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Secondary 4 Pure Physics Preliminary Examination Paper 1

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Questions

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Secondary School (AI)
PRELIMINARY EXAMINATION 2024
Version 1 of 5

Subject: Pure Physics
Level: Secondary 4
Paper: 2 (Structured Questions)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________


INSTRUCTIONS TO CANDIDATES

  1. Write your name, class, and date in the spaces provided.
  2. Answer all questions.
  3. Write your answers in the spaces provided on this question paper.
  4. The number of marks is given in brackets [ ] at the end of each question or part question.
  5. You may use an approved scientific calculator where appropriate.
  6. You may lose marks if you do not show your working or if you do not use appropriate units.
  7. Take the acceleration of free fall, g=10 m/s2g = 10 \text{ m/s}^2.

Section A: Static Electricity and Fields (Questions 1-3)

1. A student rubs a polythene rod with a woolen cloth. The rod becomes negatively charged.

(a) Explain, in terms of electron transfer, how the rod becomes negatively charged. [2]




(b) The charged rod is brought near a small, uncharged piece of paper floating on water. The paper moves towards the rod. Explain why this happens. [2]




2. Fig 2.1 shows two point charges, +Q+Q and Q-Q, separated by a distance dd.

(a) On Fig 2.1, draw the electric field lines between the two charges. Include at least four lines and indicate the direction with arrows. [2]
(Space for diagram)
<br><br><br><br>

(b) A small positive test charge is placed at point X, exactly midway between +Q+Q and Q-Q. State the direction of the resultant force acting on the test charge. [1]


3. Electrostatic precipitators are used in factory chimneys to remove ash particles from smoke.

(a) Describe how the ash particles acquire a charge in the precipitator. [1]


(b) Explain why the charged ash particles are attracted to the collecting plates. [2]




Section B: Current Electricity and D.C. Circuits (Questions 4-9)

4. A copper wire has a length of 2.0 m and a cross-sectional area of 1.0×106 m21.0 \times 10^{-6} \text{ m}^2. The resistivity of copper is 1.7×108Ω m1.7 \times 10^{-8} \Omega \text{ m}.

(a) Calculate the resistance of the wire. [2]
<br><br><br>

(b) State and explain what happens to the resistance of the wire if its temperature increases. [2]



5. Fig 5.1 shows the current-voltage (I-V) characteristic graph for a filament lamp.

(a) Describe the shape of the graph and explain why it is not a straight line through the origin. [3]




(b) Calculate the resistance of the lamp when the potential difference across it is 6.0 V and the current is 0.5 A. [2]
<br><br><br>

6. In the circuit shown in Fig 6.1, a battery of e.m.f. 12 V and negligible internal resistance is connected to two resistors, R1=4.0ΩR_1 = 4.0 \Omega and R2=8.0ΩR_2 = 8.0 \Omega, in series.

(a) Calculate the total resistance of the circuit. [1]
<br>

(b) Calculate the current flowing through the circuit. [2]
<br><br>

(c) Calculate the potential difference across R2R_2. [2]
<br><br>

7. Fig 7.1 shows a parallel circuit containing two resistors, RA=6.0ΩR_A = 6.0 \Omega and RB=3.0ΩR_B = 3.0 \Omega, connected to a 12 V supply.

(a) State the potential difference across resistor RAR_A. [1]


(b) Calculate the total current supplied by the battery. [3]
<br><br><br>

(c) Calculate the total effective resistance of the parallel combination. [2]
<br><br>

8. A potential divider circuit is used to provide a variable output voltage. Fig 8.1 shows a 12 V supply connected across a uniform resistance wire XY of length 100 cm. A sliding contact S can move along the wire.

(a) Explain how the output voltage VoutV_{out} measured between X and S changes as S moves from X to Y. [2]



(b) If the contact S is at the midpoint of the wire, calculate VoutV_{out}. [1]
<br>

9. A student investigates the relationship between the length of a wire and its resistance.

(a) Identify the independent variable and the dependent variable in this experiment. [2]
Independent: ________________________
Dependent: ________________________

(b) State one variable that must be kept constant to ensure a fair test. [1]



Section C: Practical Electricity and Magnetism (Questions 10-14)

10. An electric kettle is rated at 240 V, 2000 W.

(a) Calculate the current flowing through the kettle when it is operating normally. [2]
<br><br>

(b) Calculate the electrical energy consumed by the kettle if it is switched on for 5 minutes. Give your answer in Joules. [2]
<br><br>

(c) Suggest a suitable rating for the fuse to be used in the plug of this kettle. [1]


11. Fig 11.1 shows the wiring of a 3-pin plug.

(a) State the color of the insulation for the live wire. [1]


(b) Explain the function of the earth wire in an appliance with a metal casing. [2]



12. A transformer is used to step down the voltage from 240 V to 12 V for a low-voltage lamp. The primary coil has 1000 turns.

(a) Calculate the number of turns on the secondary coil. [2]
<br><br>

(b) The lamp draws a current of 2.0 A. Assuming the transformer is 100% efficient, calculate the current in the primary coil. [2]
<br><br>

(c) In reality, transformers are not 100% efficient. State one reason for energy loss in a transformer. [1]


13. Fig 13.1 shows a simple d.c. motor consisting of a rectangular coil placed between the poles of a magnet.

(a) State the rule used to determine the direction of the force on the current-carrying coil. [1]


(b) Explain the function of the split-ring commutator in the motor. [2]



14. A magnet is dropped through a vertical copper tube. It falls much slower than a non-magnetic object of the same mass.

(a) Explain why an e.m.f. is induced in the copper tube as the magnet falls. [2]



(b) Explain why this induced e.m.f. causes the magnet to fall slowly. [2]




Section D: Electromagnetic Induction and Applications (Questions 15-20)

15. Fig 15.1 shows a coil connected to a sensitive galvanometer. A bar magnet is pushed into the coil.

(a) State what is observed on the galvanometer when the magnet is:
(i) pushed into the coil: ________________________________________ [1]
(ii) held stationary inside the coil: _______________________________ [1]
(iii) pulled out of the coil: ______________________________________ [1]

(b) State two ways to increase the magnitude of the induced e.m.f. in this experiment. [2]



16. An a.c. generator produces an output voltage that varies with time. Fig 16.1 shows the waveform.

(a) Define the term 'frequency' of an a.c. supply. [1]


(b) If the time period of the wave is 0.02 s, calculate the frequency. [2]
<br><br>

17. High-voltage transmission lines are used to transmit electrical power over long distances.

(a) Explain why electrical power is transmitted at high voltage. [2]



(b) A power station generates 500 MW of power at 25 kV. This is stepped up to 400 kV for transmission. Calculate the current in the transmission lines at 400 kV. [2]
<br><br>

18. Fig 18.1 shows a transformer with a primary coil connected to a d.c. battery and a switch. The secondary coil is connected to a galvanometer.

(a) Describe and explain what is observed on the galvanometer when the switch is:
(i) closed: ______________________________________________________ [2]


(ii) kept closed for a few seconds: _________________________________ [1]


(iii) opened: _____________________________________________________ [1]


19. A student wants to demagnetize a permanent magnet.

(a) Describe a method using an a.c. supply to demagnetize the magnet. [2]



(b) State one property of a 'soft' magnetic material that makes it suitable for transformer cores. [1]


20. Fig 20.1 shows a circuit breaker connected in series with an appliance.

(a) State one advantage of a circuit breaker over a fuse. [1]


(b) Explain how a circuit breaker operates when the current exceeds the rated value. [2]




END OF PAPER

Answers

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4

ANSWER KEY AND MARKING SCHEME
PRELIMINARY EXAMINATION 2024
Version 1 of 5

Subject: Pure Physics
Level: Secondary 4
Paper: 2 (Structured Questions)
Total Marks: 60


Section A: Static Electricity and Fields

1.
(a) Electrons are transferred from the woolen cloth to the polythene rod. [1]
The rod gains excess electrons, giving it a net negative charge. [1]

(b) The negative rod repels electrons in the paper to the far side, leaving the near side positively charged (induction). [1]
The attractive force between the rod and the induced positive charge is greater than the repulsive force from the further negative charge, resulting in a net attraction. [1]

2.
(a) Lines curve from +Q+Q to Q-Q. [1]
Arrows point from +Q+Q towards Q-Q. [1]
(Lines should not cross; density higher near charges).

(b) Towards Q-Q (or to the right, assuming Q-Q is on the right). [1]

3.
(a) Ash particles pass through a region of high voltage/corona discharge where they gain electrons (or ions attach to them). [1]

(b) The collecting plates are earthed/positively charged. [1]
Opposite charges attract, so the negatively charged ash is attracted to the plates. [1]


Section B: Current Electricity and D.C. Circuits

4.
(a) R=ρLAR = \frac{\rho L}{A} [1]
R=1.7×108×2.01.0×106R = \frac{1.7 \times 10^{-8} \times 2.0}{1.0 \times 10^{-6}}
R=0.034ΩR = 0.034 \, \Omega [1]

(b) Resistance increases. [1]
As temperature increases, lattice ions vibrate more vigorously, increasing the frequency of collisions with free electrons, thus impeding their flow. [1]

5.
(a) The graph curves with decreasing gradient (current increases less rapidly than voltage). [1]
As current increases, the temperature of the filament increases. [1]
Higher temperature increases resistance, so a larger increase in voltage is needed for the same increase in current. [1]

(b) R=VIR = \frac{V}{I} [1]
R=6.00.5=12ΩR = \frac{6.0}{0.5} = 12 \, \Omega [1]

6.
(a) Rtotal=R1+R2=4.0+8.0=12.0ΩR_{total} = R_1 + R_2 = 4.0 + 8.0 = 12.0 \, \Omega [1]

(b) I=VR=1212=1.0 AI = \frac{V}{R} = \frac{12}{12} = 1.0 \text{ A} [2] (1 for formula/sub, 1 for ans)

(c) V2=I×R2=1.0×8.0=8.0 VV_2 = I \times R_2 = 1.0 \times 8.0 = 8.0 \text{ V} [2]

7.
(a) 12 V [1] (Voltage across parallel branches is equal to supply voltage).

(b) IA=VRA=126=2.0 AI_A = \frac{V}{R_A} = \frac{12}{6} = 2.0 \text{ A} [1]
IB=VRB=123=4.0 AI_B = \frac{V}{R_B} = \frac{12}{3} = 4.0 \text{ A} [1]
Itotal=IA+IB=2.0+4.0=6.0 AI_{total} = I_A + I_B = 2.0 + 4.0 = 6.0 \text{ A} [1]

(c) 1Rtotal=1RA+1RB\frac{1}{R_{total}} = \frac{1}{R_A} + \frac{1}{R_B} [1]
1Rtotal=16+13=16+26=36=12\frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}
Rtotal=2.0ΩR_{total} = 2.0 \, \Omega [1]
(Or R=VI=126=2.0ΩR = \frac{V}{I} = \frac{12}{6} = 2.0 \, \Omega)

8.
(a) VoutV_{out} increases linearly from 0 V to 12 V. [1]
Because the resistance of section XS is proportional to its length, and VoutV_{out} is proportional to this resistance (potential divider principle). [1]

(b) 6.0 V [1]

9.
(a) Independent: Length of wire [1]
Dependent: Resistance (or Current/Voltage to calculate R) [1]

(b) Cross-sectional area / Thickness / Material / Temperature [1]


Section C: Practical Electricity and Magnetism

10.
(a) P=IVI=PVP = IV \Rightarrow I = \frac{P}{V} [1]
I=2000240=8.33 AI = \frac{2000}{240} = 8.33 \text{ A} [1]

(b) E=PtE = Pt [1]
t=5×60=300 st = 5 \times 60 = 300 \text{ s}
E=2000×300=600,000 JE = 2000 \times 300 = 600,000 \text{ J} (or 600 kJ) [1]

(c) 10 A or 13 A [1] (Must be higher than operating current 8.33 A).

11.
(a) Brown [1]

(b) The earth wire connects the metal casing to the ground. [1]
If the live wire touches the casing, a large current flows to earth, blowing the fuse/tripping the breaker, preventing electric shock. [1]

12.
(a) NsNp=VsVp\frac{N_s}{N_p} = \frac{V_s}{V_p} [1]
Ns=1000×12240=50 turnsN_s = 1000 \times \frac{12}{240} = 50 \text{ turns} [1]

(b) VpIp=VsIsV_p I_p = V_s I_s (100% efficiency) [1]
240×Ip=12×2.0240 \times I_p = 12 \times 2.0
Ip=24240=0.1 AI_p = \frac{24}{240} = 0.1 \text{ A} [1]

(c) Heating of coils due to resistance / Eddy currents in core / Hysteresis loss / Flux leakage. [1] (Any one)

13.
(a) Fleming’s Left-Hand Rule [1]

(b) It reverses the direction of current in the coil every half rotation. [1]
This ensures the force on the coil always acts in the same rotational direction, allowing continuous rotation. [1]

14.
(a) As the magnet falls, the magnetic flux through the copper tube changes. [1]
This changing flux induces an e.m.f. (and eddy currents) in the tube (Faraday’s Law). [1]

(b) The induced eddy currents create a magnetic field that opposes the motion of the falling magnet (Lenz’s Law). [1]
This upward magnetic force opposes gravity, reducing the net downward force and thus the acceleration. [1]


Section D: Electromagnetic Induction and Applications

15.
(a) (i) Needle deflects in one direction. [1]
(ii) No deflection (returns to zero). [1]
(iii) Needle deflects in the opposite direction. [1]

(b) Move magnet faster / Use a stronger magnet / Increase number of turns in coil. [2] (Any two)

16.
(a) The number of complete cycles (or waves) per second. [1]

(b) f=1Tf = \frac{1}{T} [1]
f=10.02=50 Hzf = \frac{1}{0.02} = 50 \text{ Hz} [1]

17.
(a) High voltage reduces the current for the same power (P=IVP=IV). [1]
Lower current reduces energy loss due to heating in the cables (Ploss=I2RP_{loss} = I^2R). [1]

(b) P=IVI=PVP = IV \Rightarrow I = \frac{P}{V} [1]
I=500×106400×103=500,000400=1250 AI = \frac{500 \times 10^6}{400 \times 10^3} = \frac{500,000}{400} = 1250 \text{ A} [1]

18.
(a) (i) Needle deflects momentarily. [1]
Changing current in primary creates changing magnetic field, which cuts secondary coil, inducing e.m.f. [1]
(ii) No deflection. [1]
(Magnetic field is constant, so no change in flux).
(iii) Needle deflects momentarily in the opposite direction. [1]

19.
(a) Place magnet inside a solenoid connected to an a.c. supply. [1]
Slowly withdraw the magnet (or slowly reduce the a.c. current to zero). [1]

(b) It is easily magnetized and demagnetized. [1]

20.
(a) Can be reset / Reusable / Faster response / More precise. [1] (Any one)

(b) When current is too high, the electromagnet inside the breaker becomes strong enough. [1]
It attracts an iron armature, which trips the switch and breaks the circuit. [1]


END OF MARKING SCHEME