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

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 C.

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.

(a)(i) A machine is a device used to overcome a load at one end by the application of an effort at another end. Examples of simple machine include the screw jack, can opener, pulley system. (ii) M.A. is the ratio of the load (L) to the Effort (E). \(M.A = \frac{Load}{Effort}\) V.R. is the ratio of the distance moved by the effort tc the distarice moved by the load (at the same time). \(V.R = \frac{\text{distance moved by effort}}{\text{distance moved by load}}\) Efficiency (E) is the ratio of the useful work done by the machine (work output) to the total work put into the machine (work input). \(Efficiency = \frac{\text{work done on load}}{\text{work done by effort}} \times 100%\) \(Efficiency = \frac{M.A}{V.R} \times 100%\); \(Work = Force \times Distance\) \(Efficiency = \frac{\text{work output}}{\text{work input}} \) = \(\frac{Load \times \text{distance moved by load}}{Effort \times \text{distance moved by effort}}\) = \(M.A \times \frac{1}{V.R} \times 100%\) (b) The efficiency of a simple machine is not 100% because of the work done in overcoming friction. (c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m. (i) \(V.R = \frac{2\pi L}{pitch} = \frac{2 \pi \times 0.7}{0.0044}\) = \(2 \times \frac{22}{7} \times \frac{0.7}{44 \times 10^{-4}}\) \(V.R = 1000\) (ii) \(Efficiency = \frac{M.A}{V.R} \times 100%\) \(\frac{80}{100} = \frac{M.A}{1000} \implies M.A = 800\) (iii) \(M.A = \frac{Load}{Effort}\) \(Effort = \frac{8000 \times 10}{800} = 100N\) (iv) Work done = \(\frac{load \times \text{distance moved by load}}{effort \times \text{distance moved by effort}}\) = \(\frac{8000 \times 10 \times 0.2}{0.8} = 20,000J\)

(a) A mass of dry air is entrapped in a uniform capillary tube, sealed at one end, by means of a mercury pellet. Both the tube and thermometer are then firmly attached a half-meter rule and placed in a water bath as shown above. The water is stirred and its temperature \(\theta_{1}\) at a steady state is noted. The corresponding length, \(L_{1}\) of the entrapped air is also noted and recorded. The water is gradually heated and stirred uniformly until another steady temperature, \(\theta_{2}\) is attained and recorded. The corresponding length, \(L_{2}\) of the air column in the capillary tube is again read and recorded. The experiment is repeated to obtain more sets of readings. A graph of L against \(\theta\) is plotted as shown. Deduction: Since the volume V of the tube is proportional to its length, L the graph shows that the volume of the entrapped air increases (directly) with increase in temperature at constant pressure. Precautions: Dry air must be used. Also, error of parallax in reading thermometer and meter rule avoided. (b) \(S = \frac{V_{1} - V_{0}}{V_{0} \theta}\) = \(\frac{18.7 - 13.7}{13.7 \times 100}\) = \(\frac{5}{1370}\) = \(3.65 \times 10^{-3} K^{-1}\) (c) As temperature increases the kinetic energies the molecules increase and they bombard the wall the container at a faster rate since their velocities has increased. To keep the rate of change of momentum per unit area (i.e pressure) the same, the molecules will have to travel longer distances betweon collision This implies increase in volume.

(a) Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of the change of the flux linked with the circuit. The direction of the induced current is always such as to oppose the change producing it. As coil is rotating in the magnetic field, an e.m.f which is due to the changing flux induces between its terminals. The e.m.f. varies from maximum when the coil is horizontal to minimum when the coil is vertical at which position connections between brushes and terminals of coil are changed by commutator. This ensures that current continues to flow in the same direction. (c) – By increasing the number of turns – Winding the coil on the soft iron core – Increasing the strength of the magnet – turning the coil faster.

(a)(i) Fusion: Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy. Example: \(^{2} _{1} H + ^{2} _{1} H \to ^{4} _{2} He + Energy\) (ii) Fission: Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process. Example: \(^{235} _{92} U + ^{1} _{0} N \to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\) (iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns (b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by. (c) \(I = I_{m} \sin 2 \pi ft ... (1)\) I - Instantaneous value of current \(I_{m}\) - Peak value of current F - frequency of current \(I = 30 \sin 100 \pi t ... (2)\) Comparing (1) and (2), (i) \(2 \pi f = 100\pi \therefore f = 50 Hz\) (ii) \(I_{m} = 30A\) (iii) \(I_{r.m.s} = \frac{I_{m}}{\sqrt{2}}\) = \(\frac{30}{\sqrt{2}} = 21.2A\)

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

2KClO3(s) → 2KCl(s) + 3O2(g) From the above balanced equation, 2 moles of reactant decomposed to give 3 moles of O2 i.e. i mole of reactant will produce (3)/(2) moles O2. ∴ 2.5 moles KClO3 will give = 3.75 moles

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 A.

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

Topic is Chemical Bonding