2nd Year Physics — MCQs & Practice Questions

1. Uses E = mcΔθ with Δθ = 70 °C88 − 18, not 88[1]
2. E = 1.5 × 4200 × 70 = 4.41 × 10⁵ J[1]
3. Uses t = E/P with P = 2000 Wconverting kW to W[1]
4. t = 441000 / 2000 = 220 saccept 220.5 s[1]
5. Energy is lost to the surroundings / the container is also heatedeither reason accepted[1]

(a) 4.4 × 10⁵ J (b) 220 s (c) heat lost to surroundings

1. Temperature rises from −10 °C to 0 °C[1]
2. Horizontal section at 0 °C while the ice melts[1]
3. Temperature rises from 0 °C to 100 °C[1]
4. Longer horizontal section at 100 °C while the water boilsthe boiling plateau must be longer than the melting one[1]
5. During the flat sections the energy supplied breaks bonds between particles rather than raising kinetic energythe explanation mark[1]

1. Heating makes particles vibrate more and push further apart, so each metal expands[1]
2. The two metals expand by different amounts for the same temperature risethis is the essential point[1]
3. The strip bends toward the metal that expands less[1]
4. Used in a thermostat / fire alarm / oven switchany valid use[1]

Temperature is the average kinetic energy of the particles, measured in °C or K. Heat is the total thermal energy of all the particles, measured in joules, and depends on how many there are.

The two metals expand by different amounts for the same temperature rise, so the strip curves toward the metal that expands less.

The energy supplied is used to break the bonds holding the particles in the lattice, not to increase their kinetic energy. Since temperature measures kinetic energy, it does not change.

1. Recognises two stages: melting, then warminga single-stage answer cannot score more than two[1]
2. Melting: E₁ = mL = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J[1]
3. Warming: E₂ = mcΔθ = 0.20 × 4200 × 20[1]
4. E₂ = 1.68 × 10⁴ J[1]
5. Total = 6.68 × 10⁴ + 1.68 × 10⁴ = 8.36 × 10⁴ Jadding the two stages[1]

8.36 × 10⁴ J

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume. Both the mass and the temperature must be constant.

The particles gain kinetic energy and move faster, so they strike the walls harder and more frequently. With the same wall area, the force per unit area increases.

The gas does work pushing back the surroundings, so W is negative. With Q = 0, ΔU is negative, the internal energy falls and the temperature drops.

1. Convert: 17 °C = 290 K, 137 °C = 410 K[1]
2. Constant pressure, so V₁/T₁ = V₂/T₂Charles's law[1]
3. 0.015 / 290 = V₂ / 410[1]
4. V₂ = 0.015 × 410 / 290[1]
5. V₂ = 0.021 m³a 41% rise in kelvin gives a 41% rise in volume[1]

0.021 m³

1. Uses 1 − T_c/T_h with kelvin[1]
2. 1 − 300/500 = 0.40, so 40%[1]
3. Maximum work = 0.40 × 6000 = 2400 J[1]
4. Heat rejected = 6000 − 2400[1]
5. = 3600 Jthis cannot be avoided — the second law requires it[1]
6. Actual efficiency = 1800/6000 = 0.30, or 30%[1]
7. Lower than the maximum because the real engine is not reversible[1]
8. Friction, turbulence and heat lost to the surroundings all reduce the work obtained[1]

(a) 40% (b) 2400 J of work, 3600 J rejected (c) 30%

1. The increase in internal energy equals the heat supplied plus the work done on the system[1]
2. ΔU = Q + W[1]
3. W = +250 J (work done ON the gas), Q = −100 J (heat lost)the signs are the difficult part[1]
4. ΔU = −100 + 250 = +150 J[1]
5. The temperature rises[1]
6. Internal energy has increased, and for a gas that means greater mean kinetic energy[1]

ΔU = +150 J, so the temperature rises

The acceleration must be proportional to the displacement from a fixed equilibrium point, and it must be directed towards that point. Together these give a = −ω²x.

Speed is greatest at the equilibrium position, because all the energy is kinetic there and the net force is zero. Acceleration is greatest at maximum displacement, because the restoring force is largest there — although the body is momentarily at rest.

1. ω = 2π/T = 2π/0.80 = 7.854 rad s⁻¹Angular frequency comes first; everything else is built on it.[1]
2. v_max = ωx₀ = 7.854 × 0.040The amplitude must be converted to metres.[1]
3. v_max = 0.314 m s⁻¹Occurring at the equilibrium position.[1]
4. a_max = ω²x₀ = 61.68 × 0.040 = 2.47 m s⁻²Occurring at maximum displacement, where the speed is zero.[1]

v_max = 0.314 m s⁻¹, a_max = 2.47 m s⁻²

1. (a) T = 2π√(L/g) = 2π√(1.2/9.81) = 2π√0.12232The standard pendulum formula.[1]
2. T = 2.20 sClose to 2 s, as expected for a metre-scale pendulum.[1]
3. (b) ω = 2π/2.20 = 2.856 rad s⁻¹; v_max = ωx₀ = 2.856 × 0.050Converting 5.0 cm to 0.050 m.[1]
4. v_max = 0.143 m s⁻¹At the lowest point of the swing.[1]
5. (c) The period is unchanged, because T = 2π√(L/g) contains no amplitude termIsochronism — the property that made pendulum clocks possible.[1]
6. The total energy is quadrupled, since E ∝ x₀² and the amplitude has doubledThe squared dependence is the point of the part.[1]
7. (d) Critical damping returns the system to equilibrium in the shortest time without overshooting — used in car suspension or door closers.Definition plus a named application.[1]

(a) 2.20 s; (b) 0.143 m s⁻¹; (c) period unchanged, energy ×4; (d) fastest return without overshoot

Resonance occurs when the frequency of a driving force equals the natural frequency of the system, producing a maximum amplitude and the most efficient transfer of energy. Increasing the damping makes the peak lower and broader, and shifts it slightly to a lower frequency.

Sources with the same frequency and a constant phase difference. Without that, the phase relationship changes randomly and no stable pattern of fringes is seen.

Light is a transverse wave, so its vibrations are perpendicular to the direction of travel and one plane can be selected. Sound is longitudinal, vibrating along the direction of travel, so there is no plane to select.

Diffraction is greatest when the gap is comparable to the wavelength. Sound wavelengths are around a metre, similar to a doorway, while visible light is about 5 × 10⁻⁷ m — far too small to spread noticeably.

1. d = 1 / (300 × 10³) = 3.33 × 10⁻⁶ m300 per mm is 3 × 10⁵ per metre[1]
2. Uses d sin θ = nλ with n = 1[1]
3. sin θ = 590 × 10⁻⁹ / 3.33 × 10⁻⁶ = 0.177[1]
4. θ = 10.2°[1]
5. Highest order: n ≤ d/λ = 5.6, so n = 5sin θ cannot exceed 1[1]

θ = 10.2° for the first order; 5 orders visible

1. The two paths to the centre are equal in length[1]
2. So the path difference is zero, the waves arrive in phase and interfere constructively[1]
3. Uses x = λD/a[1]
4. x = (5.9 × 10⁻⁷ × 2.4) / 4.0 × 10⁻⁴[1]
5. x = 3.5 × 10⁻³ m, about 3.5 mm[1]
6. The fringe spacing doubles[1]
7. Because x is inversely proportional to a[1]
8. The fringes become more widely separated and easier to measure[1]

(b) 3.5 mm (c) spacing doubles

1. A laser beam is split down two perpendicular arms and recombined[1]
2. The arms are set so the returning beams cancel by destructive interference[1]
3. A tiny change in one arm's length upsets the cancellation, so light appears at the detectora fraction of a wavelength is enough[1]
4. Events such as black hole mergers emit almost no light[1]
5. But they do radiate gravitational waves, which also pass through intervening dust and gas unimpeded[1]

Electrons are transferred from the cloth onto the rod. The rod gains electrons and becomes negative; the cloth is left with a positive charge.

A region in which a charge experiences a force. The direction of the field is the direction of the force on a small positive test charge.

The rod induces a separation of charge in the paper, drawing the opposite charge to the near side. That near charge is closer to the rod, so its attraction outweighs the repulsion of the far side.

1. Converts 25 mm = 0.025 m[1]
2. Uses E = V/d = 5000 / 0.025[1]
3. E = 2.0 × 10⁵ V m⁻¹[1]
4. F = QE = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N[1]

E = 2.0 × 10⁵ V m⁻¹, F = 6.4 × 10⁻¹⁴ N

1. Friction between the tanker and the air, and between the fuel and the tank walls, transfers electrons[1]
2. The tanker is insulated by its rubber tyres, so the charge cannot escape and accumulates[1]
3. The charge raises the potential of the tanker[1]
4. Eventually a spark jumps to a nearby earthed object[1]
5. The spark could ignite the fuel vapour and cause an explosion[1]
6. The wire provides a conducting path to earth[1]
7. Electrons flow along it until the tanker is at the same potential as the earth[1]
8. So charge leaks away steadily instead of building to a spark[1]

1. The lines are parallel and equally spaced, showing a uniform field[1]
2. They run from the positive plate to the negative plate[1]
3. The force is upward, towards the positive plate[1]
4. Because the drop is negative, so the force is opposite to the field direction[1]
5. Increase the potential difference across the plates[1]
6. Move the plates closer together, or increase the charge on the dropany one[1]

(b) upward, towards the positive plate

The negative half of the cycle cancels the positive half, so the mean current is zero. Heating depends on I²R, and squaring makes the negative half positive, so energy is dissipated throughout.

The steady direct current that would deliver the same average power to a resistor as the alternating current does.

Capacitive reactance decreases, since X_C = 1/(2πfC). Inductive reactance increases, since X_L = 2πfL.

1. Uses V_rms = V_peak/√2[1]
2. V_rms = 170 / 1.414 = 120 V[1]
3. I_rms = V_rms/R = 120/40[1]
4. I_rms = 3.0 A[1]
5. P = V_rms I_rms = 120 × 3.0 = 360 Wusing peak values would double this[1]

120 V, 3.0 A, 360 W

1. A transformer works by a changing magnetic field inducing an e.m.f. in the secondary[1]
2. Direct current gives a steady field, so nothing is induced[1]
3. I = P/V = 40 × 10⁶ / 400 × 10³[1]
4. I = 100 A[1]
5. P_lost = I²R = 100² × 3.0[1]
6. = 30 kW, less than 0.1% of the output[1]
7. At 40 kV: I = 1000 Aten times the current[1]
8. P_lost = 1000² × 3.0 = 3.0 MW — a hundred times more, and 7.5% of the output[1]

(b) 100 A, 30 kW (c) 3.0 MW

1. The inductive and capacitive reactances are equal, X_L = X_C[1]
2. The current reaches its maximum[1]
3. The two reactances cancel, so the impedance falls to just the resistance — its minimum value[1]
4. Tuning a radio receiver[1]
5. Adjusting the capacitance shifts the resonant frequency to match one station, so only that signal produces a large current[1]

X_L = X_C; current is maximum; used for tuning a radio

No emission occurs below a threshold frequency, whatever the intensity. And emission is instantaneous even in very dim light, rather than requiring time for energy to build up.

Greater intensity means more photons per second, but each photon still carries the same energy hf. One photon is absorbed by one electron, so the energy per electron is unchanged — only the number emitted rises.

The minimum energy needed to remove an electron from the surface of the metal.

1. E = hf = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵[1]
2. E = 7.96 × 10⁻¹⁹ J[1]
3. Converts: 7.96 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 4.97 eV[1]
4. Uses KE_max = hf − φ = 4.97 − 3.2[1]
5. KE_max = 1.8 eV[1]

photon 4.97 eV, electrons up to 1.8 eV

1. Uses KE = eV = 1.6 × 10⁻¹⁹ × 2500[1]
2. KE = 4.0 × 10⁻¹⁶ J[1]
3. Uses KE = ½mv²[1]
4. v² = 2 × 4.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹[1]
5. v = 3.0 × 10⁷ m s⁻¹[1]
6. λ = h/mv = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 3.0 × 10⁷)[1]
7. λ = 2.4 × 10⁻¹¹ mcomparable to atomic spacing, which is why diffraction is observable[1]
8. A diffraction pattern of rings, showing that electrons behave as waves[1]

4.0 × 10⁻¹⁶ J, 3.0 × 10⁷ m s⁻¹, 2.4 × 10⁻¹¹ m, rings

1. Wave: interference or diffraction (for example the double-slit experiment)[1]
2. Particle: the photoelectric effect, or line spectra[1]
3. Its de Broglie wavelength is λ = h/mv[1]
4. Its mass and speed are enormous compared with h, so λ is around 10⁻³⁴ m[1]
5. That is far smaller than any gap it could pass through, so no diffraction is observable[1]

interference/diffraction for waves; photoelectric effect for particles

1. 92 protonsthe proton number[1]
2. 235 − 92 = 143 neutrons[1]
3. 92 electrons, since the atom is neutral[1]
4. Same number of protons (92); different number of neutrons (146 instead of 143)[1]
5. Chemical behaviour depends on the electrons[1]
6. Both have 92 electrons arranged identically, so they react in the same way[1]

(a) 92 p, 143 n, 92 e (b) same Z, different A (c) identical electron arrangement

1. Most alpha particles passed straight through[1]
2. So the atom is mostly empty space[1]
3. Some were deflected through large angles[1]
4. So there is a concentrated positive charge repelling them[1]
5. A very few were reflected almost straight back[1]
6. So the nucleus is very small and contains most of the atom's mass[1]

1. In that model the positive charge is spread thinly throughout the atom[1]
2. So the repulsive force on an alpha particle anywhere would be small[1]
3. It could not produce the large-angle deflections or backscattering that were observed[1]

The nucleon number A is the total number of protons and neutrons in the nucleus. The proton number Z is the number of protons alone.

They have the same number of protons and therefore the same number of electrons in the same arrangement. Chemical behaviour is decided by the electrons, not by the number of neutrons.

1. Almost all the alpha particles passed straight through[1]
2. A very small fraction were deflected through large angles, some straight back[1]
3. Passing straight through shows the atom is mostly empty space[1]
4. Large deflections show a concentrated region of positive charge[1]
5. Which repels the positive alpha particle[1]
6. Backward scattering shows that region is also very massive, and very small[1]
7. So each alpha particle meets at most one nucleus[1]
8. A thicker foil would cause multiple scattering and the result could not be interpreted[1]

1. 18 hours is 3 half-lives18 ÷ 6[1]
2. Halves three times: 800 → 400 → 200 → 100[1]
3. 100 counts per minute[1]
4. Each half-life removes only half of what remains, so some always remains[1]
5. Temperature or pressure[1]
6. Chemical state or physical form of the sampleany second valid factor[1]

(a) 100 counts/min (b) halving never reaches zero (c) temperature, pressure, chemical state

1. Alpha: A = 218 − 4 = 214[1]
2. Z = 84 − 2 = 82lead[1]
3. Beta: A unchanged at 214[1]
4. Z = 82 + 1 = 83bismuth[1]
5. A neutron changes into a proton and an electron, and the electron is emitted[1]

(a) ²¹⁴₈₂Pb (b) ²¹⁴₈₃ (c) a neutron becomes a proton plus an emitted electron

1. Gamma[1]
2. It is penetrating enough to leave the body and be detected, and least ionising so it does least damageboth halves wanted[1]
3. Short, so the activity falls quickly and the patient is not exposed for long[1]
4. But not so short that it decays away before the scan can be completedthe balance is the point of the question[1]

Handle it with long tongs to increase the distance from the body, and return it to its lead-lined container immediately after use. (Also acceptable: never point it at anyone; minimise exposure time.)

The low level of ionising radiation always present in the environment. Sources include radon gas from rocks, cosmic rays, medical X-rays and food.

1. The time taken for the number of undecayed nuclei in the sample to halve[1]
2. Equivalently, the time for the count rate to fall to half its value; the process is random so this is an average[1]
3. 32 days is 4 half-lives[1]
4. Uses 640 ÷ 2⁴[1]
5. = 40 counts per minute[1]
6. The extra 15 counts per minute is background radiation[1]
7. It is measured with the source removed and subtracted from every reading[1]
8. The proton number increases by 1[1]
9. The nucleon number is unchangeda neutron becomes a proton plus an electron[1]

(b) 40 counts per minute

The difference between the total mass of the separate nucleons and the mass of the assembled nucleus. The nucleus is lighter, because energy was released when it formed.

Total binding energy simply grows with the number of nucleons. Dividing by that number gives a fair measure of how tightly each nucleon is held, and therefore of stability.

The moderator slows fast neutrons down, because slow neutrons are much more readily absorbed and sustain the chain reaction. The control rods absorb neutrons, and are moved to keep the reaction steady.

1. Uses E = Δm × 931.5[1]
2. E = 0.0304 × 931.5[1]
3. E = 28.3 MeV[1]
4. Divides by the 4 nucleons[1]
5. = 7.1 MeV per nucleonhigh for such a light nucleus, which is why helium is so stable[1]

28.3 MeV total, 7.1 MeV per nucleon

1. Iron-56 has the greatest binding energy per nucleon[1]
2. So it is the most stable nucleus[1]
3. Joining light nuclei produces a nucleus higher on the curve[1]
4. The products are more tightly bound, so energy is released[1]
5. Splitting a heavy nucleus produces fragments higher on the curve[1]
6. Again the products are more tightly bound, so energy is released[1]
7. Fusing iron would produce nuclei lower on the curve, which absorbs energy rather than releasing it[1]
8. With no energy released, the outward pressure supporting the star fails and the core collapses[1]

iron is most stable; both processes move products up the curve; fusing iron would absorb energy

1. Fusion produces no long-lived radioactive waste — the product is helium[1]
2. And its fuel, hydrogen isotopes, is effectively unlimitedeither point, developed[1]
3. Both nuclei are positively charged and repel strongly[1]
4. They must be brought close enough for the strong nuclear force to act, needing temperatures of millions of kelvin[1]
5. At those temperatures the plasma cannot touch any container, so it must be confined magnetically, and holding it stable long enough to gain net energy is unsolved[1]
6. Fission[1]

fusion: no long-lived waste, unlimited fuel; difficulty is containment at millions of kelvin; fission is used today

Air has a far lower acoustic impedance than skin, so at an air–skin boundary almost all the ultrasound would be reflected and very little would enter the body. The gel has an impedance close to that of skin, which removes the air gap and allows the pulse to be transmitted.

Momentum must be conserved. The electron and positron have almost no momentum before the event, so the total afterwards must be near zero. A single photon would carry momentum in one direction, which is impossible — so two photons are emitted in almost exactly opposite directions.

1. Distance travelled = ct = 1540 × 65 × 10⁻⁶ = 0.1001 mThe total path length, there and back.[1]
2. Depth = 0.1001/2 = 0.050 m = 5.0 cmHalving for the round trip is the step most often forgotten.[1]
3. Z = ρc = 1060 × 1540Acoustic impedance is density times speed.[1]
4. Z = 1.63 × 10⁶ kg m⁻² s⁻¹Typical of soft tissue, and far greater than air.[1]

Depth 5.0 cm; Z = 1.63 × 10⁶ kg m⁻² s⁻¹

1. (a) Advantage: ultrasound is non-ionising, so it is safe for a fetus. Disadvantage: it gives poorer resolution and cannot image through bone or gas.One of each is required; both must be genuine comparisons.[1]
2. (b) I/I₀ = e^(−0.60 × 4.0) = e^(−2.4)Units of µ and x are consistent in cm.[1]
3. = 0.0907, about 9.1%Bone attenuates strongly, which is why it appears white on a radiograph.[1]
4. (c) x½ = ln2/µ = 0.693/0.60 = 1.16 cm4.0 cm is nearly 3.5 half-value thicknesses, consistent with roughly 9% transmitted.[1]
5. (d) The tracer accumulates where metabolic activity is highest, so the image shows how tissue is functioning rather than its shape.The contrast with structural imaging must be explicit.[1]
6. (e) E = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ JEach photon carries the rest energy of one electron.[1]
7. E = 8.20 × 10⁻¹⁴ / 1.60 × 10⁻¹³ = 0.51 MeVConverting joules to MeV by dividing by 1.60 × 10⁻¹³.[1]

(b) 9.1%; (c) 1.16 cm; (e) 0.51 MeV

The half-value thickness is the thickness of a material that reduces the transmitted intensity to half its incident value. It is independent of the initial intensity because the attenuation is exponential: equal thicknesses remove equal fractions, not equal amounts.

A standard candle is an astronomical object whose luminosity is known independently of its distance — Type Ia supernovae are the standard example. It is needed because a faint appearance could mean either a dim object or a distant one; knowing the true output allows the two to be separated and the distance calculated.

λ_max × T = 2.90 × 10⁻³ m K — the peak wavelength is inversely proportional to the surface temperature. A blue star peaks at a shorter wavelength, so it is hotter than a red star.

1. F = L/(4πd²) = 8.0 × 10²⁶ / (4π × (4.0 × 10¹⁷)²)The distance must be squared inside the bracket.[1]
2. 4π(1.6 × 10³⁵) = 2.011 × 10³⁶Evaluating the denominator separately reduces errors.[1]
3. F = 3.98 × 10⁻¹⁰ W m⁻²Very small, as expected at interstellar distances.[1]
4. T = 2.90 × 10⁻³ / 550 × 10⁻⁹ = 5270 KConverting nanometres to metres first.[1]

F = 3.98 × 10⁻¹⁰ W m⁻², T = 5270 K

1. (a) Δλ = 672.1 − 656.3 = 15.8 nmObserved minus laboratory wavelength.[1]
2. z = Δλ/λ = 15.8/656.3 = 0.0241A dimensionless ratio, so the nm units cancel and need not be converted.[1]
3. (b) v = zc = 0.0241 × 3.00 × 10⁸ = 7.22 × 10⁶ m s⁻¹Well below c, so the approximation z = v/c is valid.[1]
4. (c) d = v/H₀ = 7.22 × 10⁶ / 2.2 × 10⁻¹⁸Rearranging Hubble's law.[1]
5. d = 3.28 × 10²⁴ mRoughly 350 million light years.[1]
6. (d) Galaxies recede in every direction with speed proportional to distance, so space itself is expanding rather than us sitting at the centre. Extrapolating that expansion backwards implies everything was once together — the Big Bang.The all-directions point is essential; without it the data could be read as an ordinary explosion with us at its centre.[1]

(a) 0.0241; (b) 7.22 × 10⁶ m s⁻¹; (c) 3.28 × 10²⁴ m; (d) expansion in all directions run backwards

Stefan's law gives L = 4πr²σT⁴, so with the radius unchanged the luminosity depends on T⁴. Doubling the temperature multiplies the luminosity by 2⁴ = 16.