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Experiment: to demonstrate the law of floatation with a worked example

Answer Text:
Experiment: to demonstrate the law of floatation:
Procedure
1. Weigh the block in air and record its weight as W1.
2. Put water into the overflow can (eureka can) up to the level of the spout.
3. Collect displaced water in a beaker. Record the weight of the beaker first in air and record as W2. Weigh both the beaker and the displaced water and record as W3.
4. Record the same procedure with kerosene and record your results as shown below.

5. What do you notice between W1 and W3 – W2

Discussion
The weight of the displaced liquid is equal to the weight of the block in air. This is consistent with the law of floatation which states that “A
body displaces its own weight of the liquid in which it floats”. Mathematically, the following relation can be deduced
Weight = volume × density × gravity = v × ρ × g, therefore
W = v d × ρ × g where vd is the volume of displaced liquid.
NOTE – Floatation is a special case of Archimedes principle. This is because a floating body sinks until the upthrust equals the weight of the body.
Example
A wooden block of dimensions 3 cm × 3 cm × 4 cm floats vertically in methylated spirit with 4 cm of its length in the spirit. Calculate the weight of
the block. (Density of methylated spirit = 8.0 ×# 102 kgm^-3#).
Solution
Volume of the spirit displaced = (3 × 3 × 4) = 36 #cm^3# = #3.6 × 10^-5 m^3#
Weight of the block =v d × ρ × g = (3.6 × 10-5) × 8.0 × 102 × 10 = #2.88 × 10^-1 N#.