You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 ...
Question 1 Report
You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.
(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.
(ii) Evaluate \(A = 0.25\pi D^2\), where \(\pi = 3.14\).
(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.
(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside the boiling tube.
(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.
(vi) Evaluate \(h = (y - y_0)\), log h and log m.
(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.
(viii) Tabulate the results.
(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).
(b)(i) State in full the law on which the experiment in (a) is based.
(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of 0.1 m\(^2\). Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod = 720 kg m\(^{-3}\), g = 10 m s\(^{-2}\)].
Answer Details
(a) Results and graph
The external diameter of the boiling tube measured with the slide vernier caliper is:
\[D=2.41\ \text{cm}\]
Hence,
\[A=\frac{\pi D^2}{4}=\frac{3.14(2.41)^2}{4}=4.56\ \text{cm}^2.\]
The initial depth of immersion of the loaded boiling tube is:
\[y_0=3.000\ \text{cm}.\]
| Mass, \(m\) (g) | Depth, \(y\) (cm) | \(h=y-y_0\) (cm) | \(\log h\) | \(\log m\) |
|---|---|---|---|---|
| 2 | 3.439 | 0.439 | −0.358 | 0.301 |
| 5 | 4.097 | 1.097 | 0.040 | 0.699 |
| 7 | 4.535 | 1.535 | 0.186 | 0.845 |
| 10 | 5.193 | 2.193 | 0.341 | 1.000 |
| 12 | 5.632 | 2.632 | 0.420 | 1.079 |
The graph of \(\log m\) against \(\log h\) is shown below.
Using two widely separated points on the best-fit line, \((-0.358,\ 0.301)\) and \((0.420,\ 1.079)\),
\[\text{gradient}=\frac{1.079-0.301}{0.420-(-0.358)}=\frac{0.778}{0.778}=1.00.\]
Thus, \(\log m=\log h+0.659\), showing that \(m\propto h\).
(b)(i)
Law of flotation: A body floating in a fluid displaces a quantity of the fluid whose weight is equal to the weight of the body.
(b)(ii)
Mass of rod:
\[m=\rho V=720(0.1)(0.63)=45.36\ \text{kg}.\]
Weight of rod:
\[W=mg=45.36\times10=453.6\ \text{N}.\]
Density of liquid:
\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]
If \(y\) is the depth of immersion, then the upthrust is
\[U=\rho_l(0.1y)g=1260(0.1y)(10)=1260y.\]
For flotation, upthrust equals weight:
\[1260y=453.6.\]
\[y=\frac{453.6}{1260}=0.36\ \text{m}.\]
Depth of immersion = \(0.36\ \text{m}\).
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