- Using the spring balance provided, determine the weight of object of mass M = 5.0g. Record this weight as W\(_{1}\).
- Determine the weight of the object when completely immersed in water contained in a beaker as shown in the diagram. Record the weight as W\(_{2}\)
- Determine the weight of the object when it is completely immersed in the liquid labelled ‘L’. Record the weight as W\(_{3}\). Evaluate, u = (W\(_{1}\) -W\(_{3}\)) and v = (W\(_{1}\) – W\(_{3}\)).
- Repeat the procedure with the objects of masses M = 10, 15, 20, and 25g. In each case, evaluate u = (W\(_{1}\) – W\(_{3}\)) and v = (W\(_{1}\) – W\(_{3}\)) on the vertical axis against u = (W\(_{1}\) – W\(_{2}\) on the horizontal axis.
- Determine the slope, s, of the graph. (vii) State two precautions taken to ensure accurate results.
(b)i. A piece of brass of mass 20.0g is hung on a spring balance from a rigid support and completely immersed in kerosine of density 8.0 x 10\(^{2}\)kg m\(^{-3}\). Determine the reading on the spring balance. [g= 10ms\(^{-2}\)], density of brass = 8.0 x 10\(^{3}\) kg m\(^{-3}\) J
ii. State Archimede’s principle and the law of floatation.
Explanation
| 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 | 4.20 | 4.70 | 0.80 | 0.80 |
| 10.0 | 10.00 | 7.90 | 8.60 | 2.10 | 1.40 |
| 15.0 | 15.00 | 12.75 | 13.50 | 2.25 | 1.50 |
| 20.0 | 20.00 | 17.10 | 16.40 | 2.90 | 3.60 |
| 25.0 | 25.00 | 22.10 | 21.40 | 2.90 | 3.60 |
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2.35-1.00}{3.25-1.65} = \frac{1.35}{1.60}\) = 0.84
Precautions:
- I avoided parallax error in reading spring balance.
- I noted zero error of spring balance.
- l avoided splashing of liquid.
- I avoided object touching the bottom or side beaker.
- I cleaned the mass before dipping it into the liquid.
(bi) Weight of kerosine displaced = volume of object x density of kerosine x g
\(\frac{\text {Mass of object}}{\text {density of of object}}\) x density of kerosine x g
= \(\frac{20 \times 10^{-3}kg \times 8.0 \times 10^{2} kgm^{-3} \times 10ms^{-2}}{ 8.0 \times 10^{-3} kgm^{-3}}\)
= 0.02N = upthrust of the weight of object in air = mass x g
= 20 x 10\(^{-3}\) X 10 = 0.2N
Tension in the spring = weight of an object in the air - upthrust
= 0.2N - 0.02N=0.18N
The reading on the spring balance= 0.18
ii. Archimede's principle states that when a body is wholly or partially immersed in a liquid, it experiences an upthrust equal to the weight of the liquid displaced. On the other hand, the law of floatation states that a floating body displaces its own weight of the liquid in which it floats.