2025 RI H2 Physics Prelims P2 Answers

2025 Preliminary Examination H2 Physics Paper 2 Solutions 1 (a) ( )( )( ) 2 2 2 3 0.234 5.13 10 11.38 10 1.72 10 2330 kg m m abc − − − − = =    = (b) ( ) 3 0.002 0.01 0.01 0.01 23300.234 5.13 11.38 1.72 40 kg m m abc m a b c m a b c     − =     = + + +  = + + + = ( ) 32330 40 kg m −= (c) zero error / incorrect calibration of calipers / balance 2 (a) Let n be number of 5.0 g masses. By principle of moments, take moments about the edge of the table, Total anti-clockwise moments = total clockwise moments ( )( ) ( )( )( ) ( )( )( ) − = − +  = 30.101.5 9.81 0.11 9.81 0.50 0.10 5.0 10 9.81 0.802 7.75 n n Therefore, the maximum number of masses is 7. (b) Let tension in string be T. Consider forces acting on the metre rule and taking moments about the edge of the table, string holding the 1.5 kg mass will cause a reduction of anti-clockwise moments 0.10 0.0502TT==  string attached to ruler at midpoint will cause a reduction of clockwise moments ( )( )sin60 0.40 0.346TT=  = Since the reduction of clockwise moments is greater, t here will be a net anticlockwise moment acting on the metre rule. Hence more 5.0 g masses can be placed (compared to (a)) before the cantilever topples. 3 (a) The vertical displacement of the yoke is r sin(t). Hence, the acceleration is ( ) 22 sinr t x  − =− , which is the defining equation of simple harmonic motion.

2 © Raffles Institution (b) (i) 1. ( ) 00 1 2 0.0800.40 1.3 m s vy r    − = = = = 2. ( ) 2 00 2 2 2 0.0800.40 20 m s ay   − = = = or ( ) 00 2 2 1.30.40 20 m s av   − = = = (ii) 4 (a) (i) When the path difference between the waves that meet at the screen is integer multiple of wavelength of the light, they meet in phase and there is constructive interference, forming bright fringe. When the path difference between the waves that meet at the screen is odd integer multiple of half wavelength of the light, they meet in antiphase and there is destructive interference, forming dark fringe. correct path difference meeting in phase / antiphase constructive / destructive interference (ii) The single slit diffraction pattern from each slit forms the envelope of the double slit interference pattern. Where the minima of the single slit diffraction pattern coincide with the bright fringes of the double slit interference pattern, these bright fringes cannot be observed and will be seen missing from the interference pattern. (b) (i) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9  3.00 m 17.7 mm t / s F / N 5.9 −5.9 0 | | | | | | | | | |

3 © Raffles Institution [Turn over ( ) 9 3 4 sin sin sin tan as is smalltan 590 10 35.4 10 1 2 3.00 1.00 10 m 0.100 mm b b         − − − = = =   =     = = (ii) According to the Rayleigh criterion, the two diffraction patterns are just distinguishable if the central maximum of one diffraction pattern coincides with the first minimum of the other (diffraction pattern). This means the minimum angular separation min of the two central maxima for the patterns to be resolved is 9 min 3 590 10 0.0059 rad0.100 10b  − −  = =  . Since the angle between the two beams of light is smaller than 0.0059 rad, the two diffraction patterns are unresolved. 5 (a) gas that obeys pV  T for all values of p, V and T where p is pressure, V is volume and T is thermodynamic temperature (b) ( )( ) ( ) 61.6 10 0.20 8.31 22 273.15 130 mol pVn RT=  = + = (c) ( ) ( )( ) − − = = +=  = 2 r.m.s. r.m.s. 2 1 13 where is the molar mass22 3 3 8.31 22 273.15 4.2 10 420 m s Mc RT M RTc M (d) When gas reaches equilibrium with surroundings, its pressure is 3.6  104 Pa. ( )( ) ( ) 43.6 10 0.20 3.8827 mol8.31 50 273.15 pVn RT = = = −+ ( ) 23.8827 4.2 10 0.16 kgm −=  = 6 (a) The amount of energy transformed from chemical to electrical per unit charge (driven around a complete circuit).

4 © Raffles Institution (b) ( ) ( ) PR 22 7 8 8 ' ' 90 10 10 105.0 10 805.7 10 5.7 10100 8.9912 9.0 R AA  −− − − − =+   =  +     = = (c) (i) 9.0 8.9912 25 0.26 A V R= = + = I (ii) When galvanometer shows a null reading, VPQ = E. ( ) ( ) PQ 22 7 8 8 65 10 10 100.26477 5.0 10 805.7 10 5.7 10100 1.8 V ER −− − − − =   =  +     = I Alternatively, ( ) ( ) 22 7 PQ 8 8 65 10 10 105.0 10 6.7982 805.7 10 5.7 10100 R −− − − −   =  + =      PQ PR variable resistor 9.0 6.7982 9.08.9912 25 1.8 V RE RR = + = + = (d) (i) As there is no current in the solar cell, potential difference (p.d.) across its internal resistance is zero. Hence, the terminal p.d. of the solar cell is equal to E and VPQ = E. (ii) The resistance of a potentiometer of uniform cross-sectional area is smaller and therefore the potential difference across the potentiometer is now smaller. Hence, the balance length would be longer =  PQ PQ PR PR LVV L . or The potential drop across the first 75 cm of the potentiometer wire is now lower as the resistance across it is lower. Hence, balance length would be longer. 7 (a) (i) Nuclei that have the same number of protons but different number of neutrons.

5 © Raffles Institution [Turn over (ii) Time taken for the number of undecayed nuclei to be reduced to half its original number. or Time for activity to halve. (b) (i) 40 40 0 19 20 1K Ca antineutrino (or neutrino)− −→ +  + one mark each for each correct decay product including mass and atomic numbers (ii) ( ) ( )( )( ) 2 227 8 13 39.963998 39.962591 1.66 10 3.00 10 2.102058 10 J 1.31 MeV E m c − − = = −   = = (c) The ratio of potassium to argon to calcium is 2:1:9.    −=  =−  =  =   + + =  = 0 0 0 01/2 9 9 ln 1 ln lnln2 1.25 10 2 1 9lnln2 2 3.23 10 years tN N e N tN Nt N Nt N 8 (a) (i) Time taken to reach the top 900 2.5t = Let n be the number of passengers. ( )( )( )( ) rate of gain of gravitational potential energy of passengers 2 24 75 9.81 300 900 2.5 29430 29000 W P nmgh t = = =   == Alternatively, the vertical speed of the ( )sinv  , can be used instead ( ) ( ) ( )( )( ) sin sin 3002 24 75 9.81 2.5 900 29430 29000 W P Fv nmg v nmgv = == =   ==

6 © Raffles Institution (ii) Any one point from: It does not account for: 1. energy losses due to friction in the moving parts of the chair lift 2. drag forces acting on the moving chairs, which vary with wind conditions. 3. additional ski equipment which would result in the average mass being more than 75 kg (b) (i) ( )( )cos 75 9.81 cos9.0 726.69 N 730 NN mg = = = = Alternatively, from the graph 1.540ga = (acceptable range 1.525 to 1.550ga = ), ( )( )75 1.540 729.24 730 Ntan9 tan9 gmaN = = = = or ( ) ( ) ( )( ) ( )( ) 2 222 75 9.81 75 1.540 726.63 730 NgN mg ma    = − = − = =     (ii) ( )0.080 726.69 58.135 58 N fN == == (iii) ( ) 2 sin 75 9.81 sin9.0 58.135 75 0.76 m s mg f ma a a  − −= − = = Alternatively, From Fig. 8.4, when  = 9.0 and  = 0, ag = 1.54 m s−2. ( ) 2 75 1.54 58.135 75 0.76 m s gma f ma a a − −= −= = (c) (i) ( )( )( )cos 0.12 75 9.81 cos22.0 81.861 82 Nf N mg  = = = = = (ii) In order to maintain constant speed, the acceleration of the skier is zero. 2 81.861 75 1.0915 1.1 m s g g f ma fa m − = == == From Fig. 8.4, the corresponding ski angle is 72.5. (iii) The skier can decrease the ski angle. (d) (i) From Fig. 8.7, for hard snow, when  = 61, r = 9.2