2013 MJC H2 Physics Solutions

2013 Meridian JC H2 Physics Preliminary Examination Solutions 1 Answers to 2013 Prelim Paper 1 (H2 Physics) 1 D 11 A 21 B 31 C 2 A 12 C 22 C 32 A 3 B 13 A 23 B 33 B 4 B 14 A 24 D 34 A 5 D 15 B 25 A 35 A 6 A 16 D 26 B 36 C 7 A 17 C 27 B 37 C 8 C 18 A 28 C 38 D 9 A 19 C 29 B 39 C 10 B 20 B 30 D 40 C MCQ 1: (D) Reasoning: Random errors cannot be eliminated but it can be reduced by averaging repeated measurements. By nature, random errors are of varying sign and magnitude. MCQ 2: (A) Reasoning: √ MCQ 3: (B) Reasoning: Since the area under the acceleration time graph is the change in speed and the initial speed of the object is zero, the biggest area under the graph will correspond to the largest speed. At B, the area under the graph is the largest, following which the area is negative which means that the speed decreases from B. MCQ 4: (B) Reasoning: Using: 152 = 202 + 2a (70) a = -1.25 m s-2 02 = 152 + 2a(x) x = 90 m MCQ 5: (D) Reasoning: The resultant of the two forces 3 N and 4 N will be 5 N pointing in this direction. Hence the resultant of all three forces will then be 1 N pointing in the similar direction.

2013 Meridian JC H2 Physics Preliminary Examination Solutions 2 MCQ 6: (A) Reasoning: The 3 forces acting on the rod is weight, F and the hinge force For equilibrium, all 3 forces must pass through a common point (concurrent) and form a closed triangle. MCQ 7: (A) Reasoning: By Hooke’s Law, 3 00 9 81 25 2 0 500 0 0208 mg sin f ke . ( . )(sin ) .e .m θ   MCQ 8: (C) Reasoning For elastic collision, Relative speed of approach = relative speed of separation v2 – v1 = u1 – u2 (where the sign conventions of u1, u2, v1, v2 are to the right) Hence ux – (-uy) = vy – (-vx) ux + uy = vx + vy MCQ 9: (A) Reasoning: Work done by friction = 60 x 10 = 600 J = Increase in internal energy ∆Ek + ∆Ep +∆Ee = Wsupplied – Wdissipated Kinetic energy = 150(10) -600 – 100(10 sin 30) = 400 J MCQ 10: (B) Reasoning: For each jet engine, Power output 0.80Power input  power outputPower input = 0.8 Power input = 0.8 Fv where F = half of total thrust Power input = 8.0 )250(000 200 x 2 1       31.3 MW

2013 Meridian JC H2 Physics Preliminary Examination Solutions 3 MCQ 11: (A) Reasoning: ( ) MCQ 12: (C) Reasoning: Net force acting on the satellite at that instant is now zero. Since the satellite is already moving with a speed tangent to the orbit, Newton’s 1st law says that it will continue to move along the tangent to the orbit. MCQ 13: (A) Reasoning: ( ) ( ) MCQ 14: (A) Reasoning: 3 2 2 2 4() 43 3 GRGM G V G Ra R R R aR          Since  and R is doubled, a’ is 4a. MCQ 15: (B) Reasoning: Frequency = 50/47 Hz KEmax = (0.5)( 5.0 × 10-3)(2π (50/47)(150 x 10-3))2= 2.5 x 10-3 J MCQ 16: (D) Reasoning: Amplitude, x0 = 22/2 = 11 mm When the point of needle is about to move downwards through the cloth, displacement of needle from origin is 3 mm. v =  = (2π (4.5))(√ = 0.299 ms-1 MCQ 17: (C) Reasoning: The pulse reflects off the fixed support with a 180o phase change. Hence, the reflected pulse superimpose with the incoming pulse to give a resultant pulse that is of a higher amplitude than before. 22 o -xx

2013 Meridian JC H2 Physics Preliminary Examination Solutions 4 MCQ 18: (A) Reasoning: At T/4 seconds later, the wood is moving downwards at position A. The wood is oscillating vertically, not travelling to the right. MCQ 19: (C) Reasoning: Maxima closest = Fringe separation, x, is the smallest Using For same D, x ∝ Smallest ratio is option C MCQ 20: (B) Reasoning: For pipe A, L 1/4 λ1 λ1 = 4L f1 = v/4L For pipe B, L ½ λ2 λ2 = 2L f2 = v/2L = 2(v/4L) = 2f1 = 2(220) = 440 Hz MCQ 21: (B) Reasoning: 53 80 ( ) 80 (1.0 10 )(0.2 10 ) 60 J U Q W pV             MCQ 22: (C) Reasoning: The temperature at which boiling occurs for a liquid is dependent on whether the molecules of the liquid have sufficient energy to completely overcome the force of attraction between the molecules in the liquid state as well as to do work against atmospheric pressure. Hence the pressure exerted on the liquid surface will affect the boiling temperature.

2013 Meridian JC H2 Physics Preliminary Examination Solutions 5 MCQ 23: (B) Reasoning: By definition. MCQ 24: (D) Reasoning: Option A is wrong because electric charge needs to be quantized Option B is wrong because gravitation potential cannot be positive Option C is wrong because electric field strength at a point due to the two point charges should be proportional to the sum of the field strength due to the two charges. MCQ 25: (A) Reasoning: Let r be resistance of R and S; then 0.5r is the resistance of R and S. When bulb S blows, total circuit resistance increases from r/3 to r/2  total circuit current decreases from Io to (2/3)Io . Since current through each bulb unchanged, bulb brightness unchanged. MCQ 26: (B) Reasoning: MCQ 27: (B) Reasoning: The voltmeter and the 2.0 kΩ resistor has a combined equivalent resistance of 0.667 kΩ. The p.d. across the equivalent resistor can be determined using the potential divider principle, = 5.1667.0 667.0  x 10 = 3.07 V. MCQ 28: (C) Reasoning: Adding a resistor in series with the se condary cell does not alter the balance length as no current flows through the resistor when the balance length is attained. Current depends on the motion of both positive and negative charges. Total charge flowing past a point in 1 minute 15 19 15 19 4 1.5 10 1.6 10 4.4 10 1.6 10 9.44 10 C Q            Total current = 4 59.44 10 1.57 10 16 A60 Q t      

2013 Meridian JC H2 Physics Preliminary Examination Solutions 6 MCQ 29: (B) Reasoning: When the current and magnetic field are both on the horizontal plane and are perpendicular to each other, there is force and the force is vertical. When the current and magnetic field are parallel to each other, then there’s no magnetic force. Force is zero. MCQ 30: (D) Reasoning: Induced E is directly proportional to dB/dt, which is proportional to dI/dt. dI/dt is the gradient of the given graph. Consider one cycle: For the first ¼ of the cycle, gradient becomes steeper, hence induced E increases. For the second ¼ of the cycle, gradient becomes gentler, hence induced E decreases. For the third ¼ of the cycle, gradient becomes steeper but is now negative, hence induced E increases in magnitude but reverse in direction. For the fourth ¼ of the cycle, gradient becomes gentler, hence induced E decreases in magnitude. MCQ 31: (C) Reasoning: The coil is turn over 180o. The flux linkage therefore undergoes a change of 2AB. 2 ave AB tt  MCQ 32: (A) Reasoning: Since one particle has –q, the other particle must have +q charge as they separated from a neutral charge. By COM, the two particles separated with the same speed, opposite direction. But since their charges are opposite, the magnetic force acting on them is the same. So, they will move along the same circle and meet after completing half a circle (half T) Therefore, time taken, ½ T is .

2013 Meridian JC H2 Physics Preliminary Examination Solutions 7 MCQ 33: (B) Reasoning: ( ) ( ) ( ) ( ) ( ) ( ) ( ) () 22 () 120 60 7200 4500 7200 1.6 1.6 1 generated mean rms generated rms generated generated mean rms transmission rms transmiss ion rms transmission rms transmission lost rms transmission P V