RI 2023 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9749 H2 Physics 2023 GCE A-Level Suggested Solutions Paper 1 1 C The varying reaction time of a person will cause him/her to start and stop the stop-watch either early or late giving rise to random error. 2 D Displacement is the area under the v – t graph ( )1 2= + p qr 3 A ( )= = = = 2 2 211 1 22 2 2 mv pE mv mm p mE Average force, ( ) ∆= ∆ −−= = = ⇒∝ 0 2 pF t pF t Ft p Ft mE Ft E 4 C Since collision is elastic, Relative speed of approach = relative speed of separation ( )1 2 21 12 21 −− = − +=− u u vv uu v v 5 B Let the CG be a distance x below the centre of the upper square. Treating the sheet as horizontal and take moments about the CG. ( )0.5 4 24 1.33 mgx mg x xx x = − = − = Hence, it is 6.0 − 1.33 = 4.7 cm Or: ( ) ( ) 11 2 2 12 0.5 2.0 6.0 1.5 4.67 cm cm mx mxx mm mm m += + += =

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 6 B ( ) ( ) 2 2 2211 10.6 0.6 9 .7 4.0 1.522 2 40.0125 W = ×−= × − = mmP uv tt Option A: 12 1.5 18 W= ×=P Option B: 12 3.3 39.6 W= ×=P Option C: 24 0.60 14.4 W= ×=P Option D: 24 2.8 67.2 W= ×=P 7 C In half a year, Earth will complete half a circle around the Sun. Hence angular displacement is π rad. 8 B Gravitational field strength is gravitational force exerted per unit mass. 9 C For path C, the spacecraft will constantly experience gravitational force due to Earth. Hence it will tend to curve towards the Earth unless the spacecraft fires its rockets to maintain its straight path. 10 A Time duration t for each molecule to hit the same shaded wall again 2p v= ( )( )−−∆= = = = 22 2 N mv mvpF tt Nmv mNv p p v 11 B ( ) 0 273 273 273 273 273 2 ° ° ° ∝ ∴= + = C C C UT U U UU 12 C =P mc t θ For the same P supplied, gradient of temperature-time graph of P (i.e. t θ ) is steeper than that of Q. Hence, c of P will be smaller than that of Q. =Pt mL . Since P took a longer time t, L of P is larger than that of Q. 13 B ( ) 22 222 6 0 23 3 31 1 1 2.9 10 2 2.0 1022 6.02 10 3.0 10 4.2 10 J − − − −  × = × ××  = × mx πω 14 D Phase difference 1 360 1203= × °= ° 2 2 2 3 2.252 ∝ = =X Y AI I I

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 15 B After 1st filter 0=I After 2nd filter 2 0 cos 45 °=I After 3rd filter 22 00cos 45 cos 45 0.25° °==I I 16 D sinb θλ= Using small angle approximation, tan 2 22 b xb D D cDb x fx θλ λ λ ≈  ≈ ≈= 17 A ( ) 9 63 3 sin 2 500 10 2.0 10 m 2.0 10 mm60sin 2 1lines per mm 5002.0 10 − −− − = × = = ×= × °  = = × n d d λθ 18 A increase linearly with since is increased at a constant rate = ∝ ∴ Dx a xD x t D λ 19 D The beam of protons will be attracted to plate Y (negative plate) and move in a parabolic path as the electric force on them is constant (always downwards). 20 B 0 1 4= ⇒∝QVV rrπε P P 22 636 3 600 V ∴= = V V 21 B Definition of e.m.f. 22 D For balance point to be reached, both the positive polarity of the cells must be facing the same side i.e., to the left based on the potentiometer circuit. Either X or Y must be on point P so that it can be moved to find the balance length. Option D is the only possible option. 23 C When intensity is zero, the resistance of LDR is maximum. The combined resistance will be maximum. As intensity increases, the resistance of the LDR decreases. Hence the combined resistance will decrease. 24 A The magnetic field lines can be imagined as produced by a North pole on the left of the field lines. The bar magnet will be attracted to the North pole and move along the horizontal field line.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 25 D The final position of the beam is upwards and rightwards compared to the position with no fields. For the electrons to be deflected upwards, either the magnetic field must be pointing rightwards or the electric field must be pointing downwards. For the electrons to be deflected rightwards, either the magnetic field must be pointing downwards or the electric field must be pointing leftwards. From the options given, magnetic field must be pointing rightwards and electric field leftwards. 26 C Unit of magnetic flux density is tesla. Since magnetic flux is the product of magnetic flux density and area, its unit is tesla metre2. 27 D The graph shows a half -wave rectification wave. It can only be produced by having a sinusoidal source and diodes pointing in the same direction. 28 A ( )( ) ( ) λ λ −− − = ×− −− × × = × = × 8 19 34 8 3.00 1030.6 122.4 1.60 10 6.63 10 1.35 10 m E hf 29 A λ λ = = hp ph 30 C 2 0 0 ×= =rms VTVV T

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 Paper 2 1 (a) The pressure at different depths h in the water is given by ρgh + atmospheric pressure, where ρ is the density of water. Hence the pressure at the lower surface is higher than the pressure at the upper surface. Therefore, the upward force acting on the lower surface is larger than the downward force acting on the upper surface , resulting in a net upward force which is the uprhrust. [1] [1] (b) (i) By Archimedes’ principle, upthrust = weight of water displaced. Also, volume of water displaced = volume of block. ( ) 6 Upthrust 1000 27.8 10 9.81 0.2727 (shown) Vgρ − = =× ×× = ≈ 0.27 N [1] [1] (ii) Applying principle of moments to Fig. 1.1: 8.3 18.0WF×= × (1) Fig. 1.2: ( )7.8 19.0 0.27WF×= ×− (2) (2) ÷ (1): 78 19 0.27 83 18 F F −= 78 87.6 23.7FF F = − = 2.5 N [1] [1] [1] 2 (a) Minimum kinetic energy occurs at the highest point where the velocity is cos 42 0.743vv = . Hence, ( ) 2 1 2 1 0.743 652 65 2 20.7 m s0.55 0.743 mv v − = ×= =× Hence, vertical component is ( ) 120.7sin42 13.8 14 m s shown−= ≈ [1] [1] [1] (b) Let the time of flight be t. Applying v u at= + to the vertical motion, 14 14 9.81 2.85 s t t −= − = Horizontal velocity 120.7 0.743 15.4 m s −= ×= Horizontal displacement 15.4 2.85=×= 44 m [1] [1] [1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 (c) The net force and the average deceleration in the upward motion is larger in magnitude than the net force and average acceleration in the downward motion. Hence, the time taken to reach the maximum height will be shorter. [1] OR: As energy is lost continuously due to air resistance, the average speed in the upward motion is higher than the average speed in the downward motion. Since the distance travelled is the same, the time taken to reach the maximum height will be shorter. 3 (a) (i) Distance of the satellite from the centre of Earth = 36000 + 6400 = 42400 km. ( ) 11 24 22 3 6.67 10 6.0 10 6800 42400 10 GMmF r −× ×× ×= = = × 1510 N [1] [1] (ii) All geostationary satellites should have the same period T. 2 2 2 2 23 2 4 GMm mrr mr T Tr GM ω π π = =  = Hence, T is independent of the mass m of the satellite. [1] (b) (i) Magnetic force provides the centripetal force: 2 23 19 3.4 10 0.0018 1.6 10 mvBev r mvr Be − − = ×= = =×× 0.118 m [1] [1] (ii) From mvr Be= , the radius depends on the momentum of the particle too. Hence, ions with the same charge but different momentum will travel in circular path of different radii. [1] 4 (a) (i) The sound wave produced by the loudspeaker travels to the metal plate and is reflected. The reflected wave travels toward the loudspeaker and superpose with the incident wave. Since the incident wave and the reflected wave have the same amplitude, wavelength and frequency and travel in opposite directions, the resultant disturbance is a stationary wave. [1] [1] (ii) From Fig. 4.2, 3 3 2 10 0.50 10 2.5 10 s T T − − = ×× = ×