Secondary 4 Pure Physics Practice Paper 3

Answer Key - Pure Physics Secondary 4 (Version 3)

Section A

Question 1 (a) Weight (or $mg$) [1] (b) Initially, weight is the only force, causing acceleration. As speed increases, air resistance increases [1]. The resultant force ($W - R$) decreases, so acceleration decreases [1]. Eventually, air resistance equals weight, resultant force is zero, and velocity becomes constant [1]. (c) Terminal velocity decreases [1] because the parachute increases surface area, increasing air resistance for the same speed, resulting in a lower speed where $R = W$ [1].

Question 2 (a) For a body in rotational equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot [2]. (b) Anticlockwise moment = $0.1\text{kg} \times 10 \times (40-10) = 30\text{ Nm}$. Clockwise moment (Weight of rule) = $W \times (50-40) = 10W$. $10W = 30 \rightarrow W = 3\text{ N}$ [2]. (c) Total anticlockwise moment needed = $3\text{ Nm}$ (from rule's weight). $0.2\text{kg} \times 10 \times d = 3 \rightarrow 2d = 3 \rightarrow d = 1.5\text{ cm}$ from pivot. Position = $40 + 1.5 = 41.5\text{ cm}$ or $40 - 1.5 = 38.5\text{ cm}$ [2].

Question 3 (a) $P = F/A = 100 / (2.0 \times 10^{-3}) = 50,000\text{ Pa}$. $F_2 = P \times A_2 = 50,000 \times (5.0 \times 10^{-1}) = 25,000\text{ N}$ [2]. (b) Force multiplication / Mechanical advantage [1].

Question 4 (a) $Q = mc\Delta\theta = 0.8 \times 390 \times (95 - 25) = 0.8 \times 390 \times 70 = 21,840\text{ J}$ [2]. (b) $\text{Input} = \text{Output} / 0.8 = 21,840 / 0.8 = 27,300\text{ J}$ [2].

Question 5 (a) $\sin c = 1.0 / 1.5 = 0.667 \rightarrow c = 41.8^\circ$ [2]. (b) Light must travel from denser to less dense medium [1] and angle of incidence must exceed the critical angle [1].

Question 6 (a) $P_{out} = Fv = (5 \times 10) \times 0.4 = 20\text{ W}$ [2]. (b) $P_{in} = 20 / 0.7 = 28.57\text{ W}$ [2].

Question 7 (a) $1600 \rightarrow 800 \rightarrow 400 \rightarrow 200$ (3 half-lives) [1]. (b) $T_{1/2} = 12 / 3 = 4\text{ hours}$ [2].

Section B

Question 8 (a) $240 / 12 = 20:1$ [1]. (b) For ideal: $V_p I_p = V_s I_s \rightarrow 240 \times 0.5 = 12 \times I_s \rightarrow 120 = 12 I_s \rightarrow I_s = 10\text{ A}$ [3]. (c) $P_{in} = 120\text{ W}$. $P_{out} = 120 \times 0.85 = 102\text{ W}$. $I_s = 102 / 12 = 8.5\text{ A}$ [3]. (d) Use laminated iron core (reduce eddy currents) [1]; Use soft iron core (easy magnetization) [1].

Question 9 (a) Light intensity increases $\rightarrow$ Resistance of LDR ($R_2$) decreases [1]. Since $R_1$ is constant, the proportion of voltage across $R_2$ decreases [1]. Therefore, output voltage decreases [1]. (b) $V_{out} = (R_2 / (R_1 + R_2)) \times V_{in} \rightarrow 4 = (R_2 / (2000 + R_2)) \times 12$ $4(2000 + R_2) = 12 R_2 \rightarrow 8000 + 4 R_2 = 12 R_2 \rightarrow 8 R_2 = 8000 \rightarrow R_2 = 1000\text{ }\Omega$ [3]. (c)

Generated diagram for this question.

[2].

Question 10 (a) The needle of the galvanometer deflects momentarily [1]. (b) Moving magnet creates a change in magnetic flux through the coil [1]. According to Faraday's Law, this induces an electromotive force (EMF) [1], which drives a current [1]. (c) Use a stronger magnet [1], increase the number of turns in the coil [1], or move the magnet faster [1]. (d) It reverses the direction of the current in the coil every half turn [1]. This ensures the force on the coil remains in the same direction [1], allowing the motor to rotate continuously in one direction [1].

Question 11 (a) $I = P / V = 2000 / 230 = 8.70\text{ A}$ [2]. (b) No, it will not blow [1] because the operating current (8.70 A) is less than the fuse rating (13 A) [1]. (c) Provides a low-resistance path to earth [1] so that in case of a fault (live wire touching case), the current flows to earth, blowing the fuse and preventing electric shock [1].