AJC H2 PHY P1 Soln

1 9749/01/AJC2017/Prelim [Turn Over 2017 AJC JC2 H2 Physics Prelim Solutions Paper 1 (30 marks) Answer 1 2 3 4 5 6 7 8 9 10 D A B B C D B D C D 11 12 13 14 15 16 17 18 19 20 A C B D A C C C A C 21 22 23 24 25 26 27 28 29 30 A D A C A D B B B A No Answer & Solution 1 Ans: D 2000 pV = 2000 x 10-12 V = 2 x 10-9 V = 2 x 10-15 x 106 V = 2 x 10-15 x M V p.d. = work done per unit charge, hence V is equivalent to J C-1 2 Ans : A For the projectile motion, horizontal speed is constant, v. Vertical speed is zero initially and acceleration is g, downwards. Let the height of the table be h. Using s = ut + ½ at2 h = 0 + ½ gT2 T = g h2 T is independent of mass. Range, D = vT. Since both v and T are independent of mass, D will remain the same. 3 Ans : B The scale reads the contact force between the sack and scale. Since the scale reading is larger than the mass, the upward contact force must be larger than the downward weight. The sack experiences a net force and acceleration upwards. (12 – 10) x 9.81 = 10 x acceleration acceleration = 1.962 ≈ 2.0 m s -2. 4 Ans : B For elastic collision, total KE is conserved OR relative speed of approach equals to relative speed of separation. Total KE before impact = ½ mv 2 + ½ mv2 = mv2 Thus, total KE after impact = mv2. Answer C is wrong because during collision, KE is not conserved. It is converted to elastic PE then back to KE. Answer D is wrong because total momentum before impact = mv + m(-v) = 0

2 9646/01/AJC2016/Prelim Answer A is wrong because when the spheres stick together, it is already a perfectly inelastic collision. Also note that relative speed of approach = u 1 – u2 = v – (-v) = 2v (non-zero). If the spheres stick together, they will share the same speed and relative speed of separation will be zero. 5 Ans : C Work done is minimum if the force applied is just enough to overcome the barrel’s weight on its way up the plank, with no change in speed. The barrel gains GPE without any gain in KE. Minimum work = Gain in GPE = mgh = 50 x 9.81 x 1.6 = 784.8 J ≈ 780 J Alternative method : Work done = Force x distance = mg sin θ x d = 50 x 9.81 x (1.6/3.4) x 3.4 = 784. 8 J 6 Ans: D Upthrust = weight of air displaced (smallest force as density of air is very small). Viscous drag is proportional to speed which increases as ball falls. At terminal speed, downward weight of ball is a constant which is balanced by upward drag and upthrust. Hence, weight > viscous drag > upthrust. 7 Ans: B For equilibrium, all three forces must meet at a common point, which is a point below trapdoor along the line of action of W. Hence option C and D are wrong as direction of H is wrong. Tension must act away from the door, hence direction of T is wrong. So option A is also wrong. 8 Ans: D The vertical component of the acceleration is the centripetal acceleration which is present since the particle is performing circular motion. The horizontal component of the acceleration causes the speed of the object to increase. 9 Ans: C a = v 2/r, So v = ar = 0.030120 = 1.897 Since ω is constant and ω = v/r Hence ω = 1 1 r v = 2 2 r v v2 = 2 1 1 rr v  = 0.0500.030 1.897  =3.16 m s-1 OR a = ω2r ω = √(120/0.030) = 63.25 rads-1 v = r ω = 0.050 (63.25) = 3.16 ms-1

3 9749/01/AJC2017/Prelim [Turn Over 10 Ans: D The antenna will have the same speed as the space-craft and is still bound in orbit. There is still gravitational force acting on it which causes it to move in circular motion. 11 Ans: A Let the amount of added heat be Q. So, wwQm cT ……………. (1) {heat added to water} iiQm c T ……………. (2) {heat added to iron} Eqn. (1) = (2), wi iw cT cT   Since cw > ci, ∆Ti > ∆Tw  the final temperature of iron is going to be higher than the water’s final temperature. 12 Ans: C The total volume of the cylinder is not changed. Neither is the total number of molecules. Hence, the average space given to each molecule remains constant. This implies that the average distance between molecules is unchanged. 13 Ans: B WQU  in case (i), W –ve, U +ve (PV   T), Q +ve in case (ii), ∆U = 0 since T is constant. W is -ve, thus Q is positive and heat is gained. 14 Ans : D Taking the initial position at 650 mm mark, s 167.0 2 2sin5025 2sinsin 00                    t t tTxtxx   15 Ans : A At x = 0.2 m, a = -ω2x 20 = - ω2(-0.2) ω = 10 rad s-1 vmax = ωxo = 10 x 0.2 = 2.0 m s-1 600 mm 650 mm 675 mm 700 mm

4 9646/01/AJC2016/Prelim 16 Ans: C 3 x (Period of S) = 30. Hence Period of S= 10 ms At t = 4 1 x period of S = 2.5 ms, amplitude of S = displacement of R. y = yo sin t = yo sin [(2/T)t] = 8 sin [( 2/30)x2.5] = 4 cm 17 Ans: C Since particles A and B are at two sides of a node of a stationary wave, they are anti-phase. Hence phase difference is 180° Maximum KE is proportional to amplitude. Since amplitude of A < amplitude of B, K A < KB 18 Ans: C For diffraction grating: n  = d sinθ or sinθ = n  /d for n=1, sin (15.4) =  /d ……….(1) for n=2, sin 2 = 2 /d …………..(2) (2)/(1): sin 2 = 2 sin (15.4) 2 = 32.8° Angle between first and second maxima = 32.8 – 15.4 = 16.7 19 Ans : A Terminal p.d. = 1.2 x 80 = 96 V ---(1) Terminal p.d. = 120 – (1.2+0.40) r ---(2) Equating (1) and (2), r = 15 Ω Alternative method: Let r be the internal resistance of supply and X the resistance of the lower resistor. p.d. across 80  resistor = p.d. across the lower resistor,  24040.0 80x2.1X total current = 1.2 + 0.40 = 1.6 A = supply voltage/total resistance         15r 24080 240x80r 1206.1 20 Ans : C Energy = QV 21 Ans : A Let VR = 0 V. VQ = 6 V VP – VQ = 3 V  VP = 9 V VP – VS = 5 V  VS = 4 V Hence, VQS = 6 – 4 = 2 V and VSR = 4 – 0 = 4 V

5 9749/01/AJC2017/Prelim [Turn Over 22 Ans: D 2 parallel plates, so E is constant  F is constant and in the same direction as x, as shown in the diagram. Potential V increases linearly with x. Since electron is negatively charged, from U = qV, U of electron decreases linearly with x. 23 Ans: A Efield lines radiates out of positive charges and enters the negative charge. Efield lines of like charges should repel each other. 24 Ans: C As the number of coils is halved, solenoid length is also halved, hence n does not change. As I = V/R where R = L/A, since L is halved, R will be halved and I will be twice as before. Hence T will be twice. 25 Ans : A The peak value of the induced e.m.f. , E0 = (2f)BAN and T=1/f When f is halved, E0 will halve but T will double. 26 Ans : D Power dissipated across a resistor, P = Vrms2/R When Vrms is doubled with R unchanged, P will increase 4 times 27 Ans : B The number of photoelectrons emitted per second ( t N ) is directly proportional to the intensity of incident radiation. Since the intensity of radiation is constant, the number of photoelectrons varies proportional with time. 28 Ans : B eV = hc/min higher V means smaller min so Q has the higher voltage applied same characteristic X-ray indicates same target material 29 Ans : B Based on COE and COM, the kinetic energy of the daughter nuclei is negligible hence the total energy released is shared between β particle and neutrino. Since the highest possible KE of β particle is 13.4 MeV, i.e. when neutrino has zero KE, hence the total ene