Solve 3 (b) (i) Ammeter reading when key K is open.

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Step 1: Solve 3 (b) (i) Ammeter reading when key K is open. When the key K is open, the circuit branch containing the ammeter is an open circuit. No current can flow through an open circuit. The ammeter reading will be $0 \, \text{A}$.

Step 2: Solve 3 (b) (ii) Voltmeter reading when key K is open. When the key K is open, the branch with the 20 $\Omega$ resistor is still connected directly across the 12 V voltage source. Therefore, the voltage across the 20 $\Omega$ resistor is the source voltage. The voltmeter reading will be $12 \, \text{V}$.

Step 3: Solve 3 (c) Combined resistance when key K is closed. The circuit diagram shows two resistors in parallel. One resistor is 20 $\Omega$. The other resistor (in the branch with the switch and ammeter) has no value indicated. We will assume this unlabeled resistor is also $20 \, \Omega$ for the calculation. When the key K is closed, the two $20 \, \Omega$ resistors are in parallel. The combined resistance ($R_{eq}$) for two parallel resistors is given by: $R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$ Substituting the values: $R_{eq} = \frac{20 \, \Omega \times 20 \, \Omega}{20 \, \Omega + 20 \, \Omega} = \frac{400 \, \Omega^2}{40 \, \Omega} = 10 \, \Omega$ The combined resistance of the circuit is $\boxed{*10 \, \Omega*}$ (assuming the unlabeled resistor is $20 \, \Omega$).

Step 4: Solve 3 (d) Ammeter reading when key K is closed. Assuming the resistor in the ammeter branch is $20 \, \Omega$, and the key K is closed, the voltage across this parallel branch is $12 \, \text{V}$. The current ($I$) through this branch is given by Ohm's Law: $I = \frac{V}{R}$ Substituting the values: $I = \frac{12 \, V}{20 \, \Omega} = 0.6 \, \text{A}$ The ammeter reading will be $\boxed{*0.6 \, A*}$ (assuming the unlabeled resistor is $20 \, \Omega$).

Step 5: Solve 3 (e) (i) Equivalence of 1 kWh in joules. 1 kilowatt-hour (kWh) is a unit of energy. $1 \, kWh = 1 \, kW \times 1 \, \text{h}$ Convert kilowatts to watts and hours to seconds: $1 \, kW = 1000 \, \text{W}$ $1 \, h = 3600 \, \text{s}$ $1 \, kWh = 1000 \, W \times 3600 \, s = 3,600,000 \, \text{J}$ 1 \, kWh = \text{*3.6 \times 10^6 \, J*}

Step 6: Solve 3 (e) (ii) Calculate the cost of electricity. First, calculate the total power consumption: Power of cooker = $1000 \, W = 1 \, \text{kW}$ Power of immersion heater = $300 \, W = 0.3 \, \text{kW}$ Total power = $1 \, kW + 0.3 \, kW = 1.3 \, \text{kW}$ Next, calculate the total energy consumed per day: Energy = Total power $\times$ Time Energy = $1.3 \, kW \times 10 \, h = 13 \, \text{kWh}$ Finally, calculate the total cost: Cost = Energy consumed $\times$ Cost per kWh Cost = $13 \, kWh \times 75 \, frs/kWh = 975 \, \text{frs}$ The cost of electricity used is $\boxed{*975 \, frs*}$

Step 7: Solve 4 (a) (i) Two properties of plastics useful as engineering materials. Two properties of plastics that make them useful as engineering materials are: • They are lightweight with a high strength-to-weight ratio.