2025 TJC H2 Phy Prelim P1 Solution

2025 H2 Physics P1 Solutions 1 A Units of Q is C = A s Units of V is J C-1 = kg m2 s-2 / A s Hence units of C = A2 s4 m-2 kg-1 2 A Average of h A – 6.63 , B – 6.63, C – 6.65, D – 6.62 A, B and D are accurate Range of variation of h A - 0.04, B – 0.02, C – 0.07, D – 0.02 B and D are precise 3 C Δ𝑣 𝑣 × 100% = 1 2 (0.05 4.30 + 20 1450 + 0.05) × 100% = 3.8% ≈ 4% 4 A Initial velocities for both stones, u = 0 s = ut + ½ gt2 = ½ gt2 For the same time interval, Δt, both stones fall down by the same distance, s. Therefore the distance between them will always remain the same as that at the point of release. 5 D 2 6 2(27)(900)(0.4 10 )(3.0) 0.087 NF N Av −= =  = 6 B The weight of the water converted to ice is the same as that of the ice. 7 B Instantaneous power required to drive the escalator friction friction 3 3 sin 3062 20 9.81 0.75 2.5 1060 7.1 10 W 7.1 kW Fv P Mg v P =+ =  + =     +  = =

8 C Drag at 21.0 m s-1: 110 5.24 kN21 Df == Drag at 15.0 m s-1: 2 ' 15.0 ' 2.67 kN21.0 D D D f ff =  = Force of engine at 15.0 m s-1: 110 7.33 kN15 engineF == Hence resultant force = 7.33 – 2.67 = 4.66 kN 9 B Work done = mass x change in potential = (1) (final potential – initial potential) = − 3𝐺𝑀 2𝑅 − (− 𝐺𝑀 𝑅 ) = − 𝐺𝑀 2𝑅 10 C Top: Ftop +mg = mr2 Ftop = 77 [4(2/3.7)2 – 9.81] = 133 N (downwards) Bottom: : Fbottom = mg + mr2 Fbottom = 77 [4(2/3.7)2 + 9.81] = 1640 N (upwards) 11 D Resultant force towards the sun equals to mac. 12 D GPE of orbiting satellite Ep = MmG r− Gravitational force = centripetal force  2 2 Mm vGmrr =  211 22 k MmE mv G r==  2pkEE=− ' ppE E E=+  '22 kkE E E− =− +  ' 2 kk EEE=− 13 A Total number of moles of gas is constant. Since PV = nRT, we take the initial number of moles in the smaller (A) and larger bulb (B) to be n and 8n respectively. Total number of moles = 9n.

At new equilibrium, the pressure will be the same for both. ' ' (80 273.15) (8) (9 ') (9 ') (10 273.15) fA fB PV n RT n R P V n n RT n n R = = + = − = − + Dividing: '(353.15) 8 (9 ')(283.15) 9 ' 8 353.15 ' 283.15 9 8 353.15 1' 283.15 ' 0.8194 0.8194 0.18 Vn V n n nn n n n nn n n n n = − − = =+ =  = − = 14 A 71.0 )4 5sin( )8 52sin( 0 0 0 0 8 5 −= =  = =   x x T Tx x x Tt 15 C ( ) 2 21 22E m f x = ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2110.5 2 3.0 0.40 0.72 222 0.72 m f x m f x E  =  = 16 D 2 1 2 1 ' 150 2.0 1200 ' 0.71 PI rd r IA A r A A mm =   = = 17 C Unpolarised light after passing through a polarizer, its intensity is halved (therefore we eliminate options A/B). Amplitude remains at A. (The component of E field perpendicular to the polariser axis is absorbed (e.g. AY is absorbed), leaving the transmitted light having amplitude AX)

Since the polarization angle is 75° (or 105o), by resolving the electric field, ooA' =A cos75 =A sin15 18 A First minima position: b sin  = 