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

P = (80 x 5 + 100 x 3)8 = 5600W Cost = \(\frac{5600}{1000} \times 5 = N28\)

10 - P sin 30 = 0 10 - P x 0.5 = 0 P = 20N

Efficiency = \( \frac{W(120N)INPUT}{WORK OUTPUT} \times \frac{100}{1}\) And for incline plane V.R =\( \frac{1}{Sin\theta}\) = V.R = \(\frac{1}{Sin30} =\frac{1}{0.5}\) = 2 \(\text{therefore } \frac{60}{120} = \frac{M.A}{2}\) M.A = \(\frac{120}{100}\) = 1.2 \(\text{but M.A } = \frac{load}{effort} = \frac{120}{\text{E}}\) \(\text{therefore} \frac{120}{\text{E}}\) = 1.2 E = \(\frac{120}{1.2 }\) = 100N Therefore, Effort up the plane = 100N

from hooke's law F=Ke \( \frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore} \frac{4}{0.02} = \frac{15}{e_2} \) e1 = \( \frac{0.02 x 15}{4} = 0.075M \) The new length of the string is 4 + 0.075 = 4.075 = 4.08

Mathematically Boyle's law = PV = K Where P = Pressure and V = Volume By definition Pressure = \( \frac{F}{A} = \frac{F}{M \times M} \) Again Workdone = force x displacement therefore \( PV = \frac{F}{A} \times V = \frac{F}{M^2} \times \frac{M^3}{1} \\ PV = \frac{F}{x\times x} \times \frac{x \times x \times x}{1} = F \times x \) = force x displacement

Spaces and Forces: The differences observed in solids, liquids, and gases can primarily be attributed to the arrangement, spacing, and intermolecular forces acting between the molecules. Solids: Tightly packed molecules with strong intermolecular forces, giving a definite shape and volume. Liquids: Close molecules that can move past each other, resulting in a definite volume but no definite shape. Gases: Widely spaced molecules with weak intermolecular forces, filling the entire container.

Workdone = product of the charge Q, and the p.d V. i.e Work done = QV = 600 * 10-6 *2.1 * 103 =1.8j

In general, Power = IV; \( \implies 40KW = IV \\ \text{Therefore } 40000 = 1 \times 800\\ \implies I = \frac{40000}{800} = 50A,\text{tune the current through} \) Resistor = 50A power loss= I2R = 502 x 2 = 2500 x 2 = 5.0 x 103W

since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally 5 x 10-5

Lenz's lew of electrical magnetic induction is usefully referred to as the law of conservation of energy

Power of the lamp = 60W; \(I = {\frac{P}{V}}\) \(I = {\frac{60}{120}}\) = \(0.5A\) \(R = {\frac{V}{I}}\) \(R_mains\) = \(\frac{220}{0.5}\) = \(440\Omega\) \(R_lamp\) = \(\frac{120}{0.5}\) = \(240\Omega\) \(\therefore\) The non-inductive resistance to keep the lamp = \((440 - 240)\Omega\) = \(200\Omega\)

Mass of Cu deposited = Area x thickness x density = electrochemical equivalent x current x time. Thickness = \(\frac{\text{electrochemical equivalent} \times \text{current} \times \text{time}}{Area \times density}\) = \(\frac{3.3 \times 10^{-7} \times 5 \times 3600}{60 \times 10^{-4} \times 8.9 \times 10^{3}}\) = 1.1 x 10\(^{-4}\) m

(a) Stress = \(\frac{Force}{Area} = \frac{50}{3.142(1 \times 10^{-3})^2}\) (b) Strain = \(\frac{extension}{length} = \frac{2.5 \times 10^{-5}}{5}\) = 5 x 10\(^{-5}\)

(a) Factors that can affect the rate of diffusion are: — Mass — Temperature — Density. (b) Effects of surface tension are: — needle gently placed on the surface of water floats — water skaters are able to walk on the surface of water — spilled mercury forms spherical droplets — soap bubbles are spherical in shape.