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

Table of values/observation S/N R\(\Omega\) L(cm) R\(^{-1}\)(\(\Omega^{-1}\)) L\(^{-1}\)(cm\(^{-1}\)) 1 30.00 88.80 0.033 0.01126 2 20.00 84.60 0.050 0.01182 3 10.00 79.80 0.100 0.01253 4 5.00 71.50 0.200 0.01399 5 3.00 69.50 0.333 0.01439 6 1.00 42.40 1.000 0.02358 Slope (s) = \(\frac{\bigtriangleup {R^{-1}(\Omega)}}{\bigtriangleup {L^{-1}(cm^{-1})}} = \frac{1.08-(-0.96)}{0.002175-0.00175}\) = \(\frac{1.08-(-0.96)}{0.002175-0.00175} = \frac{2.04}{0.00425}\) = 4800cm\(\Omega ^{-1}\) I = -1.14(\(\Omega ^{-1}\) (i) K I\(^{-1}\) \(\frac{1}{I} = \frac{-1}{1.14}\) = 0.877\(\Omega\) (ii) C = \(\frac{L_{o}}{S} = \frac{78.50cm}{4800cm \Omega ^{-1}}\) = 0.01635\(\Omega\) Precautions: - The key was opened when readings were not taken. - I ensured tight connections. - I avoided parallax error when reading potentiometre. (b)i. When R = 15\(\Omega\), R\(^{-1}\) = \(\frac{1}{R} = \frac{1}{15}\) = 0.0666\(\Omega ^{-1}\) From the graph, L\(^{-1}\) = 0.001175, \(\frac{1}{L}\) 0.01175, L = \(\frac{1}{0.01175}\) = 85.11cm I = 0.5+y\(^{-1}\), -1.14 = 0.5 + y\(^{-1}\), y\(^{-1}\) = 1.14 - 0.5 = -0.64, \(\frac{1}{y}\) = -0.64 \(\therefore\) y = \(\frac{-1}{0.64}\) = -1.6 (iii) e.m.f. is the p.d between the terminals of a cell when an open circuit or the work done in moving unit charge round a complete circuit.

The three 2 ohms are in series, therefore the total resistance is \(R=R_1+R_2+R_3\) \(R_ {total} =2+2+2= 6\) ohms. resketching circuit; 6 ohms is used to replace the three 2 ohms which is therefore placed by the 3 ohms resistor. The 3 ohms and 6 ohms resistors are in parallel, so the combined resistance is \(R=\frac{R_1xR_2}{R_1+R_2}\) \(R= \frac{3\times6}{3+6}\) \(R=\frac{18}{9}=2\) ohms. Resketching circuit: The 2 ohms is used to replace the 3 ohms and 6 ohms laying side-by-side. Therefore, the 3 resistors (1 ohm, 1 ohm, and 2 ohms) are in series. Thus, the total resistance is \(R= (R_1+R_2+R_3)\) \(R_{total} =1+1+2=4\) ohms. Therefore, the total resistance measured at PQ is 4 ohms.

resonance frequency \(f_r\) = \(\frac{ 1}{ 2\pi \sqrt{LC}}\) \(f_r\) = \(\frac{1}{2\pi \sqrt{5\times10^{-6} \times 8\times 10^{-3}}}\) \(f_r\) = \(\frac{1}{ \pi 0.0004}\) \(f_r\) = \(\frac{ 2500}{\pi}\)

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

(a) A body is said to be moving with uniform acceleration when its time rate of increase of velocity is constant. It is measured in m/s\(^2\). It is a vector quantity. The above sketch shows the v-t graph of the ball. The velocity decreases uniformly from A to zero (p) at the top of K its flight in a time kp. P , Va It then increases in velocity uniformly as it returns to the point of projection A. (ii) The magnitude of the slope of the line AP = g or Ap\(^-\) = g, where g = acceleration of free fall due to gravity. The height attained when the ball is thrown up is equal to the area of the triangle AKP. The distance it falls from its maximum height is equal to the area of the triangle A\(^-\)PC which is equal to triangle AKP. (c)(i) The second stone moves upwards and its acceleration g = 10m/s\(^2\). From equation v = u - gt 10 = 20 - 10t -10 = -10t t = 1 The second stone has been in air for 1s before they met. (ii) The first stone has been moving for (1 + 2)s = 3s From equation v = u + gt = 2 - (10 x 3) = 20 - 30 = -10m/s The minus sign means the first stone is moving down and its velocity is 10m/s.

(a) Heat is a form of energy due to temperature gradient while temperature is the degree of hot-ness or cold-ness of a body. (b)(i) The change in volume expansion of a fixed mass of a liquid. (ii) The change in volume of a gas at constant pressure. (iii) The change in electrical resistance of a substance. (iv) The change in e.m.f generated in a thermo-couple. (Note: Any correct two) (c)(i) It does not wet glass (unlike alcohol) (ii) It is opaque and easily seen (unlike alcohol) (iii) It is a better conductor of heat. (iv) It measures higher temperature than alcohol. (Note: Any correct two) (d) Support the thermometer vertically ensuring the bulb is covered with melting pure ice shavings. Allow the mercury thread to fall till its level remains steady. Mark this position on the thermometer as the lower fixed point. Suspend the thermometer in the steam of a hypsometer. Allow the mercury thread to rise until its level remains steady. Mark this position on the thermometer as the steam point. Sub-divide the fundamental interval into a number of equal parts. Precautions (i) The thermometer bulb should not be placed in the boiling water of the hypsometer. (ii) The thermometer should be inserted in the hypsometer such that the mercury thread is visible just above the top of the apparatus. (iii) Some few minutes should be allowed to pass when the mercury thread is steady before making the thermometer. (iv) One should ensure that the steam is at atmospheric pressure. (Note: Any correct two) (e) The breezes are due to the fact that the specific heat capacity of water is greater than that of land. During the day the land is hotter than the sea. A low pressure area is created on the land and cooler air from the sea blows into the land. This wind is called sea breeze. At night the sea is warmer than the land. A low pressure area is created over the sea, resulting in cooler air blowing from the land where the pressure is higher into the sea. This is called land breeze.

(a) Description: The circuit is set up as shown above. The key (k) is closed. The rheostat (R) is adjusted so that lowest current is flowing. Ammeter and voltmeter readings (I and V) are recorded. Adjust rheostat to increase current. Ammeter and voltmeter readings are recorded for other positions of the rheostat. Values of (are plotted against values of I. Slope, which is the value of the resistance is determine. Precautions; i) Kcy should be removed when readings are not taken ii) Connections should be tight (iii) Parallax' error when reading ammeter and voltmeter should be avoided (iv) Only low current should be used v) Zero error of the meters should be avoided. Note: Any correct two) Resistance, R = P'/A where = length of wire, A = cross sectional area, p = resistivity )f wire. But slope, S of graph = R. Hence S = P'/A Measure I with a metre rule. Measure diameter with a micrometer screw gauge and determine A. Substitute p = \(\frac{SA}{I}\) c) Diagram i) Total e.m.f. = 4V Total resistance = 1 + 2 + \(\frac{3 \times 3}{ 2 \times 3}\) = 3 + 1\(\frac{1}{2}\) = \(\frac{1}{2} \Omega\) E = IR\(_T\), = E/R :.Current, I = 4/9 x 2/1 = 8/9; A = 0.9A (ii) P.d = IR = 8/9 x 3/2 = 4/3 = 1.3V (iii) Lost volts = Ir = \(\frac{8}{9}\) x 1 = \(\frac{8}{9} = 0.9V\)

(a) When EM radiation of appropriate frequency or wavelength is incident on a metal surface, it causes electrons to be emitted from the surface. This phenomenon is called photoelectric effect. It is a surface phenomenon. Emission is instantaneous and K.E. of emitted electrons depend on frequency or wavelength of incident photons. (b) A = Incident light or EM radiations B = Cathode C = Photoelectrons D = Anode E = Vacuum (Evacuated space) F = Glass tube (c)(i) Intensity of the incident radiation (ii) Energy or frequency of the incident radiation. (d) Similarities (i) In both, particles are emitted (ii) Both are surface phenomenon (iii) Both involves energy changes Difference (i) Evaporation involves molecules while photoemission involves electrons. (ii) Evaporation causes cooling while photoemission does not. (e)(i) Thermionic emission. (ii) Ionization of gases. (iii) Field emission (f) Photocells can be used. (i) To open or close doors automatically (ii) In automatic switching device. (iii) In television cameras (iv) As a burglar alarm (v) As safety devices in paper mills and related industries (vi) To produce sounds of varying intensities on film tracks.

Table of values/Observation S/N m(g) x(cm) (cm) 1 50.00 49.75 0.0202 2 70.00 39.65 0.0252 3 90.00 35.10 0.0285 4 110.00 30.75 0.0325 5 130.00 27.40 0.0365 6 150.00 25.00 0.0400 Slope (s) = \(\frac{\bigtriangleup {m(g)}}{\bigtriangleup {\frac {1}{x}(cm^{-1})}}\) = \(\frac{130-10}{0.04-0.0145}\) \(\frac{140}{0.0255}\) = 4705.88(g cm) Deduction: When \(\frac{1}{x}\) = 0, m = \(\frac{1}{x}\)66.00g Precaution: - l avoided draught by switching off the fans. - I ensure that the suspended mass did not touch the table. - I avoided error due to parallax when reading the metre rule - Zero error of the metre rule was noted. (b)i. From the graph, when m = 0, \(\frac{1}{x}\) = 0.0125 \(\therefore\) \(\frac{1}{x} = \frac{1}{0.0125}\) = 80 cm (1) Sum of upward forces = sum of downward forces. (2) Algebraic sum of all the moments about the pivot is Zero. ii. when m = 100g. X = 0.0305 \(\therefore\) \(\frac{1}{x} = \frac{1}{0.0305}\) = 32.97cm

Table of values/observation S/N X^\circ(degree) y^\circ (degree) Sin x Cos y 1 10.00 74.00 0.1736 0.2756 2 20.00 61.50 0.3420 0.4772 3 27.50 47.00 0.4617 0.6820 4 34.00 33.5 0.5592 0.8339 5 40.00 26.00 0.6428 0.8988 6 44.00 14.00 0.6947 0.9703 Slope (s) = \(\frac{\bigtriangleup {\text {sin x}}}{\bigtriangleup {\text {cos y}}} = \frac{0.69-0.03}{0.97-0.1}\) = \(\frac{0.66}{0.87}\) = 0.7586 = 0.76 K = \(\frac{1}{S} = \frac{1}{0.7586}\) = 1.318 \(\approx\) 1.32 Precautions: I ensured that the optical pins were vertical l ensured reasonable spacing of pins. I used a sharp pencil. l avoided parallax error in reading the protractor. (b)i. \(\frac{\text {sin i}}{\text {sin r}}\) = constant for a given pair of media where i = angle sin r of incidence and r = angle of retraction. (ii) Retraction is the bending of rays at the interface of two media while diffraction is the spreading out of rays of light around the edges of obstacles or aperture. (iii)(1) Light must travel from a dense medium to a less dense medium. (2) The angle of incidence in the dense medium must be greater than the critical angle.

Table of values/observation S/N R\(\Omega\) L(cm) R\(^{-1}\)(\(\Omega^{-1}\)) L\(^{-1}\)(cm\(^{-1}\)) 1 30.00 88.80 0.033 0.01126 2 20.00 84.60 0.050 0.01182 3 10.00 79.80 0.100 0.01253 4 5.00 71.50 0.200 0.01399 5 3.00 69.50 0.333 0.01439 6 1.00 42.40 1.000 0.02358 Slope (s) = \(\frac{\bigtriangleup {R^{-1}(\Omega)}}{\bigtriangleup {L^{-1}(cm^{-1})}} = \frac{1.08-(-0.96)}{0.002175-0.00175}\) = \(\frac{1.08-(-0.96)}{0.002175-0.00175} = \frac{2.04}{0.00425}\) = 4800cm\(\Omega ^{-1}\) I = -1.14(\(\Omega ^{-1}\) (i) K I\(^{-1}\) \(\frac{1}{I} = \frac{-1}{1.14}\) = 0.877\(\Omega\) (ii) C = \(\frac{L_{o}}{S} = \frac{78.50cm}{4800cm \Omega ^{-1}}\) = 0.01635\(\Omega\) Precautions: - The key was opened when readings were not taken. - I ensured tight connections. - I avoided parallax error when reading potentiometre. (b)i. When R = 15\(\Omega\), R\(^{-1}\) = \(\frac{1}{R} = \frac{1}{15}\) = 0.0666\(\Omega ^{-1}\) From the graph, L\(^{-1}\) = 0.001175, \(\frac{1}{L}\) 0.01175, L = \(\frac{1}{0.01175}\) = 85.11cm I = 0.5+y\(^{-1}\), -1.14 = 0.5 + y\(^{-1}\), y\(^{-1}\) = 1.14 - 0.5 = -0.64, \(\frac{1}{y}\) = -0.64 \(\therefore\) y = \(\frac{-1}{0.64}\) = -1.6 (iii) e.m.f. is the p.d between the terminals of a cell when an open circuit or the work done in moving unit charge round a complete circuit.

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

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

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