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Mercury does not wet glass because (force of) cohesion/attraction of mercury molecules is greater than (force of) adhesion/attraction between glass and mercury molecules. Water wets glass because (force of) adhesion/attraction between glass and water molecules is greater than (force of) cohesion/attraction of water molecules.

(a) Tensile stress is the ratio of force to cross-sectional area of a wire/rod. It is expressed in Newton per metre squared (Nm\(^{-2}\)) (b) Young’s modulus is the ratio of tensile stress to tensile strain. It is expressed in Newton per metre squared. (Nm\(^{-2}\))

(a) Properties of cathode rays: Cathode rays are negatively charged. travel in straight line in field free space. are deflected by electric/magnetic field. possess(kinetic) energy. possess momentum. (b)(i) The intensity of cathode rays may be increased by raising the temperature of the cathode/increasing the current through the heater. (ii) The energy of cathode rays may be increased by raising the potential difference between the anode and the cathode/the anode potential.

(a) PROCEDURE Make at least 3 well-spaced pin holes round the edge of the cardboard. Clamp the pin horizontally and suspend the cardboard on it through one of the pin-holes such that the cardboard can swing freely Hang the simple pendulum on the same pin and let its string be very close to the cardboard. When the whole system is at rest (or in equilibrium) trace the plumbline on the cardboard. Repeat the procedure for each of the two other pin holes. CONCLUSION The point at which the (three) traced lines intersect is the centre of gravity of the cardboard. PRECAUTIONS - Repeat procedure - Pin rigidly and firmly held by retort stand and clamp - Allow the simple pendulum to rest before tracing the shadow of the plumbline on the cardboard. - The string to be close to the cardboard b) Diagram: Both axes correctly labelled Any correct shape of graph showing acceleration segment Acceleration = gradient of AB Total distance covered = area under the graph (c) Diagram - At least one axis labelled - correct shape of graph Let V\(_1\) = maximum velocity after 20 sec. V\(_2\) = maximum velocity after 30 sec. (i) Then \(\frac{V_1}{20}\) = 6 ∴ V\(_1\) = 120 ms\(^{-1}\) Also V\(\frac{V_2 - V_1}{10}\) = 4 \(\frac{V_2 - 120}{10}\) = 4 V\(_2\) = 160 ms\(^{-1}\) Maximum speed = V\(_2\) = 160ms\(^{-1}\) (ii) Total distance covered = Area of \Delta + Area of trapezium after 1st 30 seconds = ( ½ x 20 x 120) + ½ (120 + 160) x 10 = 1200 + 1400 = 2600m (iii) Average speed = \(\frac{\text{Total distance}}{\text{ Total time}}\) = \(\frac{ 2600}{30}\) = 86.67 ms\(^{-1}\)

(a) A clinical thermometer has small temperature range. The glass will crack/burst due to excessive pressure created by expansion of mercury. (b) Increase in pressure (i) increases the boiling point; (ii) decreases the melting point. (c) At the boiling point of pure water, the saturated vapour pressure (s.v.p) of the water is equal to the external atmospheric pressure. (d) Diagram Volume V is proportional to length Ɩ stated or implied (i) P\(_1\)V\(_1\) = P\(_2\)V\(_2\) implies P\(_1\)P\(_1\) = P\(_2\)P\(_2\) (76 + 20) 15 = 76 x L\(_2\) Ɩ\(_2\) = 96 x 15 76 Ɩ\(_2\) = 18.95 cm (ii) P\(_1\)L\(_1\) = P\(_3\) L\(_3\) = (76 + 20) 15 = (76 – 20) x Ɩ\(_3\) Ɩ\(_3\) = 96 x 15 56 Ɩ\(_3\) = 25.7l cm

(a) Differences between a sound wave and a radiowave Sound wave Radio wave is a mechanical wave is an electromagnetic wave is longitudinal is transverse has speed less than speed of light. has speed equal to that of light requires a material medium for propagation/cannot travel through vacuum. does not require a material medium for propagation/can travel through vacuum. (b) The vibrating tuning fork forces the table top to vibrate. The table top has a much larger area and is therefore in contact with a larger volume of air undergoing vibration. Hence, sound heard is amplified. (c) Conditions necessary for: (i) production of stationary wave - Two waves must be traveling in opposite directions. - Incident and reflected waves must have the same frequency/wavelength. - Incident and reflected waves must have equal amplitude. - There must be superimposition. - There must be a barrier (ii) formation of interference wave pattern - Two sources must be involved. - The two sources must be coherent/have a constant phase difference as well as the same frequency and amplitude. - The waves that are interfering must have the same amplitude - The distance between the sources must be of the order of the wavelengths of the waves. (iii) total internal reflection - The wave must be traveling from a denser medium to a less dense medium - The angle of incidence of the wave in the denser medium must be greater than the critical angle for the medium. (d)(i) Diagram Correct diagram (ii) \(\frac{ Sin (A + dm)}{2}\) n = Sin (A/2) Sin (60 + 41)/2 Sin (60/2) = 1.54

a) Differences between a d. c. generator and a moving coil galvanometer. D. C. generator moving coil galvanometer converts mechanical energy to electrical energy converts electrical energy to mechanical energy. uses split rings or commutator uses hair springs. rotation of coil is continuous rotation of coil is incomplete uses carbon brushes (as terminals) uses jeweled bearings (as terminals) (b) Eddy currents are currents induced in a conductor when subjected to varying magnetic field. Any valid additional information e.g. eddy current flows in a circular path or closed loops; eddy current generates heat; eddy current cannot flow through gaps or slots; The currents move in such a direction as to oppose the change producing them. Devices in which eddy currents are applied pointers of sensitive electric meters sensitive mass balances brakes in large electric motors speedometers in automobiles detection of cracks in railway tracks detection of metals (c) Principle on which a potentiometer is based When a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences. OR The p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section. (d) Vector diagram showing V\(_L\) with arrow V\(_C\) with arrow V\(_R\) with arrow (i) V\(^2\) = V\(_R^2\) + (V\(_L\) - V\(_C\))\(^2\) 110\(^2\) = 80\(^2\) + (40 – V\(_C\))\(^2\) (40 – V\(_C\)) = \(\sqrt{(110 + 80) (110 – 80)}\) 40 – V\(_C\) = + 75.5 V\(_C\) = 40 + 75.5 V\(_C\) = 115.5V (ii) X\(_c\) \(\frac{V_c}{I} = \frac{1}{2 \pi f c}\) C = \(\frac{1}{2 \pi fV_c}\) C \(\frac{2}{2 \times 3.14 \times 60 \times 115.5} = 45.9\muF\) (iii) \(V_L = IX_L = I \times 2 \pi fL}\) L = \(\frac{V_l}{I \times 1 \times 2 \pifVc}\) C = \(\frac{2}{2 \times 3.142 \times 60 \times 2}\) = 0.053H

(a)(i) Emission line spectra consist of distinct and separate bright lines of definite wavelengths on a dark background. Any valid additional information e.g - They are obtained when light from a luminous source undergoes dispersion and is observed directly. - When an electron moves from one energy level to another energy level, line spectra are observed - The colour of the spectra is a characteristic of the source. (ii) Line absorption spectra are obtained when light passes through a cool gas and certain wavelengths of the light are absorbed. This gives series of dark lines, each corresponding to one of the wavelengths absorbed. (b) Diagram When radiation of appropriate frequency falls on the cathode (with photosensitive surface) electrons are emitted. Because the anode is positive with respect to the cathode, the electrons are attracted to the anode. The electrons flow completes the circuit and current flows. (c)(i) Wave properties of X-rays - diffraction - interference - reflection - refraction - polarization (ii) Uses of X-rays other than in medicine to study crystal structures to detect presence of cracks in welded parts in artistic works for determining the authenticity of such works at security check points to scan X-ray spectroscopy for identifying isotopes of elements. (d) \DeltaE = E\(_i\) - E\(_f\) = - 1.6 – (- 10.4) = 8.8 e V or 1.41 x 10\(^{-18}\) J E = hf = hc/\lambda or \lambda = hc/E = 6. 6 x 10\(^{-34}\) x 3 x10\(^{8}\) 8.8 x 1.6 x 10\(^{-19}\) \lambda = 1.4 x 10\(^{-7}\) m

Center of gravity C = 49.5cm Mass of metre rule (MR) = 135g, d = 80cm = 0.8m h(cm) d(m) t(sec) T=\(\frac{\text{t(sec)}}{20}\) T\(^{2}\)(Sec\(^{2}\)) 40 0.8 28.0 \(\frac{28.0}{20}\)=1.40 1.9600 50 0.8 29.0 \(\frac{29.0}{20}\)=1.45 2.1025 60 0.8 31.0 \(\frac{31.0}{20}\)=1.55 2.4025 70 0.8 34.0 \(\frac{34.0}{20}\)=1.70 2.8900 80 0.8 37.0 \(\frac{37.0}{20}\)=1.85 3.4225 Slope (s) = \(\frac{\bigtriangleup {T}^{2}}{\bigtriangleup {h(cm)}}\) = \(\frac{(3.42-0.40)sec^{2}}{(80-21)cm}\) = \(\frac{3.02sec^{2}}{59cm}\) = 0.0512sec\(^{2}\)cm\(^{-1}\) K = \(\frac{S}{Q}\), Q = \(\frac{2}{25d^{2}}\) K = \(\frac {\frac{S}{2}}{25d^{2}}\) S = 0.0512sec\(^{2}\)cm\(^{-1}\), d = 50cm = 0.8cm K = \(\frac{0.0512sec^{2}cm^{-1} \times 25 \times (0.80)^{2}}{2}\) K = \(\frac{0.0512 \times 25 \times 0.6400sec^{2}cm^{-1}}{2}\) = 0.4096sec\(^{2}\)cm\(^{-1}\)m\(^{2}\) Precautions: i - ensured smooth and regular oscillations in a horizontal plane. - ensured there was no error due to parallax when reading my stopwatch. (b)i. Couple -This can be defined as two forces of the same magnitude acting in opposite directions but not on the same line. ii. Practical application of couple include: - Opening or closing taps. - Turning a spanner. - Setting a flywheel in motion about an axis through its center - Motion of a spinning spindle. - Riding/pedaling of bicycle

S/N L(cm) Ii(A) L\(^{-1}\)(cm\(^{-1}\)) 1 20 1.00 0.050 2 35 0.90 0.029 3 50 0.85 0.020 4 65 0.80 0.015 5 80 0.75 0.013 When L\(^{-1}\) = 0, Ii = 0.75A \(I_o\) = 75A \(\frac{I_{o}}{I}= \frac{0.75A}{0.20A}\) = 37.5 SEE THE GRAPH ABOVE... Precautions: i - ensured there was no error due to parallax when reading my ammeter. - ensured that I removed the key after each reading. (b)i. E.m.f is the total work done in pushing one coulomb of electricity round an external resistor and the internal resistor of the source, or the total energy obtained per coulomb from a cell or battery ii. \(\frac{Ey}{Ex} = \frac{Iy}{Ix}\) Iy = 60cm Ex = 1.00v Ix = 40.cm Ey = ? \(\frac{Ey}{1.00}=\frac{60}{40}\) Ey = \(\frac{60 \times 1.00}{40}\) = 1.50.v

∴i,ii, iii and iv have correct units.

From V\(^2\) = U\(^2\) + 2as O = U\(^2\) - 2gH, since at maximum height H, the final velocity, v = O. when K.E = P.E ∴ H = \(\frac{(U)^2}{2g}\) Halfway is the \(\frac{\text{height}}{2}\) PE = KE at \(\frac{(1)}{2}\) H = \(\frac{(U)^2}{2g}\) x (\(\frac{1}{2}\)) = \(\frac{U^2}{4g}\)

F = \(\frac{mv^2}{\text{r}}\) = \(\frac{100 \times 5^2}{10}\) = 250N R = mg = 100 x 10 = 1000N coefficient of friction, µ = \(\frac{\text{F}}{\text{R}}\) = \(\frac{(250)}{1000}\) = 0.25.

hlpl = hwpw where, L =liquid and W = water (pl)/pw = (hw)/hl = (26.5)/20.4 = 1.299

relative density = (weight of substance )/weight of equal volume of water = (4 - 2)/5 - 2 = (2)/3