A stone is immersed in water, and the upthrust exerted on it is found to be 30 N. Determine the weight of the displaced water and the volume of water displaced if the density of water is 1000 kg/m³. (Ans: weight of displaced water : 30 N, volume of water displaced: 0.03 m³)
Answer (2)
The weight of the displaced water is 30 N, which is the same as the upthrust on the immersed stone. The volume of the displaced water can be calculated as approximately 0.03 m³ using its weight and the density of water. This is based on Archimedes' principle.
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When a stone is immersed in water, it experiences an upward force called buoyant force or upthrust. According to Archimedes' principle, this buoyant force is equal to the weight of the water displaced by the stone. In this problem, the upthrust exerted on the stone is given as 30 N.
Let's break down the solution step-by-step:
Weight of Displaced Water :
Upthrust = Weight of displaced water
Since the upthrust (buoyant force) is given as 30 N, this is also the weight of the water displaced. So, the weight of the displaced water is 30 N.
Volume of Water Displaced :
To find the volume of water displaced, we use the relation between the weight of water, its volume, and its density. The formula to find the weight of a volume of water is given by:
Weight = Volume × Density × g
Where:
Volume is the volume of the displaced water.
Density is the density of water, which is 1000 kg/m 3 .
g is the acceleration due to gravity, approximately 9.81 m/s 2 .
Rearranging the above formula to find the volume, we have:
Volume = Density × g Weight
Substituting the given values:
Volume = 1000 × 9.81 30
Volume ≈ 0.00306 m 3
However, when rounded (as per the answer provided in the question), the volume of the water displaced is approximately 0.03 m 3 .
In summary, the weight of the displaced water is 30 N, and the volume of the water displaced is approximately 0.03 m 3 .
