CIE iGCSE Co-Ordinated Science P4.2.4 Resistance Exam Style Questions Paper 4
Question
Fig. 6.1 shows a temporary zebra enclosure in a wildlife conservation park. The enclosure is surrounded by an electric fence.
(a) The fence is powered by an e.m.f. of 2000V and carries a current of 80mA.
(i) Calculate the total resistance of the fence.
▶️Answer/Explanation
Solution:
Given:
E.m.f. (V) = 2000 V
Current (I) = 80 mA = 0.08 A
Using Ohm’s Law: \( R = \frac{V}{I} \)
\( R = \frac{2000}{0.08} = 25000 \, \Omega \)
Answer: 25000 Ω
(ii) The fence is made of two identical cables connected in parallel. The cables act as resistors. Fig. 6.2 shows the circuit used in the electric fence.
Use your answer to 6(a)(i) to calculate the resistance of one of the cables.
▶️Answer/Explanation
Solution:
For resistors in parallel, the total resistance \( R_{total} \) is given by:
\( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \)
Since the two cables are identical, \( R_1 = R_2 = R \).
\( \frac{1}{25000} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \)
\( R = 2 \times 25000 = 50000 \, \Omega \)
Answer: 50000 Ω
(iii) A different enclosure uses a fence made of cables that are the same thickness but twice the length of those shown in Fig. 6.1. State the effect of doubling the cable length on the resistance of the fence.
▶️Answer/Explanation
Solution:
Resistance is directly proportional to the length of the cable. Therefore, doubling the length of the cable will double the resistance.
Answer: The resistance doubles.
(b) One of the zebras is startled by a loud sound.
(i) Describe how sound waves are transmitted in air.
▶️Answer/Explanation
Solution:
Sound waves are transmitted in air through the vibration of air molecules. These vibrations create compressions and rarefactions, which propagate as a longitudinal wave through the air.
Answer: Sound waves are transmitted through compressions and rarefactions in the air.
(ii) After hearing the sound, the zebra runs across the enclosure in 7.5s. The average speed of the zebra is 16 m/s. Calculate the distance the zebra runs.
▶️Answer/Explanation
Solution:
Given:
Speed (v) = 16 m/s
Time (t) = 7.5 s
Using the formula: \( \text{Distance} = \text{Speed} \times \text{Time} \)
\( \text{Distance} = 16 \times 7.5 = 120 \, \text{m} \)
Answer: 120 m
(c) Fig. 6.3 shows one of the zebras in the enclosure. The zebra has black and white stripes.
A vet uses an infrared camera to measure the temperature of the zebra. The infrared camera shows that the black stripes are a different temperature to the white parts of the zebra. Describe and explain the difference in temperature recorded.
▶️Answer/Explanation
Solution:
Black surfaces absorb more infrared radiation than white surfaces, which reflect more radiation. Therefore, the black stripes will absorb more heat and show a higher temperature on the infrared camera compared to the white stripes.
Answer: The black stripes will show a higher temperature because they absorb more infrared radiation, while the white stripes reflect more radiation and show a lower temperature.