2nd Year Physics — MCQs & Practice Questions

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 rate of flow of electric charge, I = Q/t. Its unit is the ampere (A), where one ampere is one coulomb per second.

The current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature stays constant.

The metal ions vibrate more strongly, so the moving electrons collide with them more often. Each collision impedes the flow, so the resistance rises.

1. Converts the area: 0.50 mm² = 0.50 × 10⁻⁶ m²the step most often dropped[1]
2. Uses ρ = RA/L[1]
3. ρ = (0.068 × 0.50 × 10⁻⁶) / 2.0[1]
4. ρ = 1.7 × 10⁻⁸ Ω mcopper[1]
5. Doubling the length doubles the resistance: 0.136 Ω[1]

ρ = 1.7 × 10⁻⁸ Ω m; R = 0.14 Ω

1. The graph passes through the origin[1]
2. It is a straight line at low potential difference[1]
3. It then curves towards the V axis, so the gradient falls[1]
4. A larger current heats the filament[1]
5. The ions vibrate more and the electrons collide with them more often[1]
6. So the resistance increases and the current no longer rises in proportion[1]
7. R = V/I = 6.0 / 0.50 = 12 Ω[1]
8. P = VI = 6.0 × 0.50 = 3.0 W[1]

(c) 12 Ω and 3.0 W

1. Total resistance = 20 + 25 = 45 Ω[1]
2. Uses I = V/R[1]
3. I = 9.0 / 45 = 0.20 A[1]
4. Uses V = IR for the thermistor[1]
5. V = 0.20 × 25 = 5.0 V[1]
6. The potential difference across the thermistor decreases[1]
7. Its resistance falls as it warms, so it takes a smaller share of the supply voltage[1]

(a) 0.20 A (b) 5.0 V (c) it falls

The force per unit current per unit length on a conductor placed at right angles to the field, B = F/(IL). Its unit is the tesla (T), equal to 1 N A⁻¹ m⁻¹.

The force qvB always acts at right angles to the velocity. A perpendicular force does no work, so the kinetic energy and therefore the speed are unchanged — only the direction alters.

The force is zero, because F = BIL sin θ and sin 0° = 0. Only the component of the wire perpendicular to the field experiences a force.

1. Uses F = BIL with sin 90° = 1[1]
2. F = 0.15 × 4.0 × 0.25[1]
3. F = 0.15 N[1]
4. At 30°: F = BIL sin 30° = 0.15 × 0.5[1]
5. F = 0.075 Nhalf, because sin 30° = 0.5[1]

0.15 N, then 0.075 N

1. Uses F = qvB[1]
2. F = 1.60 × 10⁻¹⁹ × 2.0 × 10⁶ × 0.35 = 1.12 × 10⁻¹³ N[1]
3. The force is always perpendicular to the velocity[1]
4. So it changes the direction of motion but not the speed[1]
5. A constant force at right angles to a constant speed is centripetal, giving circular motion[1]
6. Sets qvB = mv²/r[1]
7. r = mv/(qB) = (1.67 × 10⁻²⁷ × 2.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.35)[1]
8. r = 0.060 m, about 6 cm[1]

(a) 1.12 × 10⁻¹³ N (c) 0.060 m

1. They attractthe opposite of what most people expect[1]
2. Each wire sits in the magnetic field produced by the other[1]
3. The field of the first wire at the second is perpendicular to that wireright-hand grip rule[1]
4. Fleming's left-hand rule then gives a force on the second wire directed towards the first[1]
5. Reversing one current makes them repel[1]

attract; reversing one current makes them repel

The product of the magnetic flux density and the area perpendicular to the field, Φ = BA cos θ. Its unit is the weber (Wb).

The flux through the coil is not changing. An e.m.f. depends on the rate of change of flux linkage, and that rate is zero — the strength of the magnet is irrelevant.

The induced current always opposes the change producing it. It follows from conservation of energy — if the current aided the change, energy would be created from nothing.

1. Uses e.m.f. = BLv[1]
2. = 0.25 × 0.40 × 6.0[1]
3. = 0.60 V[1]
4. At 30°: e.m.f. = BLv sin 30°only the perpendicular component cuts field lines[1]
5. = 0.30 V[1]

0.60 V, then 0.30 V

1. Deflects one way as the magnet approaches[1]
2. Falls to zero at the instant the magnet is centred, where the flux is momentarily not changing[1]
3. Deflects the opposite way as it leaves[1]
4. The flux through the coil is increasing as the magnet approaches[1]
5. The induced current flows so as to oppose that increase[1]
6. So the near face of the coil becomes a north pole, repelling the magnet[1]
7. With a complete circuit an induced current flows and opposes the motion[1]
8. Work is done against that force, converting gravitational energy into electrical energy rather than kinetic[1]

deflect, zero at the centre, deflect the other way

1. Face-on, the flux through the coil is at its maximum[1]
2. At a maximum the flux is momentarily not changing, and e.m.f. depends on the rate of change[1]
3. When the coil is edge-on to the field, a quarter turn laterflux is zero but changing fastest[1]
4. They maintain electrical contact with the rotating coil without reversing the connections, preserving the a.c. output[1]
5. Rotate the coil faster[1]
6. Use a stronger field, more turns, or a larger coil areaany one[1]

zero face-on, maximum edge-on

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

A crystalline solid has a regular repeating lattice and a sharp melting point. An amorphous solid has no long-range order and softens over a range of temperature.

In elastic deformation the atoms are pulled slightly further apart but keep the same neighbours, so the material returns to its original length. In plastic deformation planes of atoms slip permanently over one another, so it does not.

In a semiconductor, thermal energy lifts more electrons across the small band gap into the conduction band, so there are more charge carriers. In a metal the number of carriers is fixed, and heating makes the ions vibrate more so the electrons collide with them more often.

1. σ = F/A = 38 / 3.5 × 10⁻⁷[1]
2. σ = 1.09 × 10⁸ Pa[1]
3. ε = e/L = 1.8 × 10⁻³ / 2.0convert mm to m[1]
4. ε = 9.0 × 10⁻⁴no units[1]
5. E = σ/ε = 1.2 × 10¹¹ Paabout right for copper[1]

σ = 1.09 × 10⁸ Pa, ε = 9.0 × 10⁻⁴, E = 1.2 × 10¹¹ Pa

1. Copper: straight line at first, then curving over into a long plastic region before breaking[1]
2. Copper reaches a maximum stress and then extends considerably before failure[1]
3. Glass: straight line all the way[1]
4. Glass breaks abruptly at the end of the straight line, with no plastic region[1]
5. Copper is ductile — it has a long plastic region and can be drawn into wire[1]
6. Glass is brittle — it breaks with almost no plastic deformation[1]
7. The gradient is the Young modulus[1]
8. It measures the stiffness of the material and is independent of the sample dimensions[1]

copper ductile with a long plastic region; glass brittle; gradient = Young modulus

1. A range of energies that electrons in the solid are not allowed to have[1]
2. It separates the valence band from the conduction band[1]
3. Conductor: no gap — the bands overlap or the upper band is part-filled[1]
4. Semiconductor: a small gap, about 1 eV, which some electrons cross at room temperature[1]
5. Insulator: a large gap of several eV that electrons cannot cross at ordinary temperatures[1]

no gap / small gap / large gap

Adding a small amount of another element to a pure semiconductor to change its conductivity. It produces n-type, where the carriers are free electrons, and p-type, where they are holes.

In a semiconductor, heat frees more charge carriers, so more current can flow. In a metal the number of carriers is fixed, and heating makes the ions vibrate more so the electrons collide with them more often.

The depletion layer widens and essentially no current flows. This one-way behaviour is used to rectify alternating current into direct current.

1. The two components are in series across the supply[1]
2. They share the supply voltage in proportion to their resistances[1]
3. As it gets darker the resistance of the LDR increases[1]
4. So the LDR takes a larger share of the supply voltage[1]
5. The base voltage, taken across the LDR, therefore rises[1]
6. When the base voltage exceeds about 0.6 V the transistor switches on[1]
7. A small base current then allows a much larger collector current[1]
8. That current flows through the lamp, lighting ita relay would be used for a mains lamp[1]

1. An OR gate is required[1]
2. Inputs 0,0 give output 0[1]
3. Inputs 0,1 and 1,0 both give output 1[1]
4. Inputs 1,1 give output 1[1]
5. An AND gate would now be needed[1]
6. The alarm would not sound if only one sensor detected the fire[1]
7. A real fire could therefore be missed, so the design is less safe[1]

(a) OR gate (b) AND gate — but it fails to warn when only one sensor triggers

1. Only the positive halves of each cycle appear[1]
2. The negative halves are missing, leaving gaps along the axis[1]
3. Half-wave rectification[1]
4. Add a capacitor in parallel with the resistor[1]
5. It charges during each pulse and discharges through the gaps, holding the voltage upa four-diode bridge would also give full-wave rectification[1]

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

One of a set of discrete energies that an electron in the atom is allowed to have. It cannot possess any energy between two levels.

Electrons fall between fixed energy levels, so the energy differences take only certain values. Since ΔE = hf, only certain frequencies are emitted.

Both arise from the same set of energy gaps. Absorption lifts an electron across a gap and emission drops it back across the same gap, so the photon energies — and therefore the wavelengths — are identical.

1. ΔE = −1.5 − (−5.4) = 3.9 eVpositive, since energy is released[1]
2. Converts: 3.9 × 1.6 × 10⁻¹⁹ = 6.24 × 10⁻¹⁹ J[1]
3. Uses f = ΔE/h = 6.24 × 10⁻¹⁹ / 6.63 × 10⁻³⁴[1]
4. f = 9.4 × 10¹⁴ Hz[1]
5. λ = c/f = 3.0 × 10⁸ / 9.4 × 10¹⁴ = 3.2 × 10⁻⁷ mjust into the ultraviolet[1]

6.24 × 10⁻¹⁹ J, 9.4 × 10¹⁴ Hz, 3.2 × 10⁻⁷ m

1. The hot interior of the Sun emits a continuous spectrum containing all wavelengths[1]
2. This light passes outward through the cooler gases of the Sun's atmosphere[1]
3. Atoms there absorb photons whose energy exactly matches a gap between their energy levels[1]
4. Those wavelengths are removed from the beam, leaving dark lines[1]
5. Each element has a unique set of energy levels and so a unique pattern of lines[1]
6. Matching the observed pattern against laboratory spectra identifies the elements[1]
7. That energy levels in atoms are discrete rather than continuous[1]
8. And that light is emitted and absorbed in quanta of energy hf[1]

absorption by cooler outer gases; unique patterns identify elements

1. The energy needed to remove the electron completely from the atom[1]
2. 13.6 eV, since the level must be raised from −13.6 eV to zero[1]
3. −3.4 − (−13.6) = 10.2 eV[1]
4. Absorption occurs only if the photon energy matches a gap between levels exactly[1]
5. 8.0 eV matches no gap from the ground state — the first is 10.2 eV — so the photon passes straight through[1]

13.6 eV; 10.2 eV; 8.0 eV matches no gap

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