Determination of the density of a liquid using a weighted test tube

• $V$: volume of the portion in the tube filled with wax
• $M$: mass of the tube a with its contents
• $A$: external area of the cross section of the cylindrical portion of the tube
• $m$: mass of additional weights inserted into the tube
• $\rho$: density of the liquid
• $l$: length of the cylindrical portion (above wax level) immersed in the liquid when the tube floats in a liquid

According to the principle of floatation:

$$(M+m)g=(V+A~l)\rho~g$$ $$l=(\frac{1}{A~\rho})m+\frac{1}{A}(\frac{M}{\rho}-V)$$

If $G$ is the gradient of the graph of $l$ against $m$,

$$G=\frac{1}{A~\rho}$$

If $d$ is the external diameter of the tube,

$$A=\frac{\pi~d^{2}}{4}$$ $$\rho=\frac{4}{\pi~d^{2}G}$$

1. Put the minimum amount of lead shots that would make the boiling tube float vertically in the liquid.
2. Pour wax into the tube until the lead shots are completely covered (the spherical portion of the tube should be filled with wax).
3. Paste the paper strip on the inner wall so that the zero of the scale begins from the lower end of the cylindrical portion (Figure 9.1).
4. Fill the tall jar with the liquid (sodium chloride), make the tube float vertically in it and record the length $l$ of the tube immersed in liquid.
5. Insert one mass unit ($m$) into the tube and record the corresponding $l$ value. Increase $m$ and obtain about six values.
6. Measure the external diameter of the tube using a vernier calliper at three different places of the tube taking three readings normal to each other at each place.