Using the spring balance provided, determine the weight of object of mass \(M = 5.0\text{g}\). Record this weight as \(W_{1}\). Determine the weight of the ...
Question 1 Report
- Using the spring balance provided, determine the weight of object of mass \(M = 5.0\text{g}\). Record this weight as \(W_{1}\).
- Determine the weight of the object when completely immersed in water contained in a beaker as shown in the diagram. Record the weight as \(W_{2}\)
- Determine the weight of the object when it is completely immersed in the liquid labelled 'L'. Record the weight as \(W_{3}\). Evaluate, \(u = (W_{1} - W_{3})\) and \(v = (W_{1} - W_{3})\).
- Repeat the procedure with the objects of masses \(M = 10, 15, 20,\) and \(25\text{g}\). In each case, evaluate \(u = (W_{1} - W_{3})\) and \(v = (W_{1} - W_{3})\) on the vertical axis against \(u = (W_{1} - W_{2}\) on the horizontal axis.
- Determine the slope, \(s\), of the graph. (vii) State two precautions taken to ensure accurate results.
(b)i. A piece of brass of mass \(20.0\text{g}\) is hung on a spring balance from a rigid support and completely immersed in kerosine of density \(8.0 \times 10^{2}\text{ kg m}^{-3}\). Determine the reading on the spring balance. \([g = 10\text{ms}^{-2}]\), density of brass = \(8.0 \times 10^{3}\text{ kg m}^{-3}\) J
ii. State Archimede's principle and the law of floatation.
Answer Details
(a) Determination of the relative density of liquid L by upthrust
The object is hung from the spring balance and its weight in air \(W_1\) is read. It is then fully immersed in water and the reading \(W_2\) taken, and finally fully immersed in liquid L and the reading \(W_3\) taken. The upthrust in water is \(U=(W_1-W_2)\) and the upthrust in L is \(V=(W_1-W_3)\). The procedure is repeated for the five masses.
The diagram of the apparatus is shown below.
Table of readings
| M /g | \(W_1\) /g | \(W_2\) /g | \(W_3\) /g | \(U=(W_1-W_2)\) /g | \(V=(W_1-W_3)\) /g |
|---|---|---|---|---|---|
| 5.0 | 5.00 | 4.00 | 4.20 | 1.00 | 0.80 |
| 10.0 | 10.00 | 8.00 | 8.40 | 2.00 | 1.60 |
| 15.0 | 15.00 | 12.00 | 12.60 | 3.00 | 2.40 |
| 20.0 | 20.00 | 16.00 | 16.80 | 4.00 | 3.20 |
| 25.0 | 25.00 | 20.00 | 21.00 | 5.00 | 4.00 |
Graph of \(V=(W_1-W_3)\) against \(U=(W_1-W_2)\)
Slope of the graph
Taking two points on the line of best fit, \((U_1,V_1)=(1.00,\,0.80)\) and \((U_2,V_2)=(5.00,\,4.00)\):
\[ s=\frac{V_2-V_1}{U_2-U_1}=\frac{4.00-0.80}{5.00-1.00}=\frac{3.20}{4.00}=0.80 \]Since the slope \(s=\dfrac{W_1-W_3}{W_1-W_2}=\dfrac{\text{upthrust in L}}{\text{upthrust in water}}\), the slope is the relative density of liquid L:
\[ \boxed{s=0.80} \](vii) Two precautions
- I ensured the object was completely immersed in the liquid and did not touch the sides or bottom of the beaker.
- I read the spring balance at eye level with the object at rest to avoid parallax error, and I noted and corrected the zero error of the balance.
(b)(i) Reading of the spring balance for the brass
Weight of the brass in air:
\[ W=mg=20\times10^{-3}\times10=0.2\ \text{N} \]Volume of the brass:
\[ V=\frac{m}{\rho_{\text{brass}}}=\frac{20\times10^{-3}}{8.0\times10^{3}}=2.5\times10^{-6}\ \text{m}^3 \]Upthrust = weight of kerosine displaced:
\[ U=\rho_{\text{k}}\,V\,g=8.0\times10^{2}\times2.5\times10^{-6}\times10=0.02\ \text{N} \]Reading on the spring balance = tension in the spring = weight in air \(-\) upthrust:
\[ =0.2-0.02=\boxed{0.18\ \text{N}} \](b)(ii) Statements
Archimedes' principle: When a body is wholly or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces.
Law of flotation: A floating body displaces its own weight of the fluid in which it floats.
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