JPJC 2024 JC2 H2 Prelim P4 Solution

2024/JPJC/Prelim/9749/04 [Turn over Suggested Solutions No. Solution Remark 1(a) R = 68  I = 121.1 mA y = 41.2 cm [1] - for correct d.p and unit in I and y 1(b) R/ y/cm I/mA R y /  m−1 IR/V 56 42.5 121.1 130 6.8 68 41.2 121.1 170 8.2 82 40.0 121.0 210 9.9 100 38.5 121.0 260 12.1 120 37.7 121.2 318 14.5 [1] - headings and units - 5 sets of data [1] - d.p. of raw data - s.f. of processed data [1] correct calculation, allow 1 slip 1(c) Refer to attached graph. [1] axes: units, scale [1] plotted points accurate to half of smallest division [1] best fit line 1(c) Given R Q R QyF=− I Graph of R y vs IR is plotted, where Q F is the gradient and −Q is the y-intercept. ( ) ( )( ) ( ) 305 137.5Gradient= 23.9 14.0 7.0 = 23.9 Substitute 14.0,305 into the equation, 305 23.9 14.0 29.6 Q F Q Q − =−  = + − = Since Q F =23.9  29.6 F = 23.9 F = 1.23 F = 1.23 V Q = 29.6  m−1 [1] - Big triangle - substitution of gradient coordinates [1] Q F calculated correctly [1] Q calculated correctly [1] F calculated correctly [1] Q and F with correct units

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3 2024/JPJC/Prelim/9749/04 [Turn over 1(d) d1 = 0.19 mm d2 = 0.19 mm Diameter of wire, d = 0.19 0.19 0.19 mm2 + = Given 2 4Q d  = 2 4 Qd = ( ) 2329.6 0.19 10 4  −   = = 8.39  10−7  m [1] - correct measurements with unit and d.p - repeated measurement [1]  calculated correctly

4 2024/JPJC/Prelim/9749/04 No. Solution Remarks 2(a)(i) 1 = 60 2 = 60  = 60 + 60 602  = [1] - correct measurements - repeat measurement - nearest degree 2(a)(ii) n1= 11 n2 = 11 n = 11 + 11 112 = [1] correct measurements [1] repeat measurement 2(b) Given n = k tan k = 11 6.4tan tan60 n  ==  [1] k calculated correctly [1] answer for k in 2 s.f 2(c) / n1 n2 n k = tan n  60 11 11 11 6.4 70 14 14 14 5.1 80 26 26 26 4.6 [1] - headings and units - 3 sets of data - no d.p for raw data - correct s.f for processed data [1] correct calculation 2(d) For  = 90, the two pendulums will make contact with the wooden rod at the same length. Hence both pendulums will always oscillate in phase. [1]

5 2024/JPJC/Prelim/9749/04 [Turn over No. Solution Remarks 3(a)(i) L = 26.2 26.2 26.2 cm2 + = h = 71.8 71.8 71.8 cm2 + = [1] - correct unit and measurement for h and L. - repeat measurement 3(a)(ii) 71.8 2.7426.2 h L == [1] - correct calculation - answer in 3 s.f 3(a)(iii) Let Y = h L Y h L Y h L   =+ Percentage uncertainty in 0.2 0.2 100%71.8 26.2 h L = +  = 1 % [1] for correct percentage uncertainty (1 or 2 s.f.) [1] Accept h & L = 0.2 or 0.3 cm 3(b) m/g h/cm L/cm h L 100 71.8 26.2 2.74 200 79.8 18.2 4.38 300 84.1 13.9 6.05 400 86.7 11.3 7.67 500 88.5 9.5 9.3 [1] - headings and units - 5 sets of data - d.p, units of raw data - s.f of processed data [1] correct calculation

6 2024/JPJC/Prelim/9749/04 3(c)(i) [1] - all points plotted correctly - best fit line drawn 3(c)(ii) 8.4 3.6Gradient= 0.0166440 150 − =− Given 2 1hm LX=+ Graph of vs h mL is plotted  gradient = 2 X  2 X = 0.0166 X = 120 g [1] - substitution of gradient coordinates - value of X calculated correctly 3(d) Mass X = 138.96 139.04 139.00 g2 + = Percentage difference = 139 120 100% 16%120 − = The percentage difference between two values of X is 16% which is higher than the percentage difference in (a)(iii), hence value in (a)(iii) does not explain the difference. [1] - calculate the % difference of X - % difference higher than (a)(iii) 3(e)(i) Mass M = 100 g = 1.00  10−3 kg Thickness t = 0.79 0.78 0.80 0.79 mm3 ++ = Distance L = 26.8 26.8 26.8 cm2 + = [1] - correct unit and measurement of t - repeat measurement - correct d.p

7 2024/JPJC/Prelim/9749/04 [Turn over Width u = 1.3 1.3 1.3 cm2 + = Distance moved down by hacksaw blade, a = 14.6 14.5 14.6 cm2 + = Given 3 3 4MgLa Yut=  ( ) ( ) 3323 33 2 2 3 4 100 10 9.81 26.8 104 14.6 10 1.3 10 0.79 10 MgLY aut −− − − −      ==      = 8.07  1010 Pa = 80.7  109 Pa = 81 GPa [1] - correct unit and measurement of u - repeat measurement for u only - correct d.p [1] - correct measurement of a - repeat measurement - correct d.p [1] Y calculated correctly 3(e)(ii) - difficulty in determining the vertical deflection a of the ruler. - slotted weights may not be securely attached to the end of the ruler which may result in movement or slipping of the weights. [1] any one 3(e)(iii) Y is the stiffness of the metal. The value of Y for wood is 12 GPa which is much lower than value of the metal in (e)(i). This mean the wood is less stiff and more prone to deformation under load. When the same load is placed at the end of the wooden beam with the same dimensions as the metal hacksaw blade, the wooden beam will bend more and break . (deflection for metal blade will be smaller and more linear). [1]

8 2024/JPJC/Prelim/9749/04 3(f) Fig. 1 1. Set up the apparatus as shown in Fig. 1. 2. Secure the metal hacksaw blade to the bench top tightly using the G -clamp. Ensure that the blade does not slip or shift during the oscillation. 3. Attach 100 g load to the end of the blade. Use Blu-Tack and rubber band to secure the load. 4. Measure the distance between the centre of the load and the edge of the surface L using ruler. 5. Set the load into oscillation by displacing the load slightly downwards. 6. Take the time taken for N oscillations using stopwatch and calculate the period 12 2 ttT N += . 7. Calculate frequency of oscillation 1f T= . 8. Repeat the step 4 to 7 using different load M and L. M/kg L/m Time for 40 oscillations T/s f2/s−2 Z/kg m3 s−2 t1/s t2/s 0.100 0.268 26.7 26.9 0.670 2.28 0.00439 0.200 0.243 24.4 24.2 0.608 2.71 0.00778 Safety Precautions: - Wear gloves to prevent cut and scratches. - Ensure the G -clamp holding the metal blade is secured and does not slip. - Maintain safe distance from the oscillating blade to avoid injury. - Oscillating with small amplitude to avoid large and uncontrolled swing. [1] workable set up & label the diagram [2] Correct and detailed procedure [1] -heading with unit [2] Z calculated correctly - Minus [1] if t1 & t2  20 s - Two sets of M and L must be different [1] any one safety precaution

9 2024/JPJC/Prelim/9749/04 [Turn over Diagram 4 Aim: To investigate how the resistance R of the LDR depends on the intensity I of the incident light and distance d from the lamp. Diagram Refer to diagram above. Experiment 1 – to determine p Independent variable: Distance d from the lamp Dependable variable: Resistance R of the LDR Controlled variable: - Intensity I of the incident light - Alignment of lamp with LDR - Ambient light intensity [2] feasible set up and labelled diagram, e.g. - correct circuit diagram - voltmeter // LDR - ammeter in series with LDR - orientation of lamp and LDR - ruler measuring distance d - intensity meter measuring intensity of source [1] correct variables for Experiment 1 rheostat V A Distance d metre rule 12 V battery ammeter LDR voltmeter resistor lamp 12 V battery Intensity meter

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