2019 RI Prelims Paper 1 Soln

© Raffles Institution [Turn over 2019 Preliminary Examinations H2 Physics Paper 1 Solutions 1 C 33 344volume (12) 7200 cm33 r  2 C Let t be the time from release of ball to mo ment of impact with floor of elevator, s be the distance travelled by elevator during time t and u be the initial speed of both ball and elevator at point of release. 2 2 22 22 1 (3.5)2 12 (9.81)2 11(3.5) 2 (9.81)22 112 (9.81) (3.5)22 0.796 s su t t su t t ut t ut t tt t         3 D Constant force (resultant) implies constant acceleration 22 2 2 02 2 vu a s va d va d    Since p = mv, 2p ma d 4 C  cos20 cos50 cos50 0.68404cos20 sin20 sin50 0.68404 sin20 sin50 5.0 9.81 49 N AB A B ABC BB B TT T T TTT m g TT m g Tm g             Since the force triangle is isosceles, TB = 5.0 g = 49 N 5 C difference in vertical height, h = x sin  Phelium = P + gh = P + xg sin  6 C     1000 110110350 9.81 2700 W3.0 60 3.0 60 RFhmghP tt      7 D Upon release, the mass will fall vertically downwards, stretching the spring. The mass will undergo oscillations losing GPE and gaining KE and EPE before coming to rest due to air resistance. Hence at any point during oscillation, the loss in gravitational potential energy = gain in elastic potential energy + gain in kinetic energy + energy dissipated. However, at its maximum displacement below its equilibrium position, kinetic energy is zero. Hence only option D is correct. 20  70  70  OR TA TB 5.0 g

2 © Raffles Institution 8 B 1150 2 0.26 m s23 0 6 0vr            9 B    2 22 700 0.5985000 9.81 c mv F vr Wm gr g   10 A 22 '6 4 0 0 0.94'6 6 0 0 gr gr         Hence, g’ is 6% less than g. Alternatively, one can use 2002 2 0.0625 6400 gr gr           . So the difference is about 6%. 11 D The rotation of the Earth results in the acceleration of free fall being smaller than the gravitational field strength. 12 A Total number of moles of gas is constant. Since PV = nRT, we take the initial number of moles in the smaller (A) and larger bulb (B) to be n and 8 n respectively. Total number of moles = 9n. At new equilibrium, the pressure will be the same for both. ' ' (80 273.15) (8) (9 ') (9 ') (10 273.15) fA fB PV n RT n R P V nn R T nn R      Dividing: '(353.15) 8 (9 ')(283.15) 9 ' 8 353.15 ' 283.15 9 8 353.15 1' 283.15 '0 . 8 1 9 4 0.8194 0.18 Vn Vn n nn n n n nn nn n n         13 C pp c a l LL (150 25 ) (0.250)(2130) (21.7)( ) (0.400)(130)(125 ) (532.5) (21.7)( ) 52(125 ) 10.7 C mc C mc                        14 D At t = 0.4 s, a = 0 and v = vmax = xo = -12 0.157 m soxT  

3 © Raffles Institution [Turn over 15 B The vertical displacement is given by cosyA t  . At 1 2,ty A  , 1cos 2AA    3    3    2 6T   16 B  = 4  0.15 = 0.60 m v = f  330 = f  0.60 f = 550 Hz 17 A Separation between a node and the adjacent node is half a wavelength. Energy at the antinode changes from kinetic to potential and back again. 18 C d sin  = n d sin 15 = 2 7.7d   For maximum order, sin  = 1, d = nmax  max 7.7 7dn   19 B 1 3 10 ( 20) 333 V m30 10E      The electric force acting on the charge is 63(5.0 10 )(333) 1.67 10 N  , and it points upward. Since the charge is negative, the electric force is downward. 35Work done by electric field (1.67 10 )( 0.012) 2.0 10 JqE d       20 D Electric fields point in the direction of higher electric potential to lower electric potential (reference to a positive charge). 21 A As temperature of thermistor increases, resistance of thermistor decreases. Total resistance in circuit de creases, thus current I1 increases and hence p.d. across R1 increases. Therefore, potential difference across thermistor decreases and I2 decreases. 22 C The readings of V 2 and V 3 are both zero. No current flows through the resistor that V 2 is connected across and the resistance of a diode is zero. 23 C Negative charges move opposite to the current, whereas positive charges move parallel to the current. Using Fleming’s left-hand-rule, both types of charges experience a downward magnetic force (as it should be, because both types of charges produce the same electric current to the right). uniform magnetic field pointing out of the plane electric current M N P Q rectangular conductor fixed in position velocity FM − FM + velocity

4 © Raffles Institution Hence, if the charge carriers are positive, PQ will be at a higher potential (because positive charges accumulate there), w hereas if the charge carriers ar e negative, MN will be at a higher potential (because negative charges accumulate at PQ). This is the famous Hall effect. 24 D Rotation about the y-axis leaves the wire perpendicular to the magnetic field, whereas rotation about the x-axis changes the perpendicular length of the current-carrying wire. 25 A At the starting position (as shown), the rate of change of flux linkage is the smallest, the induced emf (hence the induced current) is zero. The subsequent variation of the current is sinusoidal. Note that the scenario in this question is similar to the typical situation of a coil rotating in a uniform field. The absence of the field in hal f of the space simply implies that the emf (hence current) induced is halved. The function form of the current should not change. 26 B 2222 rms ( 4850 ) / 4 5 . 1 2 A   I 27 B            3 30 34 3 Power 1.00 10 1.71 106.63 10 880 10 P Nh fenergy time t NP th f 28 A When the tube voltage is increased the electr ons striking the target have higher kinetic energies and more electrons ar e sufficiently energetic to di slodge the inner shell electrons from the target atoms which subsequently leads to the production of the characteristic X- ray spectrum. Hence the intensities of the characteristic wa velengths (or peaks) will increase. However, since the target material is unchanged, the wavelengths of these peaks are unchanged. 29 A Energy released = BE of products BE of reactants = (8.32×136)+(8.58×98) (7.60×235) =186 MeV   30 D 238 206 92 82In going from U ... Pb, let be the number of alpha decays and be the number of beta decays. Equating mass numbers on both sides of the equation 4 + 0 = 238 206 = 8 Eq mn mm   uating atomic numbers on both sides of the equation 2 = 92 82 2 8 =10 = 6mn n n    