2025 RI H2 Physics Prelims P1 Answers

2025 Preliminary Examination H2 Physics Solution Paper 1 1 A The vector sum of the two components is not R. 2 A The d will increase initially as Car Y moving slower and Car X. After that, Car Y will move with speed greater than Car X and the relative speed is increasing . Hence gradient is getter steeper. 3 C Net force up the slope ( ) ( )630 15 30 sin30 15 30ga− + = + 29.095 m sa −= For the 15 kg crate, net force up the slope 15 sin30 15R g a− = ( )15 15 sin30 15 9.095 9.81sin30 210 NR a g= + = +  = 4 B Energy stored during the process is the area under the force-extension graph 5 A For translational equilibrium, the net force has to be zero . Since the forces act in the opposite direction on the same object and have the same magnitude, the net force acting on the object is zero. Option B: Even though the two forces act in opposite directions, they are spaced apart from each other which results in a torque or rotation. Hence, the object is not in equilibrium. Option C: The direction of the torque of a couple depends on the direction of the two forces and is independent of the position of the pivot. Option D: The pair of forces act on the same object, so they are not an action-reaction pair of forces. 6 D dUF dx=− Magnitude of force = gradient of graph = 1.400 0.3125 6.0 N0.380 0.200 − =− 7 D Drag force, D = 250 N Resolving forces along the slope 1.0sin 250 900 9.81 985.75 N12F D mg  = + = +   =  985.75 24 23.7 kWP Fv= =  = Alternative method: P = rate of work done against drag + rate of increase in GPE = (250  24) + 900  9.81  2 = 23.7 kW ( )( ) ( ) ( ) ( ) ( )( ) 1 2 2 1 1 2 2 1 1 2 2 1 2 1 2 1 2 T T e e T T x x x x T T x x = + − = + − − −  = + − N F W D

2 © Raffles Institution 8 B At the top of the circle, 2 22 (0.10)(6.0) (0.10)(9.81) 6.2 N0.50 mvT mg r mvT mg r += = − = − = 9 B ( )   =− =−     =−      =− 3 2 1 volume density 14 3 4 3 MGGrr Gr r Gr 10 B ( ) ( ) ( ) 32 32 32 22 2 3 3 22 2 3 3 33 3 3 3 in the same time , 3 3 90 467.7 467.7 360 107.7 Y Y Y Y Y Y Y Y Y X X X X X X X X X Y Y X XY X YY XX Y Mm MMG m R G GR R R Mm MMG m R G GR R R MG RR M RG R t            = → = → = = → = → = = = = ==  =  =  − =  Alternative method: ( ) ( ) 23 32 32 3X Y Tr T k r T k r  = = 2 33 5.196 X Y XY T T TT  =  = X takes ¼ period to move thought 90. 1 1.2994 XYTT= 0.299 360 108Y =  =  11 C ( ) ( ) 2 2 2 0 22 22 0 2 2 2 2 1 2 1 2 1 3.0 2.1 m KE m x x PE m x xxKE PE x xx x   =− = − == −= = 12 D The child on a swing is pushed at regular intervals matching the swing's natural frequency, and the amplitude of the swing increases each time. This is an example of resonance where the driving frequency matches the natural frequency of the system.

3 © Raffles Institution [Turn over 13 C Let the intensity of incident light be I0. By Malus’s law, ( ) 2 0 cos 30 ... 1=II When filter is rotated anticlockwise by an angle of 45, the transmission axis is 15 from the vertical. The new intensity I’, ( ) 2 0 cos 15 ... 2=I' I (2)  (1), 2 cos15 cos30 ' 1.2 =  = I' I II 14 C Let length of tube be L. The wavelengths  of the lowest frequencies of sound produced are: 3 5 7 2 1, , , ,... , 4 4 4 4 4 4 21 nLn L n    +−= =   = − ¢ 0 0 0 0 0 21 , 4 357, , , ,...4444 ,3 ,5 ,7 where the lowest frequency 4 vf vf n vnL v v v v LLLL vf f f f f L   + = = −= = ==  ¢ Hence, the different frequencies, 92 Hz,3 92 Hz,5 92 Hz,7 92 Hz,... 92 Hz,276 Hz,460 Hz,644 Hz,... =    = 15 A ( ) sin sin50.8 3 ... 1 dn d   = = For highest order, let  = 90 (and sin 90 = 1). ( ) ... 2dn = ( ) ( ) 12 1 , sin50.8 3 3 sin50.8 3.87 n n n =  =  = Hence, highest order is 3. 16 D Molecules move with different speeds, which can be different from the root -mean-square speed. 17 A The water molecules collide with the walls of the bottle and rebound with greater velocities. Hence, the total microscopic kinetic energy of the water molecules increases.

4 © Raffles Institution 18 B 0 1 2EPE 4 2 ACA B A D B C C D BD QQQ Q Q Q RR R Q Q Q QQQ RR R   ++ = + + + ( )( ) ( )( ) ( )( ) ( )( ) 5 6 5 5 5 6 6 6 0 1.0 10 5.0 10 1.0 10 1.0 10 2 21 22EPE 4 1.0 10 5.0 10 5.0 10 5.0 10 2 2 22 0.50 J  − − − − − − − −  −   −  −    =  −    +  + =− 19 D A negatively charged particle moves from region of lower potential to region of higher potential. Option A: This statement is correct at the mid-point between a charge of +Q and a charge of −Q. Option B: This statement is correct at the mid-point between a charge of +Q and a charge of +Q. Option C: E tends asymptotically toward zero as r increases. So, dE dr decreases as r increases. 20 C eeeNNNnAvq Ave Ave veV AL L= = = =I 22 5 194.8 10 3.2 10 1.60 10 1.2 A0.20 −−=     =I 21 B 1 11 1 0.5882 25 5.0 0.5882 2.9412 V 2.9412 1.5 A2.0 effR V − = + + =  =  = ==I 22 A v V + − P Q W X Y Z R R QA QB QC QD

5 © Raffles Institution [Turn over When temperature increases, RLDR decreases, VPY decreases, VWP increases and VP decreases. When light intensity increases, Rthermistor decreases, VXQ decreases, VQ increases. Hence, VPQ = (Vp − VQ) decreases. 23 B The magnitude of the magnetic force remains constant as the orientation of the magnetic field and direction of current does not change. The torque is not constant as the perpendicular distance between the couple will vary as the loop rotates. 24 D ( ) ( ) 73 33 950(4 10 ) (2.0) 2.985 10 T0.80 2 96 2.985 10 1.6 10 3.8 mV0.24 oBn NBA t   −− −− = =  =       = = = I 25 B The current in the wire produces a magnetic flux density into the plane of the paper in the circular coil. As the coil moves away, the magnetic flux linkage in the coil decreases. By Lenz law, the direction of the induced current is clockwise to oppose this decrease. By conservation of energy, kinetic energy decreases as current is induced. 26 C For the first half cycle: Only diode D1 is conducting and the effective resistance in the circuit is R. In the second half cycle: Only diode D2 is conducting and t he effective resistance in the circuit is 1.75R. For half of a cycle, 2 0 0 1mean power 22 VP R== ( ) 22 00 2 02 2 1.75 0.392 VV RR VP R + == 27 B According to Heisenberg’s uncertainty principle, rp r h   . Having a well-defined radius means 0r= , which means rp = . Yet, based on the model, the radial momentum is zero. 28 A The sharp characteristic lines are due to inner-shell electron transitions in the target atoms; electrons knocked out by incident electrons, followed by higher -level electrons filling the vacancies and releasing energy as X-rays. 29 C total energy of gamma photons = rest mass energy of the two particles ( )( ) 2 231 8 14 22 9.11 10 3.0 10 8.2 10 J E mc E   −− = =   =  D1 R D2 0.75R R T 0 2T

6 © Raffles Institution 30 D ( ) ( ) product reactant 6 19 11 energy released 8.4 144 8.5 90 7.6 235 10 1.6 10 3.02 10 J BE BE − − =− =  +  −     =