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(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.

K = 500Nm-1; M = 5 x 10-3kg; e = 0.02m; g = 10ms-2. The elastic potential energy stored in the compressed spring is given by = 1/2Ke2, and the energy is transformed into gravitational potential energy = mgh, thus from the principle of energy conversion, we have that mgh = 1/2Ke2 5 x 10-3 x 10 x h = 1/2 x 500 x (0.02)2 ∴ h = 1 x 500 x 0.02 x 0.022 x 5 x 10-3 x 10 = 5 x 2 x 0.02___10-1 = 10 x 0.02 x 10 = 2 ∴ h = 2m

To find the depth to which a test tube sinks when placed upright in water, we apply the principle of buoyancy. Given: - Radius of the test tube, \( r = 1.0 \, \text{cm} = 0.01 \, \text{m} \) - Mass of the test tube, \( m = 8.8 \, \text{g} = 0.0088 \, \text{kg} \) - Density of water, \( \rho = 1000 \, \text{kg/m}^3 \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) Calculate the weight of the test tube The weight \( W \) of the test tube is given by: \(W = mg = 0.0088 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.088 \, \text{N}\) Calculate the buoyant force The buoyant force \( F_b \) is given by: \(F_b = \rho g V_{\text{displaced}} = \rho g \pi r^2 h\) But, the buoyant force is equal to the weight: \(\rho g \pi r^2 h = W\) Substituting the known values: \(1000 \times 10 \times \pi \times (0.01)^2 \times h = 0.088\) Solve for \( h \) Rearranging gives: \(h = \frac{0.088}{1000 \times 10 \times \pi \times (0.01)^2}\) Calculating: \(h = \frac{0.088}{1000 \times 10 \times \pi \times 0.0001} = \frac{0.088}{3.142} \approx 0.028 \, \text{m}\) ≈ 2.8cm

Vol = cross-sectional area x length ∴ vol. of water disharged per sec = 60ms-1 x 0.5m = 30m3 5-1 => Vol. of water discharged in the said 20s = 30m3 x 20 = 600m3 ∴ vol. of the container = 600m3

Pressure P = \(\frac{F}{A}\); Pressure P = \(\frac{F}{\pi r^2}\) = \(\frac{7.7 \times 10^{-2}}{\pi \times 10^{-10}}\) = 2.45 x 10\(10^8\)Nm\(^{-2}\)

AS the stone is whirled round and te string suddenly cuts, say at A, B, C or D. The stone will fly off in the direction indicated VA, VB, VC or VD respectively which is tangential to the circular path

Since the lever is a uniform one, its center of gravity will be acting half way (i.e 45cm) ∴from the principle of moments, F x 45 = 30 x 30 ∴ F = (30 x 30) / 45 = 20N

R.H = \(\frac{Partial Pressure of Water Vapour Present}{S.V.P at temperature of atmosphere}\) x 100 = \(\frac{10}{17.50}\) x 100 = 57.14% ≈ 57%

In vibration in a closed pipe/tube as shown in the figure, fundamental note is produced at the first position of resonance. When the length of the air column L = λ/4; => λ = 4L. Since the open end of the tube is taken as an antinode; and closed end as the node. Thus the associated frequency is given by f = V/λ = V/4L ∴ f = V/4L

In general intensity of a wave is the rate of flow of energy across unit area normal to the direction of propagation of the wave intensity α1/r2 where r is the distance away from the point source. Thus if the distance from the point source of the sound wave is doubled, then the intensity will be decreased by a factor of 4