CJC.2025.H2.Phy.PRELIM.P1.Ans

CANDIDATE NAME MARK SCHEME CLASS 2T PHYSICS 9749/01 Paper 1 Multiple Choice Questions September 2025 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write your name and class in the spaces at the top of this page. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write and shade your name, NRIC / FIN number and HT group on the Answer Sheet (OMR sheet) , unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question, there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet (OMR sheet). Read the instructions on the Answer Sheet carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 23 printed pages and 1 blank page. [Turn over Catholic Junior College JC2 Preliminary Examinations Higher 2

2 DATA speed of light in free space c = 3.00 x 108 m s-1 permeability of free space 0 = 4 x 10-7 H m-1 permittivity of free space 0 = 8.85 x 10-12 F m-1 (1/(36)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2

3 [Turn over FORMULAE uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as work done on / by a gas W = p V hydrostatic pressure p = gh gravitational potential  = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin t velocity of particle in s.h.m. v = v0 cos t = 22 0 xx − electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin t magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-t) decay constant λ = 1 2 ln2 t

4 1 Which length is equal to 1 dm? A 1  100 mm B 1  101 mm C 1  100 cm D 1  101 cm Answer: D 1 dm = 1  10−1 m = 1  102 mm = 1  101 cm 2 A particle accelerates from rest. The graph shows the variation of the velocity v of the particle with time t. Which graph shows the variation of the velocity v with the acceleration a of the particle? A B C D Answer: D The gradient of the v -t graph is positive and constant. Hence, the acceleration is constant with respect to time. Hence, the v-a graph should be D such that the velocity is increasing at constant a.

5 [Turn over 3 An astronaut on the Moon, where there is no air resistance, throws a ball. The ball’s initial velocity has a vertical component of 8.00 m s−1 and a horizontal component of 4.00 m s−1, as shown. The acceleration of free fall on the Moon is 1.62 m s−2. What will be the speed of the ball 9.00 s after being thrown? A 6.60 m s−1 B 7.70 m s−1 C 10.6 m s−1 D 14.6 m s−1 Answer: B Take upwards and rightwards as positive directions. Vertical direction: 1 8.00 ( 1.62)(9.00) 6.58 m s y y yv u a t − =+ = + − =− Horizontal direction: 1 4.00 (0)(9.00) 4.00 m s x x xv u a t − =+ =+ = speed after 9.00 s = ( ) 2 2 1 6.58 (4.00) 7.70 m s− −+ = 4 A uniform square metal sheet of length x is cut into an ‘L’ shape. What is the distance of the centre of gravity of the sheet of metal from side AB? A 1.0 x B 1.2 x C 1.5 x D 1.8 x initial velocity path of ball 8.00 m s−1 4.00 m s−1 x x A B

6 Answer: B Let the weight of 1 square be W. Hence, the weight of 2 stacked squares on the right will be 2W. Let the horizontal distance of the CG of the ‘L’ shape be x from B. Hence, the equivalent weight of the combined L shape will be 3W acting downwards on this CG. Taking moments about B, Moment of the combined weight of the L shape about B = Sum of moments of the weight of the individual squares about B. ( )3W x = W(0.5 x) + 2W(1.5 x) 3 x = 3.5 x x = 1.2 x x x A B W 2W CG

7 [Turn over 5 Several identical springs, each having the same spring constant, are joined in four arrangements. A different load is applied to each arrangement. Which arrangement has the largest extension? A B C D Answer: A Let k be the spring constant of each of the identical springs. For a spring system, Fnet = keff (e) where Fnet is the load keff is the effective spring constant e is the extension of the arrangement For A: e = 2 / (0.5k) = 4 / k (largest) For B: e = 1 / (k / 3) = 3 / k For C: e = 6 / 2k = 3 / k For D: e = 8 / 3k = 2.7 / k (smallest)

8 6 The energy conversions inside a power station burning fossil fuel can be simplified as shown. chemical energy W thermal energy X electrical energy Y Which expression gives the efficiency of the power station? A Y W B ( ) Y WX+ C Y X D ( ) Y W X Y++ Answer: A Efficiency = Useful power / Input power = (Y/t) / (W/t) = Y / W 7 A metal wire is stretched to breaking point and the force–extension graph is plotted. Which graph is correctly labelled with the elastic region, the plastic region and the area representing the work done to stretch the wire until it breaks? A B C D Answer: D The force is directly proportional to the extension in the elastic region. The work done is given by the area under the force extension graph. Breaking point is at the end of the graph. Only option D fits the description.

9 [Turn over 8 The Earth takes 24 hours to complete one rotation on its axis. What is the angular velocity of the Earth as it rotates on its axis? A 1.75 x 10-3 rad s-1 B 1.99 x 10-7 rad s-1 C 4.36 x 10-3 rad s-1 D 7.27 x 10-5 rad s-1 Answer: D  = 2  / T = 2  / 24 x 60 x 60 = 7.27 x 10-5 rad s-1 9 A stone is attached to a string. The stone is then caused to swing in a vertical circular motion at a constant speed. Which of the following statements is incorrect? A The magnitude of resultant force acting on the stone is constant throughout the circular motion. B The acceleration is always directed towards the centre of the circle throughout the circular motion. C The kinetic energy of the stone is constant throughout the circular motion. D The tension in the string when the stone is at the highest point of the circular motion is higher than that when the stone is at the lowest point. Answer: D Centripetal force is constant in magnitude The centripetal acceleration is always acting towards the centre of circle K.E. is constant because speed is constant since KE = ½ m v2 At lowest poi