PHS 2024 4Exp PP Prelim P1 worked solution (shared)

1 [Turn Over PRESBYTERIAN HIGH SCHOOL SCIENCE DEPARTMENT MARKSCHEME Subject: Physics Test: PrelimExam Setter: Year: 2024 Level: Sec 4 Express Answers to Paper 1: MCQ 1 C 2 D Fall at a constant acceleration of 10 m/s2 All objects will fall at the same rate if there is no air resistance. 3 C • greater mass • greater weight • accelerates longer before air resistance balances with the weight • take longer time to teach terminal velocity • higher terminal velocity 4 A gb = 9.81 N/kg gt = 9.79 N/kg Wb = mgb 600 = m(9.81) m = 61.2 kg Wt = mgt = (61.2)(9.79) = 599 N 5 B When  = 180° , magnitude of resultant force = 4 + ( -2) = 2 N When  = 90° , magnitude of resultant force = √(22 + 42) = 4.47 N When  = 0° , magnitude of resultant force = 4 + 2 = 6 N For resultant of 4N, 90° <  < 180° 6 A Resultant force, F = ma From graph : 1st Part: F , acceleration  Since gradient of v-t graph = acceleration When acceleration , gradient of graph  From graph : 2nd Part: F is zero, acceleration is zero Since gradient of v-t graph = acceleration When acceleration is zero, gradient of graph is zero or constant speed

2 7 A Volume of water = 150 – 100 = 50 cm3 Mass of water = 50 × 1.0 = 50 g Mass of paraffin = 0.80 × 100 = 80 g Total mass = 50 + 80 = 130 g 8 A (G.P.E)A + (K.E)A = (G.P.E)B + (K.E)B + W.D against friction (2)(10)(2) = ½ (2) v2 + 2.6 × 4 v = 5.4 m/s 9 B In an area of 1.0 m2 of each panel receives 0.80 kJ of energy from the Sun in 1.0 s. Power input for 1.0 m2 = energy/time = 0.80 kJ / 1.0 s = 0.80 kW Power input for 7.2 m2 = 0.80 kW × 7.2 = 5.76 kW Efficiency = 1.0 kW/ 5.76 kW × 100% = 17% 10 A chemical energy is transferred electrically to the internal store and kinetic store. 11 A The pollen particles are being bombarded by the heavier particles of the water. 12 C P gas - P atm = 30 cm Hg P gas = 30 cm of water + 76 cm Hg = (0.3)(1000)(10) + (0.76)(13 600)(10) = 106 360 Pa = 106 000 Pa 13 D Pressure at X = Pressure at Y Since height of liquid X is greater than liquid Y, therefore density of liquid X is smaller than density of liquid Y. 14 C For the energy to be transferred only by radiation, it must be in a vacuum as conduction and convection require a medium for energy to be transferred. 15 C High Heat Capacity --→ more change in internal energy is needed to change the temperature -→ temperature increase at slower rate. Since for Y, the rate of increase in temperature is the slowest, it has the largest specific heat capacity. 16 A One bucket contains 5 kg of water at 20 °C and the other contains 1 kg of water at 80 °C. Let the final temperature be . internal energy lost by 1 kg of water = internal energy gained by 5 kg of water (1) c (80 - ) = 5 c (  - 20) 80c -  = 5 - 100  = 30 °C

3 [Turn Over 17 D Ultrasound is a longitudinal wave. 18 A v = f λ = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 6600  = 1320 4  = 0.05 m 19 C a sound that becomes quieter (smaller amplitude) with a higher pitch (higher frequency) 20 D Each coil completes 4.0 oscillations in 2.0 s. … = 2.0 oscillations in 1.0 s. Therefore frequency = 2.0 Hz wavelength = distance between 2 successive compression = YZ 21 B mobile phone (microwaves) sunbed (ultraviolet rays) television controller (infrared waves) toaster (infrared waves) 22 A v = f 50 = f (20) F = 2.5 Hz Period, T = 1/f = 0.40 s 0.30 s later, wave will have travel a distance of ¾  23 B Objects Q, R, S and T give rise to images that can be seen by the eye at E. 24 C Refractive index, n = 𝑠𝑖𝑛55 𝑠𝑖𝑛 33 = 1.50 speed of light in the glass = 3.0/1.5 x 108 = 2.0 x 108 m/s 25 A  is the critical angle as it produce angle of refraction of 90° As not all light rays are reflected, there is no total internal reflection

4 26 D Focal length = 3 cm Object distance = 2F = 6.0 cm to produce image of same size. 27 A Negative induced charges on left and positive induced charges on the right of each sphere as the negative will be attracted to the positively charged rod. 28 C V = IR Current, I = V/R = 6 ÷ 3 = 2.0 A Charge = current x time = 2.0 × 60 = 120 C 29 D Straight line… ratio of I/R is constant… resistance is constant Curved section… ratio of I/R decreases… resistance increases 30 B Lamp Z becomes dimmer as the total resistance of the circuit will increase when lamp Y is removed. 31 C p.d across LDR = 2.0 V p.d across thermistor = 4.0 V ratio of resistance of thermistor to resistance of LDR = 4/2 = 2:1 1000  (warm)/ 500  (bright) = 2 : 1 32 B Since the earth from the metal case is connected to the live terminal, the metal case will be live. 33 D Using P = V2 R For the power dissipated to be the highest, the total resistance of the circuit has to be the lowest. 34 B When electrical appliances are earthed, current leakage is channelled to the ground, preventing electric shock, while a large current may blow the fuse and disconnect the appliance due to the metal case's small resistance.

5 [Turn Over 35 D using a larger control current will increase the strength of the magnetic field. 36 D Using fleming left hand rule the direction of the force acting on the wire is upward. Based on newton 3rd Law, the wire will exert a downward force on the magnet. Therefore the reading on the balance will increase 37 D speed of rotation of the coil steadily increases, the amplitude and the frequency of the induced e.m.f will also increases. 38 B Using Lenz Law, the direction of the induced current will be in opposite direction magnet falls away from the coil. 39 D Using fleming left rule, the particles have to be negatively charged. 40 C Nuclear fusion is a process in which two light atomic nuclei combine to form one heavier atomic nucleus and releases a huge amount of energy.