Physics
WAEC 2013
You are provided with a uniform metre rule of mass, M indicated on its reverse side, a knife-edge, a graduated measuring cylinder of known mass, M\(_{1}\) marked on it and other necessary apparatus.
- Read and record with values of M and m\(_{1}\).
- Balance the metre rule horizontally on the knife edge. Read and record the balance point as G.
- Tie a loop of thread around the neck of the measuring cylinder.
- Fill the cylinder with the sand provided to the 2cm\(^{3}\) mark. Record the volume, V, of the sand.
- Hang the cylinder at the 2 cm mark of the metre rule and adjust the position of the knife edge until the rule balances horizontally.
- Read and record the new balance position K.
- Determine the value of e and f.
- Determine the mass, m\(_{2}\), of the sand in the measuring cylinder. Hint: m\(_{2}\) = (\(\frac{\text {M x f}}{e}\)) - m\(_{1}\).
- Repeat the procedure by filling the measuring cylinder to the mark V = 4,6,8 and 10 cm\(^{3}\). In each case, ensure that the measuring cylinder is kept constant at the 2 cm mark on the metre rule.
- Tabulate your readings.
- Plot a graph with m\(_{2}\) on the vertical axis and V on the horizontal axis.
- Determine the slope, s, of the graph.
- State two precautions taken to ensure accurate results.
(b)i. Determine the mass of 7.5 cm\(^{3}\) of the sand using your graph.
ii. A gold coin of mass 102.0 g has a uniform cross-sectional area of 10.0 cm\(^{2}\). Calculate its thickness. [Density of gold=19.3 g cm\(^{-3}\)]
Explanation
| volume of sand v/cm\(^{3}\) |
k/cm |
\(\theta\)=(k-p)/cm |
f=(G-K)/cm |
m\(_{2}\)=\(\frac{mxf}{e}\)-m\(_{1}\) |
| 2 |
43.5 |
41.5 |
6.9 |
1.614 |
| 4 |
42.7 |
40.7 |
7.7 |
3.443 |
| 6 |
41.9 |
39.9 |
8.5 |
6.372 |
| 8 |
40.1 |
39.1 |
9.3 |
9.416 |
| 10 |
40.4 |
38.4 |
10.0 |
12.178 |
Slope = \(\frac{\bigtriangleup {m}_{2}}{\bigtriangleup {v}}\)
= \(\frac{12-8-4}{10.4-4.4} = \frac{(8.8)9}{(6)cm^{3}}\)
= 1.4679/cm\(^{3}\)
Observation;
Mass of uniform metre rule m--> 130g
Mass of graduated measuring cylinder m\(_{1}\) --20g
Balance point G = 50.4cm
PRECAUTIONS; i
- Avoided draught
- Avoided parallax error in reading metre rule/ measuring cylinder
- Knife-edge firmly clamped
- Avoided measuring cylinder from resting on/touching the table
- Repeated readings are shown on the table
(b)i. 7.5cm\(^{3}\) correctly shown on graph m\(_{2}\) = 8.4g
ii. V = \(\frac {\text{Mass}} {\text{Density}} = \frac{102.0}{19.3}\) = 5.28
Volume = thickness x area
thickness = \(\frac {\text{Volume}}{\text{Area}} = \frac{5.28}{10}\)
OR
t = \(\frac{m}{A} = \frac{102}{19.3 \times 10}\) = 0.528cm
Report an Error
Ask A Question
Download App
Quick Questions
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}