Physics: Unit-wise Subjective Question Bank

Question bank · 28 min read

Physics: Unit-wise Subjective Question Bank

Groups B and C of the Grade 11 Physics paper are sixty-four of the seventy-five marks: eight questions of five marks and three of eight. This chapter gives you worked practice for both, arranged by the five content areas of the CDC syllabus — Mechanics, Heat and Thermodynamics, Waves and Optics, Electricity and Magnetism, and Modern Physics. Every answer is written the way you should write it in the hall: law first, substitution with units, answer boxed in words.

How to use this chapter
  • Attempt first. Cover the answer, write your own against the clock (12 minutes per 5-marker, 19 per 8-marker), then mark yourself against the model.
  • Split the marks. A 5-mark numerical is usually law (1), substitution (2), arithmetic (1), unit and answer (1). An 8-mark question combines a derivation (3–4), a diagram (1–2) and a numerical (2–3).
  • Use the stated g. Questions here say g = 10\ \text{m s}^{-2} or 9.8\ \text{m s}^{-2}; use whatever your paper states.
About years and sources

Every question below is an original practice question written to the CDC pattern and carries no past-paper year. The syllabus outline asks for "years each appeared", but Grade 11 papers are not published by NEB, so no year is claimed here. Instead, the section headings give each area's teaching hours (Mechanics 45, Heat 26, Waves and Optics 15, Electricity and Magnetism 27, Modern Physics 15 of 128), the public measure of how much of the paper to expect.

3.1 Mechanics (units 1–8, 45 hours)

Group B — Short answer (5 marks each)

  1. Using dimensional analysis, derive an expression for the time period T of a simple pendulum, assuming it depends on its length l, the mass m of the bob and the acceleration due to gravity g. State one limitation of the method. [5]
  2. Two forces of 10\ \text{N} and 6\ \text{N} act at a point with an angle of 60^{\circ} between them. Find the magnitude and direction of the resultant. [5]
  3. A ball is projected with a speed of 20\ \text{m s}^{-1} at 30^{\circ} above the horizontal (g = 10\ \text{m s}^{-2}). Find the time of flight, the maximum height and the horizontal range. [5]
  4. A gun of mass 5\ \text{kg} fires a bullet of mass 20\ \text{g} at 400\ \text{m s}^{-1}. State the principle you use and find the recoil velocity of the gun. [5]
  5. State the laws of solid friction. A 20\ \text{kg} block rests on a horizontal floor with \mu = 0.3. Find the horizontal force needed to move it at constant speed (g = 9.8\ \text{m s}^{-2}). [5]
  6. A 2\ \text{kg} body is released from rest at a height of 20\ \text{m} (g = 10\ \text{m s}^{-2}). Using energy, find its speed when it is 5\ \text{m} above the ground, and show that the total energy there equals the initial energy. [5]
  7. Derive an expression for the acceleration due to gravity at a height h above the Earth's surface. Find g at a height R/2 (g_0 = 9.8\ \text{m s}^{-2}). [5]
  8. A car of mass 1000\ \text{kg} rounds a level curve of radius 50\ \text{m} at 10\ \text{m s}^{-1}. Find the centripetal force, and the banking angle at which no friction is needed on this curve (g = 10\ \text{m s}^{-2}). [5]
  9. A steel wire 2\ \text{m} long with cross-section 1\ \text{mm}^{2} is stretched by 1\ \text{mm} by a force of 100\ \text{N}. Find the Young modulus and the elastic energy stored. [5]
Answers — Group B, Mechanics

1. Let T = k\,l^{a}m^{b}g^{c}. Dimensions: [T] = T, [l] = L, [m] = M, [g] = LT^{-2}. So T^{1}= L^{a}M^{b}L^{c}T^{-2c}. Comparing: M: b = 0; T: -2c = 1 so c = -\tfrac12; L: a + c = 0 so a = \tfrac12. Hence T = k\sqrt{l/g}, independent of mass. Limitation: the dimensionless constant k (= 2\pi) cannot be found by this method, and it fails if a quantity depends on a sum or a function such as \sin\theta.

2. R = \sqrt{P^{2} + Q^{2} + 2PQ\cos\theta} = \sqrt{100 + 36 + 2(10)(6)(0.5)} = \sqrt{196} = 14\ \text{N}. Direction from the 10\ \text{N} force: \tan\alpha = \dfrac{Q\sin\theta}{P + Q\cos\theta} = \dfrac{6(0.866)}{10 + 3} = 0.400, so \alpha \approx 21.8^{\circ}.

3. T = \dfrac{2u\sin\theta}{g} = \dfrac{2(20)(0.5)}{10} = 2.0\ \text{s}. H = \dfrac{u^{2}\sin^{2}\theta}{2g} = \dfrac{400(0.25)}{20} = 5.0\ \text{m}. R = \dfrac{u^{2}\sin2\theta}{g} = \dfrac{400(0.866)}{10} = 34.6\ \text{m}.

4. Principle: conservation of linear momentum (no external horizontal force). 0 = m_bv_b + Mv_g, so v_g = -\dfrac{m_bv_b}{M} = -\dfrac{0.020\times400}{5} = -1.6\ \text{m s}^{-1}: the gun recoils at 1.6\ \text{m s}^{-1} opposite to the bullet.

5. Laws: (i) friction opposes relative motion; (ii) limiting friction is independent of the area of contact; (iii) F \propto R, F = \mu R; (iv) kinetic friction is less than limiting friction and independent of speed. R = mg = 20\times9.8 = 196\ \text{N}; F = \mu R = 0.3\times196 = 58.8\ \text{N}.

6. Fall of 15\ \text{m}: v^{2} = 2gh = 2(10)(15) = 300, so v = 17.3\ \text{m s}^{-1}. At 5\ \text{m}: K = \tfrac12(2)(300) = 300\ \text{J}, U = mgh = 2(10)(5) = 100\ \text{J}, total 400\ \text{J}. Initially U = 2(10)(20) = 400\ \text{J}, K = 0. Equal, so mechanical energy is conserved.

7. At the surface g_0 = GM/R^{2}. At height h: g = \dfrac{GM}{(R+h)^{2}}, so g = g_0\left(\dfrac{R}{R+h}\right)^{2} = g_0\left(1 + \dfrac{h}{R}\right)^{-2}. For h = R/2: g = 9.8\times\left(\dfrac{2}{3}\right)^{2} = 9.8\times\dfrac{4}{9} = 4.36\ \text{m s}^{-2}.

8. F = \dfrac{mv^{2}}{r} = \dfrac{1000\times100}{50} = 2000\ \text{N}. With banking, \tan\theta = \dfrac{v^{2}}{rg} = \dfrac{100}{50\times10} = 0.2, so \theta = 11.3^{\circ}.

9. Y = \dfrac{FL}{A\,\Delta l} = \dfrac{100\times2}{(1\times10^{-6})(1\times10^{-3})} = 2\times10^{11}\ \text{N m}^{-2}. Energy = \tfrac12F\Delta l = \tfrac12(100)(10^{-3}) = 0.05\ \text{J}.

Group C — Long answer (8 marks each)

  1. (a) Derive the three equations of uniformly accelerated motion using the velocity–time graph. [5](b) A car accelerates uniformly from 10\ \text{m s}^{-1} to 30\ \text{m s}^{-1} in 200\ \text{m}. Find the acceleration and the time taken. [3] [8]
  2. (a) Derive an expression for the orbital speed and period of a satellite at height h above the Earth. [4](b) What is a geostationary satellite? State its conditions. [2](c) For R = 6400\ \text{km}, g = 10\ \text{m s}^{-2} and h = 3600\ \text{km}, find the orbital speed and period. [2] [8]
Answers — Group C, Mechanics

1. (a) Draw a straight line AB on a v–t graph from v = u at t = 0 to v = v at time t. Slope = a, so a = \dfrac{v-u}{t}, giving v = u + at. The displacement is the area under the line: a rectangle ut plus a triangle \tfrac12(v-u)t, so s = ut + \tfrac12(v-u)t = ut + \tfrac12at^{2}, giving s = ut + \tfrac12at^{2}. The area is also the trapezoid s = \tfrac12(u+v)t; substituting t = (v-u)/a gives s = \dfrac{v^{2}-u^{2}}{2a}, so v^{2} = u^{2} + 2as. (b) a = \dfrac{v^{2}-u^{2}}{2s} = \dfrac{900 - 100}{400} = 2.0\ \text{m s}^{-2}; t = \dfrac{v-u}{a} = \dfrac{20}{2} = 10\ \text{s}.

2. (a) Gravity supplies the centripetal force: \dfrac{mv^{2}}{R+h} = \dfrac{GMm}{(R+h)^{2}}, so v = \sqrt{\dfrac{GM}{R+h}} = R\sqrt{\dfrac{g}{R+h}} (using GM = gR^{2}). Period T = \dfrac{2\pi(R+h)}{v} = 2\pi\sqrt{\dfrac{(R+h)^{3}}{GM}}. Both are independent of the satellite's mass. (b) A satellite that appears fixed over one point of the Earth. Conditions: orbit in the equatorial plane; period 24\ \text{h}; same sense as Earth's rotation (west to east); height about 36\,000\ \text{km}. (c) R + h = 10^{7}\ \text{m}; v = 6.4\times10^{6}\sqrt{\dfrac{10}{10^{7}}} = 6.4\times10^{6}\times10^{-3} = 6400\ \text{m s}^{-1}. T = \dfrac{2\pi\times10^{7}}{6400} = 9.8\times10^{3}\ \text{s} \approx 2.7\ \text{h}.

3.2 Heat and Thermodynamics (units 9–13, 26 hours)

Group B — Short answer (5 marks each)

  1. Define linear, superficial and cubical expansivity. Show that \gamma = 3\alpha for an isotropic solid. [5]
  2. State Newton's law of cooling. A body cools from 80^{\circ}\text{C} to 60^{\circ}\text{C} in 10 minutes in a room at 20^{\circ}\text{C}. Estimate the time to cool from 60^{\circ}\text{C} to 40^{\circ}\text{C} using the average-temperature form. [5]
  3. A 0.5\ \text{kg} metal block at 100^{\circ}\text{C} is dropped into 0.4\ \text{kg} of water at 20^{\circ}\text{C} and the mixture reaches 30^{\circ}\text{C}. Find the specific heat capacity of the metal (c_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}; neglect the container). [5]
  4. Find the heat required to change 10\ \text{g} of ice at -10^{\circ}\text{C} into steam at 100^{\circ}\text{C}. (c_{ice} = 2100, c_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}, L_f = 3.36\times10^{5}, L_v = 2.26\times10^{6}\ \text{J kg}^{-1}) [5]
  5. Define thermal conductivity. Find the rate of heat flow through a glass pane of area 2\ \text{m}^{2} and thickness 0.1\ \text{m}... use k = 0.8\ \text{W m}^{-1}\text{K}^{-1} with faces 20\ \text{K} apart. [5]
  6. State Stefan–Boltzmann's law. A black body of surface area 0.01\ \text{m}^{2} is at 727^{\circ}\text{C}. Find the power radiated (\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}). [5]
  7. State the zeroth law of thermodynamics and explain how it underlies the use of a thermometer. [5]
  8. Calculate the volume of 2 mol of an ideal gas at 27^{\circ}\text{C} and 1.013\times10^{5}\ \text{Pa} (R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}). [5]
Answers — Group B, Heat

1. \alpha = \dfrac{\Delta l}{l\,\Delta\theta} (per unit length per kelvin); \beta = \dfrac{\Delta A}{A\,\Delta\theta}; \gamma = \dfrac{\Delta V}{V\,\Delta\theta}. A cube of side l_0: V_0 = l_0^{3}; at \theta, l = l_0(1+\alpha\theta), so V = l_0^{3}(1+\alpha\theta)^{3} \approx V_0(1 + 3\alpha\theta) (neglecting \alpha^{2}, \alpha^{3}). Comparing with V = V_0(1+\gamma\theta): \gamma = 3\alpha. Similarly \beta = 2\alpha.

2. The rate of loss of heat of a body is proportional to the excess of its temperature over the surroundings (for small differences). First interval: mean excess = 70 - 20 = 50; fall = 20 in 10 min, so rate = 2.0\ \text{K min}^{-1} at 50\ \text{K} excess, i.e. 0.04\ \text{min}^{-1} per kelvin. Second interval: mean = 50^{\circ}\text{C}, excess 30; rate = 0.04\times30 = 1.2\ \text{K min}^{-1}; time = 20/1.2 = 16.7 min.

3. Heat lost by metal = heat gained by water: 0.5\,c\,(100-30) = 0.4(4200)(30-20), so 35c = 16800, c = 480\ \text{J kg}^{-1}\text{K}^{-1}.

4. Ice warms -10 \to 0: 0.010\times2100\times10 = 210\ \text{J}. Melts: 0.010\times3.36\times10^{5} = 3360\ \text{J}. Water 0 \to 100: 0.010\times4200\times100 = 4200\ \text{J}. Boils: 0.010\times2.26\times10^{6} = 22\,600\ \text{J}. Total = 30\,370\ \text{J} \approx 3.0\times10^{4}\ \text{J}.

5. Thermal conductivity k: the heat flowing per second through unit area of a slab of unit thickness when the faces differ by 1\ \text{K} (steady state), unit \text{W m}^{-1}\text{K}^{-1}. H = \dfrac{kA\Delta\theta}{d} = \dfrac{0.8\times2\times20}{0.1} = 320\ \text{W}.

6. The energy radiated per second by a black body is proportional to the fourth power of its absolute temperature and its area: P = \sigma AT^{4}. T = 727 + 273 = 1000\ \text{K}. P = 5.67\times10^{-8}\times0.01\times10^{12} = 567\ \text{W}.

7. If two bodies are each in thermal equilibrium with a third, they are in thermal equilibrium with each other. A thermometer is the third body: when it reads the same for two bodies they have the same temperature and no net heat flows between them. In a mercury thermometer, the mercury reaches equilibrium with the body and its column length marks the common temperature.

8. T = 300\ \text{K}. V = \dfrac{nRT}{P} = \dfrac{2\times8.31\times300}{1.013\times10^{5}} = 4.92\times10^{-2}\ \text{m}^{3} \approx 49\ \text{litres}.

Group C — Long answer (8 marks each)

  1. (a) Derive the expression P = \tfrac13\rho\overline{c^{2}} for the pressure of an ideal gas using the kinetic theory. [5](b) Deduce that the mean kinetic energy of a molecule is \tfrac32kT. [1](c) Find the r.m.s. speed of nitrogen (M = 28\times10^{-3}\ \text{kg mol}^{-1}) at 27^{\circ}\text{C}. [2] [8]
Answers — Group C, Heat

1. (a) Take N molecules, each of mass m, in a cubical box of side l. A molecule with velocity components u_x, u_y, u_z strikes a wall perpendicular to x and rebounds elastically; momentum change = 2mu_x. It returns to the same wall after time 2l/u_x, so the rate of change of momentum (force) is \dfrac{2mu_x}{2l/u_x} = \dfrac{mu_x^{2}}{l}. For all molecules, F = \dfrac{m}{l}\sum u_x^{2} and the pressure on the wall of area l^{2} is P = \dfrac{m}{l^{3}}\sum u_x^{2} = \dfrac{m}{V}\,N\,\overline{u_x^{2}}. By symmetry \overline{u_x^{2}} = \overline{u_y^{2}} = \overline{u_z^{2}} = \tfrac13\overline{c^{2}}. So P = \dfrac{mN}{3V}\overline{c^{2}} = \tfrac13\rho\overline{c^{2}}. (b) For one mole PV = RT and P = \dfrac{1}{3}\dfrac{N_Am}{V}\overline{c^{2}}, so RT = \tfrac13N_Am\overline{c^{2}} and \tfrac12m\overline{c^{2}} = \dfrac{3RT}{2N_A} = \tfrac32kT with k = R/N_A. (c) c_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3\times8.31\times300}{0.028}} = \sqrt{2.67\times10^{5}} \approx 517\ \text{m s}^{-1}.

3.3 Waves and Optics (units 14–18, 15 hours)

Group B — Short answer (5 marks each)

  1. An object is placed 10\ \text{cm} from a concave mirror of focal length 15\ \text{cm}. Find the position and magnification of the image and state its nature. [5]
  2. A coin lies at the bottom of a water tank 12\ \text{cm} deep. If n_{water} = 4/3, find the apparent depth. Explain why a pool looks shallower than it is. [5]
  3. Define the critical angle and total internal reflection. Find the critical angle for glass of refractive index 1.5. [5]
  4. A prism of angle 60^{\circ} and refractive index \sqrt{2} is used. Find the angle of minimum deviation. [5]
  5. An object is 30\ \text{cm} from a convex lens of focal length 20\ \text{cm}. Find the image distance and magnification. Also find the power of a concave lens of focal length 25\ \text{cm}. [5]
  6. Define the dispersive power of a material. Find it for glass with n_v = 1.532, n_r = 1.514 and n_y = 1.522. [5]
Answers — Group B, Waves and Optics

1. Real-is-positive convention: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} with u = 10, f = 15 gives \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{10} = -\dfrac{1}{30}, so v = -30\ \text{cm}: a virtual image 30\ \text{cm} behind the mirror. m = -\dfrac{v}{u} = +3: erect and three times larger (as in a shaving mirror).

2. n = \dfrac{\text{real depth}}{\text{apparent depth}}, so apparent depth = \dfrac{12}{4/3} = 9\ \text{cm}. Rays leaving the coin bend away from the normal on entering air, so to the eye they appear to come from a point nearer the surface.

3. The critical angle is the angle of incidence in the denser medium for which the refracted ray grazes the interface (angle of refraction 90^{\circ}). Total internal reflection occurs when light travelling in the denser medium meets the boundary at an angle greater than C. \sin C = \dfrac{1}{n} = \dfrac{1}{1.5} = 0.667, so C = 41.8^{\circ}.

4. n = \dfrac{\sin\frac{A+D}{2}}{\sin\frac{A}{2}}. \sin\dfrac{60+D}{2} = \sqrt{2}\sin30^{\circ} = 0.7071, so \dfrac{60+D}{2} = 45^{\circ} and D = 30^{\circ}.

5. \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{1}{60}, so v = 60\ \text{cm} (real). m = -\dfrac{v}{u} = -2 (inverted, twice the size). Power P = \dfrac{1}{f} = \dfrac{1}{-0.25} = -4\ \text{D}.

6. Dispersive power \omega = \dfrac{n_v - n_r}{n_y - 1}: the ratio of the angular dispersion to the mean deviation. \omega = \dfrac{1.532 - 1.514}{1.522 - 1} = \dfrac{0.018}{0.522} = 0.0345.

Group C — Long answer (8 marks each)

  1. (a) Draw the path of a ray through a prism and derive n = \dfrac{\sin\frac{A+D}{2}}{\sin\frac{A}{2}}. [5](b) A ray falls on a prism of angle 60^{\circ} at 50^{\circ} and emerges at 50^{\circ}. Find the deviation and the refractive index. [3] [8]
  2. Derive the mirror formula \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f} for a concave mirror and show that f = R/2. [6] Find R if an object 20\ \text{cm} away gives a real image 60\ \text{cm} away. [2] [8]
Answers — Group C, Waves and Optics

1. (a) (Diagram: ray PQ meets face AB at Q, bends towards the normal, meets face AC at R, emerges as RS bending away; angles i_1, r_1 at Q and r_2, i_2 at R; the deviation D is between the incident ray PQ extended and RS.) Deviation at the first face = i_1 - r_1; at the second = i_2 - r_2, so D = (i_1 + i_2) - (r_1 + r_2). In the quadrilateral AQNR (normals meeting at N), r_1 + r_2 = A. So D = i_1 + i_2 - A. At minimum deviation the ray passes symmetrically: i_1 = i_2 = i and r_1 = r_2 = r = A/2, so D_m = 2i - A, or i = \dfrac{A + D_m}{2}. Snell's law n = \dfrac{\sin i}{\sin r} gives n = \dfrac{\sin\frac{A+D_m}{2}}{\sin\frac{A}{2}}. (b) Emergent angle equals incident angle, so this is minimum deviation: D = i_1 + i_2 - A = 50 + 50 - 60 = 40^{\circ}. n = \dfrac{\sin50^{\circ}}{\sin30^{\circ}} = \dfrac{0.766}{0.5} = 1.53.

2. (Diagram: object AB beyond C; ray AD parallel to the axis reflects through F; ray AP strikes the pole and reflects with equal angle; image A'B' between F and C.) Use the real-is-positive convention: PB = u, PB' = v, PF = f. The ray AP hits the pole at angle \theta and reflects at the same angle, so right triangles ABP and A'B'P are similar: \dfrac{AB}{A'B'} = \dfrac{u}{v}. The ray AD (with DP = AB) passes through F, so right triangles DPF and A'B'F are similar: \dfrac{AB}{A'B'} = \dfrac{DP}{A'B'} = \dfrac{PF}{FB'} = \dfrac{f}{v - f}. Equating, \dfrac{u}{v} = \dfrac{f}{v-f}, so uv - uf = vf. Dividing by uvf gives \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}. For a distant object u = \infty and the image forms at F: v = f. A ray parallel to the axis meets the surface at D, and the radius CD is the normal, so the angle of incidence \angle CDA equals the angle of reflection \angle CDF, which makes \triangle CDF isosceles with CF = FD \approx FP for a small aperture. Hence f = R/2. Numerical: \dfrac{1}{f} = \dfrac{1}{20} + \dfrac{1}{60} = \dfrac{4}{60}, so f = 15\ \text{cm} and R = 2f = 30\ \text{cm}.

3.4 Electricity and Magnetism (units 19–23, 27 hours)

Group B — Short answer (5 marks each)

  1. State Coulomb's law. Find the force between two 2\ \mu\text{C} charges placed 3\ \text{cm} apart in air (k = 9\times10^{9}\ \text{N m}^{2}\text{C}^{-2}). [5]
  2. Define electric field intensity. Find the field at 0.30\ \text{m} from a point charge of 5\ \text{nC}, and the force on a 2\ \text{nC} charge there. [5]
  3. Define electric potential and the volt. Find the work done in taking a 2\ \mu\text{C} charge through a potential difference of 100\ \text{V}. [5]
  4. Capacitors of 2\ \mu\text{F}, 3\ \mu\text{F} and 6\ \mu\text{F} are connected(a) in series and(b) in parallel. Find the equivalent capacitance in each case. [5]
  5. Derive the energy stored in a capacitor. Find it for 10\ \mu\text{F} charged to 100\ \text{V}. [5]
  6. A cell of emf 12\ \text{V} and internal resistance 1\ \Omega is connected to a 5\ \Omega resistor. Find the current, the terminal potential difference and the power dissipated in the external resistor. [5]
  7. State Ohm's law and explain why it fails for a filament lamp. Find the resistance of a copper wire 2\ \text{m} long, 0.5\ \text{mm}^{2} in area (\rho = 1.7\times10^{-8}\ \Omega\,\text{m}). [5]
Answers — Group B, Electricity and Magnetism

1. The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, along the line joining them: F = k\dfrac{q_1q_2}{r^{2}}. F = \dfrac{9\times10^{9}\times(2\times10^{-6})^{2}}{(0.03)^{2}} = \dfrac{9\times10^{9}\times4\times10^{-12}}{9\times10^{-4}} = 40\ \text{N} (repulsive).

2. The force per unit positive test charge at that point, E = F/q (unit \text{N C}^{-1}). E = \dfrac{kq}{r^{2}} = \dfrac{9\times10^{9}\times5\times10^{-9}}{0.09} = 500\ \text{N C}^{-1}. F = qE = 2\times10^{-9}\times500 = 1.0\times10^{-6}\ \text{N}.

3. Potential at a point is the work done per unit positive charge in bringing it from infinity to that point. One volt is one joule per coulomb. W = qV = 2\times10^{-6}\times100 = 2\times10^{-4}\ \text{J}.

4. (a) Series: \dfrac{1}{C} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1, so C = 1\ \mu\text{F}. (b) Parallel: C = 2 + 3 + 6 = 11\ \mu\text{F}.

5. Raising the charge on the plates from q to q + dq against the potential V = q/C needs work dW = \dfrac{q}{C}\,dq. Total: W = \int_0^Q\dfrac{q}{C}\,dq = \dfrac{Q^{2}}{2C} = \tfrac12CV^{2} = \tfrac12QV. U = \tfrac12(10\times10^{-6})(100)^{2} = 0.050\ \text{J}.

6. I = \dfrac{E}{R + r} = \dfrac{12}{6} = 2.0\ \text{A}. Terminal p.d. V = E - Ir = 12 - 2 = 10\ \text{V}. P = I^{2}R = 4\times5 = 20\ \text{W}.

7. Ohm's law: at constant temperature the current through a conductor is proportional to the potential difference across it. A filament heats up, so its resistance rises with current and the V–I graph is not a straight line. R = \dfrac{\rho l}{A} = \dfrac{1.7\times10^{-8}\times2}{0.5\times10^{-6}} = 0.068\ \Omega.

Group C — Long answer (8 marks each)

  1. (a) Derive the capacitance of a parallel-plate capacitor, C = \dfrac{\varepsilon_0A}{d}. [4](b) Show that the energy density is \tfrac12\varepsilon_0E^{2}. [2](c) Plates of area 0.02\ \text{m}^{2} are 1\ \text{mm} apart in air; find C and the energy at 100\ \text{V} (\varepsilon_0 = 8.85\times10^{-12}\ \text{F m}^{-1}). [2] [8]
  2. (a) State Kirchhoff's current law and voltage law and name the principle each expresses. [3](b) A 12\ \text{V} battery (negligible internal resistance) is connected to a 4\ \Omega resistor in series with a parallel pair of 6\ \Omega and 3\ \Omega. Find the current in each resistor. [5] [8]
Answers — Group C, Electricity and Magnetism

1. (a) For large plates the field between them (area A, charge \pm Q) is E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0A} (Gauss's law). The potential difference is V = Ed = \dfrac{Qd}{\varepsilon_0A}. By definition C = \dfrac{Q}{V} = \dfrac{\varepsilon_0A}{d}. (b) U = \tfrac12CV^{2} = \tfrac12\dfrac{\varepsilon_0A}{d}(Ed)^{2} = \tfrac12\varepsilon_0E^{2}\,(Ad). The volume between the plates is Ad, so the energy per unit volume is \tfrac12\varepsilon_0E^{2}. (c) C = \dfrac{8.85\times10^{-12}\times0.02}{1\times10^{-3}} = 1.77\times10^{-10}\ \text{F} = 177\ \text{pF}. U = \tfrac12CV^{2} = \tfrac12(1.77\times10^{-10})(10^{4}) = 8.85\times10^{-7}\ \text{J}.

2. (a) Current law: the algebraic sum of currents at a junction is zero, \sum I = 0 — conservation of charge. Voltage law: the algebraic sum of emfs and potential drops round a closed loop is zero, \sum E = \sum IR — conservation of energy. (b) The parallel pair: \dfrac{1}{R}_p = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}, so R_p = 2\ \Omega. Total R = 4 + 2 = 6\ \Omega. I = \dfrac{12}{6} = 2.0\ \text{A} in the 4\ \Omega resistor. Voltage across the pair: 2\times2 = 4\ \text{V}. Current in 6\ \Omega: \dfrac{4}{6} = 0.67\ \text{A}; in 3\ \Omega: \dfrac{4}{3} = 1.33\ \text{A}. Check: 0.67 + 1.33 = 2.0\ \text{A} ✓.

3.5 Modern Physics (units 24–26, 15 hours)

Group B — Short answer (5 marks each)

  1. Define half-life and decay constant, and derive the relation between them. [5]
  2. A radioactive sample of mass 80\ \text{g} has half-life 4 days. How much remains after 12 days? [5]
  3. Calculate the binding energy and the binding energy per nucleon of ^{4}_{2}\text{He} (m_p = 1.007825\ \text{u} for hydrogen atom, m_n = 1.008665\ \text{u}, m_{He} = 4.002603\ \text{u}, 1\ \text{u} = 931.5\ \text{MeV}). [5]
  4. Distinguish between nuclear fission and nuclear fusion. [5]
  5. Using band theory, distinguish between a conductor, a semiconductor and an insulator. [5]
  6. Explain what happens to the resistance of a semiconductor and a metal as temperature rises, and why. [5]
Answers — Modern Physics, Group B

1. Half-life T_{1/2}: the time in which half of the radioactive nuclei in a sample decay. Decay constant \lambda: the probability of decay of a nucleus per unit time (fractional decay rate). From N = N_0e^{-\lambda t}, at t = T_{1/2}, N = N_0/2: \tfrac12 = e^{-\lambda T_{1/2}}, so \ln2 = \lambda T_{1/2} and T_{1/2} = \dfrac{0.693}{\lambda}.

2. 12/4 = 3 half-lives, so N = 80\times\left(\tfrac12\right)^{3} = 10\ \text{g}.

3. Mass of constituents: 2(1.007825) + 2(1.008665) = 2.015650 + 2.017330 = 4.032980\ \text{u}. Mass defect = 4.032980 - 4.002603 = 0.030377\ \text{u}. B = 0.030377\times931.5 = 28.3\ \text{MeV}; per nucleon = 28.3/4 = 7.07\ \text{MeV}.

4.

Point Fission Fusion
Process a heavy nucleus (U-235) splits into two medium nuclei two light nuclei (hydrogen isotopes) join into a heavier one
Trigger a slow neutron temperature of about 10^{7}\ \text{K}, so nuclei overcome their repulsion
Energy about 200\ \text{MeV} per fission about 17.6\ \text{MeV} per D–T fusion, which is more per unit mass
Waste radioactive fission products little radioactive waste
Use reactors, atom bomb the Sun, hydrogen bomb; not yet controlled commercially
5. In a conductor the conduction and valence bands overlap or the conduction band is partly filled, so electrons move freely. In an insulator the forbidden gap is large (above about 5\ \text{eV}) and no electrons reach the conduction band. In a semiconductor the gap is small (\approx1.1\ \text{eV} for Si, 0.7\ \text{eV} for Ge), so at room temperature a few electrons cross the gap, leaving holes.
6. Metal: resistance rises, because the ion lattice vibrates more and scatters electrons more often; the number of carriers hardly changes. Semiconductor: resistance falls, because heat frees many more carriers across the gap, which outweighs increased scattering.

Group C — Long answer (8 marks each)

  1. (a) State the law of radioactive decay and derive N = N_0e^{-\lambda t}. [4](b) Draw the decay curve and mark T_{1/2}. [1](c) A sample of 64\ \text{g} is reduced to 8\ \text{g} in 18 days. Find the half-life and the decay constant. [3] [8]
Answers — Modern Physics, Group C

1. (a) Law: the rate of decay of a radioactive sample is proportional to the number of undecayed nuclei present. -\dfrac{dN}{dt} = \lambda N. Separating variables, \dfrac{dN}{N} = -\lambda\,dt. Integrating from N_0 at t = 0 to N at t: \ln\dfrac{N}{N_0} = -\lambda t, so N = N_0e^{-\lambda t}. (b) (Graph: an exponential curve falling from N_0 at t = 0; mark N_0/2 at t = T_{1/2}, N_0/4 at 2T_{1/2}, N_0/8 at 3T_{1/2}.) (c) \dfrac{64}{8} = 8 = 2^{3}, so 18\ \text{days} = 3T_{1/2} and T_{1/2} = 6\ \text{days}. \lambda = \dfrac{0.693}{6} = 0.1155\ \text{day}^{-1}.

3.6 Examiner's note: where the five- and eight-mark marks are lost

Area Common loss Remedy
Mechanics Wrong sign for recoil velocity; using g = 9.8 when 10 is given; mixing \text{cm} and \text{m} in Y Define directions first; state which g you use; convert units before substituting
Heat Using ^{\circ}\text{C} in Stefan's law; forgetting the latent heat step in a phase-change numerical Convert to kelvin; list every stage (warm, melt, warm, boil)
Waves and Optics Sign convention muddle in mirror and lens formulae State the convention once, then keep to it; check real/virtual from sign of v
Electricity Series/parallel mix-up; treating V = IR with internal resistance omitted Draw the circuit; reduce step by step; add r to the total
Modern Using mass defect in kg with MeV; counting half-lives wrongly Use 1\ \text{u} = 931.5\ \text{MeV}; compute t/T_{1/2} first

Chapter summary

  • Group B and C are 64 of 75 marks; a numerical is marked in steps (law, substitution, arithmetic, unit), so show every line.
  • Mechanics: v = u + at, s = ut + \tfrac12at^{2}, v^{2} = u^{2} + 2as; T = 2u\sin\theta/g, H = u^{2}\sin^{2}\theta/2g, R = u^{2}\sin2\theta/g; g_h = g_0\left(R/(R+h)\right)^{2}; v_{orbit} = R\sqrt{g/(R+h)}.
  • Heat: \gamma = 3\alpha, \beta = 2\alpha; H = kA\Delta\theta/d; P = \sigma AT^{4}; PV = nRT; P = \tfrac13\rho\overline{c^{2}}, \tfrac12m\overline{c^{2}} = \tfrac32kT.
  • Optics: \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}; n = \sin\frac{A+D}{2}/\sin\frac{A}{2}; \sin C = 1/n; \omega = (n_v - n_r)/(n_y - 1).
  • Electricity: F = kq_1q_2/r^{2}; C = \varepsilon_0A/d; U = \tfrac12CV^{2}; I = E/(R + r); Kirchhoff's laws are conservation of charge and of energy.
  • Modern: N = N_0e^{-\lambda t}, T_{1/2} = 0.693/\lambda; B = \Delta m\times931.5\ \text{MeV}.

Practice questions

Group A — Multiple choice (1 mark each)

  1. The dimensions of the Young modulus are [1]a[MLT^{-2}]b[ML^{-1}T^{-2}]c[ML^{2}T^{-2}]d[ML^{-2}T^{-2}]
  2. A projectile has its maximum range when the angle of projection is [1]a30^{\circ}b45^{\circ}c60^{\circ}d90^{\circ}
  3. Three capacitors of 3\ \mu\text{F} each are in series. The equivalent capacitance is [1]a1\ \mu\text{F}b3\ \mu\text{F}c6\ \mu\text{F}d9\ \mu\text{F}
  4. After four half-lives the fraction of a radioactive sample left is [1]a1/4b1/8c1/16d1/32
Answers to Group A

1. (b) — stress has the dimensions of pressure.

2. (b) — R \propto \sin2\theta, maximum at \theta = 45^{\circ}.

3. (a) — 1/C = 3/3 = 1.

4. (c) — (\tfrac12)^{4} = 1/16.

Group B — Short answer (5 marks each)

  1. A body of mass 4\ \text{kg} moving at 6\ \text{m s}^{-1} collides inelastically with a 2\ \text{kg} body at rest and they move together. Find the common velocity and the loss of kinetic energy. [5]
  2. Find the equivalent resistance of 4\ \Omega and 12\ \Omega in parallel, and the current from a 6\ \text{V} source through it. [5]
  3. Find the temperature rise of 0.2\ \text{kg} of water when it absorbs 4200\ \text{J} of heat (c = 4200\ \text{J kg}^{-1}\text{K}^{-1}). [5]
  4. State Snell's law and find the refractive index if the angle of incidence is 45^{\circ} and of refraction 30^{\circ}. [5]
Answers to Group B

1. 4\times6 = 6v, v = 4\ \text{m s}^{-1}. Initial K = \tfrac12(4)(36) = 72\ \text{J}; final K = \tfrac12(6)(16) = 48\ \text{J}; loss = 24\ \text{J}.

2. R = \dfrac{4\times12}{16} = 3\ \Omega; I = 6/3 = 2\ \text{A}.

3. \Delta\theta = \dfrac{Q}{mc} = \dfrac{4200}{0.2\times4200} = 5\ \text{K}.

4. The ratio \sin i/\sin r is constant for a given pair of media, and the incident ray, refracted ray and normal lie in one plane. n = \dfrac{\sin45^{\circ}}{\sin30^{\circ}} = \dfrac{0.7071}{0.5} = 1.41.

Group C — Long answer (8 marks each)

  1. Derive the work–energy theorem for a constant force and use it to find the stopping distance of a 1000\ \text{kg} car moving at 20\ \text{m s}^{-1} when a constant braking force of 5000\ \text{N} is applied. [8]
  2. Define the principal focus of a convex lens, derive the thin-lens formula and find f if an object at 40\ \text{cm} gives a real image at 24\ \text{cm}. [8]
Answers to Group C

1. For a constant force F producing acceleration a over displacement s: v^{2} = u^{2} + 2as with a = F/m gives Fs = \tfrac12mv^{2} - \tfrac12mu^{2}: the work done by the net force equals the change in kinetic energy. Braking: -5000\,s = 0 - \tfrac12(1000)(400) = -2.0\times10^{5}, so s = 40\ \text{m}.

2. The principal focus is the point on the axis at which rays parallel to the axis converge after refraction. Derivation: for an object AB at u, ray AO through the optical centre is undeviated; the ray parallel to the axis refracts through F. From similar triangles ABO \sim A'B'O: \dfrac{h_o}{h_i} = \dfrac{u}{v}; from COF \sim A'B'F: \dfrac{h_o}{h_i} = \dfrac{f}{v - f}. So \dfrac{u}{v} = \dfrac{f}{v-f}, giving uv - uf = vf and \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} (real-is-positive). Numerical: \dfrac{1}{f} = \dfrac{1}{40} + \dfrac{1}{24} = \dfrac{3 + 5}{120} = \dfrac{1}{15}, so f = 15\ \text{cm}.