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

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

(a) Moment of a force is the product of force and perpendicular distance from the turning point. Moment = F x D The S.I unit is Nm. (b) Conditions: (i) The sum of forces in one direction must be equal to sum of forces in the opposite direction. (ii) Sum of clockwise forces must be equal to sum of anti-clockwise forces about the same point. Sum of clockwise moment = Sum of anti- clockwise moment. \(M \times 18 = 60 \times 24\) Mass of metre rule \(M = \frac{60 \times 24}{18} = 80g\) Clockwise moment = anti- clockwise moment \(80 \times 48 - (13 + x) = 60 \times x\) \(80 \times 35 - x = 60x\) \(2800 - 80x = 60x \implies 2800 = 60x + 80x\) \(2800 = 140x \implies x = 20cm\) Distance of balance point from the zero end = 20cm + 13cm = 33cm. (c) \(Efficiency = \frac{\text{Work output}}{\text{Work input}} \times 100%\) = \(\frac{\text{Load x Distance moved by load}}{\text{Effort x Distance moved by effort}}\) = \(M.A \times \frac{1}{V.R} \times 100%\) (ii) \(80% = \frac{200 \times 3.0}{\text{Work done by effort}} \times100% \) \(0.8 = \frac{600}{\text{Work done by effort}} \implies \text{Work done by effort} = \frac{600}{0.8} = 750J\)

(a) Vacuum between the double walls prevents heat loss by conduction and convection. The walls are silvered to prevent heat loss by radiation. The cover is made of poor conductor (cork) and it prevents evaporation. (b) Boyle's law states that the volume of a given mass of gas is inversely proportional to its pressure provided its temperature is constant. Mathematically, \(V = \frac{K}{P}\) where K is a constant of proportionality; P = Pressure ; V = Volume. \(\therefore PV = K \implies P_{1} V_{1} = P_{2} V_{2}\) (i) Atmospheric pressure (76cm Hg) \(V_{1} = 20cm, V_{2} = ? , P_{1} = (15cm + 76cm) = 91 cmHg\) \(P_{2} = 76 cmHg\) Using Boyle's law, \(P_{1} V_{1} = P_{2} V_{2}\) \(V_{2} = \frac{P_{1} V_{1}}{P_{2}}\) \(V_{2} = \frac{91 \times 20}{76} = \frac{1820}{76}\) = \(23.95 cm \approxeq 24cm\) Note: We have assumed that volume is proportional to length. Also, \(P_{2} = 76 cmHg\) because only the atmospheric pressure is acting in this case. (ii) In this case, \(V_{1} = 20cm\), \(P_{1} = 91 cmHg \) as above ; In this case, \(V_{2} = ?\) but \(P_{2} = (76 - 15) cmHg = 61 cmHg\) \(V_{2} = \frac{91 \times 20}{61} = \frac{1820}{61}\) = \(29.84 cmHg \approxeq 30 cmHg\) Clinical thermometer is used to determine the temperature of human body which is about 37^\circC for a healthy person. It is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure because (1) The range of the clinical thermometer is usually between 40^\circC and 50^\circC. Hence when put in boiling water it over expands since the temperature of boiling water is 100^\circC. (2) The boiling water temperature will be too high for the glass material thus cracking it.

(a)(i) Electric Field Intensity: An electric field is a region in which a force of electrical origin is experienced. Electric Field Intensity (E) at any point is the force per unit charge which the field exerts at that point. The S.I unit is NC\(^{-1}\). It is a vector quantity and its direction is the same as that of the force or a small positive test charge placed in the field. Mathematically, \(E = \frac{F}{q}\), F - force and q - charge. (ii) Electric Lines of Force: Electric lines of force are imaginary lines used to represent electric field patterns. They have the following characteristics: (1) At any point, the tangent to a line of force gives the direction of E at that point. (2) The number of lines per unit cross-sectional area is proportional to the magnitude of electric field. The closer the lines, the larger the electric field. (b) Field due to -q at p: \(E_{1} = \frac{q}{4 \pi \varepsilon _{0} r^{2}}\) = \(\frac{-5.0 \times 10^{-8}}{0.05^{2}} \times 9 \times 10^{9}\) = \(- 180000 NC^{-1}\) \(E_{2} = \frac{+ q}{4 \pi \varepsilon _{0} r^{2}}\) = \(\frac{5.0 \times 10^{-8}}{0.03^{2}} \times 9 \times 10^{9}\) = \(+ 500000 NC^{-1}\) \(\therefore\) Resultant of the electric field intensity E at the point P is \(E = E_{1} + E_{2} = 500000 - 180000 = 320000 NC^{-1}\) In the direction of the positive charge. (c) (i) Reactance of the capacitor \(X_{c} = \frac{1}{\omega c} = \frac{1}{2 \pi f c}\) = \(\frac{1}{2 \times \pi \times 25 \times 10 \times 10^{-6}} = \frac{2000}{\pi} \Omega\) (ii) Impedance of the circuit, \(Z = \sqrt{X_{c} ^{2} + R^{2}}\) = \(\sqrt{(\frac{2000}{\pi})^{2} + 1000^{2}}\) = \(\sqrt{(636.62)^{2} + (1000)^{2}}\) = \(\sqrt{1405285.0244}\) = \(1185.45 \Omega\) (iii) Current through the circuit, \(I = \frac{E}{Z}\) = \(\frac{90}{1185.45} = 0.0759 A_{rms}\) (iv) Voltage across the capacitor \(V = IX_{c}\) = \(0.0759 \times 636.62\) = \(48.319 V_{rms}\) (v) Average power used in the circuit \(P = IV\) = \(I^{2} Z\) = \((0.0759)^{2} \times 1185.45\) = \(6.829 W\)

Thrust = force x time thrust = \(\frac{m(v - u)}{t}\) x \(\frac{t}{1}\) 270 = m(300 - 0) m = 0.90kg

a = \(\frac{F}{m}\) = \(\frac{60000 -(4800 \times10)}{4800}\) = 2.5ms-2

X = ut + \(\frac{1}{2}\)at2 = (0 x 4) + (\(\frac{1}{2}\) x 0.1 x 4 x 4) = 0.8m workdone = force x distance = 40 x 0.8 = 32J

P.E = mgh = 1000m joules K.E = \(\frac{1}{2}\)mv2 = 200m joules %converted = \(\frac{200m}{1000m}\) x 100 = 20%

momentum force = Rsin\(\theta\) - \(\mu\)Rcos\(\theta\) Fm = (40 x 10 x sin 45) - (40 x 10 x 0.25 x cos 45) = 212N

Height below liquid = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\) volume of cube = 10\(^3\)cm\(^3\) volume of liquid displaced = \(\frac{4}{5}\) x 1000 = 800cm\(^3\) density of liquid displaced; = \(\frac{500}{800}\) = 0.625

V = F\(\lambda\) = 2 x 0.3 = 0.6ms-1

Distance between the boat and crest = 100m time = \(\frac{100}{25}\) = 4sec

Electrical energy = IVT = 0.5 x 40 x 2 x 60 = 2400J

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