TEST OF PRACTICAL KNOWLEDGE QUESTION Study the diagrams above and use them as guides in carrying out the following instructions. Using the spring balance pr...

Answer Details

Practical: upthrust with a spring balance

For each object of mass M you record its weight in air \( W_1 \), its weight fully immersed in water \( W_2 \), and its weight fully immersed in the liquid L \( W_3 \), then evaluate the upthrusts

\[ U = W_1 - W_2 \qquad V = W_1 - W_3 \]

Graph: plotting V (vertical) against U (horizontal) gives a straight line through the origin, and its slope \( s = \dfrac{V}{U} \) equals the ratio of the density of liquid L to that of water, i.e. the relative density of L.

Two precautions

• Ensure the object is fully immersed without touching the sides or bottom of the beaker, and that no air bubbles cling to it.
• Read the spring balance at eye level to avoid parallax, and allow it to come to rest before reading.

(b)(i) Archimedes' principle

When a body is wholly or partially immersed in a fluid, it experiences an upthrust (upward force) equal to the weight of the fluid it displaces.

(b)(ii) Reading of the spring balance

Mass of brass \( = 20.0\,\text{g} = 0.020\,\text{kg} \); density of brass \( = 8.0 \times 10^{3}\,\text{kg m}^{-3} \); density of kerosene \( = 8.0 \times 10^{2}\,\text{kg m}^{-3} \); \( g = 10\,\text{m s}^{-2} \).

Weight in air:

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

Volume of the brass:

\[ V = \frac{m}{\rho_{brass}} = \frac{0.020}{8.0 \times 10^{3}} = 2.5 \times 10^{-6}\,\text{m}^3 \]

Upthrust from the kerosene:

\[ U = \rho_{k} V g = (8.0 \times 10^{2}) \times (2.5 \times 10^{-6}) \times 10 = 0.020\,\text{N} \]

Spring-balance reading (apparent weight):

\[ W - U = 0.20 - 0.020 = 0.18\,\text{N} \]

The spring balance reads 0.18 N.