CIE iGCSE Co-Ordinated Science P4.2.4 Resistance Exam Style Questions Paper 3
Question
A room in a house has an electric heater.
(a) Fig. 12.1 shows part of the circuit containing the heater.
Complete the circuit diagram in Fig. 12.1 by adding the correct electrical symbol for a fuse.
(b) When the circuit is switched on, the current in the heater is 3A and the supply voltage is 240V.
(i) Calculate the resistance of the heater.
State the unit of your answer.
resistance = ……………………………….. unit ………..
(ii) The fuse in the circuit needs to be replaced.
Explain why a 5A fuse is used and not a 3A fuse.
(c) Fig. 12.2 shows the heater as part of an underfloor heating system.
(i) When the heater is switched on, thermal energy passes through the solid floor to heat the air in the room.
The temperature of the air in the room increases slowly.
State the method of thermal energy transfer through the solid floor.
(ii) State the method of thermal energy transfer that heats all the air in the room.
(d) Some water spills onto the floor and evaporates.
Describe evaporation in terms of the motion of water molecules.
(e) There are solar cells on the roof of the house.
State one advantage and one disadvantage of generating electricity using solar cells. Do not include the cost.
advantage ………………………………………………………………………
disadvantage ………………………………………………………………….
▶️Answer/Explanation
Ans : 12(a) fuse inserted in gap using the correct symbol ;
12(b)(i) evidence of resistance = voltage / current (in any form) or 240 / 3 ;
= 80 ;
Ω / ohms ;
12(b)(ii) fuse needs to be slightly greater than maximum current (or it would blow in normal use) ;
12(c)(i) conduction ;
12(c)(ii) convection ;
12(d) faster moving / most energetic molecules (leave) ;
(leave) from the surface of the liquid ;
12(e) renewable energy / does not produce CO2 etc ;
need sunlight / only works during the day etc ;
Question
(a) A car has two headlamps, connected in parallel, across a 12V battery. There is one switch in the circuit which controls both lamps.
(i) Complete the circuit diagram in Fig. 9.1 to show how the two lamps and the switch are connected to the battery.
Fig. 9.1
(ii) The current passing through each lamp is 4.0A.
The potential difference across each lamp is 12V.
Calculate the resistance of each lamp.
State the unit of your answer.
resistance = …………………….. unit ……………….
(b) The car is crossing a long bridge.
Fig. 9.2 shows a gap in the road surface in the middle of the bridge.
On a hot day the temperature of the road surface increases.
(i) State what happens to the gap as the temperature increases.
…………………………………………………………………………………………………………………….
(ii) Explain why the gap is needed.
……………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………….
(c) The driver of the car notices that the sound from the engine is louder and has a higher pitch when the car accelerates up a hill.
(i) State how the amplitude of the sound wave changes when the car accelerates up the hill.
…………………………………………………………………………………………………………………….
(ii) State how the frequency of the sound wave changes when the car accelerates up the hill.
…………………………………………………………………………………………………………………….
(iii) The car gains thermal energy as it accelerates up the hill.
State two other forms of energy gained by the car as it accelerates up the hill.
1 ……………………………………………………………………………………………………………………
2 ……………………………………………………………………………………………………………………….
▶️Answer/Explanation
Ans: 9(a)(i) all symbols correct ;
two lamps in parallel with battery ;
switch to control both lamps ;
9(a)(ii) R = V/I (in any form symbols or words) or 12/4 ;
= 3 ;
ohms / Ω ;
9(b)(i) gap is smaller/closes ;
9(b)(ii) road/bridge needs to expand on hot day/
road/bridge could be damaged by expansion ;
9(c)(i) amplitude increases ;
9(c)(ii) frequency increases ;
9(c)(iii) kinetic (energy);
gravitational potential (energy) ;
Question
Fig. 9.1 shows an aircraft at rest on a runway.
(a) The mass of the aircraft is 400 000 kg.
Calculate the weight of the aircraft. The gravitational field strength, \( g \), is 10 N/kg. (Topic – P1.3)
▶️Answer/Explanation
Weight = 4,000,000 N
Explanation: Weight is calculated using the formula \( W = m \times g \), where \( m \) is mass and \( g \) is gravitational field strength. Substituting the values, \( W = 400,000 \, \text{kg} \times 10 \, \text{N/kg} = 4,000,000 \, \text{N} \).
(b) The aircraft starts from rest and accelerates along the straight runway. The aircraft engines produce a constant horizontal thrust force of 1,200,000 N. A constant frictional force of 500,000 N acts on the aircraft. (Topic – P1.5)
(i) Calculate the resultant horizontal force acting on the aircraft.
▶️Answer/Explanation
Resultant force = 700,000 N
Explanation: The resultant force is calculated by subtracting the frictional force from the thrust force: \( F_{\text{resultant}} = 1,200,000 \, \text{N} – 500,000 \, \text{N} = 700,000 \, \text{N} \).
(ii) Explain why the aircraft accelerates.
▶️Answer/Explanation
The aircraft accelerates because there is a resultant force acting on it.
Explanation: According to Newton’s Second Law of Motion, an object accelerates when there is a resultant force acting on it. In this case, the resultant force of 700,000 N causes the aircraft to accelerate.
(c) Fig. 9.2 shows a TV monitor in the cabin of the aircraft and the energy transferred each second by the monitor.
(i) The number of joules of sound energy transferred per second is shown as XJ. Calculate the value of X. (Topic – P1.6)
▶️Answer/Explanation
X = 1 J
Explanation: The total energy input is 200 J, and the energy outputs are 119 J (light) and 80 J (thermal). Therefore, the sound energy \( X = 200 \, \text{J} – (119 \, \text{J} + 80 \, \text{J}) = 1 \, \text{J} \).
(ii) The monitor has a resistance of 1900 Ω. The current passing through the monitor when in use is 0.060 A.
Calculate the potential difference across the monitor. State the unit of your answer. (Topic – P4.2)
▶️Answer/Explanation
Potential difference = 114 V
Explanation: The potential difference (V) is calculated using Ohm’s Law: \( V = I \times R \). Substituting the values, \( V = 0.060 \, \text{A} \times 1900 \, \Omega = 114 \, \text{V} \).
(iii) The current of 0.060 A is the same as 60 mA. The fuse in the electrical supply to the monitor has to be replaced. Several fuse ratings are available.
10 mA 50 mA 100 mA 250 mA
State which fuse is the correct choice. Explain your answer. (Topic – P4.4)
▶️Answer/Explanation
Fuse = 100 mA
Explanation: The fuse rating should be higher than the maximum current (60 mA) but not too much higher to ensure safety. A 100 mA fuse is the most appropriate choice as it is the smallest value above 60 mA.
Question
(a) An electric heater is used to heat a classroom in a school.
The arrows on Fig. 12.1 show the circulation of air around the classroom.
P and Q are two positions within the air circulation.
Complete the sentences using words from the list.
conduction convection cooled radiation warmed
Position P shows ………………………………. air.
Position Q shows ………………………………. air.
This method of thermal energy transfer is called ………………………………. .
(b) In the classroom, a student draws diagrams to represent the three states of matter.
Fig. 12.2 shows the diagrams drawn. Box X shows the arrangement of particles in a solid.
Box Y shows the arrangement of particles in a liquid.
(i) In box Z, draw the arrangement of particles in a gas.
(ii) Complete the sentences below using only the words solid, liquid and gas.
Solidification occurs when a …………………………….. turns into a …………………………….. .
Condensation occurs when a …………………………….. turns into a …………………………….. .
(iii) State the melting point and the boiling point of water at standard atmospheric pressure.
melting point = ……………….. °C
boiling point = ………………… °C
(c) In another lesson, the student builds an electric circuit.
Fig. 12.3 shows the circuit diagram.
(i) State the name of the components represented by the symbols in Table 12.1.
(ii) When there is a potential difference of 6V across the lamp, a current of 0.3A passes through the lamp.
Calculate the resistance of the lamp.
resistance = …………………………………………….. Ω
▶️Answer/Explanation
Ans: 12(a) cooled ;
warmed ;
convection ;
all correct ;
12(b)(i)
random arrangement and widely spaced ;
12(b)(ii) liquid and
solid ;
gas and
liquid ;
12(b)(iii) 0 °C and 100 °C ;
12(c)(i) ammeter ;
voltmeter ;
12(c)(ii) R = V ÷ I (symbols or words) or 6 ÷ 0.3 ;
R = 20 (Ω) ;