RI 2024 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9749 H2 Physics 2024 GCE A-Level Suggested Solutions Paper 1 1 A ( ) ( ) π − − = ×× = × = ×× ≈× 22 6 17 2 217 9 2 12 Approximate area 4 10 10 5.0265 10 m 5.0265 10 10 Gm 5 10 Gm 2 B ≤≤ 10 tt Acceleration has a positive and constant value. Object’s velocity increases linearly in the positive direction. Gradient of v – t graph is positive. ≤≤12t tt At t = t1, acceleration immediately switches to a negative value and remains constant until t = t2. Object’s velocity decreases linearly in the positive direction. Gradient of v – t graph is negative. 3 C Taking upward as positive, ( ) ( ) = + = °+ − = 2 2 1 2 10 15sin40 1.62 2 11.9 s yy ys ut at tt t 4 B As the cylinder is in equilibrium and acted upon by three forces, the vector triangle formed by the three forces forms a close loop in one direction. Options (A) and (C) are out. 5 D = = 3 ...... (1)3 W kx Wk x = = = ∴= ' ' [ substitute eqn(1) here] 3 '3 W kx WWx Wk x xx 6 B When specimen is stretched under the tensile load from zero to a maximum value, the work done by the tensile force on the specimen is area under OPQT. When relaxed, the work done by the specimen is area under SRQT. Hence the net work done on the specimen = area OPQT − area SRQT = area OPRS.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 7 C Since the car is moving up the hill at a constant speed, Forward driving force of the engine uphill = F + component of weight downhill = + × × °=engine 600 900 9.81 sin10 2133.14 NF = ×= ≈ engine 2133.14 16 34130 W 34 kW P 8 D Work done on the gas is when the gas is compressed by an external force and its volume decreases i.e. RS and SP. 9 A dAngular velocity d r tt rt θω = = = l l 10 A ω π − =  × =  × = 2 27 2 1.3 10 2 2 24 3600 0.0344 m s towards the centre of the Ear th. car 11 A As GM rφ =− where M is the mass of the object and r is the distance between the object and points P or Q, the magnitudes of their individual gravitational potentials decrease as r increases. Q P PQ GM GM GMd r r rrφ ∆= − −− =   where d =rQ − rP. As P and Q move away from the object while maintaining the same separation, d is unchanged but rp rQ increases, hence ∆φ decreases. 12 A The change in momentum of the particle after collision with the wall (shaded) ( ) 2p mv mv mv∆= −− = , taking leftward as the positive direction. The average force that the particle exerts on the shaded wall = 22 2 p mv mv xtx v ∆ = =∆ . 13 D Heat loss by volume V of water = Heat gained by V/4 of water (60 ) ( 20) 4 52 C VVc cρ θ ρθ θ −= − = ° 14 C At t = 0, x = −x0, EK = 0 J. 0 0 22 2 K0 cos sin 1 sin2 xx t vx t E mx t ω ωω ωω =− = = 15 C In critical damping, the mass will reach equilibrium in the shortest time without oscillating. 16 C 2 1 P IA I r = ∝

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 2 mars earth earth mars 211 mars 11 1.5 10 1300 552.931.02.3 10 Ir Ir I =   × = ×= ×  Area of solar panel to receive the same power if satellite orbits around Mars 2 1300 552.93 2.35 2.4 m A A = × = ≈ 17 D 0 0 00 cos cos15 0.966 0.97 AA AA A A φ= = °= ≈ 18 D 19 C 2 d 4 d DI nAv q n v q π= = For the same material, number density n is the same. For equal current in both wires, drift velocity is higher for the wire with the smaller diameter and vice versa. 20 B Since LR A ρ= , resistance R is minimum if L is minimum and A is maximum. Hence min .xR yz ρ= 21 A E.m.f. of the cell connected across PQ, E = P.D. across the 2.0 Ω resistor, 2.0 0.60 0.40 V3.0E = ×= Polarity connected to P is negative. 22 A As temperature of the thermistor rises, its resistance decreases. The total resistance of the circuit decreases. Current A1 increases. The p.d. across the thermistor decreases, hence current A2 decreases. 23 C Option A: electric force on electron acts perpendicularly out of plane of paper may cancel out the magnetic force acting perpendicularly into the page. It is possible that the electron moves horizontally straight from left to right. Option B: electric force on electron acts in the opposite direction to the electric field strength. Its vertical component acts vertically downward which may be cancelled by vertically upward magnetic force on it. The horizontal component of electric force will cause the electron to accelerate horizontally from left to right. Option D: electric force on electron accelerates it to the right whereas there is no magnetic force on it. It is possible that the electron moves horizontally straight from left to right. *Note: Option B is also not possible. The magnetic force is greater than the electric force in the vertical direction since the velocity of the ion increases. Net vertical force is not zero. 24 B 22rrTv vT ππ= ⇒= . Magnetic force on the particle provides for its centripetal force. 2 2 2 4 2 . Bqv mr mr T mT Bq πω π = =   ∴= 25 D Induced e.m.f. across the wing of the aircraft E = Blv = 1.0 × 10−5 × 80 × 150 = 0.12 V

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 0.12 0.12 60 7.2 Wb t φ φ ∆ =∆ ∆= × = 26 B Magnitude of a.c. voltage changes by 2V0 from T/4 to 3T/4 for a sinusoidal voltage supply defined by V = V0sinωt. Fraction of a cycle = 3T/4 − T/4 = T/2. 27 A Total resistance of the overhead cables 2 0.030 100 6.0 R = × ×=Ω 00 6100 10 400000 250 A P IV I I = ×= × = Power loss Ploss = 2502 × 6 = 375 000 W Power input to substation Pi = 100 × 106 − 375 000 = 99.625 ≈ 99.63 MW 28 D From De Broglie’s relationship hp λ= . Photon energy hc hcE pchpλ= = = where c is the speed of light in vacuum. 29 B From Heisenberg’s uncertainty principle xp h∆∆ ≈ ⇒ ( )xmv h∆ ∆≈ . There is a trade-off in the uncertainties of position (or displacement) and momentum (or velocity). Options A, C and D are incorrect. 30 D Mass of an Fe-56 nucleus = 55.93493 28 0.00055 55.92063u uu−× = ( ) 28 Mass of Fe-56 nucleus 26 1.00728 30 1.00867 55.92063 0.52875 8.78 10 kg u u uu − = × +× − = = ×

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9749 H2 Physics 2024 GCE A-Level Suggested Solutions Paper 2 1 (a) Gain in 27 ( 0) 27 27 (0.43)(9.81) 6.4 m pE mg h h = −= = = [1] [1] (b) At maximum height: 0yv = 2 1 02 2 2(9.81)(6.5) 11.21 11 m s (shown) y y u gh u gh − = − = = = = [1] (c) (i) 1 cos32sin32 11.2 tan32 17.9 m s y x uu − = °° = ° = By conservation of momentum 1 12 2 12 ()mu mu m m v+= + Since 2 0u = 11 22 12 1 (0.43)(17.9) 0.43 0.67 7.0 m s mu muv mm − += + = + = [1] [1] [1] [1] (ii) As the horizontal velocity of the original object as it rises is greater than the horizontal velocity of the combined object as it falls , it will travel a greater horizontal distance given by the product of its horizontal velocity and time. Since both objects travel the same vertical height, the time taken for the original object to rise is equal to the time taken for the combined object to fall. [1] [1] 2 (a) The molecules exert no intermolecular forces on one another except during collisions. [1] (b) The internal energy of ideal gas consists only of the microscopic kinetic energy due to random motion of its molecules since the microscopic potential energy of ideal gas is zero. [1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 (c) (i) 23 23 3Total 2 3 (2.50 10 )(1.38 10 )(21 273.15)2 1520 J kE NkT − = = × ×+ = [1] [1] (ii) 54(1.0 10 )(6.3 10 ) 63 J W pV − = −∆ = −× × =− By the First Law of Thermodynamics, UQW∆=+ 23 23 3 2 3(2.50 10 )(1.38 10 )(15) ( 63)2 141 J Q UW Nk T W − = ∆− = ∆− = × × −− = Note: The values of ∆T and ∆V given in the question are inconsistent with the ideal gas equati