RI 2020 Y5 H2 Physics TP Soln

© Raffles Institution [Turn over 2020 Year 5 Term 3 Timed Practice H2 Physics Solutions 1 D Specific heat capacity of water is about 4 200 J kg−1 K−1 and it takes about 5 min for the water in a kettle to boil P × t = m C ∆θ (O-level syllabus) P × 300 = 1 × 4200 × (100 − 30) P = 980 = 1000 W 2 C Average of results of method P = 9.76 m s−2 Average of results of method Q = 9.56 m s−2 Range of results of method P = 10.12 − 9.51 = 0.61 m s−2 Range of results of method Q = 9.61 − 9.51 = 0.10 m s−2 3 A ( ) ( ) 2 tot 22 22 tot tot let be the time taken to travel 1 2 10.50 2 1 2 2 time taken for remaining 0.5 2 1 0.41 ts s at s at s at at at tt s tt = = ⇒= = = = −= 4 A gradient of the velocity-time graph is acceleration 5 B Assuming that A moves up the slope − −= → −= → AA BB T mg f ma mg T ma s in 3 0 [1] [2] [1] + [2]: ( ) ( )( ) − −= + − −=  =−= B A ABmg mg f m m a gg f fg sin30 13.0 4.0 7 .0 1.02 7 2.81 N 6 B ( ) ( ) ( ) 1 12 21 12 2 1 1.0 3.5 7.0 1.0 1.0 7.0 1.6 0.70 m s mu m u mv mv − += + += + =− v v 7 C 4TF m a ma= = 11 43xF T m a T ma ma ma−= ⇒= − = 22 xTm a Tm a= ⇒= 1 2 3 1 T T =

2 © Raffles Institution 8 B 22 7313 1.3 10=−=A Taking moments about corner of platform: 731(4.3) (9.0)( ) 10 5.7 N = = F F 9 D Pressure in a water column is dependent only on the depth of liquid + atmospheric pressure. 10 B By COE, loss in gain in KPEE= Since they are launched with the same kinetic energy, both stones will gain the same amount of potential energy when they reach their highest point. 12 11 22 2 2 Gain in Gain in 2 1 2 PPEE m gh m gh mgh mgh hh = = = = 11 D ( )θ= + = × +× =  0sin 240 1500 9.81sin3.2 240 17 18.04 kW = 18 kW P mg v 12 (a) (i) unit of t = s 2 mmmunit of s 2 m m m s m Lwh A gd − ×× ×× = = ×× × B1 B1 (ii) 1. There may be wrong coefficients. 2. There may be missing or additional terms. B1 B1 (b) (i) Micrometer screw gauge / travelling microscope / digital vernier caliper B1 (ii) ( ) 23 6 0.32 10 1.60 4 0.500 0.23 1.12 10 m ρ − − π× × = ×× =×Ω M1 A1 (iii) 2 dVL dV L ∆ ∆∆∆∆= +++ I I ρ ρ 6 0.01 0.02 0.1 0.042 0.32 1.60 50.0 0.231.119 10 − ∆ = +++×  ρ ∆ρ = 2.81 × 10−7 = 3 × 10−7 Ω m M1 A1 (iv) ρ = (1.1 ± 0.3) × 10−6 Ω m B1 platform A 1.7 cm 6.0 cm F 1.3 cm 4.3 cm

3 © Raffles Institution [Turn over (v) Any 1: Use an ammeter/voltmeter with greater precision / resolution / least count Use a wire of thicker diameter / longer length Increase potential difference/current B1 Markers’ Comments (a) (i) Students should present their workings clearly. (ii) Many students have tried to state what is wrong with the derivation of the equation. Only those explanations with the correct physics concepts will be accepted. (b) (iii) The most common mistake made is not