O Level Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics O-Level (Answers)

TuitionGoWhere Practice Paper (AI) - Version 2

Section A: General Physics & Mechanics

Question 1 (a) $F_{net} = ma \Rightarrow 15 - F_{friction} = 2.0 \times 3.5$ $15 - F_{friction} = 7.0$ $F_{friction} = 8.0\text{ N}$ [2 marks] (b) $v = u + at \Rightarrow v = 0 + (3.5 \times 4.0) = 14\text{ m/s}$ [2 marks] (c) $s = ut + \frac{1}{2}at^2 \Rightarrow s = 0 + 0.5 \times 3.5 \times 4^2 = 28\text{ m}$ $W = F \times s = 15 \times 28 = 420\text{ J}$ [3 marks]

Question 2 (a) Pressure is transmitted equally in all directions in an incompressible fluid. A small force on a small area creates a pressure that exerts a much larger force on a larger area. [2 marks] (b) $A_A = \pi(0.025)^2, A_B = \pi(0.125)^2$ $F_A/A_A = F_B/A_B \Rightarrow F_A = 2000 \times (5/25)^2 = 2000 \times (1/25) = 80\text{ N}$ [3 marks]

Question 3 (a) $10\text{ m/s}^2$ [1 mark] (b) $v^2 = u^2 + 2as \Rightarrow v^2 = 0 + 2 \times 10 \times 20 = 400 \Rightarrow v = 20\text{ m/s}$ [3 marks] (c) As the ball falls, speed increases, so air resistance (drag) increases. Eventually, drag equals weight. Resultant force becomes zero, acceleration becomes zero, and the ball moves at a constant terminal velocity. [3 marks]

Question 4 (a) For a body in rotational equilibrium, the sum of clockwise moments about a pivot equals the sum of counterclockwise moments about the same pivot. [2 marks] (b) Distance of 100g mass from pivot = $40 - 10 = 30\text{ cm}$. Moment = $0.1\text{ kg} \times 10 \times 0.3\text{ m} = 0.3\text{ Nm}$. Weight of rule acts at 50 cm mark. Distance = $50 - 40 = 10\text{ cm} = 0.1\text{ m}$. $W \times 0.1 = 0.3 \Rightarrow W = 3.0\text{ N}$ [3 marks]

Section B: Thermal Physics & Waves

Question 5 (a) $E = P \times t = 50 \times 120 = 6000\text{ J}$ [2 marks] (b) $Q = mc\Delta\theta \Rightarrow 6000 = 0.2 \times c \times (70 - 20)$ $6000 = 0.2 \times c \times 50 \Rightarrow 6000 = 10c \Rightarrow c = 600\text{ J/kg}^\circ\text{C}$ [3 marks]

Question 6 (a) $n = \sin i / \sin r \Rightarrow 1.6 = \sin 30 / \sin r \Rightarrow \sin r = 0.5 / 1.6 = 0.3125$ $r = \arcsin(0.3125) = 18.2^\circ$ [2 marks] (b) $\sin c = 1/n = 1/1.6 = 0.625 \Rightarrow c = 38.7^\circ$ [2 marks]

(c) Since the angle of incidence (50°) is greater than the critical angle (38.7°), total internal reflection occurs. The light ray is reflected back into the plastic block. [2 marks]

**Question 7**
(a) $v = f\lambda \Rightarrow \lambda = 1540 / (2.0 \times 10^6) = 7.7 \times 10^{-4}\text{ m}$ (or 0.77 mm) [2 marks]
(b) Application: Fetal scanning/imaging of internal organs. Reason: Ultrasound is non-ionizing and does not damage cells/DNA, unlike X-rays which can be harmful to a developing fetus. [3 marks]

---

### Section C: Electricity & Magnetism

**Question 8**
(a) As temperature increases, the resistance of the thermistor decreases. This leads to an increase in the total current in the circuit. [2 marks]
(b) $R_{total} = 4.0 + 10 = 14\text{ }\Omega$. $I = V/R = 12 / 14 = 0.86\text{ A}$ [2 marks]
(c) $R_{total} = 4.0 + 2 = 6\text{ }\Omega$. $I = 12 / 6 = 2.0\text{ A}$.
$V = I \times R = 2.0 \times 4.0 = 8.0\text{ V}$ [3 marks]

**Question 9**
(a) $1/R_p = 1/6 + 1/3 = 3/6 \Rightarrow R_p = 2.0\text{ }\Omega$ [2 marks]
(b) $R_{total} = 2.0 + 2.0 = 4.0\text{ }\Omega$ [2 marks]
(c) Total current $I = 12 / 4 = 3.0\text{ A}$.
Voltage across parallel section $V_p = I \times R_p = 3.0 \times 2.0 = 6.0\text{ V}$.
Current through $3.0\text{ }\Omega$ resistor $I_2 = 6.0 / 3.0 = 2.0\text{ A}$ [4 marks]

**Question 10**
(a) $V_s/V_p = N_s/N_p \Rightarrow V_s = 240 \times (50/1000) = 12\text{ V}$ [2 marks]
(b) $P_{in} = P_{out} \Rightarrow V_p I_p = V_s I_s \Rightarrow 240 \times I_p = 12 \times 4.0$
$I_p = 48 / 240 = 0.2\text{ A}$ [3 marks]
(c) To increase voltage for transmission. This reduces the current, which minimizes energy loss as heat ($I^2 R$) in the transmission cables. [3 marks]

**Question 11**
(a) Fleming's Left-Hand Rule. [1 mark]
(b) It reverses the direction of the current in the coil every half turn. This ensures that the force on the sides of the coil always acts in the same direction, maintaining a continuous rotation in one direction. [3 marks]
(c) 1. Increase the current flowing through the coil. 2. Increase the strength of the magnetic field (using stronger magnets). [2 marks]