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(a)(i) Coulomb's law of electrostatics The electrostatic force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. (ii) Calculation of the magnitude of the electrostatic force F = \(\frac{kq_{1}q_{2}}{d^{2}}\) F = \(\frac{9.0 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{(5.2 \times 10^{-11})^{2}}\) = 8.52 x 10\(^{-8}\) N (b)(I) Proof of V\(_{out}\) Let the current through the circuit be I V\(_{in}\) = I(\(R_1\) + \(R_2\)) OR I = \(\frac{V_{in}}{R_{1} + R_{2}}\) V\(_{out}\) = I\(R_2\) II. Calculation of \(R_1\); V\(_{out}\) = V\(_{in}\)(\(\frac{R_{1}}{R_{1} + R_{2}}\)) OR \(\frac{V_{out}}{V_{in}} = \frac{R_{1}}{R_{1} + R_{2}}\) \(\frac{1}{2.5} = \frac{R_{1}}{R_{1} + R_{30}}\) \(R_1\) = 20\(\Omega\) (ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points, (c) Why wood is not suitable for use as the Core of transformers Wood is a non-magnetic material and would not provide magnetic coupling between the coils for electromagnetic induction. (d) Applications of cathode ray tube i. For studying alternating current waveform ii. For measuring frequency/voltage/ reactance/power factor and resistance of an electronic device iii. Used to monitor signals/check FM/AM characteristics

(a) Conditions necessary for: Constructive interference of waves i. lt must involve at least two waves. ii. Waves must vibrate in phase. iii. The distance between the sources of the wave must be of the order of the wavelength. Total internal reflection i. The waves must travel from a denser medium to a less dense medium. ii. The angle of incidence (in the denser medium) must be greater than the critical angle. iii. It must involve two media of different optical densities. Production of beats 1. The amplitude of the two interfering waves should be the same. ii. The difference between the frequencies of interfering waves should be small. iii. The two sources of waves must be sounded together. (b) Calculation of velocity of sound in air V = f\(\lambda\) V = 2f(\(l_2\) - \(l_1\)) = 2 x 256(100 - 35) = 33280 cm s\(^{-1}\) = 332.80 m s\(^{-1}\) (c)(i) Classification of musical instruments i. Wind ii. Stringed iii. Percussion (ii) Examples of classified instruments saxophone I. Wind: flute, trumpet, pipe, organ, iii. Stringed: piano, guitar, violin iii. Percussion: piano, drum, bell, gong. xylophone (d) Calculation of critical angle \(\eta\) = \(\frac{1}{\text {Sin C}}\) 1.5 = \(\frac{1}{\text {Sin C}}\) Sin C = 0.667 C = 41.80º (e) Calculation of temperature \(\frac{V_{1}}{\sqrt T_{1}} = \frac{V_{2}}{\sqrt T_{2}}\) \(\frac{240} {\sqrt 375} = \frac {310} {\sqrt T_{2}}\) 240 \(\sqrt T_{2}\) = 310 x 19.36 \(\sqrt T_{2}\) = 625 K = 352 ºC

(a) Factors affecting the rate i. Surface area of the liquid (exposed surface area) ii. Temperature of the liquid iii. Wind/ drought. iv. Humidity/ amount of water vapour v. Nature of the liquid vi. Air pressure (b) Explanation of latent heat; The energy needed to change the state of a substance without a change in temperature. (c)(i) Water in a clay pot cooler than in a rubber container; A clay pot has pores which allows evaporation to occur while a plastic container has no pores. (ii) Pressure cooker cooks faster than ordinary pot; Pressure cooker raises the pressure within it, which raises the temperature to/above the boiling point thus cooks faster. (d)(i) Calculation of latent heat of fusion of the body Ivt = mL mL = 12 x 40 x 1400 = 6.72 x 10\^{5}\)j (ii) Calculation of specific heat capacity of the body C = \(\frac{Ivt}{m \bigtriangleup \theta}\) = \(\frac{12 \times 40 \times 72}{1.5 \times 60}\) = 384 jkg\(^{-1}\) K\(^{-1}\) (e) difference between evaporation and boiling Evaporation Boiling It takes place at the surface of the liquid It occurs throughout the entire volume of the liquid It takes place across all temperatures It takes place at specific temperatures It produces a cooling effect It produces no cooling effect It is wind aided It is not wind aided

The primary colors of pigment are red, blue, and green. Mixing these primary colors of pigment gives us the three secondary colors: red+blue=magenta, red+ green = yellow, and blue + green = cyan

The refractive index of water is greater than that of air. So, A person legs appear to be short while standing in the tank due to refraction.

(a)() Meaning of artificial radioactivity; The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy. Emission Nature Charge Ionizing Beta (\(\beta\)) High speed electron Negative Moderately ionizing Gamma (\(\gamma\)) Electro-magnetic radiation Neutral Negligible ionizing ability Alpha particles Helium nucleus Positive Highly ionizing (b) Calculation of kinetic energy K.E = eV = 1.6 x 10\(^{-19}\) x 30 x 10\(^{3}\) = 4.8 \times 10\(^{-15}\)J (c)(i) graph of K.E versus f i. Both axes correctly distinguished ii. Reasonable scale on both axes (ii) (I) Planck's constant. h = slope of the graph Slope i. \(\bigtriangleup\)K.E and \(\bigtriangleup\)f correctly determined ii. \(\frac{\bigtriangleup K.E}{\bigtriangleup f}\) corectly determined and recorded within range. (II) Threshold frequency, \(f_0\) . The intercept on the horizontal or frequency axis. (III) Threshold frequency, \(W_o\) Negative intercept on K.E axis.

(a) OBSERVATIONS (i) Value of \(M_0\) correctly determined and recorded to at least 1 d.p in grammes (ii) Value of m correctly determined and recorded to at least 1 d.p in grammes (iii) Value of d correctly read and recorded to at least 1 d.p in cm (iv) Five values of M correctly recorded in grammes (v) Five values of Z = AK correctly read and recorded to at least1 d.p in cm and in trend. Trend: As M increases Z decreases. (vi) Five values of y = (Z - 5) correctly evaluated and recorded. (vii) Five values of x = (d - Z) correctly evaluated and recorded (vii) Five values of v = \(\frac{x}{y}\) correctly evaluated and recorded to at least 3 d.p (ix) Composite table showing at least M, Z,y, x and v. (b)(i) Condition for a body to be in stable - The algebraic sum of all external forces on the object is zero OR - The algebraic sum of all external torques from the external forces acting on the object is zero. OR - The sum of clockwise moments about a point in the object is equal to the sum of anticlockwise moment about the same point. OR - The algebraic sum of all moments about a point on the object is zero. (ii) If the distance is y, 100y = 50(6 - y) 100y = 300 - 50y 150y = 300 y = 2m

(a) OBSERVATIONS (i) Value of D correctly measure and recorded to at least 2 d.p in cm (ii) Value of A = 0.25\(\pi\)D\(^{2}\) correctly evaluated (iii) Value of \(y_0\) correctly read and recorded to at least 1 d.p in cm. (iv) Five values of m correctly recorded in grammes. (v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend. (vi) Five values of h = (y - yo) correctly evaluated. (vii) Five values of log h correctly evaluated to at least 3 d.p (viii) Five values of log m correctly evaluated to 3d. (ix) Composite table showing at least m, y, h, log h and log m. (b)(i) The Law of Flotation; A floating body displaces its own weight of fluid in which it floats. (ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6 Let y be the depth of immersion \(\therefore\) Weight of water displaced = (0.1y) x 1.26 x 1000 x 10 1260y = 453.6 y=453.6=9.36 = 1260y = 453.6 y = \(\frac{453.6}{1260}\) = 9.36

(a) OBSERVATIONS (i) Value of E correctly read and recorded to at least 1 d.p in volts. (ii) Value of i correctly read and recorded to at least 1 d.pin amperes. (iii) Five values of d correctly determined and recorded to at least 1 d.p in cm. (iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend. Trend; As d increases, I decreases (v) Five values of I\(^{1}\) correctly evaluated to at least 3 s.f. (vi) Composite table showing at least d, I, and I\(^{-1}\). (ii) P = IV Power P = \(\frac{240 \times 0.75}{100}\) 0.18kw Energy consumed in 10hr = 0.18 x 10 = 1.8kWh Cost of energy = 1.8 x 0.5 = $0.9 OR Cost of energy = \(\frac{\text {P x cost x time}}{100}\) \(\frac{0.7 \times 240 \times 10}{1000}\) = $0.9

Water is most anomalous at low temperatures, with a remarkable density maximum at 4^\circC. water is not used as a thermometric liquid because it has low thermal expansivity, high specific capacity, high vapor pressure and wets glass.

(a) OBSERVATIONS (i) Value of E correctly read and recorded to at least 1 d.p in volts. (ii) Value of i correctly read and recorded to at least 1 d.pin amperes. (iii) Five values of d correctly determined and recorded to at least 1 d.p in cm. (iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend. Trend; As d increases, I decreases (v) Five values of I\(^{1}\) correctly evaluated to at least 3 s.f. (vi) Composite table showing at least d, I, and I\(^{-1}\). (ii) P = IV Power P = \(\frac{240 \times 0.75}{100}\) 0.18kw Energy consumed in 10hr = 0.18 x 10 = 1.8kWh Cost of energy = 1.8 x 0.5 = $0.9 OR Cost of energy = \(\frac{\text {P x cost x time}}{100}\) \(\frac{0.7 \times 240 \times 10}{1000}\) = $0.9

A body sliding down a plane inclined at 30o to the horizontal, Has coefficient of friction = tan 30º or \(\frac{1}{\sqrt3}\)

(a)(i) Is the energy required to remove one mole of an electron from a gaseous atom. OR Is the minimum energy required to convert one mole of a gas of atom into one mole of a gaseous plus 1 ions (ii) \(_{4}\)Be: Is\(^{2}\) 2s\(^{2}\) \(_{5}\)B :Is\(^{2}\) 2s\(^{2}\) 2p\(^{1}\) The electron to be removed from Be is in the 2s orbital which is closer to the nucleus, nuclear attraction is greater hence first ionization energy is greater but the electron to be removed from B is in the 2p orbital which is farther away from the nucleus, hence nuclear attraction is weaker hence first ionization energy is smaller. OR 2p has less energy than 2s So it is easier to remove an electron from 2p as the nuclear charge is weaker in 2p because 2p is further away from the nucleus. Hence, first ionization energy is small. (b) Q - It Q = 0.12 x 500 = 60 C 96500 = 1F = 60 C liberates 0.015g \(\therefore\) 96500 will liberate \(\frac{96500}{60}\) x 0.015 = 24 g n((\(T_1\)) = \(\frac{m}{M}\) = \(\frac{24}{48}\) = 0.5mol 1F = 0.5mol 2F = 1mol hence charge = +2 Alternative A Q = lt = 0.12 x 500 = 60 C 0.015 g of Ti is deposited by 60 C . \(\therefore\) 48 g will be deposited by = \(\frac{60}{0.015}\) x 45 = 1920000 96500 C\(\equiv\) i mole of electron \(\therefore\) 192000 C \(\frac{1}{96500}\) x 192 = 1.989 \(\approx\) 2.0 moles of electron \(\therefore\) charge on Ti ion is +2 Alternate B \(\frac{m}{M} = \frac{It}{nF}\) n = \(\frac{Mlt}{mF}\) = \(\frac{48 \times 0.12 \times 500}{0.015 \times 96500}\) = 1.989 \(\equiv\) 2 \(\therefore\) Change on Ti ion is +2 (c)(i) Alumina (A\(l_2\)\(_{3}\)) mixed with molten cryolite (N\(a_3\)AI\(F_6\)) (i) - abundant deposits of bauxite - cheap source of electricity (d)(i) Is the ability of a substance to be attracted strongly into a magnetic field. (ii) I. \(_{24}\)Cr\(^{2+}\) - Is\(^{2}\) 2P\(^{6}\) 3S\(^{2}\) 3P\(^{6}\) 3d\(^{4}\) unpaired electrons = 4 \(_{24}\)Cr\(^{6}\) - Is\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3d\(^{6}\) unpaired electrons = 0 (II) \(_{24}\)Cr\(^{2+}\) III. The greater the number of unpaired electrons in the 3d - orbital, the| greater-the paramagnetism.

(a)(i) This can be defined as the number of atoms or molecules in one mole of a substance which is equal to 6.02 x 10\(^{23}\) (ii) Mass/m.m = no. of molecules / Avogadro's number (2.30 / m.m) x (3.01x10\(^{22}\) / 6.02 x 10\(^{23}\)) m.m = (2.30 x 6.02 x 10\(^{23}\)) / 3.01 x 10\(^{22}\) m.m = 46g / mol \(\therefore\) 14+('16 x 2) 14 + 32 = 46g /mol \(\therefore\) N\(o_2\) (b)(i) I. Two layers are formed / no reaction / immiscible II. there is a reaction / hydrolysis with steamy fumes /white fumes (ii) The reaction occurs by interaction of the tones pair of electrons of water binding to the central atom (silicon or carbon) / attaching to the central atom In silicon, the 3p - orbital is available to accept the pairs of electrons but in carbon, there is no vacant orbitals for bonding. OR Water is polar hence the reaction with SiC\(I_4\). But CC\(I_4\) is non-polar hence there is no reaction (c) (i) - Purification of metals/ copper - Electroplating

Amines possess a characteristic ammonia smell, liquid amines have a distinctive "fishy" and foul smell. -Amine. R-NH2