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(i) Elasticity is the ability of a material to regain its original shape and size after undergoing a stretching or compression. (ii) Young's modulus is the ratio of tensile stress to tensile strain. (iii) Force constant is the amount of force that causes unit extension.

Stress = \(\frac{F}{A} = \frac{40}{3.14(1 \times 10^{-3})^2} = 12.74 \times 10^{6} Nm^{-2}\) (b) Strain = \(\frac{e}{I} = \frac{24 \times 10^{-5}}{4} = 6 \times 10^{-5}\)

(a)(i) Linear momentum is the product of the mass of a body and its velocity. (ii) Impulse is the product of a large force and short time for which the force acts. (b) The principle of conservation of linear momentum states that in system of colliding bodies the total momentum of the system is constant provided no external forces act on the system. (c) (i) \(m_1u_1 + m_2U_2 = v(m_1 + m_2)\) 5 x 10\(^{3}\) x 3 + 0 = v(5 x 10\(^{3}\) + 2.5 x 10\(^{3}\)) v = \(\frac{15 \times 10^3}{7.5 \times 10^3}\) = 2m/s (ii) Initial K.E = \(\frac{1}{2}mv^2 = \frac{1}{2} MV^2 = \frac{1}{2} times 5 \times 10^{3} \times 3^2\) = 2.25 x 10\(^4\) Final K.E = \(\frac{1}{2}mv^2 = \frac{1}{2} (5 + 2.5) \times 10^3 \times 2^2\) = 1.5 x 10\(^{4}\)J Loss in K.E of the couple = (2.25 - 1.5) x 10\(^4\) = 75 x 10\(^3\)J (iii) Impulse = (\(mu_1U_1 + m_2U_2\)) - v (\(mu_1 +m_2) (0 + 5 \times 10^3) - 2 (2.25 + 5) 10^3\) = 0Ns

(a)(i) Liquids consist of molecules in random motion having different velocities, hence near the surface molecules with higher velocities and kinetic energy escape from the surface. Therefore the average kinetic energy of the remaining liquid is reduced. Since the average kinetic energy is a measure of temperature, therefore temperature is lowered. Hence evaporation causes cooling. (ii) The heat supplied to boiling water increases the kinetic energy of the molecules and also weaken the bonds between the molecules. Hence at boiling the heat supplied is used in breaking the bonds, therefore molecules escaped as steam without change in temperature. (b) (i) Boyle's law states that the volume of a fixed mass of gas at constant temperature varies inversely as the pressure upon it. (ii) Introduce dry air into burette (B) and close the tap. Read and record the barometer reading H i.e. a.p., keeping B steady, raise the reservoir R, read the mercury head h from the scale, at the same time the corresponding volume V of air from the Repeat burette. Repeat the experiment with various values of H, and determine the corresponding values of h and v. Evaluate P = H \pm h and V. Graph of P against V\(^{-1}\) The graph P against V\(^{-1}\) is a straight line graph passing through the origin, showing that pressure is inversely proportional to volume. (iii) Precautions; – I ensured that dry air is used. – I avoided parallax error when reading metre rule.

(a) (i) Neutral point is a point at which the resultant magnetic flux density is zero. (ii) (b)(i) (ii) An ammeter is accurate if the current it mea-sures is close to the true value of the current flowing through the instrument. (c) The material is surrounded by a thick wall of ferro-magnetic material which does not allow magnetism to pass through them. This shields or protects the material from the earth's magnetic field. (ii) F = qvBsin\(\theta\) = 1.6 x 10\(^{-19}\) x 2.5 x 10\(^{7}\) x 2 x 0.5 = 4 x 10\(^{2}\) N (d) Electromagnetic induction laws are - Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit. - The direction of the induced current is always such as to oppose the change producing it.

(a)(i) Photoelectric emission is the ejection of electrons from the surface of a metal when light of appropriate wavelength and frequency is incident on the metal. (ii) Applications of photoelectric emission i.e. photo cells: - Burglar alarms - Photo meters - Automatic doors - Sound production from film tracks - T.V. camera etc (b) When light of a set wavelength and frequency falls on the cathode, electrons are emitted. The electrons flow toward the anode. The flow of these electrons constitute current. (c) (i) For energy of light with frequency less than threshold frequency, complete absorption takes place. The energy absorbed is used to raise the energy of electrons or overcome the potential barrier in the photoelectric material. (ii) The number of emitted electrons are affected by: - intensity - Number of photons.

\(M_1\) = 20.0g \(M_2\) = 50.0g C.G = 50CM S/N L=P - Y (cm) d = =Q - P (cm) Y (cm) 1 40.00 37.00 20.00 2 42.00 37.50 18.00 3 44.00 38.30 16.00 4 46.00 38.70 14.00 5 48.00 39.30 12.00 Slope (s) = \(\frac{\bigtriangleup {L}}{\bigtriangleup {d}} = \frac{48-4}{39.2-28.2} = \frac{44}{10.4}\) = 4.23 Precautions: Ensure that there is no air interference when determining the position of the centre of gravity (G) of the metre-rule. Ensure that the suspended mass did not touch the table Avoid error due to parallax when reading the metre rule. Zero error of the metre rule must be noted. (b)i. ii. 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. It is also defined as the turning effect of the force about a point. It is measured in Nm - SI unit.

Table of values/observation \(l_o\) = 0.8A S/N L(cm) l(A) (L\(^{-1}\)cm\(^{-1}\)) 1 25.0 1.20 0.040 2 40.0 1.00 0.025 3 55.0 0.95 0.018 4 70.0 0.85 0.014 5 85.0 0.8 0.012 When 1\(^{-1}\) =0, l = 0.66A \(\frac{|_{o}}{|} = \frac{0.8}{0.66}\) = 1.212 Precautions: The Key was open when readings were not being taken. Tight connections were ensured. Parallax errors were avoided in reading the meter-rule Zero error was noted in reading metre-rule The jockey was not rolled on or allowed to scratch the wire (b) The p.d between two points is the work done (in joules) in moving an electric charge of one coulomb across the two points. ii. R = \(\frac{PL}{A} = \frac{4pL}{\pi d^{2}}\) d = 0.2m, \(\pi\) = \(\frac{22}{7}\), R = 7\(\Omega\), p = 8.8 x 10\(^{-7} \pi \), L = ? \(\therefore\) 7 = \(\frac{4 \times 8.8 \times 10^{-7} \times L}{\frac{22}{7} \times (0.2)^{2}}\) 7 = \(\frac{4 \times 8.8 \times 10^{-7} \times L \times 7}{22 \times 0.04} = \frac{28 \times 8.8 \times 10^{-7} \times L}{0.88}\) L = \(\frac{7 \times 0.88}{28 \times 8.8 \times 10^{-7}}\) L = \(\frac{6.16}{246.4}\) x 10\(^{-7}\) L = 0.025 x 10\(^{7}\) L = 2.5 x 10\(^{-2}\) x 10\(^{7}\) L = 2.5 x 10\(^{5}\)

[ Density of blood = 1.1 x 103 kgm-3, g = 10ms-2] Pressure = pgh, where p = density, h = the height, g = acceleration of free fall due to gravity 1.6.5 x 104 = 1.1 x 103 x 10 x h therefore h = 1.65 x 104/1.1 x 103 x 10 = 1.5m

W = Mg man's weight = 80N then his mass = \(\frac{80}{10}\) = 8kg T = Mg + Ma = M( g + a) = 8(10 + 5) = 8 x 15 = 120N.

Let the distance apart the two masses = m. From the equation F & M1 m2/ϒ2 F1 α 10 x 10/ϒ2 = F α 100/ ϒ2 F2 α 10 X 10/(3ϒ)2 => F2 α 100/9ϒ2 therefore F2/F1 = 100/9ϒ2/100/ϒ2 = 100/9ϒ2 ϒ2/100 = 1/9 Therefore F 2 : F 1 = 1 : 9

From Pythagoras, R2 = 102 + 102 = 100 + 100 = 200 therefore R2 = √200 = √2 x 100 = 10√2, in the direction 45°E

Angular velocity w = 2π f. But f = frequency no. of oscillations per second. = 30/6 = 5Hz. therefore W = 2π x 5 = 10π rad. s-1

No expanded working has been attached for this question yet. The verified answer is Option B.

Air pressure has no effect on speed of sound in an ideal gas because pressure and density both contribute to sound velocity equally, the two effects cancel each other out