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ICSE Class 9 Answered

Please answer the 26th question.
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Asked by nishabr7 | 11 Sep, 2018, 08:22: PM
answered-by-expert Expert Answer
Take a solid(say, a metallic piece). Suspend it by a thin thread from the hook of a spring balance as shown in figure.
Note its weight. Now take a eureka can and fill it with water up to its spout.
Arragane a measuring cylinder below the spout of the eureka can as shown in figure.
Now immerse the solid gently into water of the eureka can. The water displaced by it gets collected in the measuring cylinder.
When water stops dripping through the spout, note the weight of the solid and the volume collected in the measuring cylinder.
Since density of water is 1g/cm3, the volume of water collected in the measuring cylinder is numerically equal to the mass (in g)
or weight (in gf) of water displaced by the solid when it is completely immersed in water.
 
Let us assume the solid weighs 300 gf in air and 200 gf when it is completely immersed in water.
Let the volume of water collected in the measuring cylinder is 100 cm3.
 
Hence loss in weight = 300 - 200 = 100 gf
volume of water displaced = volume of solid = 100 cm3.
Since density of water = 1 g/cm3
 
weight of water displaced = 100 gf = upthrust or loss in weight.
 
It is found that the weight of water displaced by solid is equal to the difference in weight of the solid in air
and in water( i.e., loss in weight of the solid). This verifies the Archimedes principle
Answered by Thiyagarajan K | 12 Sep, 2018, 04:27: PM

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