Prelim H2P2 MS

1 Anderson Serangoon Junior College 2021 H2 Physics Prelim Mark Scheme Paper 2 (80 marks) 1a similarity: lines are radial / greater separation of lines with increased distance from the sphere / lines are perpendicular to the surface of sphere difference: gravitational field lines directed towards sphere and electric field lines directed away from sphere B1 B1 1bi The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. B1 B1 1bii (gravitational) potential is taken to be zero at infinity (gravitational) force is attractive work done by the external agent on the point mass moving it from infinity is negative B1 B1 B1 1ci M1 A1 1cii  magnitude of force decreases non-linearly with distance  decreasing gradient, never infinite nor zero B1 B1 1ciii It represents the increase in potential energy of the satellite. B1 2a total volume of molecules is negligible compared with volume occupied by the gas B1 2b pV = NkT 4.60 × 105 × 2.40 × 10–2 = N × 1.38 × 10–23 × (273 + 23) C1 9749/02/ASRJC/2021PRELIM [Turn Over force distance planet’s surface final position of satellite Penalise [1] if curve is drawn at distances less than planet’s surface. No penalty if curve is drawn beyond final position.

2 N = 2.7 × 1024 volume of one molecule = (4 / 3)r3 (= 1.41 × 10–29 m3) volume of all molecules = 2.7 × 1024 × 1.41 × 10–29 = 4 × 10 –5 m3 (1 s.f.; as this is an estimation question the uncertainty of final answer cannot be more precise than that of the given data.) or volume of one molecule = d3 (= 2.7 × 10–29 m3) volume of all molecules = 2.7 × 10–29 × 2.7 × 1024 = 7 × 10 –5 m3 (1 s.f.; as this is an estimation question the uncertainty of final answer cannot be more precise than that of the given data.) C1 A1 2c Since volume of all atoms (4 × 10 –5 m3 or 7 × 10 –5 m3) is 3 orders of magnitude less than volume occupied by the gas (2.4 × 10–2 m3), so assumption in (a) is justified. B1 2d work done on gas (P → Q) : 0 increase in internal energy (P → Q) : (+)97.0 J increase in internal energy (Q → R) : –42.5 J work done on gas (R→P) W = p ∆V = 2.10 × 105 × (1125 – 950) × 10–6 = 36.8 J increase in internal energy (R→P) since total change in internal energy is zero, 97 + (–42.5) + ∆U = 0 → ∆U = –54.5 J thermal energy supplied (R→P) Q = ∆U – W = −54.5 – 36.8 = –91.3 J A1 A1 A1 A1 A1 3a amplitude = 0.020 m f = 1/T = 1/0.60 = 1.7 Hz a = (–)2x and [ = 2f or  = 2/ T] = (2/0.60)2  2.0  10–2 = 2.2 m s–2 [accept –ve a] E C1 C1 A1 3b Resultant between upthrust and weight Upthrust increases with depth of immersion A B1 B1 3c wave starting with a peak at (0,6), and same period peak height decreasing successively exponentially wrt time. A B1 B1 B1 4a Polarisation is where the oscillations in a wave are confined to one direction only in a E B1 9749/02/ASRJC/2021PRELIM

3 plane normal to the direction of transfer of energy of the wave. OR Oscillations occur only in one plane, parallel to the direction of energy transfer. 4b (A polarised wave vibrates in a single plane in space.) Since longitudinal waves vibrate along their axis of propagation, it is not possible to polarise a longitudinal wave. Hence, only transverse waves can be polarised as it vibrates perpendicular to its axis of propagation. A B1 B1 4c Angle between axis of polarisation between images is 90°. Angle between axis of polarisation between left and right lenses is 90°. Angle of polarisation of each image needs to match angle of polarising axis of the lens for the appropriate eye. D B1 B1 B1 4d (Since the plane of the images are perpendicular to each other), the two images are never added together in either eye. D B1 5ai The incident wave reflects at the top of the tube. The incident and reflected wave interfere / superpose to form the stationary wave. A B1 B1 5aii Maximum value at h = 0, 0.30 m, 0.60 m, and 0 amplitude at h = 0.15 m, 0.45m. The line should be a modulus of a sinusoidal function. D B1 B1 5aiii1 vertical/along length of tube/along axis of tube E B1 5aiii2 phase difference = 180° E B1 5aiv When next stationary wave is formed: 0.60 m = 1.5  v = f , so f = v/ = 340 / [0.60/1.5] = 850 Hz A C1 A1 5bi The waves spread as they pass through the slits. A B1 5bii d sin θ = n sin θ = n/d D 9749/02/ASRJC/2021PRELIM [Turn Over

4 since sin θ ≤ 1, n/d ≤ 1 when n=3 and  = 540 ×  10 –9 m 3(540 ×10–9)/d ≤ 1 d ≥ 1.62 ×10–6 m when n=3 and  = 630 ×  10 –9 m 3(630 ×10–9)/d ≤ 1 d ≥ 1.89 ×10–6 m to satisfy both conditions above, minimum d is 1.89 ×10–6 m C1 A1 5biii wavelength of blue light is shorter (than both given lights) so the third order diffraction maximum is produced. A B1 6ai arrow from –0.85 eV level to –1.50 eV level E B1 6aii ΔE = (1.50 – 0.85) × 1.60 × 10–19 = 1.04 × 10–19 J Since ΔE = hc /  = (6.63 × 10–34 × 3.00 × 108)/(1.04 × 10–19) = 1.91 × 10–6 m A C1 A1 6b Spectrum appears as dark background crossed by two bright lines. Electrons in gas de-excite, emitting photons with specific energies equal to the energy difference of two levels. These photons have specific frequencies (or wavelengths) which correspond to the lines. D B1 B1 B1 6ci e.m. radiation produced whenever charged particle is accelerated electrons hitting target have distribution of accelerations A M1 A1 6cii all electron energy given up in one collision/converted to a single photon (since λmin = hc/ Emax,) maximum photon energy corresponds to minimum wavelength A B1 B1 7ai para 2 To prevent microwave/(em) radiation from leaking/escaping out /exiting of the cage/microwave oven. E A1 7aii para 1, 2 & 5 From the passage, by international convention, microwave ovens operate at frequencies at around 2.45 GHz. Wavelength of the microwaves = 3.00  10 8 / 2.45  10 9 Hz = 0.122 m 100 times smaller than 0.122 m or 12.2 cm = 1.22  10–3 m or 0.00122 m. Hence, estimated spacing of holes is 1  10 –3 m or 1.2  10 –3 m . A C1 A1 9749/02/ASRJC/2021PRELIM

5 [-1] Powers of Ten error for conversion of GHz 7bi para 3 Potential difference across the electrodes = 5000.00 V By conservation of energy, Kinetic energy gained by an electron = electrical potential energy loss by the electron = (5000.00)(1.60  10-19) = 8.00  10–16 J = 8.0  10–16 J E M1 A0 7bii para 3 Power output, P = (energy of an electron)  (n/t) Each electron has available max. 8.0 x 10–16 J of energy (assuming that the electrons start off from cathode with negligible kinetic energy) to be converted to microwaves. Least number of electrons per second A A1 7biii para 3 Not all the (kinetic) energy of the electrons is converted into microwave energy as:  electrons give off e.m. radiation of varying wavelengths as it accelerates towards the anode and hence actual energy possessed by electrons are lower when reaching the anode  some electrons hit the anode, some of its kinetic energy is also converted to thermal energy / passed to the molecules (or atoms) in the anode causing thermal agitation so less energy is available for conversion to microwave energy Not all the microwaves generated from the energy is fed into the cavity resulting in energy losses due to:  the walls in the cavity of the food chamber absorbing some of the microwaves  microwaves may be fed back / coupled back to the magnetron (resulting in actual useful power of microwaves less than the actual energy that can be supplied by the electrons) Answer must be related back to the effi