²/(2 C). With Q =100 ×10⁻⁶ C and C =10 ×10⁻⁶ F, U =(10⁻⁴)²/(2 ×10⁻⁵) =5 ×10⁻⁴ J.

A: 15 Ω

Total resistance is 12/2 =6 Ω. The parallel pair therefore has 6 −2 −1.5 =2.5 Ω. Solving 1/2.5 =1/3 +1/X gives 1/X =1/15, so X =15 Ω.

B: Neutron

A neutron absorbed by a suitable fissile nucleus can trigger fission. The resulting fission releases more neutrons, which can sustain the chain reaction in a reactor.

D: 800/π Hz

Capacitive reactance is X_C =1/(2πfC). Thus f =1/(2π ×250 ×2.5 ×10⁻⁶) =800/π Hz.

B: 3 %

After five half-lives, the fraction of original undecayed nuclei is(1/2)⁵ =1/32. As a percentage this is 3.125%, which rounds to 3% to the nearest whole percent.

A: Neutron

Conservation of nucleon number gives 23 +A_X =20 +4, so A_X =1. Conservation of charge gives 11 +Z_X =9 +2, so Z_X =0. A particle with mass number 1 and charge number 0 is a neutron.

B: A strong permanent magnet producing a strong field

A stronger magnetic field gives a larger turning effect on the current-carrying coil. This improves current sensitivity when coil turns, area and spring stiffness are unchanged.

B: Trivalent atoms

Trivalent dopants have one fewer valence electron than silicon. They introduce acceptor states and holes, making holes the majority carriers in p-type silicon.

A: Resistive heating in the transmission conductors

For the same transmitted power, increasing voltage reduces current because I =P/V. Resistive heating in the transmission conductors is I²R, so reducing current sharply reduces this loss.

B: 2.0 ×10⁻²⁸ J

Photon energy is E =hf =(6.63 ×10⁻³⁴)(3.0 ×10⁵) =1.989 ×10⁻²⁸ J. To two significant figures this is 2.0 ×10⁻²⁸ J.

D: Resistance

Sound pressure changes the contact between carbon granules, varying their electrical resistance. With an applied bias, this resistance change produces a corresponding electrical signal.

A: T

For small oscillations, a simple pendulum’s natural period depends on its length, T =2π√(L/g). T has the same length asP, so it has the same natural frequency and resonates withP.

D: 6 cm

The displaced volume equals the sphere volume:πR² ×1 =(4/3)π ×3³. Cancellingπ gives R² =36, hence the cylinder radius is 6 cm.

D: Velocity ratio exceeds mechanical advantage

For a machine, efficiency =MA/VR. Energy losses make efficiency less than one, so mechanical advantage MA must be less than velocity ratio VR.

C: Conduction and convection

Heat reaches the ice by conduction at contacting surfaces and by convection as water circulates and carries thermal energy. Conduction through the glass also helps it cool toward the water temperature.

D: Erection of the image

A simple astronomical telescope gives an inverted view. The extra erecting lens in a traditional terrestrial telescope inverts that intermediate image again, producing an upright view.

B: 4/9 m

A convex mirror has focal length−R/2 =−0.5 m. From 1/f =1/u +1/v with u =4 m,1/v =−2 −1/4 =−9/4, so v =−4/9 m: the image is 4/9 m behind the mirror.

A: cosα/sinβ

The incident angle from the normal is 90° −α, becauseα is measured from the surface. Snell’s law for air gives n =sin(90° −α)/sinβ =cosα/sinβ.

C: Helium

Helium gas is used in primary gas thermometry because its pressure, volume or sound speed can be related to thermodynamic temperature. It remains gaseous over a wide range and can be studied in the dilute-gas limit.

B: 9/11 A

The 10 Ω and 5 Ω branches have equivalent resistance 10/3 Ω. In series with 3 Ω and 1 Ω, the total is 22/3 Ω. Thus I =6/(22/3) =9/11 A.

D: 1 : 2

Both parallel capacitors have the same voltage. Since energy is½CV², their energy ratio is C_P/C_Q =2/4 =1/2, or 1:2.

D: 6 V, 2 V and 4 V

In an initially uncharged series chain, all three capacitors acquire equal charge. The equivalent capacitance is 1 μF, giving charge 12 μC. Their voltages are 12/2 =6 V,12/6 =2 V and 12/3 =4 V.

A: Copper sulfate, copper and zinc

In a Daniell cell, zinc is oxidized at the negative electrode and copper ions are reduced at the positive copper electrode. Copper sulfate supplies those ions and is traditionally described as the depolarizer.

D: 0.0 E

For a uniform electric field, equal flux enters and leaves opposite faces of a closed cube. The signed contributions cancel, while the other faces have zero normal component. Net outward flux is zero.

A: Increase

Glass polarizes in the electric field. For a given plate charge this reduces the potential difference, so C =Q/V increases when glass replaces air between plates at unchanged geometry.

D: e²/(4πε₀Gm²)

The electric force magnitude is e²/(4πε₀d²), while gravity gives Gm²/d². Dividing cancels d², leaving F_E/F_G =e²/(4πε₀Gm²).

B: 48 A

The series resistance is 4 Ω and inductive reactance 3 Ω, so impedance magnitude is √(4² +3²) =5 Ω. The RMS current is 240/5 =48 A.

D: √2 A

The d.c. heating equivalent is the RMS current. For i =2 sin(ωt) A, the peak is 2 A, so I_RMS =2/√2 =√2 A.

B: Nuclear fusion

The Sun’s core releases energy mainly by nuclear fusion, combining hydrogen nuclei into helium. The products have slightly less total rest mass, with the difference released as energy.

D: At right angles to the field

For a straight conductor in a uniform magnetic field, F =BIL sinθ. With B,I andL fixed, the force is largest at θ =90°, where sinθ =1.

A: I, II and IV

In the simple dry-sliding-friction model, F =μN. The coefficient depends on the contacting materials; force is approximately independent of apparent area and, over the model’s range, sliding speed.

C: Decrease

A floating ship displaces water whose weight equals its own. Fresh water is less dense than salt water, so a larger submerged volume is needed. The volume remaining above the surface decreases.

D: II and IV

At constant speed the kinetic energy is unchanged, so the net work is zero. Velocity still changes direction, and the net centripetal force points toward the centre. Therefore II and IV are correct.

C: 20 m/s²

Instantaneous acceleration is the derivative of velocity: a =dv/dt =4 t. At t =5 s this is a =4 ×5 =20 m/s². The constant 10 contributes no acceleration.

C: Reduced to a quarter

Mechanical power is P =Fv for force parallel to motion. Halving both quantities gives P_new =(F/2)(v/2) =P/4.

B: 40 N

Efficiency equals mechanical advantage divided by velocity ratio. Thus MA =0.75 ×5 =3.75, and effort =load/MA =150/3.75 =40 N.

A: Alcohol, mercury and alcohol

In the traditional mercury-and-alcohol Six thermometer, alcohol in the sensing side expands and displaces the mercury in the U-bend. The liquid above the mercury on the other side is also alcohol; the expansion space is farther up in its bulb.

A: Blue, yellow and red

In a normally dispersive medium, blue light has a higher refractive index than yellow or red, so it bends closest to the normal. Yellow is intermediate and red is farthest from the normal.

B: 100 μC and 200 μC

Parallel capacitors each have the full 20 V. Using Q =CV gives Q_P =5 ×20 =100 μC and Q_Q =10 ×20 =200 μC.

C: Astronomical telescope

A refracting astronomical telescope uses a long-focal-length converging objective and a short-focal-length converging eyepiece. The 500 cm and 5 cm lenses fit that arrangement.

D: 1 V and 2/3 V

Total resistance is 3 +2 +1 =6 Ω including the cell. Current is 2/6 =1/3 A, giving V₁ =(1/3) ×3 =1 V and V₂ =(1/3) ×2 =2/3 V.

A: 200 Ω

Inductive reactance is X_L =2πfL. With f =50/π Hz and L =2 H, X_L =2π ×(50/π) ×2 =200 Ω.

C: 20 years

The remaining fraction 1/64 equals(1/2)⁶, so six half-lives have elapsed. The half-life is 120/6 =20 years.

A: Equal to the number of holes in the valence band

In an intrinsic semiconductor, thermal excitation of one valence electron into the conduction band leaves one hole behind. Thus conduction-electron and valence-hole populations are equal.

C: 2.0 V

The maximum photoelectron energy is photon energy minus work function:5 −3 =2 eV. A stopping potential of 2 V removes 2 eV from an electron, bringing the fastest electrons to rest.

B: 16.0 V

Faraday’s law gives emf magnitude N|dΦ/dt|. With 200 turns and a flux change of 0.08 Wb/s through each turn, the induced emf is 200 ×0.08 =16 V.

C: P

The fuse belongs in series with the live conductor. If it opens, it disconnects the device from the live supply. A fuse in the neutral can leave the device live, while the protective earth must remain continuous.

D: 80.0 W

The line current is I =P/V =3000/1500 =2 A. The total cable loss is I²R =2² ×20 =80 W.

A: It tips if the vertical line through G falls outside its base of support

If the vertical line of weight falls outside the base, gravity produces a turning moment about the downhill edge that tips the solid. Inside the base, a supporting reaction can balance its weight.

C: 2 : 3

Newton’s second law gives F₁/F₂ =(m₁/m₂)(a₁/a₂) =3 ×2/9 =2/3. The forces therefore have ratio 2:3.

C: 4 m/s², 2 m/s²

The rising slope is (80 −0)/(20 −0) =4 m/s². The falling slope is (0 −80)/(90 −50) =−2 m/s², so the retardation magnitude is 2 m/s².

D: Vernier calipers with inside jaws

The inside jaws of vernier calipers can contact opposite inner walls of the tube and measure the separation directly. An ordinary outside micrometer cannot measure that internal diameter.

B: P is open while Q is closed

On the downstroke, water pressure below the piston closes the inlet valve Q and opens the piston valve P. Water passes upward through the piston instead of returning down the inlet.

D: 2.0 ×10⁹ Pa

The strain is (10.05 −10.00)/10.00 =0.005. Young’s modulus is stress divided by strain: E =10⁷/0.005 =2.0 ×10⁹ Pa.

A: I, II and III

Boiling occurs when saturated vapour pressure reaches the external pressure. Raising external pressure raises boiling temperature. A nonvolatile solute lowers solvent vapour pressure and raises the solution’s boiling point.

A: Decreases

For a pure liquid such as water, surface tension generally decreases as temperature rises. Greater thermal motion reduces the free-energy cost per unit area of forming its liquid surface.

A: Increases

Sound travels faster in hotter air of the same composition. Its frequency remains fixed by the source across a stationary boundary, so λ =v/f increases when the speed increases.

C: 5.00 ×10² rad/s

Compare y =A sin(ωt −kx) with the given equation. The coefficient of time in the phase is the angular frequency, so ω =500 rad/s =5.00 ×10² rad/s.

A: 8/33 kg

The copper releases Q =mcΔT =2 ×400 ×100 =80,000 J while cooling to 0°C. Melting needs mᵢL =Q, so mᵢ =80,000/330,000 =8/33 kg.

D: 3.0 ×10⁹ Hz

Three complete cycles occupy 0.30 m, so the wavelength is 0.30/3 =0.10 m. Hence f =v/λ =(3.0 ×10⁸)/0.10 =3.0 ×10⁹ Hz.

B: 3.0 ×10³ K/m

The temperature difference is 80 −20 =60 K. The magnitude of the gradient is ΔT/Δx =60/0.02 =3000 K/m =3.0 ×10³ K/m.

D: 1.5 V, 1 Ω

Identical cells connected in parallel with matching polarities retain the emf of one cell, 1.5 V. Their internal resistances act in parallel, giving r =4/4 =1 Ω.

D: Final image is at infinity.

Normal adjustment places the intermediate image at the eyepiece’s focal plane, so rays from each image point emerge parallel. The final image is therefore at infinity and a relaxed eye can view it.

D: Retains magnetism more strongly

Hardened steel retains substantial magnetization after the magnetizing field is removed and resists demagnetization. Soft iron is easier to magnetize and demagnetize, making it better suited to temporary magnetic cores.

C: Periscope

Two right-angle prisms can reflect a light path down from a higher viewing point and then horizontally to a lower observer. This displaced viewing path is the basic arrangement of a prism periscope.

D: 0.75

Snell’s law gives sin i/sin r =v_air/v_glass =3/2. With r =30° and sin 30° =0.5, sin i =(3/2) ×0.5 =0.75.

B: 10 V

Resistor voltage is in phase with current, while ideal-inductor voltage leads current by 90°. The supply RMS voltage is their phasor sum: √(6² +8²) =10 V.

D: Reduce the heat produced by reducing the eddy current.

Insulated laminations interrupt large circulating paths in the metal core. This reduces eddy currents and their resistive heating, improving transformer efficiency.

A: Motion, magnetic field and induced current

For Fleming’s generator right-hand rule, the thumb indicates conductor motion, the first finger the magnetic field, and the second finger the induced conventional current. The three directions are mutually perpendicular.

C: 2.5 ×10⁻² J

Both parallel capacitors have 100 V across them. Their total stored energy is ½(C₁ +C₂)V² =½(2 +3) ×10⁻⁶ ×100² =0.025 J.

C: 100/π Hz

Inductive reactance is X_L =2πfL. Rearranging gives f =2000/(2π ×10) =100/π Hz, approximately 31.8 Hz.

A: R₂/R₁

The resistors are in parallel, so each has the cell voltage E across it. Since P =V²/R, the ratio is P₁/P₂ =(E²/R₁)/(E²/R₂) =R₂/R₁.

B: Electrons and holes

Conduction-band electrons and mobile holes in the valence band both carry current in ordinary semiconductors. A hole represents a missing valence electron and behaves as a positive charge carrier.

B: 66.30 m/s

De Broglie’s relation is λ =h/p. At this low speed p =mv, so v =h/(mλ) =(6.63 ×10⁻³⁴)/(10⁻²⁷ ×10⁻⁸) =66.3 m/s.

B: 40 V

At series resonance, X_L =X_C. The same current passes through both components, so their voltage magnitudes are equal and opposite in phase. Thus the capacitor voltage is 40 V.

B: An excited nucleus changes to a lower energy state

In nuclear gamma emission, an excited nucleus changes to a lower energy state and releases the energy difference as a photon. This is distinct from transitions among atomic electron energy levels.

B: Momentum

Impulse is force multiplied by time, with dimensions MLT⁻² ×T =MLT⁻¹. Momentum is mass multiplied by velocity and has the same dimensions.

D: 9.0 ×10⁻⁴ N

Gravitational force varies inversely with separation squared. Reducing distance from 0.18 m to 0.06 m multiplies the force by(0.18/0.06)² =9, giving 9.0 ×10⁻⁴ N.

A: III and IV only

For fixed force, a =F/m. Thus a plotted against m is a decreasing inverse curve, while a plotted against 1/m is a straight increasing line through the origin.

B: 25 N

Velocity ratio is wheel diameter divided by axle diameter:40/8 =5. Efficiency equals mechanical advantage divided by velocity ratio, so MA =0.8 ×5 =4. Effort is 100/4 =25 N.

C: 2 ×10⁻³

Strain is extension divided by original length. Thus strain =(2.0 ×10⁻⁵)/0.01 =2.0 ×10⁻³. It has no unit because both lengths use metres.

B: I, II and IV only

For static liquids, pressure at a point is isotropic. The pressure rise below a free surface isρgh: it increases with depth, is proportional to density at a fixed depth, and does not depend on container shape.

B: 1.4 ×10⁶ Pa

The water addsρgh =1000 ×10 ×10 =100000 Pa. Adding the specified surface pressure gives 1.3 ×10⁶ +0.1 ×10⁶ =1.4 ×10⁶ Pa.

C: 15.09 cm³

For small isotropic expansion, volume expansivity is 3α. Thus V =15.00[1 +3(2 ×10⁻⁵)(100)] =15.00(1.006) =15.09 cm³.

A: 200 g

Melting the ice requires 0.042 ×3.4 ×10⁵ =14280 J. Water cooling from 17 °C to 0 °C supplies 71400 J per kilogram, so the minimum mass is 14280/71400 =0.200 kg =200 g.

D: Molecules are identical and very small compared with their separation

In the elementary kinetic model of a pure ideal gas, particles are identical and their individual size is negligible compared with their separation. Their collision rate can change with temperature and density.

A: Reflection and diffraction

Ordinary sound in air is longitudinal. It reflects from boundaries and diffracts around obstacles, but it cannot be polarized like a transverse wave.

C: 240√3 m/s

For the same ideal gas, sound speed is proportional to√T. The kelvin temperatures are 300 and 400, so v₂ =360√(400/300) =240√3 m/s.

B: The open end

At the open end, air can move most freely, giving a displacement antinode. The closed end prevents air movement and is a displacement node; pressure nodes and antinodes are reversed.

B: Magnified and erect

An object between a concave mirror and its focus forms an upright virtual image behind the mirror. The image is larger than the object.

D: Parabolic mirror.

A paraboloid sends rays from a small source at its focus into parallel rays, avoiding the spherical aberration of a spherical reflector. This makes the parabolic shape suitable for a narrow searchlight beam.

C: t(1 − 1/n)

Near normal viewing through glass gives apparent depth t/n. The apparent upward shift is actual depth minus apparent depth: d =t −t/n =t(1 −1/n).

B: I, II and IV only

Infrared is outside human visible wavelengths and is commonly associated with thermal radiation. It is a transverse electromagnetic wave, with lower frequency than blue light.

C: 9.0 ×10² N

Coulomb’s law gives F =kq₁q₂/r². Thus F =(9 ×10⁹)(5 ×10⁻⁶)(8 ×10⁻⁶)/(0.02)² =900 N. Both charges are positive, so the force is repulsive.

A: Ebonite has a negative charge while glass has a positive charge

In the usual dry charging experiment, ebonite gains electrons from fur and becomes negative; glass loses electrons to silk and becomes positive. Opposite charges attract.

C: 4.0 Ω

The three 2 Ω resistors make a 6 Ω series path. This is in parallel with 3 Ω, giving 2 Ω. The two 1 Ω lead resistors are in series with it, so total resistance is 1 +2 +1 =4 Ω.

C: 200 J

Work needed per charge is the potential rise. Moving 2 C through 100 V requires W =qΔV =2 ×100 =200 J.

A: Minimize eddy-currents

Separate insulated iron wires interrupt large circulating current paths in the core. This reduces eddy currents and the heating losses they would cause as magnetic flux changes.

D: 1.8 ×10⁷ J

At 100 A, cable heating power is I²R =100² ×0.5 =5000 W. Sustaining this for 3600 s gives energy 5000 ×3600 =1.8 ×10⁷ J.

D: 2500/π Hz

In a series RLC circuit, current resonance occurs when ωL =1/(ωC). Here LC =(8 ×10⁻³)(5 ×10⁻⁶) =4 ×10⁻⁸, so f =1/(2π√LC) =2500/π Hz.

D: 2.6 ×10⁻³ kg

Deposited mass equals electrochemical equivalent multiplied by charge: m =zQ =(1.3 ×10⁻⁷)(2.0 ×10⁴) =2.6 ×10⁻³ kg.

D: 9.0 ×10¹⁰ J

Released energy is the decrease in rest mass multiplied by c². Thus E =10⁻⁶(3 ×10⁸)² =9 ×10¹⁰ J.

B: 1, 1, −2

Force equals mass times acceleration. Mass contributes M and acceleration LT⁻², so force has dimensions M¹L¹T⁻². The exponents are 1,1,−2.

C: 84 m/s

At 2 s, x =10 +12(2²) =58 m; at 5 s, x =310 m. Position increases throughout, so distance travelled is 252 m and average speed is 252/3 =84 m/s.

A: 5.0 m/s²

The downslope component of weight is mg sin 30°. Dividing by the mass gives acceleration g sin 30° =10 ×0.5 =5.0 m/s².

D: 3.0 m/s

Horizontal momentum is conserved as the arrow embeds in the block. Thus 0.1 ×15 =(0.1 +0.4)v, giving v =1.5/0.5 =3.0 m/s.

D: Net force and acceleration

For a fixed mass, net force F =ma. Force and acceleration therefore have the same sign and reach their extrema together. In SHM, displacement is opposite in phase to acceleration, while velocity differs by a quarter cycle.

D: 60°

Vertical equilibrium requires the upward component 12 cosθ to equal 6 N. Thus cosθ =1/2 and θ =60°. The horizontal component is 12 sin 60° =6√3 N, matching the rightward force.

B: mg sinθ and friction

Weight resolves into mg sinθ down the slope and mg cosθ perpendicular to it. Friction acts up the slope against sliding; the normal reaction is perpendicular, so it contributes no force along the plane.

C: 30 J

At 10 m height, kinetic energy is½ ×0.1 ×20² =20 J and gained potential energy is 0.1 ×10 ×10 =10 J. Initial kinetic energy is their sum,30 J.

C: 80.0%

Useful output energy is mgh =200 ×10 ×6 =12000 J. Electrical input is 1500 ×10 =15000 J. Efficiency is 12000/15000 ×100% =80%.

A: tanθ = coefficient of static friction

At the point of slipping, downslope weight W sinθ equals limiting friction μₛW cosθ. Dividing by W cosθ gives tanθ =μₛ.

D: 53.2 kg

The fuel volume is 60 L =0.060 m³. Its density is 0.72 ×1000 =720 kg/m³, so fuel mass is 43.2 kg. Adding the 10 kg empty tank gives 53.2 kg.

C: 75.0 °C

The 100 °C interval corresponds to 225 −25 =200 mm. A 175 mm reading is 150 mm above zero, so temperature is 150/200 ×100 =75.0 °C.

D: 127 °C

At fixed volume, absolute pressure is proportional to absolute temperature. Initially T =27 +273 =300 K. The final temperature is 300 ×4/3 =400 K, or 127 °C.

A: Latent heat is absorbed, the mass remains constant and the volume decreases.

Melting ice absorbs latent heat while conserving its mass. Ordinary ice is less dense than liquid water near 0 °C, so the same mass occupies less volume after melting.

D: Temperature falls, remains constant for an interval, then falls again

A pure molten substance first cools to its freezing point. Temperature remains constant while latent heat is released during solidification, then falls again once the solid cools.

B: can be refracted and diffracted

Water ripples and light can both bend when propagation conditions change and spread around edges or through openings. These are refraction and diffraction.

A: 3 m

With tension and mass per unit length fixed, the same string mode has frequency inversely proportional to length. Hence 500 ×0.90 =150 L, giving L =3 m.

B: Interference

Light reflected from the upper and lower film surfaces follows different optical paths. Their waves reinforce some wavelengths and cancel others, producing the visible colours.

C: constructively to produce a wave of a larger amplitude

When two wave profiles have crests, troughs and zero crossings aligned, their displacements have the same sign at each position. Adding them produces a larger amplitude: constructive interference.

A: Decrease the vibrating length of the wire

Immersing the hanging load adds upthrust and reduces the tension in the sonometer wire. Since f =√(T/μ)/(2 L), shortening the vibrating length can offset the tension reduction and restore the original frequency.

D: 1.25

Refractive index is the ratio of vacuum light speed to speed in the material: n =3.0 ×10⁸/(2.4 ×10⁸) =1.25.

D: Infrared

Objects near ordinary environmental temperatures emit much of their thermal radiation in the infrared. Infrared detectors can measure changes in that emission to detect temperature differences.

C: 1.5 A

Identical cells in parallel retain emf 1.5 V and have combined internal resistance 1/3 Ω. With the 2/3 Ω load, total resistance is 1 Ω, so current is 1.5/1 =1.5 A.

C: 8.0 ×10² V

Potential difference is work per charge. Convert 100 μC to 1.00 ×10⁻⁴ C, then V =0.08/(1.00 ×10⁻⁴) =800 V.

B: 40 cm

At balance, resistance ratio equals uniform-wire length ratio:2/3 =l/(100 −l). Thus 2(100 −l) =3 l, giving l =40 cm from the left.

B: 2.50 ×10⁻³ C

Two equal 50 μF capacitors in series have equivalent capacitance 25 μF. The series charge is 25 ×10⁻⁶ ×100 =2.50 ×10⁻³ C. Each capacitor has that charge magnitude and a 50 V drop.

C: 1.8 ×10⁷ J

Five hours is 5 ×3600 =18000 seconds. At constant 1000 W, energy used is Pt =1000 ×18000 =1.8 ×10⁷ J.

A: 50.0 Hz

In V =V₀ sinωt, angular frequency is 314.2 rad/s. Ordinary frequency isω/(2π) =314.2/(2 ×3.142) =50.0 Hz.

A: Lags the voltage by 90°

For an ideal inductor, v =L di/dt. Differentiating a sinusoidal current advances its phase by 90°, so voltage leads current by 90° and current lags voltage by 90°.

B: Sustain a chain reaction under suitable conditions

Neutron-induced fission can release neutrons that trigger further fissions, allowing a chain reaction under suitable conditions. The released energy comes from a decrease in total rest mass.

C: Slow fast neutrons to help sustain the fission chain reaction

Graphite acts as a moderator. Collisions with carbon nuclei slow fast neutrons, making them more effective at causing fission in uranium-235 and helping sustain the chain reaction.

A: ML⁻¹T⁻²

Pressure is force divided by area. Force has dimensions MLT⁻² and area L², so pressure has dimensions ML⁻¹T⁻².

A: 0.25 m

A pendulum crosses its lowest point twice per complete oscillation, so two crossings per second mean period 1 s. From T =2π√(L/g), L =10/(4π²) =0.253… m, or 0.25 m to two decimal places.

A: P = Q + R

Following Q from O to B and then R from B to A reaches the same endpoint as P from O to A. Therefore P =Q +R.

C: 20.0 N

The two equal tensions each make 30° above horizontal. Horizontal components cancel, while vertical balance gives 2 T sin 30° =20 N. Since sin 30° =1/2, T =20 N.

A: Translational

The centre of a rolling ball moves along the ground as the ball changes position. This is translational motion; rotation describes the ball turning about its centre.

B: 4 m/s²

The hanging weight is 30 ×10 =300 N. The 20 kg block’s downslope weight component is 20 ×10 sin 30° =100 N. The net driving force is 200 N on 50 kg, so a =4 m/s².

C: 1.67 m/s²

Using the usual approximation that lunar surface gravity is one-sixth of Earth’s gives gmoon =10/6 =1.666… m/s², or 1.67 m/s².

A: 5 J

For a Hooke’s-law wire, force rises linearly from zero to 500 N over the 0.02 m extension. Stored energy is the triangular area under the force-extension graph:½ ×500 ×0.02 =5 J.

B: Pressure increases with depth of the lake.

Water pressure is higher at greater depth. As the bubble rises, external pressure falls. At constant temperature a fixed amount of gas expands as its pressure decreases, according to Boyle’s law.

B: Mean translational kinetic energy per molecule

For a classical ideal gas, average translational kinetic energy per molecule is 3 kT/2. Temperature therefore measures this average, rather than the energy of a particular molecule or total heat transfer.

B: 336 s

The temperature rise is 100 −20 =80 K. Heating needs Q =mcΔT =2.0 ×4200 ×80 =672000 J. At 2000 W, the time is 672000/2000 =336 s.

D: Polarization

Polarization selects the direction of oscillation in the plane perpendicular to travel. That directional freedom belongs to transverse waves; a purely longitudinal wave oscillates along its travel direction.

B: 300 m

The wave speed is fλ =200000 ×1500 =3 ×10⁸ m/s. At 1000000 Hz, the wavelength is 3 ×10⁸/10⁶ =300 m.

B: Require a medium to travel while light waves do not.

Sound consists of mechanical disturbances of a material medium. Light is electromagnetic radiation and can propagate through vacuum, so it does not require matter along its path.

A: Increasing the frequency

For a steady simple tone, a higher vibration frequency is heard as a higher pitch. Changing amplitude mainly changes loudness rather than the tone’s frequency.

C: Moon is between the Sun and Earth

During a total solar eclipse, the Moon lies between the Sun and Earth. An observer in the Moon’s umbra sees the entire bright solar disc covered; the alignment and apparent size must also be suitable.

A: Red and violet

Red and violet occupy opposite ends of the visible spectrum. Their separation spans the full listed colour range, exceeding any pair with an endpoint nearer the middle.

D: 75000.0 V/m

Away from the edges, the uniform field between parallel plates has magnitude E =V/d. Thus E =600/(8.0 ×10⁻³) =75000 V/m. It points from the positive plate to the earthed negative plate.

C: Different in both magnitude and direction

Earth’s magnetic field varies geographically in strength and direction. Its inclination and declination, as well as its magnitude, depend on location.

D: 6:1

Ohm’s law gives I =V/R. Tripling voltage and halving resistance gives Inew =3 V/(R/2) =6 V/R, so the new-to-old current ratio is 6:1.

B: 8.0 A

The battery’s internal resistance is in series with the cable. Total resistance is 0.5 +1.0 =1.5 Ω, giving I =12/1.5 =8.0 A.

D: 0.8 μF

The parallel pair gives 2 +4 =6 μF. The four series sections are 2,3,6 and 4 μF, so 1/C =1/2 +1/3 +1/6 +1/4 =1.25 μF⁻¹. Therefore C =0.8 μF.

B: 4.00 W

Instantaneous resistor power is p =v²/R. When the voltage magnitude reaches 20 V, its maximum power is 20²/100 =4 W.

A: 3 V

An ideal transformer has voltage ratio equal to turns ratio. The secondary voltage is 12 ×100/400 =3 V, making this a step-down transformer.

B: 80 m

The initial upward velocity is 80 sin 30° =40 m/s. At maximum height the vertical velocity is zero, so h =40²/(2 ×10) =80 m.

A: Kinetic energy is constant but linear momentum varies

At constant angular velocity on a fixed circle, speed is constant, so kinetic energy stays constant. The direction of velocity changes continuously, making linear momentum change direction.

D: 200 N

Taking moments about the pivot, the cord contributes T sin 30° ×L while the uniform beam’s weight contributes 200 ×L/2. Thus T ×0.5 =100, giving T =200 N.

B: 1 000 J

Neglecting air resistance, gravitational potential energy becomes kinetic energy. Thus K =mgh =50 ×10 ×2 =1000 J just before impact.

B: 40.0 N

Just before sliding starts, the limiting friction is μₛN. The horizontal force leaves N = mg, so F = 0.2 × 20 × 10 = 40 N.

A: I and III only.

At a point in a liquid at rest, pressure acts equally in every direction. In a connected liquid of uniform density, equal depths have equal pressure. Pressure increases with depth and does not depend on tube area.

D: 0.80

The kerosene mass is 24.8 − 15.2 = 9.6 g, and the water mass is 27.2 − 15.2 = 12.0 g. Equal bottle volumes give relative density 9.6/12.0 = 0.80.

D: Volume of P displaced is less than that of Q

A floating object displaces its own weight of either liquid. Since P is denser, less of it is needed to provide that weight: Vₚ/Vq = 1.5/2.0 = 0.75.

B: 351 K

Convert Celsius to kelvin by adding 273.15. Thus 78 °C is 351.15 K, which rounds to 351 K to the nearest kelvin.

C: 150 g

Heat lost equals heat gained: 100 c(78 − 66) = X(c/2)(66 − 50). Therefore 1200 c = 8 Xc, so X = 150 g.

C: 560 W

Melting requires Q = mL = 100 × 336 = 33600 J. One minute is 60 s, so the required power is Q/t = 33600/60 = 560 W.

D: 0.3 cm

In y = A sin(kx − ωt), the sine varies between −1 and 1. The largest displacement magnitude is therefore A = 0.3 cm.

B: 3000 m

The echo makes a return trip, travelling twice the sea depth. Thus d = vt/2 = 1500 × 4/2 = 3000 m.

A: Double its vibrating length at fixed tension and mass per unit length

For a stretched wire, f = (1/2 L)√(T/μ), where μ is mass per unit length. Doubling the vibrating length while holding tension and μ fixed halves the fundamental frequency.

B: 400 Hz

Fundamental frequency is proportional to √T/L at fixed mass per unit length. The new frequency is 400 × √4/2 = 400 Hz.

A: One focal length from the mirror

A small lamp at a concave mirror’s principal focus sends rays that reflect parallel to the principal axis in the paraxial approximation. Its distance is therefore the focal length.

B: 6.0 × 10¹⁴ Hz

A stationary boundary does not change the light’s frequency. In glass, speed and wavelength decrease together, leaving frequency at 6.0 × 10¹⁴ Hz.

B: 10 cm

For an upright virtual image of magnification 2, v = −2 u. The thin-lens equation gives 1/20 =1/u −1/(2 u) =1/(2 u), so u =10 cm.

A: 1.0 cm

For near-normal viewing through a plane glass surface, apparent depth is real depth divided by refractive index:3.0/1.5 =2.0 cm. The object appears raised by 3.0 −2.0 =1.0 cm.

D: f < u < 2 f

A projector needs a real enlarged image on a screen. A converging lens produces that image when the object is farther than f but nearer than 2 f.

D: Both the pen and the cloth are charged.

Rubbing transfers electric charge between the pen and cloth, leaving opposite charges on them. The charged pen polarizes nearby neutral paper, and the nearer opposite charge is attracted more strongly.

B: 5:4

Parallel resistors have the same voltage. Since P = V²/R, their power ratio is P₁/P₂ = R₂/R₁ =5/4, giving 5:4.

A: 0.8 A

The parallel equivalent is 12 ×6/(12 +6) =4 Ω. Including internal resistance gives 5 Ω, so total current is 12/5 =2.4 A. Terminal voltage is 2.4 ×4 =9.6 V, giving 9.6/12 =0.8 A in the 12 Ω resistor.

A: Attract each other

Each conductor produces a magnetic field at the other. Applying the magnetic-force direction rule to parallel currents in the same direction gives a force on each wire toward the other.

C: 1/100 s

Consecutive zero crossings with opposite slopes are half a cycle apart. At 50 Hz, a cycle lasts 1/50 s, so the interval is 1/100 s.

A: No photoelectron is emitted

A photon must supply at least the work function to liberate an electron. Here 2 eV is less than 3 eV, so ordinary single-photon photoemission does not occur.

D: Force x velocity

A watt is a joule per second. Force multiplied by velocity has units N·m/s =J/s =W, so it has the unit of power.

C: tan θ =3/7

The resultant has components 7 N along the larger force and 3 N perpendicular to it. Therefore tan θ =opposite/adjacent =3/7.

C: 750 m

The landing speed is 180/3.6 =50 m/s. Uniform deceleration gives average speed(50 +0)/2 =25 m/s, so distance is 25 ×30 =750 m.

B: Gravitational potential → mechanical → electrical → light

Water stored above a turbine has gravitational potential energy. Flowing water turns the turbine, which drives a generator. Electrical energy from the generator is then converted to light by a lamp.

A: 1.0 N

The object weighs 0.4 ×10 =4 N. Half its volume displaces paraffin of mass 0.5 ×0.4 ×900/600 =0.3 kg, giving 3 N upthrust. The string tension is 4 −3 =1 N.

A: e.m.f. changes with temperature

With the reference junction held at a known temperature, a thermocouple’s emf changes with the measuring-junction temperature. Calibration relates that voltage to temperature.

B: Increases by a factor of 4

For a fixed amount of ideal gas, P is proportional toT/V. Doubling absolute temperature while halving volume multiplies pressure by 2/(1/2) =4.

C: 70 °C

Ignoring heat loss, heat lost by hot water equals heat gained by cool water:200(90 −T) =100(T −30). Solving gives 300 T =21000, so T =70 °C.

C: 1.00 ×10⁷ J/kg

At the melting point, heat goes into the phase change. The mass is 10 g =0.010 kg, so specific latent heat is Q/m =10⁵/0.010 =1.00 ×10⁷ J/kg.

B: 30 cm

Consecutive antinodes of a stationary wave are half a wavelength apart. Their separation is therefore 60/2 =30 cm.

D: Infra-red waves

Infrared radiation is electromagnetic and can travel through vacuum. Sound, ultrasound and other acoustic waves need a material medium for their pressure disturbances.

A: 6.0 ×10² km

The measured time includes the outward and return journeys. One-way distance is ct/2 =(3 ×10⁸)(4 ×10⁻³)/2 =6 ×10⁵ m =6 ×10² km.

B: 2 v/r −1

For a real image, magnification magnitude is v/u. Since 1/u +1/v =2/r, multiplying byv gives v/u +1 =2 v/r. Therefore magnification magnitude is 2 v/r −1.

C: 25 cm

Using 1/f =1/u +1/v gives 1/v =1/20 −1/100 =1/25. The film must therefore be 25 cm from the lens to receive the sharp real image.

D: Ciliary muscles

Ciliary muscles change tension in the supporting fibres of the lens. This changes lens curvature and optical power, allowing the eye to focus on objects at different distances.

C: I and II only

Soft iron magnetizes readily when current flows and loses most of that magnetism when current stops. These properties make the electromagnet easy to switch on and off. Steel retains more magnetism.

C: 2.0 Ω

Terminal voltage is V =E −Ir. The voltage falls by 4 −2 =2 V when current rises by 3 −2 =1 A. Thus r =2/1 =2 Ω; both measurements then give emf 8 V.

A: 4.00 A

Each parallel branch has two 2 Ω resistors in series, giving 4 Ω per branch. Their parallel equivalent is 2 Ω. Adding the 1 Ω series resistor gives 3 Ω, so battery current is 12/3 =4 A.

D: 18 V

The 5 μF and 15 μF capacitors are in series across 24 V. Their voltage drops are inversely proportional to capacitance, so the 5 μF capacitor gets 15/(5 +15) ×24 =18 V.

A: 60 V

For a resistive instrument, P =V²/R. Therefore V =√(PR) =√(40 ×90) =√3600 =60 V.

A: Minimize electrical energy losses due to heat production

For the same transmitted power, raising voltage reduces current. Since transmission-wire heating is I²R, the smaller current reduces wasted energy.

C: 245.0 Ω

Inductive reactance is Xᴸ =2πfL. Converting 30.0 mH to 0.0300 H gives 2π ×1300 ×0.0300 =245.04… Ω, or 245.0 Ω to one decimal place.

A: I only

Cathode rays are streams of electrons, which carry negative charge. An electric field can deflect them, and a magnetic field can deflect them when they have a velocity component perpendicular to it.

B: ²²²₈₈Z

Each beta-minus decay raises atomic number by 1 without changing mass number. An alpha decay lowers mass number by 4 and atomic number by 2. The final numbers are A =226 −4 =222 and Z =88 +2 −2 =88.

D: Frequency of incident radiation

For a fixed metal surface, maximum photoelectron kinetic energy is Kmax =hf −φ. Increasing incident frequency above threshold raises the photon energy and therefore the maximum electron energy.

B: 0.5 cm

When a reading is rounded to the nearest 1 cm division, it can differ from the true value by up to half a division. The rounding-error bound is therefore±0.5 cm.

B: II and III only

For small oscillations, T =2π√(L/g). The period depends on length and gravitational acceleration, but not on bob mass. Thus II and III are the relevant quantities.

B: Load and radius of the wire

Tensile stress is force divided by cross-sectional area. For a circular wire, stress =F/(πr²), so the load and radius determine it. Extension and original length instead determine strain.

C: 1 250 kg m-3

Floating equilibrium gives sphere density =1000 ×0.5 =500 kg/m³. In the second liquid,500 =ρ ×0.4, so its density is 500/0.4 =1250 kg/m³.

C: Inversely proportional to its absolute temperature

At constant pressure, a fixed amount of ideal gas satisfies V/T =constant. Since density is mass divided by volume, density is inversely proportional to absolute temperature.

B: 9.75 ×10² J

The temperature rise is 70 −20 =50 K. Heat absorbed is mcΔT =0.05 ×390 ×50 =975 J =9.75 ×10² J.

D: Reflection

Echo sounding sends a sound pulse downward and measures its return after reflection from the seabed. With sound speed v and round-trip time t, depth is vt/2.

C: 800 Hz

Each revolution lets 32 holes pass the air jet, producing 32 pressure pulses. At 25 revolutions per second, the pulse frequency is 32 ×25 =800 Hz.

C: II and III

For objects in front of it, a convex mirror gives virtual, upright, diminished images. Its wide field of view lets a driver see more of the surrounding traffic. Therefore II and III apply.

D: Enlarged and upright.

An object between a concave mirror and its focus produces a virtual image behind the mirror. The image is upright and larger than the object.

C: 2.0 ×10⁸ m/s

The refractive index is the speed in air divided by the speed in glass. Thus v =3.0 ×10⁸/1.50 =2.0 ×10⁸ m/s.

B: Gamma rays, x-rays, ultraviolet rays, infra-red rays

In the conventional electromagnetic-spectrum order, gamma rays have the shortest typical wavelengths, followed by X-rays, ultraviolet and infrared.

D: Angle of dip

Magnetic dip, also called inclination, is the angle the magnetic field makes with the horizontal. Declination instead measures the horizontal difference between geographic and magnetic north.

A: 1.6 ×10⁻¹¹ N

Force magnitude is |q|E =(1.6 ×10⁻¹⁹)(10⁸) =1.6 ×10⁻¹¹ N. Because an electron is negatively charged, the force points opposite the electric field.

B: Changing the cell’s physical size

Cell emf depends on electrode reactions, electrolyte conditions and temperature. Scaling up an otherwise identical cell increases available charge and can change internal resistance, but does not change its equilibrium emf.

B: 10.0 μC

Charge magnitude on either plate is Q =CV =(2.0 ×10⁻⁶) ×5 =10 ×10⁻⁶ C =10 μC. The two plates carry equal and opposite charges.

A: 60 W, 240 V

Operating current is P/V. The four currents are 0.25 A,1.5 A,3.33 A and 8 A. The 60 W,240 V device draws the smallest current and needs the lowest current rating when the same protection margin is used.

D: Directly proportional to the square of the current, the resistance and the time.

Joule heating is H =I²Rt for a steady current through a constant resistance. Heat is proportional to current squared, resistance and elapsed time.

C: 1990 Ω in series

The total required resistance is V/I =100/0.05 =2000 Ω. Subtracting the galvanometer’s 10 Ω gives R =1990 Ω. It is connected in series to limit the current through the meter.

B: Ohm, volt and hertz

Impedance is voltage divided by current, so its unit is the ohm. RMS voltage is measured in volts, and resonance frequency in hertz, or cycles per second.

B: Iᵣₘₛ = I₀/√2

For sinusoidal current, the mean of sin² over a cycle is 1/2. Thus the mean square current is I₀²/2, and its square root is Iᵣₘₛ =I₀/√2.

C: Material of the photocathode

The threshold frequency is f₀ =φ/h, whereφ is the photocathode’s work function. It is therefore determined by the emitting material and its surface condition.

D: 7/8

After three half-lives, the fraction remaining is(1/2)³ =1/8. The fraction decayed is 1 −1/8 =7/8.

B: Constant magnitude and varying direction.

Centripetal acceleration has magnitude v²/r. Both speed and radius are constant, so this magnitude stays constant. Its direction continually changes as it points from the moving object toward the centre.

C: 133.3 N

Efficiency is useful work divided by input work. For motion through distance s along the slope, 0.75 = mg(s sin 30°)/(Fs). Thus F = 20 × 10 × 0.5/0.75 = 133.3 N to one decimal place.

D: 23.0cm

An extra 5 N extends the spring by 5 cm, giving compliance 1 cm/N. Its unloaded length is 25 − 5 = 20 cm. With 3 N attached, its length is 20 + 3 = 23 cm.

C: 20.0g

Let b be the mass-equivalent buoyancy in water and m the true mass. Then m − b = 15 g and m − 2b = 10 g. Subtracting gives b = 5 g, so m = 20 g.

B: The floating of a dry needle in water

Surface tension can support a carefully placed dry needle at the water surface. A ship and a balloon are mainly supported by buoyancy, while membrane diffusion is driven by concentration differences.

C: 9,040,000J

Both samples release 4 × 4200 × 20 = 336000 J as water cools from 100 °C to 80 °C. The steam also releases 4 × 2260000 = 9040000 J on condensing. That latent heat is the difference.

C: 168s

Required heat is mcΔT = 1 × 4200 × (50 − 20) = 126000 J. At 750 J/s, the heater needs 126000/750 = 168 s if no heat is lost.

C: Temperature of the liquid increases

Saturated vapour pressure rises with temperature. Warmer liquid molecules have greater energy, and equilibrium is reached at a higher vapour pressure. At a fixed temperature it does not depend on the liquid amount while both phases remain.

C: Translational kinetic energy of the molecules

For an ideal gas, the mean translational kinetic energy per molecule is 3kT/2. It is therefore directly proportional to the absolute temperature T.

A: They both transmit energy

Both light and sound transfer energy. Light can travel through vacuum, while sound needs matter. In air, sound is longitudinal and much slower than light.

D: Inverted and formed by the light from each point traveling in a straight line.

Rays from the top of an object pass through the small hole to the bottom of the screen, and rays from the bottom reach the top. Straight-line propagation therefore produces an inverted real image.

C: 2Q

With the incident ray fixed, rotating the mirror also rotates its normal by Q. The equal-angle reflection law makes the reflected direction change by twice that angle, or 2Q.

A: Pass through undeviated

In the thin-lens approximation, a ray through the optical centre is treated as undeviated. The small lateral displacement through a real lens is neglected in the ray diagram.

A: The image is virtual, 3cm in front of the lens

Using 1/f = 1/u + 1/v gives 1/v = −1/12 − 1/4 = −1/3, so v = −3 cm. The negative image distance means a virtual image on the same side as the object.

B: 4 Hz

Beats arise from alternating reinforcement and cancellation of nearby frequencies. Their frequency is the absolute difference: |260 − 256| = 4 Hz.

B: The angle of declination is the angle between the magnetic meridian and the geographic meridian

Magnetic declination is the horizontal angle between geographic north and magnetic north. Dip, or inclination, is instead the angle of the magnetic field relative to the horizontal.

C: Prevent polarization in the cell

Manganese(IV) oxide is reduced in the cell’s cathode reaction and serves as the depolarizer. Carbon improves electrical conduction in the mixture, helping the cell deliver current.

B: 2.00 A

For a resistive device, P = I²R. Hence I = √(P/R) = √(1500/375) = √4 = 2.00 A.

C: Slip rings are replaced with a split-ring commutator

A split-ring commutator reverses the coil’s connection to the external circuit every half-turn. The external current then flows in one direction, giving pulsating direct current rather than alternating current.

A: 5.6 hours

Using m = ZIt, time is 1/(5.0 × 10⁻⁷ × 100) = 20000 s. Dividing by 3600 gives 5.56 hours, or 5.6 hours to one decimal place.

D: I, II and III.

Isotopes have the same proton number. Their neutral atoms have the same electron configuration and very similar chemistry. Their different masses allow separation by physical methods such as mass separation.

C: 30 days

If 3.5 g decays,0.5 g of the original radioactive isotope remains. The sequence is 4 →2 →1 →0.5 g: three half-lives. Therefore the time is 3 ×10 =30 days.

A: 4 Mp > Mx

The proton-proton fusion chain produces a helium nucleus with less rest mass than the four initial protons. The rest-mass difference supplies the released energy and other reaction products, so 4 Mp > Mx.

A: I and II only

An alpha particle contains two protons and two neutrons. Its emission reduces the parent’s mass number by 4 and atomic number by 2. Thus I and II are true, while III is false.

D: Second

The second is the SI base unit of time. Newton, watt and joule are derived units built from kilogram, metre and second.

A: P/v

Mechanical power for a force along the direction of motion is P = Fv. Dividing by speed gives the engine’s driving force F = P/v.

C: 3.92m

Hydrostatic pressure is p = ρgh. Thus h = 102000/(2600 × 10) = 3.923… m, or 3.92 m to two decimal places.

B: 50.0 °C

The 0–100 °C resistance change is 1.05 − 0.75 = 0.30 Ω. The measured rise is 0.90 − 0.75 = 0.15 Ω, half the interval. Temperature is therefore 50.0 °C.

A: Collide less frequently with the walls of the container.

Cooling lowers the molecules’ average kinetic energy and speed. At fixed volume they strike the walls less often and with less momentum change per collision, reducing the gas pressure.

C: 120g

The copper loses Q = mcΔT = 1 × 400 × 100 = 40000 J. This melts ice of mass Q/L = 40000/333000 = 0.12012… kg, or 120 g to the nearest gram.

A: Increase the external pressure

Pure water boils near 100 °C at about 760 mmHg. At 750 mmHg it boils slightly below 100 °C. Raising the external pressure to about 760 mmHg raises the boiling point to 100 °C.

A: I and II only

Land and sea breezes arise from natural convection. In ordinary heat-transfer experiments, convection is bulk motion of a liquid or gas. A vacuum suppresses both conduction and convection across the gap, so III is false.

B: Rectilinear propagation of light

In a uniform medium, light travels approximately in straight lines. An opaque object blocks those paths and produces a shadow; an eclipse occurs when a celestial body blocks the light reaching another body or observer.

D: 12.00 cm

The magnitude of magnification is image distance divided by object distance: 40/10 = 4. Therefore the image height has magnitude 4 × 3 = 12 cm.

C: 10cm

An erect image from a converging lens is virtual: v = −2u. Substituting in 1/f = 1/u + 1/v gives 1/20 = 1/(2u). Hence u = 10 cm, inside the focal length.

A: In front of the retina

In an uncorrected short-sighted eye, parallel rays from a distant object converge before reaching the retina. A diverging corrective lens reduces the eye system’s convergence so the focus moves back onto the retina.

B: Green only

An ideal magenta surface reflects red and blue light and absorbs green. Under white illumination, the reflected red and blue combine to appear magenta.

D: 384 m/s

At the first resonance of a tube closed at one end, L = λ/4. Thus λ = 4 × 0.20 = 0.80 m and v = fλ = 480 × 0.80 = 384 m/s.

A: 106 Hz

For the fundamental of a pipe closed at one end, λ = 4L = 3.20 m. Frequency is 340/3.20 = 106.25 Hz, or 106 Hz to the nearest hertz.

C: I, II and III only

Positive net charge means there is more positive than negative charge, usually through an electron deficit. The charged body creates a field and stores electrostatic energy. At rest in electrostatic equilibrium, it carries no steady current.

B: Electric field

Electric field is force per unit positive test charge, so it has both magnitude and direction. Electric charge, potential difference and capacitance are scalars.

C: 0.50A

Identical cells in parallel retain emf 1.5 V and have internal resistance 1/3 Ω. The load current is 1.5/(2.67 + 1/3) = 0.4994… A, or 0.50 A to two decimal places.

B: 1.64 Ohms

For a uniform metre bridge, resistance ratio equals wire-length ratio at balance. With 2.0 Ω beside the 0 cm end, 2.0/R = 55/45. Thus R = 2.0 × 45/55 = 1.64 Ω to two decimal places.

C: Electromotive force

Emf is the energy supplied by a source per unit charge passing through it. That energy is transferred around the complete circuit, including the source’s internal resistance. Its unit is the volt, or joule per coulomb.

D: Low melting point.

A fuse element must melt when excessive current heats it sufficiently. Melting opens the circuit and stops the sustained overcurrent. A suitably low melting point supports this protective action.

A: Heat dissipation

For the same transmitted power, increasing voltage reduces current because P = VI. Heating loss in the wires is I²R, so lowering current reduces heat dissipation.

B: 0.03 ohms shunt resistor

The full-scale coil voltage is 0.010 × 30 = 0.30 V. The shunt carries 10 − 0.010 = 9.99 A, so its resistance is 0.30/9.99 ≈ 0.03 Ω. It must be parallel to the coil.

B: Make the magnetic core from thin, electrically insulated metal sheets

A transformer core made from thin, electrically insulated metal laminations restricts large circulating eddy currents. This reduces unwanted heating in the core.

B: Emission of electrons by incident radiation

A photoemissive cell uses incident photons to release electrons from a photosensitive surface. Those electrons can be collected to produce a measurable current.

B: Acceleration, velocity and moment of force

Acceleration and velocity have magnitude and direction. Moment of force, or torque, is also a vector with direction given by the right-hand rule. Mass, volume and density are scalars.

C: 12.0 N

Perpendicular forces obey R² = F₁² + F₂². Therefore F₂ = √(13² − 5²) = √144 = 12 N.

A: 0.67 m/s²

Mean acceleration is change in velocity divided by elapsed time: (20 − 10)/(20 − 5) = 10/15 = 0.67 m/s² to two decimal places.

D: I and IV.

An aneroid barometer measures atmospheric pressure and a manometer measures pressure or pressure difference. A hydrometer measures liquid density, while a hygrometer measures humidity.

B: 35.0 min

Heat needed is mL = 1.5 × 336000 = 504000 J. Heater power is VI = 12 × 20 = 240 W. Time = 504000/240 = 2100 s = 35.0 min.

D: Physical property proportional to the average translational kinetic energy of its molecules

Absolute temperature is a physical property, not a quantity of energy. For an ideal gas, the mean translational kinetic energy per molecule is 3kT/2, so it is proportional to kelvin temperature.

C: 1 000 Hz

A compression and the adjacent rarefaction are half a wavelength apart. Thus λ = 2 × 17 cm = 0.34 m. Frequency f = v/λ = 340/0.34 = 1000 Hz.

A: 3333 × 10⁻⁸ cm

Frequency stays unchanged on entering glass, while speed becomes c/n. Therefore wavelength becomes λ/n = (5000/1.5) × 10⁻⁸ cm ≈ 3333 × 10⁻⁸ cm.

D: 32 cm

For a convex mirror, f = −12 cm and v = −u/3. The mirror equation gives −1/12 = 1/u − 3/u = −2/u, so u = 24 cm and v = −8 cm. Their separation is 24 + 8 = 32 cm.

A: At 60°

A common three-mirror kaleidoscope uses mirrors forming an equilateral triangular tube. Adjacent mirrors meet at 60°, creating repeated symmetric images by reflection.

A: Black

In the ideal colour model, a yellow card reflects red and green light. A blue filter transmits blue and blocks red and green, leaving essentially no light from the card, so it appears black.

B: Different speeds of the various colours in glass

Glass has a refractive index that depends on wavelength. Different colours therefore travel at different speeds in it and refract by different amounts, separating the white light.

D: II and IV.

Alpha radiation consists of helium nuclei and beta radiation of electrons or positrons. They are particles with mass. Radio waves and gamma rays are electromagnetic radiation.

B: Are at the same potential.

Charges redistribute through the connecting conductor until there is no potential difference between the spheres. At electrostatic equilibrium the connected conductor is equipotential.

D: 1.70 × 10⁻⁸ Ω·m

For a uniform wire R = ρL/A. Thus ρ = RA/L = 0.425 × (0.2 × 10⁻⁶)/5 = 1.70 × 10⁻⁸ Ω·m.

B: 0.86V

Series resistance is 250 + 500 + 1000 = 1750 Ω. The 250 Ω resistor gets the fraction 250/1750 of the voltage: 6 × 250/1750 = 0.86 V to two decimal places.

C: 500 mA

At full scale, the meter voltage is 0.050 × 0.1 = 0.005 V. The shunt carries 0.005/0.0111 ≈ 0.45045 A. Adding the meter’s 0.050 A gives 0.50045 A, or 500 mA to the nearest milliampere.

B: 100 A

Deposited mass is m = ZIt. With time 5 × 60 = 300 s, I = 36/(0.0012 × 300) = 100 A.

C: Wavelength of the radiation

For a fixed metal in the ordinary photoelectric effect, Kmax = hc/λ − φ. The wavelength sets photon energy and therefore the maximum emitted-electron energy above the work function.

D: I and III only.

Force is measured in newtons and power in watts, so I and III are correct. Torque is measured in N·m, not N/m. Momentum is kg·m/s, not kg·m/s².

D: 86.6 N

The force making 30° with the resultant is the adjacent component: F = R cos 30° = 100 × √3/2 = 86.6 N to one decimal place.

D: I and IV only

Electric potential and kinetic energy are scalars. Torque and momentum are vectors, so the scalar quantities are I and IV.

C: 20.00m s-1

Impulse equals change in momentum. Starting from rest, mv = Ft = 30.0 × 0.05 = 1.5 N·s. Hence v = 1.5/0.075 = 20.0 m/s.

B: 40 rev s-1

Meshing gears pass the same number of teeth per second. Therefore 20 × 120 = 60 × n, giving the larger gear’s speed n = 40 rev/s.

D: 52.8 s

Useful energy for the climb is mgh = 60 × 10 × 22 = 13200 J. At 0.25 kW = 250 W, time = energy/power = 13200/250 = 52.8 s.

C: Energy

Pressure × volume has units (N/m²) × m³ = N·m = J. A joule is the SI unit of energy.

B: Heat capacity

Heat capacity refers to the whole body: C = Q/ΔT. The heat required for this 10 kg piece to rise by 1 K corresponds to its heat capacity. Specific heat capacity is per kilogram.

C: Conduction

Conduction transfers thermal energy through the rod without bulk motion of the metal. In metals, mobile electrons and lattice vibrations both contribute to this transfer.

D: I, II and III.

Sound is a mechanical disturbance that can propagate through gases, liquids and solids. All three provide matter whose particles can pass the disturbance onward.

D: I, II and III.

Mechanical and electromagnetic waves can diffract around openings, refract when their propagation speed changes, and interfere when they overlap. All three are characteristic wave behaviours.

A: I only

For air treated as an ideal gas, sound speed is √(γRT/M). At fixed composition it depends on temperature. Pressure has no independent effect at fixed temperature, and frequency does not enter this model.

D: 680m.

The delay is d/340 − d/(3 × 10⁸) = 2 s. Solving gives d ≈ 680.0008 m, which rounds to 680 m. Neglecting the tiny light travel time gives the same metre-rounded result.

D: 1:16

For equal length and tension in the same vibration mode, f is proportional to 1/√m. Thus f₁/f₂ = 4 implies m₁/m₂ = (1/4)² = 1/16.

A: 30 cm behind the mirror, virtual

With real object distance positive, 1/v = 1/15 − 1/10 = −1/30. The image is therefore virtual and lies 30 cm behind the mirror. The object is inside the focal length.

C: Violet

Violet is at the short-wavelength end of the visible spectrum. Green, orange and red all have longer wavelengths than violet.

B: Ultraviolet rays have shorter wavelengths than infrared rays and can produce fluorescence

Ultraviolet has shorter wavelengths than visible light and infrared. It can excite fluorescent materials, which then emit light, often at visible wavelengths.

B: R/2

With external resistance R and internal resistance r, terminal voltage is V = ER/(R + r). Setting V/E = 2/3 gives 3R = 2(R + r), hence r = R/2.

C: Torch light 6V 30W

Using I = P/V: iron = 1000/250 = 4 A; television = 110/220 = 0.5 A; torch = 30/6 = 5 A; heater = 500/110 ≈ 4.55 A. The torch draws the highest current.

B: II and III only

Moving-iron meters respond to magnetization irrespective of current direction. Hot-wire meters respond to I²R heating. Both can measure AC. An unrectified permanent-magnet moving-coil meter averages opposing torques over an AC cycle.

C: Faraday’s law

An induction coil uses a changing current to change magnetic flux through another winding. Faraday’s law relates that changing flux to the induced emf.

D: ⁶⁵₂₂X

For a neutral atom, 22 electrons mean 22 protons, so atomic number Z = 22. Mass number A = protons + neutrons = 22 + 43 = 65. The notation is ⁶⁵₂₂X.

C: N·s

One newton is kg·m·s⁻². Multiplying by one second gives N·s = kg·m·s⁻¹, the unit of momentum.

A: 2.5 m/s²

For upward acceleration, T − mg = ma. At the cable limit, a = 60000/4800 − 10 = 2.5 m/s². Any greater acceleration would exceed the stated tension limit.

C: Surface tension

The curved meniscus results from surface tension together with adhesion between water and glass. Water wets the glass, so its surface rises near the wall.

D: The ratio of its length to its height

For an ideal frictionless inclined plane, effort × length = load × height. Mechanical advantage is load/effort = length/height.

A: 22.5 °C

The 20 cm interval represents 100 °C. A rise of 4.5 cm therefore gives T = (4.5/20) × 100 = 22.5 °C.

A: 126.50 cm³

Initial volume is 5³ = 125 cm³. To first order, volume expansivity is 3α, so V = 125[1 + 3(4 × 10⁻⁵)(120 − 20)] = 126.50 cm³.

D: 76.0 cm

Initially the trapped air pressure is 75.0 − 73.5 = 1.5 cmHg. At constant temperature, p₂ = 1.5 × 8/6 = 2.0 cmHg. True atmospheric pressure is then 74.0 + 2.0 = 76.0 cmHg.

C: 50 °C

Neglecting heat loss, hot-water heat loss equals cold-water heat gain: mc(T − 20) = 3mc(20 − 10). Thus T − 20 = 30 and T = 50 °C.

D: 674360 J

First warm the ice by 2 °C: Q₁ = 2 × 2090 × 2 = 8360 J. Then melt it: Q₂ = 2 × 333000 = 666000 J. Total heat is 674360 J.

A: Copper, steel, glass

Copper conducts heat much better than steel, and steel conducts much better than glass. Their descending order of thermal conductivity is copper, steel, glass.

B: Convection and conduction

The vacuum greatly reduces conduction and convection across the space between the flask walls because there is almost no matter there. Electromagnetic radiation can still cross that space.

D: Reflection.

A white coating generally reflects more incoming sunlight than a black coating. With less solar energy absorbed, the roof gains less heat under otherwise similar conditions.

C: Tuning fork

A tuning fork is designed to vibrate mainly at one frequency and gives a nearly pure tone after brief higher-frequency transients fade. Guitar and ordinary string sounds contain strong harmonics.

C: Three

Two plane mirrors at right angles form one image by reflection in each mirror and a third by successive reflections in both. Thus there are three images.

B: 1.5 cm

The lens formula gives 1/v = 1/20 − 1/60 = 1/30, so v = 30 cm. Magnification magnitude is v/u = 30/60 = 0.5. Image height is 0.5 × 3.0 = 1.5 cm.

B: +5mm

A microscope objective uses a converging lens with a short focal length. Positive 5 mm provides stronger convergence than positive 5 cm, making it the suitable choice here.

A: N50°E

Magnetic north is 10° east of true north. A compass direction 40° east of magnetic north is therefore 10° + 40° = 50° east of true north: N50°E.

C: 24 μF

For series capacitors, reciprocals add: 1/3 = 1/6 + 1/8 + 1/C. Thus 1/C = 1/24 and the additional capacitance is 24 μF.

A: Has a smaller secondary current than primary current

An ideal transformer with more secondary turns steps voltage up. Conservation of power gives VsIs = VpIp, so the secondary current is smaller than the primary current.

D: In the direction the north pole of a compass needle would point.

The direction of a magnetic field is defined as the direction in which the north-seeking pole of a small compass points at that location.

A: Away from the other wire

Parallel conductors carrying currents in opposite directions repel. Each wire’s magnetic field exerts a force on the current in the other wire, directed away from it.

C: The maximum emitted-electron energy depends on the incident wavelength

In the ordinary single-photon photoelectric effect, maximum kinetic energy is Kmax = hc/λ − φ. It depends on wavelength and work function. Increasing intensity at fixed wavelength chiefly increases the number emitted.

C: The same atomic number but different number of neutrons

Isotopes have the same atomic number, meaning the same number of protons, but different neutron numbers. Their mass numbers therefore differ.

D: Newton

The newton is defined as kg·m·s⁻², combining base units to measure force. Kilogram, metre and kelvin are SI base units.

D: Acceleration.

Without air resistance, both objects accelerate at g and have the same velocity at each instant. Momentum, kinetic energy and gravitational potential energy also depend on mass, so these differ by a factor of three.

C: 70 g

The pivot is at 100 − 55 = 45 cm from Q. The 10 g mass is 35 cm left of the pivot, while the uniform rule’s centre is 5 cm right of it. Balancing moments gives 10 × 35 = M × 5, so M = 70 g.

C: 13.2 N

F₃ must oppose the resultant of F₁ and F₂. The resultant’s squared magnitude is 10² + 5² + 2 × 10 × 5 cos 60° = 175 N². Hence |F₃| = √175 ≈ 13.2 N; its direction is opposite the resultant.

B: 5/12

Work done against friction equals the initial kinetic energy: μmg × 3 = ½m × 5². Cancelling mass gives μ = 25/(2 × 10 × 3) = 5/12.

C: 9000.0 N

Taking downward as positive, mg − T = ma. Therefore T = m(g − a) = 1000(10 − 1) = 9000 N. The downward acceleration makes tension less than weight.

B: Radiation

Thermal radiation travels as electromagnetic waves and can cross a vacuum. Conduction requires interacting matter, and convection requires moving fluid.

A: Balance wheel of a watch

Thermal expansion changes the dimensions and moment of inertia of a watch’s balance wheel, affecting its oscillation period and timekeeping. Thermal compensation is needed to reduce this error.

C: Change unit mass of lead from solid to liquid at its melting point

Specific latent heat of fusion is the energy required to melt unit mass at the melting point without changing temperature. Both the unit mass and the solid-to-liquid change are essential.

C: Involve change of state

Both processes change liquid into gas. Evaporation occurs at the surface, while boiling forms vapour throughout the liquid when its vapour pressure reaches the surrounding pressure.

C: Is several times denser than water

For a barometer, h = p/(ρg). Mercury’s high density means a much shorter column can balance atmospheric pressure than a water column, making the instrument practical.

B: Good conduction of heat

Good thermal conduction lets a metal utensil transfer heat efficiently from the heat source to its contents. That is the relevant property for cooking.

C: 1 : 1

Equal heat gives mPcPΔTP = mQcQΔTQ. Since mP = mQ/2 and ΔTP = 2ΔTQ, the factors cancel: cP = cQ. The ratio is 1:1.

A: 1.73

The reflected angle equals the 60° incidence angle. Perpendicular reflected and refracted rays imply i + r =90°, so r =30°. Snell’s law gives n = sin 60°/sin 30° = √3 ≈1.73.

B: Gamma rays, X-rays, infra-red rays, radiowaves.

In the electromagnetic spectrum, gamma rays have shorter wavelengths than X-rays, followed here by infrared and then radio waves. This gives gamma → X-ray → infrared → radio.

B: sin⁻¹(2/3)

At the critical angle, the refracted ray in air is at 90°. Snell’s law gives 1.5 sin c = 1, so c = sin⁻¹(2/3). The incident ray must be travelling from glass toward air.

A: Gives a blurred image.

In the ray model, enlarging the pinhole lets each object point illuminate a wider patch on the screen. These patches overlap more, so the image becomes brighter but more blurred.

A: Depends on the square of the amplitude of the vibrating body.

For otherwise unchanged conditions, sound intensity varies with amplitude squared. Greater intensity is perceived as greater loudness. Loudness also depends on frequency and the listener, so it is not identical to physical intensity.

C: Flute

A flute sets the air column inside its tube into vibration. A piano and guitar primarily use vibrating strings, while a handbell uses the vibration of the bell itself.

C: 1.0 A

The parallel resistance is(5 ×20)/(5 +20) =4 Ω. Adding the series 5 Ω gives 9 Ω. Since the accumulator’s internal resistance is negligible, its current is I =9.0/9 =1.0 A.

D: Transformer

A simple brushed DC motor has a field magnet, an armature and a commutator. A transformer is a separate device for transferring AC electrical energy between circuits.

D: Chemical energy.

A Leclanché cell stores chemical energy in its reactants. During discharge, chemical reactions provide electrical energy to the external circuit.

A: Iron filings cling mainly around the ends of a bar magnet

The magnetic field is strongest near the poles of a bar magnet, so iron filings gather most densely near its ends. Like poles repel, and lodestone is magnetic.

D: Angle of declination of the locality.

Magnetic declination is the horizontal angle between magnetic north and true geographic north. Inclination instead measures the magnetic field’s tilt relative to the horizontal.

D: 7.5 × 10¹⁴ Hz

Convert 400 nm to 400 × 10⁻⁹ m. Frequency f = c/λ = (3 × 10⁸)/(400 × 10⁻⁹) = 7.5 × 10¹⁴ Hz.

E: Radian

The radian measures plane angle and is a dimensionless derived unit. Metre, ampere, kelvin and second are SI base units.

A: 95 m

From rest, distance after t seconds is ½at². The final second covers ½ × 10 × (10² − 9²) = 95 m.

E: Impulse.

Impulse is the change in momentum, equal to force multiplied by time for a constant force. Its unit is N·s. Friction, tension, upthrust and weight are forces measured in newtons.

C: 40cm from the heavier mass

Measure x from the 40 g mass. The centre of mass is x = (40 × 0 + 60 × 100)/(40 + 60) = 60 cm. It is therefore 40 cm from the heavier mass.

C: 1.67

Absolute pressure includes atmospheric pressure. Thus P₁ = ρg(10 + 20) and P₂ = ρg(10 + 40). Their ratio is 50/30 = 1.67 to two decimal places.

D: 1 000 N

Pascal’s principle gives equal pressure in the two pistons: F₁/A₁ = F₂/A₂. Therefore F₂ = 100 × 5/0.5 = 1000 N.

C: 56 °C

Heat lost by Q equals heat gained by P: 1.5(80 − T) = 1.0(T − 20). Hence 140 = 2.5T, giving T = 56 °C. Equal masses cancel from the equation.

C: Work

Slope is pressure divided by reciprocal volume, so its units are pressure × volume: (N/m²) × m³ = N·m = J. These are the units of work.

B: Between 0 °C and 4 °C

Between 0 °C and 4 °C, water contracts as it warms and expands as it cools. Its density is greatest at about 4 °C. This reversal of ordinary thermal expansion is its anomalous behaviour.

B: Both its boiling point and its saturated vapour pressure at a fixed temperature depend on external pressure.

At a fixed temperature, a pure liquid’s saturated vapour pressure is determined by the liquid and its temperature. External pressure sets the boiling temperature by the condition that vapour pressure equals external pressure.

D: Radiation

Radiation transfers energy by electromagnetic waves and can cross a vacuum. A simple kinetic molecular model alone does not describe this mechanism; electromagnetic theory is also needed.

A: 0.75km

The echo time includes travel down to the seabed and back. Depth = vt/2 = 1500 × 1/2 = 750 m = 0.75 km.

B: Speed and wavelength increase but its frequency remains the same

Sound travels faster in water than in air. Its frequency stays fixed by the source as it crosses the boundary. Since λ = v/f, the wavelength increases with speed.

A: 50cm

For the fundamental of a pipe closed at one end, L = λ/4 = v/(4f). Thus L = 340/(4 × 170) = 0.50 m = 50 cm, neglecting end correction.

A: Virtual and 60cm behind the mirror

The focal length is R/2 = 20 cm. With real distances positive, 1/v = 1/20 − 1/15 = −1/60. Thus the image is virtual, 60 cm behind the mirror; the object lies inside the focus.

E: The critical angle for glass is less than 45°

In a right-angled 45° glass prism, a ray entering normally at a short face reaches the long face at 45°. If the glass–air critical angle is smaller, total internal reflection occurs there.

D: 83.6°

Each ray is 30° from the normal in water. Snell’s law gives sin r = (4/3)sin 30° = 2/3, so r = 41.81°. The emergent rays are separated by 2r = 83.6° to one decimal place.

C: High series resistance

A high resistance in series limits the current through the meter. The resulting combination measures a larger voltage because V = I(Rseries + Rmeter).

B: 28.8 ohms

Five minutes is 300 s. Power = E/t = 600000/300 = 2000 W. Using P = V²/R gives R = 240²/2000 = 28.8 Ω.

D: II and III

A current-carrying coil experiences magnetic forces that produce a turning effect in the meter’s magnetic field. These are the magnetic effect of current and the force on a current-carrying conductor, so II and III apply.

B: 20 W

Parallel resistors have the same voltage across them. From the 5 Ω branch, V² = PR = 40 × 5 = 200 V². The power in the 10 Ω branch is therefore V²/10 = 200/10 = 20 W.

E: The mass of gas set free or metal deposited is proportional to the square of the current.

Faraday’s law gives deposited mass m = MIt/(nF). For the same material and duration, mass is proportional to current, not to its square.

D: Half the acceleration

Starting from rest with constant acceleration, s = ½at². A graph of s against t² therefore has slope ½a. Equivalently, if s = kt², differentiating twice gives a = 2k.

A: 0.25 N

The uniform rule’s weight W acts at 50 cm, 20 cm left of the pivot. Taking moments about 70 cm gives 0.4 × 15 = 0.1 × 10 + W × 20, with distances in centimetres. Thus W = (6 − 1)/20 = 0.25 N.

D: I and II

Force has dimensions MLT⁻². Both moment of force and work multiply force by a distance, giving ML²T⁻². Acceleration has dimensions LT⁻², so I and II are correct.

B: 3.6 m³

The petrol mass is 720 × 4.2 = 3024 kg. For alcohol of density 840 kg/m³, volume = mass/density = 3024/840 = 3.6 m³.

A: Converging lens

A long-sighted eye does not converge light from close objects sufficiently. A converging lens adds focusing power so that the image forms on the retina.

E: At a distance greater than the radius of curvature

A concave mirror forms a real, smaller image when the object is beyond its centre of curvature. The image then lies between the focus and the centre of curvature.

C: The coulomb

Electric charge is measured in coulombs. One coulomb is the charge transferred by a current of one ampere flowing for one second: Q = It.

E: All of the above.

For a uniform wire, R = ρL/A. Resistance depends on its length, cross-sectional area (and therefore diameter), and resistivity. Temperature also affects resistivity, so all four factors matter.

A: 0.75 A

Identical cells in parallel retain emf 1.5 V, while their combined internal resistance is 2/2 = 1 Ω. Including the external resistor, I = 1.5/(1 + 1) = 0.75 A.

E: Copper rod

A zinc-carbon dry cell contains a carbon rod, a zinc casing and a paste containing ammonium chloride and manganese dioxide. A copper rod is not one of these components.

B: Low current and high voltage

High-tension transmission means high-voltage transmission. For the same power, using a high voltage permits a low current and reduces I²R heating losses in the conductors.

C: 120 J/g

The heat supplied is I²Rt = 2² × 5 × 30 = 600 J. Specific latent heat is energy per unit mass: L = 600/5 = 120 J/g.

A: One-quarter of the wavelength

In a standing wave, adjacent nodes are half a wavelength apart. An antinode is midway between them, so a node and its nearest antinode are one-quarter wavelength apart.

C: III only

At fixed temperature and composition, pressure changes do not change the ideal-gas speed of sound. Sound speed increases with temperature. The quality or timbre of a note depends on its mixture of frequencies and their relative amplitudes, so only III is incorrect.

B: 252 Hz

Adding a small load to a tuning fork lowers its frequency. The beat frequency is the difference between the two frequencies, so the loaded fork vibrates at 256 − 4 = 252 Hz.

C: First condenses into liquid form

The dew point is the temperature to which air must be cooled, without changing its water-vapour content, to become saturated and start forming liquid droplets.

B: 23.3 °C

The 210 mm interval represents 100 °C. A rise of 49 mm therefore corresponds to (49/210) × 100 = 23.33… °C, or 23.3 °C to one decimal place.

C: Sublimation

Sublimation is the direct change from solid to gas without passing through a liquid phase. Dry ice is a familiar example.

C: 4.0 × 10⁴ J/(kg·K)

Heat supplied is Pt = 100 × 20 = 2000 J. The temperature rise is 5 K, so c = Q/(mΔT) = 2000/(0.01 × 5) = 4.0 × 10⁴ J/(kg·K).

B: 2 10/13 μF

The three parallel capacitors add to 4 + 3 + 2 = 9 μF. This 9 μF combination is in series with the first 4 μF capacitor, giving C = (4 × 9)/(4 + 9) = 36/13 μF = 2 10/13 μF.

C: 10 000 m

For the stated uniform-density model, p = ρgh. Hence h = 10⁵/(1 × 10) = 10000 m. This is the height of the simplified model atmosphere.

E: Reluctance to start moving and its reluctance to stop moving once it has begun to move.

Inertia is resistance to a change in velocity. A body tends to stay at rest or continue moving with constant velocity unless a resultant external force acts.

B: 10.0 kW

The gain in gravitational potential energy is mgh = 1000 × 10 × 10 = 100000 J. Dividing by 10 s gives 10000 W = 10 kW of useful lifting power.

D: Black

In the ideal colour model, a red shirt reflects red light and absorbs green light. Under pure green illumination there is no red light to reflect, so it appears black.

B: Red, orange, yellow, green, blue, indigo, violet.

From longest to shortest visible wavelength, the usual order is red, orange, yellow, green, blue, indigo and violet. Frequency increases in the reverse direction to wavelength.

C: Streams of electrons

Cathode rays are streams of electrons. They carry negative charge and can be deflected by electric and magnetic fields.

A: 375.0 m

Convert 800 kHz to 800000 Hz. Wavelength λ = v/f = (3 × 10⁸)/800000 = 375 m.

B: ₦0.13

Total power is 5 × 50 + 4 × 100 = 650 W = 0.65 kW. In 10 hours the lamps use 6.5 kWh. At the stated rate, the cost is 6.5 × 2 = 13 kobo = ₦0.13.

B: Altitude

Altitude is a height measured from a reference level, so it has magnitude without a vector direction. Force, weight, displacement and acceleration are vectors.

D: II and III only

Horizontal equilibrium gives P₃ = P₁ cos θ₁ + P₂ cos θ₂. Vertical equilibrium gives P₁ sin θ₁ = P₂ sin θ₂. Statement I is not a general equilibrium identity: the two angles need not be equal. Therefore II and III must hold.

A: It increases if the pull increases while the brick remains at rest

While the brick remains at rest, static friction acts leftwards and matches the rightward pull. Increasing that pull increases static friction until its limiting value is reached. The limiting value can change with normal force, but the actual static friction still balances the applied pull.

E: Weight.

Weight is the gravitational force acting on an object: W = mg. Mass measures the amount of matter and is measured in kilograms; weight is measured in newtons.

B: 2.0 × 10⁸ m/s

Refractive index n = c/v, so v = c/n = (3.0 × 10⁸)/1.5 = 2.0 × 10⁸ m/s. Light travels more slowly in this liquid than in a vacuum.

A: 380g

Relative density 19 means a density of 19 g/cm³. Mass = density × volume = 19 × 20 = 380 g. Immersion changes apparent weight, not the mass of the metal.

E: At a particular depth depends on the shape of the vessel.

For a liquid at rest, pressure due to the liquid is p = ρgh. At a given depth it depends on density and depth, not on the shape of the container.

C: I and II only

Long-sightedness makes close objects difficult to focus on; short-sightedness makes distant objects blurred. Short-sightedness is corrected using diverging lenses, so I and II are correct but III is false.

D: Shorter than 400nm

Gamma rays have much shorter wavelengths than visible light. Their wavelengths are therefore shorter than 400 nm.

A: 25 cm³

For a fixed amount of ideal gas, PV/T is constant. V₂ = V₁(P₁/P₂)(T₂/T₁) = 100 × ½ × ½ = 25 cm³. Both changes reduce the volume.

A: Source, slit, converging lens, prism, converging lens, screen.

The slit provides a narrow source. The first converging lens makes the rays parallel before the prism separates the colours; the second converging lens focuses them onto the screen to form a pure spectrum.

B: I and II only

For a given transmitted power, a higher voltage requires a smaller current. Heating loss is I²R, so a smaller current reduces heating and allows a smaller conductor cross-section for the same loss limit. Higher voltage does not mean that power travels faster.

C: 10.06 cm³

For a solid, volume expansivity is approximately 3α. V = V₀(1 + 3αΔT) = 10(1 + 3 × 2 × 10⁻⁵ × 100) = 10.06 cm³.

B: 4 A

Each parallel branch has the full 24 V across it. The current in each 6 Ω resistor is I = V/R = 24/6 = 4 A. The total current through all four branches is 16 A.

E: 2.268 × 10⁻³ kg

The charge passed is Q = It = 5 × 3600 = 18000 C. Deposited mass m = ZQ = 0.126 × 10⁻⁶ × 18000 = 2.268 × 10⁻³ kg.

E: Can be refracted and diffracted.

Both water ripples and light can change direction through refraction and spread around edges or openings through diffraction. They do not need to have the same speed or wavelength to show these wave behaviours.

B: 2 × 10⁻⁹ m

One drop has mass (5 × 10⁻⁴)/1000 = 5 × 10⁻⁷ kg. Its volume is m/ρ = 10⁻⁹ m³. Film thickness = volume/area = 10⁻⁹/0.5 = 2 × 10⁻⁹ m.

B: 10.00 cm

Hooke’s law makes extension proportional to load. An extra 10 g increases the length by 2 cm, so the 20 g load produces a 4 cm extension. The natural length is 14 − 4 = 10 cm.

B: II only

Atomic number is the number of protons, so II is correct. Mass number counts protons plus neutrons. A neutral atom has equal numbers of electrons and protons, not protons plus neutrons.

C: 0.18 J/(g·°C)

Electrical energy supplied is I²Rt = 3² × 20 × 10 = 1800 J. Using Q = mcΔT, c = 1800/(1000 × 10) = 0.18 J/(g·°C), equivalent to 180 J/(kg·°C).

C: 0.40 ohms

The current through the external resistor is I = 2/4 = 0.5 A. The internal voltage drop is 2.2 − 2.0 = 0.2 V, so internal resistance r = 0.2/0.5 = 0.40 Ω.

E: East

Point your right thumb upwards along the current. Your curled fingers show the magnetic-field direction around the wire. At a point south of the wire the tangent to this circle points east, so the field at P is eastwards.

C: Heat → kinetic energy → electricity → heat and light

The steam engine changes thermal energy into mechanical motion. The generator changes that motion into electrical energy, and the lamp changes electrical energy into light and heat.

D: I and III

Copper expands more than steel, so it becomes the outside of the bend. With copper above steel, the free end bends downwards towards the steel side, away from the contact above. Reversing the layers makes it bend upwards, allowing contact. Thus I and III are correct.

B: II and III only

A wet-and-dry bulb hygrometer uses evaporative cooling to determine humidity. A hair hygrometer uses the change in hair length with humidity. A manometer measures pressure; a hydrometer measures liquid density.

D: Vertical

Reflection in a plane mirror preserves length and reverses the component perpendicular to the mirror. Reflecting a horizontal line in a line inclined at 45° changes its orientation to vertical. The image is finite and the same size as the rod.

B: 2.25 × 10⁸ m/s

The refractive index is n = c/v. Therefore v = (3.0 × 10⁸)/(4/3) = 2.25 × 10⁸ m/s.

C: 20.0 N/C

Electric field strength is force per unit charge: E = F/q = 4/0.2 = 20 N/C.

D: Polarization

Polarization requires vibrations perpendicular to the direction of travel. Longitudinal waves vibrate along their direction of travel, so they cannot be polarized.

D: The tube is completely filled with water and P is lower than W

A continuous water column must first fill the tube, and the outlet P must be below the tank’s water surface W for gravity to drive outward flow. The inlet O only needs to remain submerged; it need not reach the bottom, and P need not be lower than O.