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 ...

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.