YIJC 2024 JC2 PRELIM H2 Phy P1 Soln

©YIJC 9749/01/YIJC/24 2024 YIJC JC2 Preliminary Exam H2 Physics Paper 1 Solution Question Answer Question Answer Question Answer 1 A 11 A 21 D 2 C 12 D 22 B 3 B 13 A 23 C 4 C 14 B 24 B 5 B 15 D 25 C 6 A 16 C 26 A 7 D 17 C 27 C 8 B 18 A 28 B 9 A 19 D 29 C 10 A 20 B 30 C MCQs Solutions Qn Answer Explanation 1 A Surface area of cylinder A = 2rL = 282.7 cm2 20.1 0.1 282.7 10 cm (to 1 s.f.)3.0 15.0 A r L A r L rLAA rL   =+     = + = + =       Therefore A = (280  10) cm2 2 C During free fall, 2 =+ = 0 + (-9.81)(1.0) = 9.81 m s v u at v v −− On hitting the sand, 22 23 32 2 0 ( 9.81) 2 (8.0 10 ) 6.0 10 m s v u as a a − − =+ = − +  =− 

©YIJC 9749/01/YIJC/24 2 3 B Considering car and trailer as a single system to find common acceleration -2 5600 1200 400 4000 N 4000 2.0 m s1500 500 net D car trailerF F f f a = − − = − − = == + Considering only the trailer: ( )500 2.0 1000 N 1000 400 1400 N net trailer net trailer F m a F T f T T = = = =− =− = 4 C Taking moments at P, force at Q must be upwards to balance the larger clockwise moment due to weight. Taking moments at Q, force at P must be downwards to balance the larger clockwise moment due to tension. 5 B ( )( )11 70 0.040 0.020 0.70 J22EPE Fx= = − = 6 A Component of the weight parallel to the slope = W sin30o = (1.0×103)(0.50) = 0.50×103 N Because the barrel is moving at constant v, the applied F = W sin30o WD = Fx = (0.50×103)(5.0) = 2500 J 7 D Useful WD in lifting the weight = (480)(3.5) = 1680 J WD by man in pulling rope = (200)(10.5) = 2100 J 1680Efficiency 100% 80%2100=  = 8 B period of sound wave, 1T f= ( ) 1 rad8 2200 864 rad s8 2 16 fT   − == = = = =  9 A Using 2 GMmKE R= and 2 GMmTE R=− , both satellites will have the same TE if their KE is equal. Hence A is false. Since KE is same, mass of satellite is proportional to R. So, if X has a larger mass, it will have a larger radius of orbit. Option B is valid. Kepler’s 3rd law would suggest that X has a larger period. Since angular velocity is inversely proportional to period, the former is smaller.

©YIJC 9749/01/YIJC/24 3 10 A Work done in moving from X to Z ( ) 1 100 120 20 J 20 2000 1980 J 990 J kg2.0 2.0 X Z X Y Y Z X X Z Z XX X W W W W W W WW → → → → → → −→ → = + = + − =− = + =− + = −= = =− 11 A ( )( ) ( )( )( ) ( )( ) 3 3 3 4 4 Heat supplied by kettle heat absorbed by water heat loss 1.2 10 30 60 0.80 4.2 10 100 25 0.80 2300 10 heat lo ss heat loss 6.8 10 J 6.8 10power loss 38 W30 60 =+   =  − +  + = ==  12 D ( ) ( ) ( ) ( ) ( ) ( ) 2 2 22 22 31 22 constant 330 120% 475 K rms rms rms rms initial final final rms rms final U NkT Nm c T c TT cc T cc T == =         =       =  = 13 A With damping, the amplitude will be lower for all frequencies due to loss in energy. 14 B Look for minimum cycle (pattern) that repeats. Distance between two successive tall peaks = 5 cm ( )( ) ( ) 1 3 Period 5 cm 2.00 ms cm 10 ms 11 100 Hz 10 10 T f T − − == = = =  15 D Each emergent light beam has the amplitude ' cos45 oAA= As there are two incident beams incident on the polarizer in phase, due to superposition the total emergent light beam has amplitude = 2'A = 2 cos45 2oAA = Since 2AI 2 2 2E A A == I I IE = 2I

©YIJC 9749/01/YIJC/24 4 16 C ( ) 5 60 cm 48 cm 0.48 m4 330 0.48 688 Hz vf f f   =  = = = = = 17 C ( ) 51 42 12 2 sin sin 5 sin 4 54 5 480 600 nm4 dn d d      = = = = == 18 A For the net force to be acting to the left, the charge Q needs to exert a force on X to counter the downward force on X by Y. Thus Q is negative. The vertical component of force on X by Q = force on X by Y 22 cos(45 ) ( 2 ) 2.8 oQq qqkk rr Qq = = Thus Q = −2.8q 19 D ( ) ( ) 21 12 Work done on the charge by external agent change in potential energy final EPE initial EPE qV qV q V V = =− =− − − =− 20 B Wire P: 22 4 ( / 2) LR A dd    = = = ll 2 4d R  = l Wire Q: ( )1 3 22 (21 2 4 43 ( / 2)QQ R dd   == l) l 22 8 4 8 33 Qdd R  == l dQ = 1.6d 21 D E = 3r + 3(1) E = 2r + 2(2) solving, E = 6.0 V

©YIJC 9749/01/YIJC/24 5 22 B Since the galvanometer reads zero when PS = 40.0 cm p.d. across the wire PS = 5.00 mV. A length of 40.0 cm of wire PS has a resistance of 40.0 (5.00) 100.0 = 2.00 . Using potential divider rule, 32.0 5.0 (2.000) 5.00 10 795 R R −= +  = 23 C 7 0 7 0 6 1 4 10 15 0.20 4 10 0.90 2 2 .6 3.4 10 S C resultant S C Bn NB r B B B    − − − = =       == = − = I I 24 B For parallel wires with current flowing in the opposite directions, the wires will repel each other. If the currents are flowing in the same direction, the wires will attract each other. Since side PS is nearer than side QR, force of repulsion is greater than attraction and thus the loop will move to the right. 25 C Using Fleming’s Left Hand Rule, electrons in the conductor are moving to the right and will experience a force upwards in a magnetic field pointing out of the plane of the page. Hence, S will be at a higher potential as electrons will gather at R. ( )( )( ) 32 3 2.8 10 15 10 sin60 10 3.6 10 V 3.6 mV Bv −− − = =     = = l (Each quantity in the equation should be components that are perpendicular to each other.) 26 A ( )( ) ( ) max max max max max 150 7.5 A20 7.5I 5.3 A 22 11average Power 7.5 15022 560 W 2sf rms VI R I IV = = = = = = == = 27 C Option A: electrons are ejected only if the wavelength is smaller than some maximum value. Option B: de Broglie wavelength of the ejected electrons is not the same as the wavelength of the incident light. Option D: The maximum energy depends on the type of metal.

©YIJC 9749/01/YIJC/24 6 28 B Option A is wrong. Most of the energy of the electrons goes towards heating the metal target Option C is wrong. “ hc eV ” is the minimum wavelength Option D is wrong. The intensity of the X-ray spectrum increases. Option B is correct. The wavelengths of the characteristic X-ray spectral lines depend on the material used to make the target (and not the potential difference applied to the electrons). 29 C As the nucleus is very small compared to the atom, there are a lot of space, hence the alpha particles can pass through. When the -particles come very close to the gold nucleus, the experience a large repulsive force and hence can deviate at large angle 30 C A beta decay will result in no change of nucleon number and +1 in proton number. A alpha decay will result in −4 in nucleon number and +2 in proton number. Hence the new particle will have nucleon number of 234 and proton number of 91. Hence its neutron number will be 143.