Study the diagrams above and use them as guides in carrying out the following instructions. Using the spring balance provided, determine the weight of the o...

Answer Details

Test of practical knowledge: measurement of upthrust with a spring balance

For each object of mass M the spring balance is read three times: the weight in air \(W_1\), the weight when the object is completely immersed in water \(W_2\), and the weight when it is completely immersed in the liquid labelled L, \(W_3\). The two upthrusts are then evaluated from

\[ U = W_1 - W_2 \qquad\text{and}\qquad V = W_1 - W_3 . \]

The apparatus is set up as shown below.

Table of readings (all weights in newtons N; \(g = 10\ \text{m s}^{-2}\)):

M /g	\(W_1\) /N	\(W_2\) /N	\(W_3\) /N	\(U=W_1-W_2\) /N	\(V=W_1-W_3\) /N
50.0	0.50	0.42	0.43	0.08	0.07
100.0	1.00	0.84	0.86	0.16	0.14
150.0	1.50	1.26	1.29	0.24	0.21
200.0	2.00	1.68	1.72	0.32	0.28
250.0	2.50	2.10	2.15	0.40	0.35

Graph of V against U

V (vertical axis) is plotted against U (horizontal axis). The points lie on a straight line passing through the origin.

Slope of the graph

Taking two widely separated points on the line of best fit, \((U_1,\,V_1) = (0.08,\,0.07)\) and \((U_2,\,V_2) = (0.40,\,0.35)\):

\[ s = \frac{V_2 - V_1}{U_2 - U_1} = \frac{0.35 - 0.07}{0.40 - 0.08} = \frac{0.28}{0.32} = 0.88 . \]

The slope \(s = 0.88\) has no unit. Since \(U\) is the upthrust in water and \(V\) is the upthrust in liquid L for the same object, the slope equals the relative density of liquid L, \(s = \dfrac{\rho_L}{\rho_{water}} = 0.88\).

Two precautions

• I read the spring balance pointer with my eye level with the scale to avoid parallax error, and allowed the pointer to come to rest before reading.
• I ensured that the object was completely immersed in the liquid without touching the sides or bottom of the beaker, and that no air bubbles clung to it.

(b)(i) Archimedes' principle

When a body is wholly or partially immersed in a fluid (a liquid or a gas), it experiences an upthrust that is equal to the weight of the fluid displaced by the body.

(b)(ii) Reading of the spring balance

Data: mass of brass \(m = 20.0\ \text{g} = 0.020\ \text{kg}\); density of brass \(\rho_b = 8.0\times10^{3}\ \text{kg m}^{-3}\); density of kerosene \(\rho_k = 8.0\times10^{2}\ \text{kg m}^{-3}\); \(g = 10\ \text{m s}^{-2}\).

Weight of the brass in air:

\[ W = mg = 0.020 \times 10 = 0.20\ \text{N} . \]

Volume of the brass:

\[ V_b = \frac{m}{\rho_b} = \frac{0.020}{8.0\times10^{3}} = 2.5\times10^{-6}\ \text{m}^3 . \]

Upthrust from the kerosene (weight of kerosene displaced):

\[ F_u = \rho_k\, V_b\, g = (8.0\times10^{2})(2.5\times10^{-6})(10) = 0.020\ \text{N} . \]

The spring balance reads the apparent weight, which is the weight in air minus the upthrust:

\[ R = W - F_u = 0.20 - 0.020 = 0.18\ \text{N} . \]

The reading of the spring balance is 0.18 N.