Fig. 3.2 shows where the pivot is placed beneath the metre rule when the rule is balanced. Read off the pivot position on the metre rule when balance is achieved. Enter the reading in centimetres, to the nearest millimetre, in Table 3.1 on page 10.
Calculate and enter, in Table 3.1: 1 the distance $a$ from the 5.0 cm mark to the pivot; 2 the distance $b$ from the pivot to the 50.0 cm mark.
The student carries out the procedure in (a) again for $m=40\,\text{g}, 60\,\text{g}, 80\,\text{g}$ and $100\,\text{g}$. The outcomes are listed in Table 3.1. Describe how the student ensures that the centre of each mass on the metre rule is positioned directly above the 5.0 cm mark.
Calculate the ratio $r=\frac{b}{a}$ for $m=20\,\text{g}$. Enter, in Table 3.1, your answer to two significant figures.
On the grid provided in Fig. 3.3 on page 11, plot a graph of $r$ on the y-axis against $m$ on the x-axis. Begin at the origin (0,0). Draw the straight line of best fit.
Calculate the gradient $G$ of your graph. Make it clear on the graph how you found the numbers used in your calculation.
The mass $M$ of the metre rule in grams is given by the equation $M=\frac{1}{G}$. Find the mass of the metre rule to the nearest gram.
Name a piece of apparatus that the student can use to measure the mass of the metre rule directly.
A student says that the centre of mass of the metre rule is at the 50.0 cm mark. Describe how you use the apparatus provided to check that this statement is correct.