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(a) Specific latent heat of a substance is the quantity of heat energy required to change a unit mass of a substance from one state to another without change in temperature. The unit specific latent heat is Jkg\(^{-1}\). (b)(i) Mass of empty calorimeter = M; Mass of cal. + water = M\(_2\) Mass of cal. + water + ice = M\(_3\); Temp. of warm water = \(\theta_1\) Final temp. when all the is has melted = \(\theta_2\); Sp. heat capacity of cal = Cc Sp. heat capacity of water = C\(_w\) latent heat of ice = L Mass of water = M\(_2\) - M\(_1\) Mass of ice = M\(_3\) — M\(_2\) Heat lost by cal. and water = (m\(_1\)Cc + (M\(_2\) - M\(_1\))C x ( \(\theta_1\) - \(\theta_2\)) Heat gained by ice in melting to water of 0^\circC = (M\(_3 - M_2\))L Heat gained by ice water from 0^\circC = (M\(^3\) - M\(_2\))C\(_ w \theta_2\), Total heat gained = Heat lost (M\(_3\) - M\(_2\)L + (M\(_3\) - M\(_2\)) C\(_w \theta_2\) = (M\(_1\)Cc + (M\(_2\) - M\(_1\)) C\(_w\) (\(\theta_2 - \theta_2\) L = \(\frac{M_1C_c + (M_2 - M_1) C_w(\theta_1 - \theta_2) - (M_3 M_2)C_w\theta_2}{(M_3 - M_2)}\) (ii) Precautions - Stir gently - Ensure that ice is dry before adding to water - Avoid parallax error in reading the balance and thermometer. (c) If M\(_s\) = mass of steam ; Heat lost by steam at 100^\circC = M\(_s\)l\(_v\) = M\(_s\) x 2.3 x 10\(^3\)(J) Heat lost by hot water from 100^\circC = 0^\circC ; = \(M_sC_w \Delta \theta\) = \(M_s\) x 4.2 x 100 (J) Heat gained by ice at 0\(^oC\) = M\(_{\text{ice}}\) L\(_f\) = 20 x 3.4 x 10\(^2\) = 6800J; Heat lost by steam + hot water = Heat gained by ice; M\(_sl_v\) + M\(_s\)C\(_w\)\(\Delta\) = M\(_{\text{ice}}\)L\(_f\); M\(_s\) (2.3 x 10\(^3\) + 4.2 x 100) = 6800; M\(_s\) = \(\frac{6800}{2720}\) = 2.5g Mass of water in container = mass of melted ice + mass of water + mass of condensed steam. = 20 + 100 + 2.5 = 122.5g

(a)(i) Refraction of a wave is the change in the direction of the wave as it moves from one medium to another of different density. (ii) Critical angle is the angle of incidence in the denser medium beyond which total internal reflection occurs. (b) (i) The conditions necessary for total internal reflection of wave to occur are:- - The wave must be travelling from a denser medium to a less dense medium - The angle of incidence of the wave in the denser medium must be greater than the critical angle of the medium. (ii) The conditions necessary for interference pattern to be formed are:- - Two or more sources must be involved. - The distance between the sources must be of the order of the wave length of the waves (c) V = \(f \lambda\) F = \(\frac{v}{\lambda} = \frac{3.6}{0.45}\) = 8.0Hz (d) (i) (ii) n = \(\frac{sin i}{sin r}\) r = sin\(^{-1} (\frac{sin i}{n})\) = sin\(^{-1} (\frac{sin 30}{1.5})\) = 19.5\(^o\) Deviation = i - r = 30 - 19.5 = 10.5\(^o\)

(a)(i) Electric potential is the work done in moving a unit positive charge from infinity to a point in an electric field. H is measured in volts or J/C. (ii) Electric potential energy is the work done or energy required to move a charged particle from infinity to a point in an electric field. The unit is Joules. (ii) Electric potential unit is in volts v. Electric potential energy is measured in Joules (J) (b) (i) Electric field intensity E = \(\frac{Q}{4 \pi \varepsilon_or^2}\) (ii) Electric potential V = \(\frac{Q}{4 \pi \varepsilon_or}\) (iii) \(\frac{E}{v} = \frac{I}{r}\) OR E = \(\frac{Y}{4 \pi \varepsilon_o r}\) (c) E = \(\frac{v}{d} = \frac{5 \times 10^4}{5 \times 10^{-3}} = 10^7 v/m\) (ii) F = qE = 1.6 x 10\(^{-19}\) x 10\(^{7}\) = 1.6 x 10\(^{-12}\) N (iii) F = ma a = \(\frac{F}{m}\) = \(\frac{1.6 \times 10^{-12}}{9.1 \times 10^{-31}}\) = 1.76 x 10\(^{18}\) m/s\(^2\)

(a)(i) In nuclear fission reaction results in splitting a heavy nucleus into two parts of nearly equal masses while in nuclear fusion process two light nuclei combine together to form a heavier nucleus. (ii) Nuclear fission releases less energy per nuclear while nuclear fusion releases more energy per nuclear. (iii) Nuclear fission can take place at room temperature while nuclear fusion takes place at very high temperature. (ii) Peaceful uses of atomic energy are: - Production of laser beams - Production of x-rays - Production of energy in discharge tubes. (b)(i) The neutrons produced as a result of fission of a nucleus can cause further fission if they strike other nuclei. This rapid multiplication of a fission process by fissile neutrons is called chain reaction. (ii) (i) Conditions necessary for chain reaction to occur are: - The mass of fissile material must exceed the critical mass - In addition to the two nuclei produced during the fission, one to three neutrons must be produced per parent nucleus. (ii) Components in a nuclear reactor used to control chain reactions are; - Boron steel - Heavy water or denterium oxide - Graphite or carbon rod. (c)(i) The type of nuclear reaction is fusion. T\(_{\frac{1}{2}} = \frac{0.693}{\lambda}\); \(\lambda = \frac{0.693}{T_{\frac{1}{2}}}\) = \(\frac{0.693}{5.40 \times 10^{3}}\) = 1.28 x 10\(^{-4}\)S\(^{-1}\)

Table of values/observation Number of oscillations (n) = 20 S/N M(g) \(t_1\)(s) \(t_2\)(s) t = \(\frac{t_{1}+t_{2}}{2(s)}\) T = \(\frac{t}{n}\) T\(^{2}\)(sec\(^{2}\)) T\(^{2}\)(sec\(^{2}\))x10\(^{-3}\) 1 50.0 5.80 5.80 5.80 0.290 0.084 84 2 70.0 6.70 6.90 6.80 0.340 0.116 116 3 90.0 7.70 7.90 7.80 0.395 0.156 156 4 110.0 8.50 8.70 8.60 0.430 0.185 185 5 130.0 9.20 9.20 9.20 0.460 0.212 212 (7)The graph of T\(^{2}\)(sec\(^{2}\)) Vs m(g) axis Scale: T\(^{2}\)(sec\(^{2}\)) axis: 2cm = 20 units m(g) axis: 2cm = 20 units (8) Slope/gradient = \(\frac{\bigtriangleup {T}^{2}(sec^{2})}{\bigtriangleup {m(s)}}\) \(\frac{234-10}{140-6} = \frac{224}{134}\) = 1.672 sec\(^{2}\)/g l on the vertical axis is (0) (9) K = \(\frac{4\pi ^{2}}{s} = \frac{4 \times ({\frac{22}{7})^{2}}}{1.672}\) = \(\frac{4 \times (22)^{2}}{7^{2} \times 1.672} = \frac{4 \times 484}{49 \times 1.672}\) = \(\frac{1936}{81.928}\) = 23.61 g/sec\(^{2}\) Precautions are: I ensured that I took care of zero error when taking the period of oscillation. I ensured that there was no conical oscillation when the mass hanger was loaded with different slotted masses. I ensured that I took repeated readings in order to avoid random errors. (b)i. Young modulus is defined as the ratio of tensile or Compressive stress to tensile or compressive strain. It is denoted by E and measured in N/m\(^{2}\). Thus, Young modulus = \(\frac{\text {tensile/compressive stress}}{\text{tensile/compressive strain}}\) E = \(\frac{F/A}{e/L} =\frac{FL}{Ae}\) Force constant is defined as the amount of force that causes a unit extension of an elastic material or the ratio of F/e of an elastic material. It is denoted by K and measured in N/m or N/m\(^{-1}\). Thus, F = Ke, K = \(\frac{f}{e}\) ii. F = 500N, K = 1.0 x 10\(^{4}\) Nm\(^{-1}\), F = ke 500 = 1 x 10\(^{4}\) x e e = \(\frac{500}{1 \times 10^{4}} = \frac{5 \times 10^{2}}{1 \times 10^{4}}\) = 5 x 10\(^{2-4}\) = 5 x 10\(^{-2}\) (0.05)m W = ½Fe = ½ x 500 x 0.05 = 12.5J

Tables of value/observation S/N L 9cm) V (volts) L\(^{-1}\)(cm\(^{-1}\)) V\(^{-1}\)(volts\(^{-1}\)) L\(^{-1}\)(cm\(^{-1}\))x10\(^{-3}\) 1 15.0 0.60 0.067 1.667 67.0 2 25.0 0.80 0.040 1.250 40.0 3 35.0 1.10 0.029 0.909 29.0 4 45.0 1.30 0.022 0.909 22.0 5 55.5 1.60 0.018 0.625 18.0 6 65.0 1.80 0.015 0.556 15.0 (6) The graph of V\(^{-1}\)(volts\(^{-1}\)) vs \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\) Scale: V\(^{-1}\)(volts\(^{-1}\)) axis: 2cm = 0.15units \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\)) axis: 2cm = 7units (7) Slope/gradient = \(\frac{\bigtriangleup {V}^{-1}(volts ^{-1})}{\bigtriangleup {L}^{-1}(cm^{-1})}\) = \(\frac{1.575-0.405}{63-4.2} = \frac{1.17}{58.8}\) = 0.0199cm/volts (8) K = \(\frac{1}{s} = \frac{1}{0.0199}\) = 50.25 volts/cm precautions are: I would ensure that the key is opened when the circuit is not in use to avoid the cell being used up. I would ensure that the electrical wires are tightly connected I would ensure that I take note of the zero error in the pointer instruments to avoid systematic error I would ensure that I take pointer readings within the vertical eye level to avoid parallax I used short connecting wires to ensure that the required energy to dive the electrical charges was not used up in overcoming the resistance of the wire. (b)i. The factors affecting the resistance of a wire are: - length of wire (R&L) - cross-sectional area of the (R \(\propto\)) - Nature of material - Temperature of the wire ii. R = 100cm x 0.02\(\Omega\)cm = 2\(\Omega\), v =1.5v t = 1min = 60s H =p x t = \(\frac{V^{2}t}{R} = \frac{(1.5)^{2}}{2}\) x 60 = 67.5J OR I = \(\frac{V}{R} = \frac{1.5}{2}\) = 0.75A H = P x t = I\(^{2}\)Rt = (0.7)\(^{2}\) x 60 x 2 = \(\frac{9}{16}\)x120 = 67.5J

Option D is the graph of stopping potential Vo against the frequency of the incident light I. The straight line graph cuts the intercept I = threshold frequency. As the frequency of the incident light increases above this point(threshold frequency), the stopping potential corresponding to the kinetic energy of the photoelectron also increases.

To calculate the amount of energy released when 0.5 kg of uranium is completely burnt, we use Einstein's mass-energy equivalence formula: \(E = mc^2\) Given: Mass of uranium (\( m \)) = 0.5 kg, Speed of light (\( c \)) = \( 3 \times 10^8 \, \text{m/s} \) Substituting the values into the formula: \(E = 0.5 \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2\) Calculating \( c^2 \): \((3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2\) Now, substituting back: \(E = 0.5 \times 9 \times 10^{16} = 4.5 \times 10^{16} \, \text{J}\) The amount of energy released when 0.5 kg of uranium is burnt completely is \(E = 4.5 \times 10^{16} \, \text{J}\)

To determine the deceleration of the motorcyclist, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance: \(v^2 = u^2 + 2as\) Where: \( v \) = final velocity (0 m/s, since he stops at the traffic light), \( u \) = initial velocity (30 m/s), \( a \) = acceleration (deceleration, which will be negative) \( s \) = distance (50 m) Substituting the values into the equation, we rearrange it to solve for \( a \): \(0 = (30)^2 + 2a(50)\) Calculating \( (30)^2 \): \(0 = 900 + 100a\) Now, rearranging for \( a \): \(100a = -900\) \(a = -\frac{900}{100} = -9 \, \text{m/s}^2\) The deceleration of the motorcyclist is \( \mathbf{9 \, m/s^2} \).

I wave - particle duality of matter, the momentum of sub-atomic particle is given by P = h/\lambda where h = planck constant ; \lambda = wave length => \lambda = h/P

The binding energy of a nucleus is the work done or energy needed to separate the nucleons (protons and neutrons) apart. Again, experiment shows that the total mass of all individual nucleons is greater than that of the nucleus in which they are together. This difference in mass is the measure of the binding energy of the particular nucleus. For helium\(^4_2\)He It has 4 nucleons, 2 protons and 2 neutrons 2:H = 2 x 1.00783 = 2.01566 2:n = 2 x 1.00867 = 2.01734 ∴ 2:H + 2:n = 4.03300u But the helium nucleus has a mass of 4.00260u ∴ Binding energy = difference between nucleons and nucleus = 4.03300U - 4.00260U = 0.03040U

Energy stored in a coil of inductance, L, and carry a current, I is given by E = 1/2 L I2 = 1/2 x 5 x 10-3 x 6 x 6 = 9.0 x 10-2J

Energy stored = Ivt = current x back e.m.f x time = 2 x 3 x 0.4 = 2.4J

Pressure developed in the effort piston = F/a ∴F/a = 40/0.4 = 400/4 = 100Nm2 this pressure is transmitted to the larger piston of area A; => P = F/A ∴ 100 = 400/A => A = 400/100 = 4m2

Coefficient of friction μ = F/R = F/mg = 500/1000 = 0.5 ∴ μ = 0.5