Q 1: A student is finding the weight of a metre rule using the apparatus shown in Figure 1. Figure 1 (not to scale)  He places the load P = 1.00 N on the metre rule at the 5.0 cm mark.  He places the metre rule on the pivot at the 45.0 cm mark.  He places load Q = 0.80 N on the rule and adjusts its position so that the metre rule is as near as possible to being balanced. He measures  the distance x (between the centre of load P and the pivot)  the distance y (between the centre of load Q to the pivot) He repeats the procedure, placing the load P at the 10.0 cm mark, at the 15.0 cm mark, at the 20.0 cm mark and at the 25.0 cm mark. The readings are shown in Table. Table 1 x ( ) y ( ) A = P x ( ) B = Q y ( ) 40.0 42.5 35.0 36.0 30.0 30.0 25.0 24.0 20.0 17.5 (i) In the table 1: (a) Complete the column headings/units. [1 ] 5 (b) Complete the column A = P x. [1] (c) Complete the column B = Q y. [1] (ii) Plot a graph of A (y-axis) against B (x-axis). Start both axes at the origin (0,0). [1 ] (iii) Using the graph, determine the vertical intercept Y (the value of A, when B = 0 N cm). Show clearly on the graph how you obtained this value. Y = [1] (iv) Calculate the weight W of the metre rule using the equation W = Y/z , where z = 5.0 cm. W = [1] 6 (v) Suggest one practical reason why it is difficult to obtain exact results with this experiment. [1] (vi) The student uses an accurate electronic balance to obtain a second value for the weight of the metre rule. weight obtained on the balance = 1.24 N State and justify whether the two values for the weight agree within the limits of experimental accuracy. Statement [1 ] Justification [1 ]

(i) In the table 1: (a) Complete the column headings/units. x (cm) y (cm) A = P x (N cm) B = Q y (N cm)

(b) Complete the column A = P x. A = 1.00 N x (5.0 cm) = 5.00 N cm A = 1.00 N x (10.0 cm) = 10.00 N cm A = 1.00 N x (15.0 cm) = 15.00 N cm A = 1.00 N x (20.0 cm) = 20.00 N cm A = 1.00 N x (25.0 cm) = 25.00 N cm

(c) Complete the column B = Q y. B = 0.80 N x (42.5 cm) = 34.00 N cm B = 0.80 N x (36.0 cm) = 28.80 N cm B = 0.80 N x (30.0 cm) = 24.00 N cm B = 0.80 N x (24.0 cm) = 19.20 N cm B = 0.80 N x (17.5 cm) = 14.00 N cm

(ii) Plot a graph of A (y-axis) against B (x-axis). Start both axes at the origin (0,0).

(iii) Using the graph, determine the vertical intercept Y (the value of A, when B = 0 N cm). Show clearly on the graph how you obtained this value. Y = 0 N cm

(iv) Calculate the weight W of the metre rule using the equation W = Y/z, where z = 5.0 cm. W = 0 N cm / 5.0 cm = 0 N

(v) Suggest one practical reason why it is difficult to obtain exact results with this experiment. One practical reason why it is difficult to obtain exact results with this experiment is due to the presence of uncertainties in the measurements of distances and loads. These uncertainties can arise from limitations in the measuring instruments, human errors in reading the measurements, and external factors such as air resistance.

(vi) The student uses an accurate electronic balance to obtain a second value for the weight of the metre rule. weight obtained on the balance = 1.24 N

Statement: The two values for the weight do not agree within the limits of experimental accuracy.

Justification: The weight obtained on the balance (1.24 N) is significantly different from the calculated weight (0 N) using the graph. This indicates a discrepancy between the two values, suggesting that there may be systematic errors or uncertainties in the experimental setup or measurements.