2025 EJC H2 Prelim P1 MS

©EJC 2025 9749/J2H2PRELIM/2025 EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 9749 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions 1 Answer: C The average value is 47.78 mm which is not close to the true value so not accurate. There is a variation in measurements. So it is not precise. 2 Answer: D distance 90time taken to travel first 90 km = 1.12 5 hour<speed> 80== time taken for remaining 110 km = 0.875 hour 1total distance 110 km 126 km htotal time 0.875 hourspeed − = = = 3 Answer: A 3 (1) 2 ' (2) (2) 2 ':1(1) 3 3' 2 3' 0.5 0.5(0.40) 0.20 cm2 kx mg kx mg x x xx x x x x x = −−− = −−− = = − = − = = =

2 ©EJC 2025 9749/J2H2PRELIM/2025 4 Answer: A 2 (0.50)(9.81) (0.50) (1) ' 3.5 (0.20) (2) Solving (1) and (2): 2.0 m s− −= − = −−− −= − = −−− = mg T ma Ta T friction m a Ta a 5 Answer: B Change in velocity = final velocity – initial velocity 2 2 10.40 0.30 0.50 m s −= + = Change in momentum = m v = (0.20)(0.50) = 0.10 kg m s–1 6 Answer: C ( ) 1 4 0.4004 0 1.20(9.81)1.00 7.36 m s Thrust weight m v mgt v v − = =  −= = 7 Answer: B (3m)(4)= mvsinθ (2m)(6) = mv cosθ tanθ = 1 θ = 45°

3 ©EJC 2025 9749/J2H2PRELIM/2025 8 Answer: B Constant speed, so no change in Ek. hence constant gradient p p E mgh dE hmg mgvdt t = == 9 Answer: A sin 60 sin100 60 80000 (1000)(9.81)(sin45 )100 6.5 450 N drive resistive resistive o resistive resistive F F W P F mgv F F   =+ =+  = + = 10 Answer: A Vertical equilibrium: coskx mg = --- (1) Vertical component of tension = weight Horizontal component of tension provides centripetal force 2 2sinkx mr T  =  --- (2) 2 2 (2) 4 (0.108sin27 ): tan27(1) (9.81) o o T = T = 0.62 s 11 Answer: A For satellite in orbit, TE = ‒ KE, so option A is correct. Since 2 GMmKE r= , larger m with same KE requires larger r, so option B is incorrect. Since 2 23 4Tr GM = , larger r will result in larger T, so option C is incorrect. Since 2 T  = , larger T means smaller ω, so option D is incorrect.

4 ©EJC 2025 9749/J2H2PRELIM/2025 12 Answer: C “It has a property that varies linearly with temperature” is not strictly essential but rather an assumption used in constructing empirical centigrade scales. 13 Answer: D 11 0.67 s1.5T f= = =   0 cos 0.12cos (2 )(1.5)(0.75) 0.0848 m 85 mm x x t x x   = = == Distance = 120 × 4 + (120 – 85) = 515 mm 14 Answer: C Particles are displaced away from the equilibrium positions towards the centre of the compression. Right is taken to be positive (distance from P) Particles displaced rightwards have positive displacements and leftwards have negative displacements. 15 Answer: C ( )( ) 22 0 0 cos (30 ) cos (60 ) 0.19 = = ooII II

5 ©EJC 2025 9749/J2H2PRELIM/2025 16 Answer: C 22 0 22 0 2 2 2 (2 ) (1) 23 ' (3 ) (2) (2) ' (3 ): ' 2.3(1) (2 ) net P Q net P Q A A A A A A I kx I k A A A A A A A I kx I k A I k A III k A = + = + = = → = −−− = + = + = = → = −−− = → = 17 Answer: C 8 12 3 3.00 10 4.0 10 mm1.5 10 80 10 Dx a  −   = = =    2.5x = 25 mm 18 Answer: A 7 6 7 6 sin 4 orders are observed and for max angle of 90 , sin90 (4)(5.50 10 ) 2.20 10 m If 5 orders were to be observed, sin90 (5)(5.50 10 ) 2.75 10 m Larger the value of results in more or de o o o dn d d d d d  − − − − = = = = = rs observed. OR o 66 sin9045 45 2.20 10 2.75 10 d n d d   −−  =      

6 ©EJC 2025 9749/J2H2PRELIM/2025 19 Answer: C s rb = 9 8 550 10 20 3.00 10 365 24 60 60 3.0 s −=      s = 3.47  1010 m 20 Answer: A 𝐼 = 𝑛𝐴𝑣𝑒, 𝑤ℎ𝑒𝑟𝑒 𝑛 𝑖𝑠 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝐼 = 𝑁 𝐴𝐿 𝐴𝑣𝑒 , 𝑤ℎ𝑒𝑟𝑒 𝑁 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑣 = 𝐼𝐿 𝑁𝑒 = 5 × 2 1.5 × 1024 × 1.6 × 10−19 = 4.17 × 10−5 𝑚𝑠−1 21 Answer: D After overcoming the threshold voltage, the diode’s resistance become negligible. The resistance of the fixed resistor is shown by having a constant V to I ratio, i.e. a straight line graph that cuts through the origin. 22 Answer: C 1 3 21 33 lVR A VV   = = ==

7 ©EJC 2025 9749/J2H2PRELIM/2025 23 Answer: A Since both branches have the same ratio vertically downwards, there is no current flowing through the middle horizontal wire. 1 1 1 R 10k 30k 6.0k 18k R 15 k eff eff =+ ++ = OR Since p.d. across the 10 k and 6 k resis tors are the same, they can be treated as being a parallel connection. Similarly the 30 k and 18 k resistors ca n be viewed as a parallel connection also. Reff   11 1 1 1 1 10 6.0 30 18 15 k −−    = + + +       = The 1st method only works if resistors in both branches have the same ratio. The 2nd method always works. 24 Answer: A 15 2.0 15 and using (0.48)(15) (0.45) (0.45)(2.0) 14 (0.48)(15) 10 (0.45 0.48) 3.0 R r V V V V IR R R V E V r r   = + = =+ = =− = − + =

8 ©EJC 2025 9749/J2H2PRELIM/2025 25 Answer: B For particle Y to be undeflected, electric force and magnetic force acting on Y must be in opposite direction. E field direction must be upwards for the electric force acting on particle Y to be in the opposite direction as the magnetic force (applying Fleming’s Left Hand Rule) regardless of the polarity of particle Y. 26 Answer: C Increasing current IS increases the magnetic force on XY. This increases the moment from the magnetic force pushing XY into the paper. Sliding the weight towards WZ increases the the perpedicular distance between the weight and pivot KL. This helps to counter-balance the increased moment from the magnetic force. 27 Answer: A By Faraday’s Law, ( )( )50 0.40 0.08 0.05 0.20 0.40 V N BAE t =  = = By Lenz’s Law, current flows in a direction to oppose the decrease in magnetic flux density (into page). This means that it should flow to produce more magnetic flux into page. By right hand grip rule, the induced current flows clockwise.

9 ©EJC 2025 9749/J2H2PRELIM/2025 28 Answer: C 2 _1 2 _2 For sine wave, 2 2 For half-wave square wave, 2 2 2Ratio required = 1 2 o o rms o o rms o o V T VV T TV VV T V V  ==  == = 29 Answer: D The is not enough energy between the highest and lowest energy levels of a Hydrogen atom to emit X-rays. 30 Answer: C This is because most of the alpha particles go through the empty space undeflected. Most of the atom’s mass must be concentrated.