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(i) Elasticity is the ability of a material to regain its original shape or size after deformations. OR after removal of stress force, OR after it has been 1 compressed. (ii) Young's Modulus is the ratio of tensile I or compressive stress to tensile or compressive strain (iii) Force constant = the amount of force that causes a unit extension of an elastic material. OR \(\frac{Force (F)}{ Extension(e)}\) of an elastic material.

(i) Stress = \(\frac{force}{\text{cross sectiona area}}\) = \(\frac{40}{3.142 (1 \times 10^{-3})^2}\) = 12.74 x 10\(^{6}\)Nm\(^{-2}\) (ii) Strain = \(\frac{e}{I}\) = \(\frac{24 \times 10^{-5}}{4}\) = 6.0 x 10\(^{-5}\)

a)(i) Interference occurs in all kinds of waves i.e. longitudinal or transverse, while polarisation occurs only in transverse waves. (ii) - In sun glasses to reduce the intensity of light. - To produce polarised light. - To produce three dimensional pictures.

(a) Angle of contact - the angle, measured through the liquid between the wall of the container and the tangent of the meniscus. (b)

M = Zit Z = \(\frac{M}{It}\) = \(\frac{0.48 \times 10^{-3}}{4 \times 10^2 \times 3600}\) = 3.3 x 10\(^{-10}\)kgc\(^{-1}\)

P = mv = 9.1 x 10\(^{31}\) x 4.2 x 10\(^{-7}\) = 9.1 x 4.2 x 10\(^{-24}\) Kgms\(^{-1}\) \(\lambda = \frac{h}{p} = \frac{6.6 \times 10^{-34}}{9.1 \times 4.2 \times 10^{-24}}\) = 1.73 x 10\(^{-11}\)m or \(\lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 4.2 \times 10^7}\) = 1.73 x 10\(^{-11}\)

(a)(i) The momentum of a body is the product of its mass and its velocity. (ii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts. (b) Mp Up + Mq Uq = Mp Vp + Mq Vq 0.25 Up + 0 = 0.25 x \(\frac{2}{3}\) Up + 0.25 x 2; 0.25 (Up - 2/3 Up) = 0.50; 1/3 Up = \(\frac{0.50}{0.25}\) = 2m/s Up = 2 x 3 = 6m/s (c)(i) Newton's second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force (ii) Let the momentum before collision = mu and momentum after collision = my, change in momentum = my - mu, Change in momentum in unit time = \(\frac{mv - mu}{t}\) From second law: \(\frac{mv - mu}{t}\) = KF \(\frac{m(v - u)}{t}\) = KF If m = 1kg, a = 1ms\(^{-2}\) and F = 1N, K = 1 ma = F (iii) Resultant force on the vehicle = 9600 - 200 = 7400N. But F = ma a = \(\frac{F}{m} = \frac{7400}{3400}\) = 2.18ms\(^{-2}\) or 2.2ms\(^{-2}\)

(a)(i) Conduction and Radiation. (ii) The specific heat capacity of land is less than that of sea. Land Breeze: At night land cools faster than sea, warm air from above the sea rises upward. It's place is taken by cooler air from above the land. Sea Breeze: During the day the land is hotter than the sea. Warm air above land rises, and its place is taken by cooler air from the sea. (b) Increase in length (L\(_2\) - L\(_1\) = L\(_1 \alpha \Delta \theta = 30 \times 1.2 \times 10^{-5} \times 50 = 1.8 \times 10^{-2}\) cm (c) Ivt = mc\(\Delta \theta\) Power x time = mc\(\Delta \theta\) (i) Iv x 360 = m x c 60 ----(i) Ivt = \(\frac{1}{2}\)mL (ii) Ivx 25 x 60 = \(\frac{1}{2}\)mL ------(ii) \(\frac{Iv \times 25 \times 60}{Iv \times 360} = \frac{\frac{1}{2}mL}{mc \times 60}\) L = \(\frac{25 \times 60 \times C \times 60 \times 2}{360}\) = \(\frac{25 \times 60 \times 4200 \times 60 \times 2}{360}\) = 2.1 x 10\(^5\) Jkg\(^{-1}\)

(a)(i) The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane. OR The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant \(\frac{\text{Sin i}}{\text{Sin r}}\) = Constant Place the triangular prism on the drawing sheet and trace the outline. Arrange two pins on one side vertically and using two other pins, determine the emergent ray on the other side. Remove the prism and. join the position of pins to meet the outline of the prism. Draw the normal at the point of incidence. Produce both the incident and the emergent rays to intersect and measure the angles of incidence I^\circ and the deviation d^\circ. Repeat the procedure for at least four other values of i. Plot a graph of d on the vertical axis against i on the horizontal axis. Obtain minimum deviation D as below. n = \(\frac{sin \frac{1}{2}(A +D)}{sin \frac{1}{2} A}\) = \(\frac{sin \frac{1}{2} (60 + D)}{sin 30}\) Precautions - neat traces - pins fixed vertically - Avoid parallax in reading protector (b)(i) (ii) n = \(\frac{\text{real depth}}{\text{apparent depth}}\) apparent depth = \(\frac{\text{real depth}}{n}\) = \(\frac{12}{1.5}\) = 8cm Apparent displacement = real depth - apparent depth = 12 - 8 = 4cm

(a)(i) Electric field intensity = Electric force per unit charge OR As negative potential i.e.E = \(\frac{-V}{d}\) OR E = \(\frac{F}{Q}\). where F is the force, Q is the charge and E electric field intensity. (ii) Electric potential =- work done per unit charge to bring a (positive) charge from infinity to a point in an electric field. (b) (c) (i) A dielectric substance is any insulating material placed between the plates of a capacitor (ii) Factors Effect on capacitor common area of plates Distance between plates Nature of dielectric capacitance increases with area capacitance decreased with increasing distance capacitance increases with dielectric constant C = \(\frac{KE_oA}{A}\) (iii) C\(_{2.3}\) = C\(_2\) + C\(_3\) = 20 + 20 = 40\(\mu\) F \(\frac{1}{C_T} = \frac{1}{C_{2.3}} + \frac{1}{C_1}\) \(C_T = \frac{\frac{C_{2.3} \times C_1}{C_2.3 + C_1} = \frac{40 \times 40}{40 \times 40}\) = 20\(\mu F\)

(i) Electrical \(to\) thermal \(to\) mechanical (kinetic energy) \(to\) thermal \(to\) electromagnetic (ii) Distinction between hard and soft X-rays Hard X-rays Soft x-rays (i) High (target) voltage (ii) short wavelength (iii) High penetration (iv) produced by high voltage Low (target) voltage Long wavelength Low penetration Produced by low voltage (iii) Uses: - Used to detect hidden cracks in materials - Used to detect alteration to work of art - To show broken bones in human body - used in crystallography i.e. to study crystals - used in diagnosis of ulcer (iv) Hazards: genetic mutations - skin burn - leukemia - Tissue damage - Tissue damage - baldness - graying - cataracts (v) Safety precautions: use of dosimeter, badges of GM counters - wearing of lead coats - Periodic medical counters - wearing of lead coats - periodic medical check up (b) \(^2_1H + ^2_1H \to ^3_1H + ^1_1H + Q\) Let atomic mass unit of tritium \(^3_1H\) be X 2.01410 x 2 x 931 = (x + 1.00783) 931 + 4.03 \(\frac{3750.254 - 4.03}{931}\) = x + 1.00783 4.0238 - 1.0078 = x 3.106 = x

M(g) \(W_1\)(g) \(W_2\)(g) \(W_3\)(g) U=(\(W_1\) - \(W_2\))g V=(\(W_1\)-\(W_3\))g 5.0 5.00 4.20 4.70 0.80 0.80 10.0 10.00 7.90 8.60 2.10 1.40 15.0 15.00 12.75 13.50 2.25 1.50 20.0 20.00 17.10 16.40 2.90 3.60 25.0 25.00 22.10 21.40 2.90 3.60 Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2.35-1.00}{3.25-1.65} = \frac{1.35}{1.60}\) = 0.84 Precautions: I avoided parallax error in reading spring balance. I noted zero error of spring balance. l avoided splashing of liquid. I avoided object touching the bottom or side beaker. I cleaned the mass before dipping it into the liquid. (bi) Weight of kerosine displaced = volume of object x density of kerosine x g \(\frac{\text {Mass of object}}{\text {density of of object}}\) x density of kerosine x g = \(\frac{20 \times 10^{-3}kg \times 8.0 \times 10^{2} kgm^{-3} \times 10ms^{-2}}{ 8.0 \times 10^{-3} kgm^{-3}}\) = 0.02N = upthrust of the weight of object in air = mass x g = 20 x 10\(^{-3}\) X 10 = 0.2N Tension in the spring = weight of an object in the air - upthrust = 0.2N - 0.02N=0.18N The reading on the spring balance= 0.18 ii. Archimede's principle states that when a body is wholly or partially immersed in a liquid, it experiences an upthrust equal to the weight of the liquid displaced. On the other hand, the law of floatation states that a floating body displaces its own weight of the liquid in which it floats.

Table os values/observation D(cm) D\(^{2}\)cm \(X_1\)(cm) \(X_2\)(cm) L=(\(x_1\)-\(x_2\))cm L\(^{2}\)cm D\(^{2}\)-L\(^{2}\)cm 100 10000 80.50 18.30 62.20 3868.8 6131.16 90 8100 70.20 19.00 51.20 2621.4 5478.56 80 6400 59.10 20.00 39.10 1528.8 4871.19 70 4900 46.70 22.00 24.70 610.09 4289.91 60 3600 29.00 7.60 21.40 457.96 3142.04 Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{6000-3000}{100-50} = \frac{300}{50}\) = 60 Evaluate K = \(\frac{8}{4} = \frac{60}{4}\) = 15 Precaution: I avoided parallax when reading the metre rule. I ensured that a sharp image is formed on the Screen. (b) Real image is formed by the actual intersection of rays whereas a virtual image is formed by the apparent intersection of rays when their directions have been produced backwards, or a Real image is formed at the front of the mirror while a virtual image is formed at the back of the mirror. ii An object beyond 2F, the image is real, inverted, smaller than an object, between F and 2F, ray parallel to the principal axis and ray passes through the optical centre.

The e.m.f of the accumulator = 1.5v Table of value/observations R(\(\Omega\)) 1\(_{amp}\) 1\(^{-1}_{amp}\) 0 0.78 1.28 1 0.50 2.00 2 0.38 2.63 3 0.30 3.33 4 0.25 4.00 5 0.22 4.55 Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{4-1}{4-2} = \frac{3}{2}\) The intercept on the vertical axis is =-1.85\(\Omega\) Precautions: I tightened connections. I avoided errors due to parallax in reading ammeter or voltmeter. I noted zero error of ammeter/ voltmeter. I removed key when readings were not taken. I took repeated readings from the table, i.e. averaging the reading. (b)i. - The lead-acid accumulator can be recharged while the Leclanche cell cannot. - The lead-acid accumulator has the capacity to supply large currents for a long time while the leclanche cell does not. ii. The two resistors 3\(\Omega\) and 4\(\Omega\) are connected in parallel. \(\therefore\) The equivalent resistance (\(\Omega\) = \(\frac{1}{R} = \frac{1}{3} = \frac{1}{4} = \frac{4+3}{12}\) = \(\frac{7}{12}\) = 1.71 The effective resistance in the circuit = 1.71 + 4 = 5.71\(\Omega\)

A mixture of iodine and sulphur can be separated by adding Carbon disulphide that is (CS₂) then stir the solution and filter it . As sulphur dissolved in CS₂, so it come into filter and got separated from the mixture.