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The observations are: (i) Photoelectric effect (ii) Electrons show refraction or reflection (iii) They have wavelength or frequency (iv) Electrons show diffraction patterns. (v) Compton effect (vi) Determination of molecular velocities in gases.

(a)(i) Properties of X-rays are:; They (a) penetrates substance (b) travel in straight lines (c) are able to ionise gases. (d) affect photographic plate or paper making them to fogge (e) undergo reflection, or refraction, or polarization or diffraction etc. (ii) X-rays are electromagnetic waves with wave properties such as; - reflection - refraction - diffraction - polarisation - travel in straight line - cannot be deflected by magnetic or electric field. (iii) X - rays (a) used to ionise gases causing them to become conducting. (b) used in industries to locate fractures in machines and metal castings. (c) used to analyse crystal structure i.e. crystallography. (iv) X - rays are hazardous because they - kill mall organisms - cause leukaemia can damage riving tissues - cause genetic mutation - cause damage to cells of large animals. (b)(i) hf = eV f = \(\frac{eV}{h}\) f = 1% of eV = \(\frac{1 \times 1.60 \times 10^{-17} \times 5.00 \times 10^4}{100 \times 6.63 \times 10^{-34}}\) = 1.207 x 10\(^7\) Hz. (ii) Times rate at which heat is removed = 99% x I x V = \(\frac{99}{100}\) x 2.00 x 10\(^{-2}\) x 5.00 x 10\(^4\) = 990W

(a) The gravitational field intensity at any point on the earth's field is the gravitational force acting on a unit mass at that point (b)(i) (ii) A simple Pendulum of length L, swinging in a vertical plane at a small angle of displacement has a period T given as \(T = 2 \pi \sqrt{\frac{L}{g}}\) \(T^{2} = \frac{4 \pi^2L}{g}\) L = \(\frac{gT^2}{4 \pi^2}\) Slope S of the graph I against \(T^2 = \frac{g}{4 \pi^2}\) (iii) g = 4\(\pi^2\)S = 4(3.142)\(^2\) x 0.25 = 9.87m/s\(^{2}\) (c)(i) Time t = \(\frac{distance}{speed} = \frac{45}{15}\) = 3(s) (ii) Height h = \(\frac{1}{2} gt^2 = \frac{1}{2} \times 10 \times 3^3 = 45m\)

(a) Upper fixed point is the temperature of steam from boiling water at standard atmospheric pressure of 760 mmHg. Lower fixed point is the temperature of pure melting ice. (b) 75.000 \(\Omega\) - 1000\(o\)C \(\frac{64.992 - 63.000}{75.000 - 63.000} = \frac{R - 0}{100 - 0}; 64.992 \Omega - R\) \(\frac{1.992}{12.000 } = \frac{R_1}{100}; 63.000 \Omega - 0^oC\) \(R_1 = \frac{1992 \times 100}{12.000} = 16.6^oC\) (c) (i) Boyle's law states that the volume of a fixed unit varies inversely as the pressure provided the temperature remains constant. (ii) At horizontal position pressure = A.P = H Gas volume = 4.5 x 10\(^{-2}\) x cross sectional area of the tube A. But PV = Constant (Boyle's law) H x 5.0 x 10\(^{-2}\) x A = (H + 8.5 x 10\(^{-2}\)) x 4.5 x 10\(^{-2}\) x A H = \(\frac{8.5 \times 10^{-2} \times 4.5}{0.5} = 0.765mHg\) A.P = hpg = 0.765 x 1.36 x 10\(^4\) x 10 = 1.04 x 10\(^{5}\)pa

a) (i) A wave motion is a disturbance which travels through a medium and transfers energy from one point to another without causing any permanent displacement of the medium itself. (ii) A stationary wave is a wave got when two progressive waves of equal amplitude and frequency or wavelength travelling in opposite direction are superimposed. In stationary wave nodes are pendulum of permanently at rest while antinodes are at maximum amplitude Stationary wave along a string (b)(i) Physical properties of wave are: diffraction, refraction, reflection and.interference. (ii) Amplitude is the maximum displacement of a oscillating object from its equilibrium position. A node is a point on a stationary wave at zero amplitude while an antinode is a point at maximum amplitude. (iii) Frequency of vibration on a stretch string depends on; - Tension on the string - Length of stretch string - Mass per unit length and density of the string (c) y = 5 sin (3x - 4t); y = A sin (\(\frac{2 \pi x}{\pi} \times 2 \pi ft\)) \(\frac{2 \pi x}{\pi}\) = 3x \(\pi = \frac{2 \pi}{3}\)m = 2\(\pi ft\) = 4t f = \(\frac{2}{\pi }\) (i) = \(\frac{2}{\pi}\) = 0.64Hz (ii) T = \(\frac{1}{f}\) = \(\frac{1}{0.64}\) = 1.56(s) (iii) V = f\(\pi\) = 0.64 x \(\frac{2 \pi}{3} = 1.34\)m/s

(i) E.m.f is the work doone in taking a unit electric charge round a closed circuit. H is measured in volts. (ii) Other sources of e.m.f. are; solar cells, fuel cells, dynamo or electric generators, thermocouple etc. (b) (c) (i) In open circuit Vo e.m.f of the cell. In closed circuit V terminal p.d of the cell. (ii) (a) I = \(\frac{E}{R + r}\) (b) Maximum power dissipated in the cell occurs when the external resistance = the internal resistance of the cell i.e. R = r. Maximum power dissipated in the cell = IE = \(\frac{E}{R + r} \times E = \frac{E^2}{R + r} = \frac{E^2}{2R}\) Maximuim power dissipated in the external load = I\(^2\)R = \(\frac{E^2R}{(R + r)^2}\) = \(\frac{E^2R}{4R^2} = \frac{E^2}{4R}\) (iii) Efficiency of the call = \(\frac{\text{power output}}{\text{power input}}\) x 100% = \(\frac{2R}{E^2} \times \frac{E^2}{4R}\) x 100% = 50%

1. 49.50cm 5. 82cm from 0 6. a = 19.50cm d = 32.50cm 7. Table of values/observation S/N M(g) d(/cm) d\(^{-1}\) d\(^{-1}\)(cm\(^{-1}\))x10\(^{-2}\) 1 30.0 32.50 0.0308 3.08 2 40.0 24.38 0.0410 4.10 3 50.0 19.50 0.0513 5.13 4 60.0 16.25 0.0615 6.15 5 70.0 13.93 0.0718 7.18 8. 9. Slope = \(\frac{\bigtriangleup {m(g)}}{\bigtriangleup {d}^{-1}(cm^{-1})}\) = \(\frac{60-10.8}{(6.16-1.04) \times 10^{-2}}\) slope (s) = \(\frac{49.2}{5.12 \times 10^{-2}}\) K = \(\frac{S}{Q} = \frac{960.94}{50}\) =19.2g Precautions taken to ensure results are as follows: Avoid the error of parallax in reading the meter-rule Avoid draught-free wind. Repeated readings must be shown on the table of values/observations. (b)i. The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. ii. For a body to be in equilibrium under the action of parallel forces, The forces must all act in the same plane The sum of the forces in one direction must be equal to the sum of the forces in the opposite direction ie Total upward force is equal to the total downward force. The algebraic sum of the moments about any point is equal to zero. i.e. The sum of the clockwise moment about a point is equal to the sum of the anti-clockwise moment about the same point.

1. \(f_o\) = 15cm 5. a = 60.0cm, b = 20.0cm Hence, L = \(\frac{a}{b} = \frac{60}{20}\) = 3 6. Tables of values/observation S/N b(cm) a(cm) L=\(\frac{a}{b}\) 1 20.0 60.00 3.00 2 25.0 37.00 1.50 3 30.0 30.00 1.00 4 35.0 26.00 0.75 5 40.0 24.00 0.6 7. 8. Slope (s) = \(\frac{\bigtriangleup {L }}{\bigtriangleup {a}} = \frac{3-0.25}{60-18.6cm}\) = \(\frac{2.75}{41.4}\) = 0.066425cm\(^{-1}\) S\(^{-1}\) = \(\frac{1}{s} = \frac{1}{0.066425(cm)^{-1}}\) = 15.05 = 15 N.B: S\(^{-1}\) = \(f_o\) = 15cm Precautions taken to ensure accurate results are as follows: Parallax error in reading meter rule avoided Readings repeated must De shown on the table Collinear alignment of apparatus ensured (b)i. u = 10cm, f = +15cm, V= ? \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f }\) \(\frac{1}{v} + \frac{1}{10} = \frac{1}{15}\) \(\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} =\frac{1}{30}\) = The characteristics of the image formed are: - it is virtual - it is enlarged or magnified i.e twice or two times as big as the object (m = 2)

1. \(v_o\) = 3.0v 2. \(V_R\) = 0.20 volts 3. l = 0.20, 1\(^{-1}\) = \(\frac{1}{1} = \frac{1}{0.2}\) = 5 V = 0.2, V\(^{-1}\) = \(\frac{1}{V} =\frac{1}{0.2}\) = 5 Table of values/observations S/N IA I\(^{-1}\)(A\(^{-1}\)) V(volts) V\(^{-1}\)(volts\(^{-1}\)) 1 0.20 5.00 0.19 5.26 2 0.25 4.00 0.23 4.35 3 0.30 3.33 0.28 3.57 4 0.35 2.86 0.33 3.03 5 0.40 2.50 0..38 2.63 Slope (s) = \(\frac{\bigtriangleup {V}^{-1}}{\bigtriangleup {I}^{-1}}\) S = \(\frac{5.1-1.0}{4.55-12} = \frac{4.1}{3.35}\) = 1.224 S\(^{-1}\) = \(\frac{1}{s} = \frac{1}{1.224}\) = 0.817 N.B: R = S\(^{-1}\) = 1.0 Precautions taken to ensure accurate results are as follows Tight connections ensured/clean terminals Repeated readings have been shown on the table Parallax error in reading voltmeter/ammeter avoided Key open when reading is not being taken (b)i. An ohmic conductor is s Conductor in which the current passing through it is proportional to the p.d across the conductor, provided temperature and other physical conditions are constant. ii. The resistivity of the material of a wire is the resistance of a unit length of the wire of unit Crofs sectional area. It is denoted by p. and measured in ohm-metre (L-m).

Evaporation is a change of state from a liquid to a gas. Distillation is a physical process that involves a phase change from liquid to gas (vapour) and then back to liquid. Crystallization is a physical change that occurs when a substance turns from a liquid to a solid crystal. Sublimation is a change of state from a solid to a gas without passing through a liquid state.

Chemical reactions always involve the formation of new substances. This is because chemical reactions change the arrangement of atoms in a substance, which creates new substances with different chemical properties.

Equation of the reaction H\(_2\)SO\(_4\)(g) + CuO(s) → CuSO\(_4\)(aq) + H\(_2\)O Relative molecular mass of H\(_2\)SO\(_4\) = 1 + 2 + 32 + 16 + 4 = 98 Relative molecular mass of CuO = 64 + 16 =80 From the above equation, 98g of H\(_2\)SO\(_4\) reacts with 80g of CuO. ∴ 4.9g of H\(_2\)SO\(_4\) will react with Xg of CuO = (4.9 * 80)/98 of CuO = 392/98 = 4.0g Alternatively, H\(_2\)SO\(_4\)(g) + CuO(s) → CuSO\(_4\)(aq) + H\(_2\)O 1 : 1 Since the combining ratio of H\(_2\)SO\(_4\) and CuO is in ratio 1:1, the number of moles will also be equal. Hence, we can first calculate the amount of H\(_2\)SO\(_4\), since we were given the mass of H\(_2\)SO\(_4\) that reacted as 4g and the Molecular mass is 98g/mol Amount of moles of H\(_2\)SO\(_4\) = \(\frac{mass}{Molecular mass}\) = \(\frac{4.9}{98}\) = 0.05 mole Amount of moles of CuO = \(\frac{mass}{Molecular mass}\) 0.05 = \(\frac{mass}{64 + 16}\) Mass = 0.05 x 80 = 4.0g

According to Charles's Law, the volume of a gas would be zero at absolute zero, which is - 273^\circC or 0 Kelvin. The Volume is of gas is zero at absolute zero of temperature. On the Celsius scale, zero is the temperature at which water freezes.

Applying the Graham's law definition; i.e, \(\frac{RSO_2}{RCH_4}\) = \(\sqrt{\frac{MCH_4}{MSO_2}}\) \(\frac{tSO_2}{tCH_4}\) = \(\sqrt{\frac{MSO_2}{MCH_4}}\) (rate of diffusion ∝ \(\frac{1}{time taken}\)) Where t(SO\(_2\)) = ?, M(SO\(_2\)) = 64 t(CH\(_4\)) = 20 sec, M(CH\(_4\)) = 16 Therefore, \(\frac{t(SO_2)}{20}\) = \(\sqrt{\frac{64}{16}}\) \(\frac{t(SO_2)}{20}\) = \(\frac{8}{4}\) t(SO\(_2\)) = 8 x \(\frac{20}{4}\) = 40 sec.

When an electron is added to a halogen atom, it will have eight valence electrons. Halogens have seven valence electrons in their outermost shell. When an electron is added to a halogen atom, it fills its outermost shell, resulting in a full valence shell of 8 electrons.