2025 RI H2 Physics Prelims P3 Section B Answers

13 © Raffles Institution [Turn over 8 (a) the force per unit length per unit current exerted on a long straight conductor placed perpendicular to the magnetic field. (b) (i) M1 − all forces shown A1 − all arrows labelled (ii) 6.0 5.0 1.2 A VR= = = I I I (iii) 3 0 sin cos 9.81 sin30 0.021 1.2 1.5 cos30 6.67 10 kg netF mg B L m m  − = =   =     = I B1 − component of weight along slope B1 − component of magnetic force along slope (iv) When the cross-sectional area is doubled, the resistance will be halved, and the current will be doubled. Hence, the magnetic force will be doubled. However, the volume will also be doubled, and the mass will be doubled. There is no net force acting on rod XY and the rod remains at rest. (c) (i) As the rod move down the slope, there is magnetic flux cutting, resulting in an induced e.m.f. across the rod. (ii) 2 2 1 1 2 19.81 0.020 sin30 2 0.44294 m s mgh mv m mv v − =    = = cos (0.021)(1.5)(0.44294)cos30 0.0121 V BLv= = = (iii) 1. As the rod moves down the slope, the area and magnetic flux linkage decreases. By Lenz’s law, the direction of the induced current will be from P to Q to increase the magnetic flux linkage. 2. When the switch is closed, there will an induced current in the rod. By conservation of energy, the loss in gravitational potential energy of the rod is equal to the gain in kinetic energy of the rod and heat generated by resistive heating. Hence, there is a smaller gain in kinetic energy and a lower speed, resulting in smaller e.m.f. and the answer in (c)(ii) to be smaller. weight normal contact force magnetic force

14 © Raffles Institution 9 (a) (i) Diffraction of light when passes through a single slit / Double-slit interference / Diffraction of light through a diffraction grating. (ii) Photoelectric effect / Compton effect (not in syllabus) (b) (i) The energies of the hydrogen atoms are quantized into discrete levels. When an atom transits from a higher energy level to a lower energy level, photons of energy equal to the difference in the two energy levels are emitted. Energy of photon is given by hcE = . Hence, only photons of specific wavelengths are emitted. (ii) 34 8 7 19 Energy of photon of blue light 6.63 10 3.00 10 4.86 10 4.093 10 J 2.6 eV hc  − − − =   =  = = Since the energy of the photon is higher than the work function energy photoemission will be observed. (iii) 1. Power of red light = Intensity  Area = 6.80  103  3.00  10−4 = 2.04 W 2. 7 34 8 18 PowerNumber of photons per second Energy per photon 2.04 2.04 4.86 10 6.63 10 3.0 10 4.98 10 hc  − − = = =    = 34 27 1 7 6.63 10Momentum of each photon 1.364 10 kg m s4.86 10 h  − −− − = = =   27 18 9 Total force 1.364 10 4.98 10 6.80 10 N − − =    = (c) (i) 1. 2 7 72 1 1 1 4 1 1 1 1.097 10 46.56 10 3 n R n n n  − =−  =  −   = 2. 2 7 72 1 1 1 4 1 1 1 1.097 10 44.86 10 4 n R n n n  − =−  =  −   =

15 © Raffles Institution [Turn over (ii) For shortest wavelength, n= . min 7 min 1 1 1 4 3.65 10 m R  − =−   = (iii) 2 2 22 2 22 2 22 222 11 2 2 2 & 2 n n n n n n hcEE E E hcR n hcR hcREE n hcR hcREE n hcR hcREE n −= − = −  − = − − = − − −   − = − ( ) ( ) ( ) 18 34 8 7 22 1 2.18 106.63 10 3.0 10 1.097 10nE nn − − =−      =− Hence, the energy values are: n = 2: −5.45  10−19 J n = 3: −2.42  10−19 J n = 4: −1.36  10−19 J Alternative method From E to E2, 34 8 19 7 6.63 10 3.0 10 5.455 10 J 3.646 10 hcE  − − −    =− =− =−   From E3 to E2, 34 8 19 7 6.63 10 3.0 10 3.032 10 J 6.56 10 E − − −    =− =−   From E4 to E2, E = −4.093  10−19 J n = 2 (−5.45  10−19 J) n = 3 (−2.42  10−19 J) n = 4 (−1.36  10−19 J) n =  (0 J)