TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a uniform metre rule of mass, M indicated on its reverse side, a knife-edge, a graduated measurin...

Answer Details

Test of Practical Knowledge (moments: mass of sand by balancing a metre rule)

Observations

• Mass of the uniform metre rule, \(M = 130\,\text{g}\)
• Mass of the empty measuring cylinder, \(m_1 = 20\,\text{g}\)
• Balance point of the bare rule, \(G = 50.4\,\text{cm}\)

The cylinder is hung at the 2 cm mark of the rule and the knife-edge is moved until the rule balances at position \(K\). Then \(e = K - 2\) is the distance from the knife-edge to the load, and \(f = G - K\) is the distance from the knife-edge to the rule's centre of gravity. Taking moments about the knife-edge:

\[ (m_1 + m_2)\,e = M\,f \quad\Rightarrow\quad m_2 = \frac{M f}{e} - m_1 \]

For example, at \(V = 2\,\text{cm}^3\): \(K = 43.5\), \(e = 41.5\), \(f = 6.9\), so \(m_2 = \dfrac{130 \times 6.9}{41.5} - 20 = 21.61 - 20 = 1.61\,\text{g}\).

Table of readings

\(V\)/cm\(^3\)	\(K\)/cm	\(e = K-2\)/cm	\(f = G-K\)/cm	\(m_2 = \frac{Mf}{e}-m_1\)/g
2	43.5	41.5	6.9	1.61
4	42.7	40.7	7.7	3.44
6	41.9	39.9	8.5	6.37
8	40.1	39.1	9.3	9.42
10	40.4	38.4	10.0	12.18

Graph of \(m_2\) against \(V\)

Plotting \(m_2\) (vertical axis) against \(V\) (horizontal axis) and drawing the best straight line through the points:

Slope of the graph

Reading two well-separated points off the line of best fit, \((V_1, m_2{}_1) = (4.4,\,4.0)\) and \((V_2, m_2{}_2) = (10.4,\,12.8)\):

\[ s = \frac{\Delta m_2}{\Delta V} = \frac{12.8 - 4.0}{10.4 - 4.4} = \frac{8.8}{6.0} = 1.47\ \text{g cm}^{-3} \]

The slope is the mass per unit volume (density) of the sand.

Two precautions

• Draught was avoided and the knife-edge was firmly clamped so that the rule balanced steadily and horizontally before each reading.
• Parallax error was avoided by reading the balance point on the rule and the sand volume in the cylinder with the eye placed directly above the scale.

(b)(i) Mass of 7.5 cm\(^3\) of sand

Reading the best-fit line at \(V = 7.5\,\text{cm}^3\) (shown by the dashed guide lines on the graph) gives

\[ m_2 = 8.4\,\text{g} \]

(b)(ii) Thickness of the gold coin

Mass \(= 102.0\,\text{g}\), cross-sectional area \(= 10.0\,\text{cm}^2\), density \(= 19.3\,\text{g cm}^{-3}\).

\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{102.0}{19.3} = 5.28\,\text{cm}^3 \]\[ \text{thickness} = \frac{\text{Volume}}{\text{Area}} = \frac{5.28}{10.0} = 0.528\,\text{cm} \]