HCI 2022 Prelim P1 Guide

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 © Hwa Chong Institution 1 2022 C2 H2 Physics Prelim Exams Paper 1 Suggested Solutions 1 D 6 B 11 B 16 D 21 C 26 C 2 B 7 A 12 D 17 D 22 A 27 B 3 B 8 C 13 B 18 C 23 C 28 D 4 B 9 B 14 B 19 C 24 A 29 B 5 B 10 A 15 C 20 A 25 D 30 D 1 D 550 GHz = 5.50 x 1011 Hz This corresponds to radiation with wavelength 𝜆 = 3.0 × 108/(5.50 × 1011) ≈ 10-3 m Wavelength of green light is of order of magnitude 10 -7 m. 2 B Conceptual Question. Mean value should be close to the true value, while experimental values should have a large spread. 3 B The gradient of v-t graph give the acceleration, and initial gradient must be ½ of the original. The area under the v-t graph gives the distance between the two train stations, so it must the same for the second train. 4 B By Newton’s first law, the objects can continue at the state of constant velocity. 5 B 𝑀𝑏𝑢𝑏 + 0 = 𝑀𝑏𝑣𝑏 + 𝑀𝑝𝑣𝑝 (5)𝑢𝑏 + 0 = (5)𝑣𝑏 + (1)𝑣𝑝 (1) 𝑢𝑏 − 0 = 𝑣𝑝 − 𝑣𝑏 𝑢𝑏 = 3.33 − 𝑣𝑏 (2) Using (1) and (2) to solve for 𝑢𝑏 = 1.998 𝑚 𝑠−1 6 B Take moments about the pivot. Torque = 5.0(2)(0.5 sin 40º) = 3.2 N m 7 A Consider the vertical component of the tension, and letting the angle that the cord makes with the metal sheet be θ, 2Tsin(θ) = 10 → sin(θ)=0.25 → θ = 14.5º Length of cord required = 2(10/cos(θ)) = 20.7 = 21 cm 8 C Option A and D have the wrong units and can be ruled out immediately. Option B is wrong as vt does not give the total distance traveled as the object is not moving at constant speed. 9 B 2 22 3000 8.02 7897 790060 100a r ms −   = =  =       

2 © Hwa Chong Institution 2 10 A As the Earth is taken to be a uniform sphere, the gravitational potential at any point on the Earth’s surface is the same (as GM R =− , where M is the mass of the Earth and R is the radius of the Earth), hence the gravitational potential energy of the Earth-rocket system is the same whether the rocket is near Equator or near the North Pole. As such the escape speed is the same whether the launch is near the Equator or near the North Pole. 11 B Those who chose C or D did not know that gra vitational potential energy is negative. Those who chose A or C did not know that orbital radius is the sum of the radius of planet and altitude of the satellite. 12 D If both gases are at the same temperature, their average microscopic kinetic energy and root-mean-square speed will be the same. Since the volume of container B is larger than that of A, the gas pressure and density will be smaller in B. 13 B Net work done on the gas = – area enclosed = – area of circle = – = – 3.14 J By the first law of thermodynamics, there is a net heat transfer of 3.14 J into the system. 14 B The frequency of the driver determines the oscillating frequency. So, f = 1/2 = 0.50 Hz 15 C Intensity of beam after passing Polaroid Q = I cos2(30°) Intensity of beam after passing Polaroid R = I cos2(30°) x cos2(60°-30°) = 0.56 I (2 s.f) 16 D 2 4 source sourcePPIntensity Area x  ==I (As the source is a point source, the Area through which energy pass through is the surface area of a sphere with radius x.) As Psource is constant, 2 1 x I 2212 2 2 1 2 0.5( ) ( ) 4x x xx =  =  =I I I I I I As received reciever P Area I = CP =I 2 2 2 2 = 84received recieverP Area C P == I I 17 D Using Rayleigh’s Criterion min a  =

3 © Hwa Chong Institution 3 If angular seperation of light sources  is greater than min , the two images will be resolved (distinguished on the screen). To reduce min , one can reduce  or increase a so options A and C will help to distinguish the two images. For option C, by reducing distance D, the actual angular separation of the light sources  gets larger so it is larger than min . This change also helps to improve the ability to distinguish the two images. For option D, changing L has no impact on  or min so it is the correct answer. 18 C Note that the vertical nylon lines result in the horizontal diffraction images while the horizontal nylon lines results in the vertical diffraction images. From the equation for diffraction grating, dsinθ = nλ For monochromatic light of fixed λ, For the same order of maxima, when d is larger, sinθ is smaller and hence θ is smaller. This means that the separation of the straight through image and the image of the particular order is closer together. Since the separation of the straight through image and the vertical images are smaller compared to the separation of the straight through image and the horizontal images, the d (slit separation) for the horizontal lines must be larger than the d for the vertical lines. 19 C Option A : 2 0 2 4 qF r= . r decreases and hence, force increase. Option A is incorrect. Option B : Electric field strength is a vector quantity. Sketching the individual E due to each charge at O and summing vectorially to find the resultant shows the E has decreased. Option B is incorrect. Option D: Electric potential is a scalar quantity and hence resultant potential at O remains zero even when the change is made. Hence option D is incorrect. Option C : 2 04 qU r=− . r decrease, U becomes more negative. Hence U decreases. Option C is correct. (Alternatively you can think of the fact that the charges being closer now experience a stronger attractive force and hence will be at lower potential energy and will require a greater amount of energy to separate them to infinity.) 20 A (gradient of graph)dVE V x dx=− =− − Arrangement 1 : E-field is uniform and hence the gradient of V-x graph should be constant. Arrangement 2 : E-field decreases as we move from P to Q, hence the gradient of V-x graph should be become gentler as we move from P to Q. 21 C Bulb X : Rx = 102 / 20 = 5.0  Bulb Y : Ry = 52 / 2 = 12.5 

4 © Hwa Chong Institution 4 p.d. across X = 5.0 / (12.5 + 5.0) x 15.0 V = 4.3 V power dissipated in X < 20 W p.d. across Y = 12.0 / (12.5 + 5.0) x 15.0 V = 10.7 V power dissipated in Y > 2 W 22 A 𝑅 = 𝜌𝑙 𝐴 and 𝑙 𝐴 = constant 𝑅 = 𝜌𝑙2 / constant 23 C Points A, B, C and D have the same potentials. Thus, no current flows through all 4 R’ resistors. The above circuit can be re-drawn as shown below. The effective resistance between P and Q is the twice of the net resistance of two resistors (across points P & D and P & A) in parallel. The effective resistance between P and Q = R. P R R R R Q A B D C P R R’ R’ R R R’ R’ R Q A B D C

5 © Hwa Chong Institution 5 24 A Apply RHGR to determine the direction of B-field at O due to the current flowing in a straight wire at P, Q, R and S respectively. 25 D Using a high direct current supply to coil X will result in high flux linkage in coil Y. But the flux linkage is not changing. Thus no e.m.f. will be induced in coil Y to produce a reading. 26 C For Loop 3 and loop 4, as the loops did not experience any change in magnetic flux linkage, there were no e.m.f. induced in loop 3 and loop 4. Thus no induced currents flow in loop 3 and loop 4 so the loops did not experien