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) Evaluated \( A = 0.25\pi D^2 \), 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 \( 0.1\ \mathrm{m}^2 \). Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod =\(720\ \mathrm{kg\ m}^{-3}\), g = \(10\ \mathrm{m\ s}^{-2}\)].
Answer Details
(a) Observations, table and graph
The external diameter of the boiling tube is
\[D=1.20\ \text{cm}\]
Hence, the cross-sectional area is
\[A=\frac{\pi D^2}{4}=\frac{3.14(1.20)^2}{4}=1.13\ \text{cm}^2.\]
The initial depth of immersion of the loaded boiling tube is
\[y_0=3.0\ \text{cm}.\]
| Mass added, \(m\) (g) | Depth of immersion, \(y\) (cm) | \(h=y-y_0\) (cm) | \(\log h\) | \(\log m\) |
|---|---|---|---|---|
| 2 | 4.8 | 1.8 | 0.255 | 0.301 |
| 5 | 7.4 | 4.4 | 0.643 | 0.699 |
| 7 | 9.2 | 6.2 | 0.792 | 0.845 |
| 10 | 11.9 | 8.9 | 0.949 | 1.000 |
| 12 | 13.6 | 10.6 | 1.025 | 1.079 |
The graph of \(\log m\) against \(\log h\) is shown below.
From the line of best fit,
\[\text{gradient}=\frac{\Delta(\log m)}{\Delta(\log h)}\approx1.01\approx1.\]
Thus, \(m\propto h\). This agrees with the flotation relation \(m=\rho Ah\), where \(\rho\) and \(A\) are constant.
(b)(i) Law of flotation
A floating body displaces a weight of fluid equal to its own weight.
(b)(ii) Depth of immersion of the rod
Density of liquid:
\[\rho_l=1.26\times1000=1260\ \text{kg m}^{-3}.\]
For equilibrium of the floating rod,
\[\rho_rALg=\rho_lAyg,\]
where \(y\) is the immersed length. Therefore,
\[y=\frac{\rho_rL}{\rho_l}=\frac{720\times0.63}{1260}=0.36\ \text{m}.\]
Depth of immersion = \(0.36\ \text{m}\).
Download The App On Google Playstore
Everything you need to excel in your exams
