TJC 2023 H2 Prelim Paper 3 Solutions

2023 JC2 H2 Prelim Paper 3 Solutions 1 (a) The electric field lines, along which the electric forces exerted by the charged sphere on other charges act, are always perpendicular to the surface of the sphere (i.e. the electric field lines are radial). Hence the field lines appear to originate from the centre of the sphere. M1 A0 (b) (i) arrow pointing to the left labelled ‘electric force’ (FE) and arrow pointing downwards labelled ‘weight’(W) C1 (ii) 1. Electric force on the particle FE = QE FE = VQ / d = (2.0 × 102 × 8.0 × 10-19) / 4.0 × 10-2 = 4.0 × 10-15 N resultant force ( ) ( ) 2215 15 153.9 10 4.0 10 5.6 10 N− − −=  +  =  A1 C1 A1 2. let  be the angle to horizontal tan = weight /electric force  = tan-1 (3.9 × 10-15 / 4.0 × 10-15) = 44° A1 (c) (i) downward sloping line from (0, 2.0) M1 magnitude of gradient of line increases with time and line ends at (T, 0) A1 (ii) Horizontally: s = ½ at2 2.0 x 10-2 = ½ (4.0 x 10-15 /(3.9 x 10-15 / 9.81) T2 T = 0.063 s C1 A1 2 (a) (i) gravitational field strength is equal to the negative gravitational potential gradient Or state equation g = - d/dr B1 (b) (i) G = GM/r2 gA = G(3M)/(2R)2 = ¾ GM/R2 gB = G(M)/(R)2 = GM/R2 so conclude gA< gB alternative soln: by calculating the gradient at surface and using gradient = field strength deduce gradient at surface of A < gradient at surface of B C1 A1 (ii) M is the point/position in which net gravitational field strength is zero B1 (iii) sketch: one curve, starting with gradient of decreasing magnitude at 2R and finishing with gradient of increasing magnitude at D – R field strength shown as zero near the point of maximum potential (position of M correct), with gravitation field strength at surface of A< B B1 B1 Substitution must be shown

(c) (i) The gravitational force of attraction between the stars are of the same magnitude. This force provides the centripetal force for each star to orbit about their common centre of mass. B1 (ii) There is no external force acting on the binary stars, so the c.m. should not change position. B1 (iii) centripetal force = MAdω2 = MB(2.8 × 108 –d)ω2 Or MAdA = MBdB MA / MB = 3.0 = (2.8 × 108 – d) / d d = 7.0 × 107 km C1 A1 3 (a) (i) Molecules collide with one another in a haphazard manner hence they possess different kinetic energies at any time. (word “random” does not get marks in answer) B1 (ii) No intermolecular forces in ideal gas, hence no potential energy (Do not accept specific forces eg. “attraction”; vague description eg. “interaction” B1 (iii) p = 1/3 ρ <c2> => pV = 1/3 M <c2> since ρ = M/V For ideal gas, pV = n RT Hence, 1/3 M <c2> = n RT Total KE = ½ M <c2> = 3/2 n RT (Must start with “Use the equation” given, other methods not accepted) M1 B1 (iv) Low pressure High temperature (Low pressure implies molecules are far apart, high temperature implies high KE) B2 (b) (i) pV = nRT (6.0 x 105) (0.10) = (10) (8.31) T T = 722 K M1 A1 (ii) The products of p and V are not constant eg. the product of p and V is 5.0 x 104 J at B, but 6.0 x 104 J at A. (or calculate temperature at B = 600 K, lower than 722 K at A) B1 (iii) Work done in one cycle estimated from the area ABC enclosed = 8.8 × 104 J (Note positive sign since net work is done by the gas Do not accept area “under curve”) M1 A1

(iv) In one cycle 0=+= WqU => q = 8.8 × 104 J (working or otherwise clear explanation required) A1 4 (a) (i) 1 cycle ( ) 3 1.90 ms 11 526.3 526 Hz 1.90 10 T f T − = = = = =  (Students are allowed to use one or more cycles.) C1 A1 (ii) ( ) 2 4 25 42 1.90 0.76 ms5 2 5 t T t T Tt     = =     = = =       1.50 0.76 2.26 ms+= C1 A1 (iii) ( ) 22 2 11 2 1 1 22 1 1 1 4 for 0.25 0.25 120 60 cm PPII rr IP r IP r II Prr P Prr P =   = = = = = = C1 A1 (b) (i) All the particles in a progressive wave oscillate with the same amplitude. The particles in a stationary wave oscillate with amplitudes that range from zero at the nodes to a maximum at the antinodes. B1 (ii) All the particles within a wavelength of a progressive wave have different phases. All the particles between two adjacent nodes of a stationary wave have the same phase. Particles in adjacent segments have a phase difference of  radians. B1

5 (a) (i) waves at (each) slit/aperture spread hence wave(s) able to superpose/meet/interfere in overlap region B1 B1 (ii) Unable to achieve constant phase difference/coherence for two separate light source(s) or to ensure waves/light from the double slit are coherent/have constant phase difference B1 (iii) Greater intensity of bright fringes Narrower bright fringes B1 B1 (b) x = λD / a λ = (36 × 10–3 × 0.48 × 10–3) / (16 × 2.4) = 4.5 × 10–7 m C1 C1 A1 (c) (i) water is flat/still/no disturbance destructive interference as the path difference is 2.5λ when sources start in phase B1 B1 (c) (ii) 1. surface/water/P vibrates/oscillates with max amplitude waves from the two dippers arrive in phase 2. surface/water/P vibrates/ oscillates as a non -zero minima due to incomplete destructive interference B1 B1

6 (a)(i) Iron core prevent flux losses or improve flux linkage B1 (a)(ii) Primary a.c. current causes changing flux in iron core e.m.f. / current (induced) in core induced/eddy current in core causes (joule)heating B1 B1 (b)(i) diode correctly drawn to give polarity across load. B1 (b)(ii) ss pp NV=NV V0 = √2 × Vrms = √2 × 240 ratio = 9.0 / (√2 × 240) = 1/38 or 0.027 C1 C1 A1 (b)(iii) Vrms = 9.0/2 = 4.5 V Mean power output = Vrms2/R = (4.5)2/4.5 = 4.5 W C1 A1

7 (a) Magnetic flux density - force per unit current per unit length when conductor is placed perpendicular to the magnetic field magnetic flux – product of area and magnetic flux density normal to the area B2 B1 (b) Φ = BA sinθ B1 (c) (i) Φ = 1.8 × 52 × 10–2 × 95 × 10–2 sin 90o = 0.89 Wb C1 A1 (ii) As frame rotates, there is a rate of change of magnetic flux/cutting of flux, By Faraday’s law, (induced) e.m.f. is proportional to rate of change/cutting of (magnetic) flux (linkage) B1 B1 (iii) Sides PS and QR(both correct) B1 (iv)1 greatest rate of change of flux occurs when θ = 0°,180o and 360 or when plane of frame is parallel to field, so induced emf is maximum least rate of change of flux occurs when θ=90° and 270o or when plane of frame is perpendicular to field, so induced emf is zero Magnitude of e.m.f. determined by rate of change of flux B1 B1 B1 (iv)2 Since flux  = BA sin  = BA sint which is sinusoidal Induced emf = d/dt = BA cos t is also sinusoidal B1 B1 (iv