MI 2024 PU3 H2 PHYSICS PRELIM P1 ANS

1 H2 Physics PU3 Preliminary Examination Paper 1 Answers S/N Answer Explanation 1 D [𝐸] = |𝑄|𝑛 |𝑘| kg m2 s-2 = 𝐴𝑛 s𝑛 𝐴2 s4 kg-1 m-2 n = 2 2 C a = b2c c = 2 a b 2c a b c a b   =+ 3 C At point A: object just reached the ground for the first time At point B: object just about the leave the ground after first bounce At point C: object reached the maximum height after first bounce At point D: object just reached the ground for the second time 4 C 21 2=+s ut at u = o since the stone falls from rest. 212 2= at 24 = ta 2=t a ------------ (1) where t is the time taken for stone to fall on Earth Acceleration of free fall of moon = 6 a , where a is the acceleration of free fall on Earth. Sub 6 a into equation (1) ( ) 22 66 6 = = =moontt aa 5 B Area under the force-time graph = change in momentum = impulse Total area = (5 x 0.5) + (15 x 0.5) = 10 p = m v = 10 v =v - 0 ➔ v = 10 / 4 = 2.5 m s–1 2.5 m s–1 + 2.5 m s–1 = 5.0 m s–1

2 6 B A and D obviously wrong. Try taking moments about C, clockwise moment = (3m)(1.5x) = 4.5mx anti-clockwise moment = (3m)(1.5x) + (m)(0.5x) = 4.5mx + 0.5mx = 5 mx Therefore C is also wrong and centre of gravity is slightly to the left side of C. 7 C Since the balloon is in equilibrium, upthrust on balloon = weight of balloon and helium + force by spring 𝜌𝑎𝑖𝑟𝑉𝑏𝑎𝑙𝑙𝑜𝑜𝑛𝑔 = 𝜌ℎ𝑒𝑙𝑖𝑢𝑚𝑉𝑏𝑎𝑙𝑙𝑜𝑜𝑛𝑔 + 𝑚𝑏𝑎𝑙𝑙𝑜𝑜𝑛𝑔 + 𝑘𝑥 (1.29)(5.0)𝑔 = (0.180)(5.0)𝑔 + ( 3.5 1000)𝑔 + (100)𝑥 x = 0.544 m 8 C Force constant, 𝑘 = 5.0 0.1 = 50 N m-1 Initial extension = 𝐹 𝑘 = 3 50 = 0.06 m and final extension = 𝐹 𝑘 = 2.5 50 = 0.05 m Change in E.P.E.= 1 2 𝑘𝑥𝑓2 − 1 2 𝑘𝑥𝑖2 = 1 2 (50)(0.05)2 − 1 2 (50)(0.06)2 = −0.028 J 9 D Frictional force on the object provides the centripetal force mr2. Both objects have the same angular velocity and same mass, but the centripetal force required for Q is larger due to larger radius. When the centripetal force required exceeds the frictional force available, Q starts to slide. 10 D Using COE: 2 2 1 2 2 ------------- (1) if if EE GPE KE mgr mv mgr mv = = = = 2 bottom 2 bottom sub (1) into eqn 2 3 mvT mg r mvT mg r mgr mgr mg −= =+ =+ =

3 11 C 2= GMg r 2 Mg r Let the quantities with subscript ‘J’ represent that of Jupiter while those with subscript ‘E’ represent that of Earth. 2 JJ E 2 E E J =   gM r g M r ( ) 23 J 7 E 6370 10318 7.15 10  =   g g Since gE = 9.81 N kg-1, ( ) ( ) 23 J 7 6370 10318 9.817.15 10  =   g = 24.8 N kg-1 12 B 1 3.05.0 20 3.0 7.0 4.0 J kg BC BC C AB AB C Um U U     −  =     + −= = =  + − + = = − 13 A When two objects A and B are placed in thermal contact, heat flows from the hotter object to the colder object B, until they reached thermal equilibrium. At thermal equilibrium, both objects A and B are at the same temperature. 14 B Average K.E. ( )( ) 23 2133 1.38 10 27 273.15 6.2 10 J22 −−= =  + = kT 15 D () 4 40 160 mmc Ltt L tct    = = = = 16 C Amplitude, xo = 0.36/2 = 0.18 m Vmax = xo = (2/T) xo = (2/0.60)(0.18) = 1.9 m s-1 17 B Damping force opposes motion. 18 C Since I is proportional to A2, AR = A + (2A) = 3A Hence, 9 I

4 19 D At the fundamental mode, 1 m2  = ,  = 2 m → Option A is incorrect. The mid -point of the rope is not stationary at fundamental mode. Option B is incorrect. 2nd Harmonic-  = 1 m 3rd Harmonic-  = 0.667 m 4th Harmonic-  = 0.5 m 5th Harmonic-  = 0.4 m When the fundamental frequency is doubled, there are 3 nodes on the rope. Option C is incorrect. At the 5th harmonic, the wavelength is 0.4 m. Option D is correct. 20 B Using Rayleigh’s criterion for resolution:  = b For small angle, θ = sin θ = tan θ (700  10-9) / (30  10-3) = x / 5.0 x = 1.2  10-4 m = 0.12 mm 21 B E = − r V d d = 0 where V = constant If the electric field strength is zero at a point, it only means that the potential gradient is zero at that point. But the value of the potential at that point need not be zero. 22 B From graph, when current is 40 mA, p.d. is approximately 1.2 V. Therefore, p.d. across resistor = 6.0 – 1.2 = 4.8 V R = V/I = (4.8)/(0.040) = 120 Ω

5 23 D At balance, VAB = VDE Since no current flows through the bottom cell at balance, VDE = E2 Also, 2AB AB DE AC AC AC AC L V V E L V V V=== Then, VAC = E1 1 w W R Rr + (apply potential divider principle to the top loop) = 3.0 x 6.0 6.0 1.0+ = 2.571 V So balance length, LAB = LAC 2 AC E V = 1.0 x 1.5 2.571 = 0.58 m 24 A The electric force on a positively charged particle is directed towards SR. To pass through undeviated, the magnetic force must be directed towards PQ. When the strength of the magnetic field is increased, the magnetic force increases but the electric force remains the same. So the particles will now be deflected towards PQ. 25 B ε = - ∆(flux linkage) / ∆t ε = - N B ∆A / ∆t ε = - (1) (1.2) ( 6 x 10-3 – 0 ) / ( 3 – 0 ) ε = - 0.0024 V

6 26 A 1 120 60 2.0 V ss pp s s VN VN V V = = = Then Is = Vs/R = 2.0/3.0 = 0.667 A For ideal transformer, 120 2.0 0.667 0.011 A p p s s p p V I V I I I =  =  = 27 C The range of wavelengths for visible light is 400 nm to 700 nm. Since hcE = , the energies of these photons range from ( )( ) ( )( ) 34 8 9 19 6.63 10 3.00 10 1.7759 eV 700 10 1.60 10 − −−  =  to ( )( ) ( )( ) 34 8 9 19 6.63 10 3.00 10 3.1078 eV 400 10 1.60 10 − −−  =  . Only 3 transitions will result in emissions of such photons: 6.12 4.28 1.84 eV−= 6.81 4.28 2.53 eV−= 7.02 4.28 2.74 eV−= 28 D For the cut-off wavelength, min hceV = Since the min for graph 2 is halved of graph 1, that means that the accelerating potential for graph 2 is doubled that of graph 1. Since the characteristic wavelength remain the same for graph 1 and graph 2, this means that the target metal is the same. 29 A Difference in binding energies of products and reactants = Gain in kinetic energies of products + Energy of gamma ray (39.25 + 28.48 – 64.94) = 2.31 + E E = 0.48 MeV

7 30 A A range of (kinetic) energies indicates a range of speeds for the β particles. Since beta particles are emitted with a range of speeds, the products of a beta decay process cannot just consist of the daughter nuclide (product nuclide) and the beta particle as this would imply definite speeds for both products, in order for linear momentum to be conserved. Option B is wrong. There is no such observation. Neutrino is chargeless. The total charge of the decay products is equal to the charge of the parent nuclide. Option C is wrong. It is a true observation, but the loss in mass during β decay is due to conversion to energy released, and not the existence of the neutrino. Option D is wrong. There is no such observation. Neutrino is chargeless so has no ionising power, and therefore cannot be observed in a cloud chamber.