An object of volume 400cm3 and 2.5gcm-3 is suspended from a spring balance with half its volume immersed in water. Determine the reading on the spring balan...
Question 1 Report
An object of volume 400cm3 and 2.5gcm-3 is suspended from a spring balance with half its volume immersed in water. Determine the reading on the spring balance. (Density of water = 1gcm-3)
Answer Details
The reading on the spring balance is 800g. To understand why, we need to consider the principle of buoyancy, which states that an object partially or completely submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by the object. In this case, the object has a volume of 400cm3 and a density of 2.5gcm-3, which means its weight is: 400cm3 x 2.5gcm-3 = 1000g When the object is partially submerged in water, it displaces a volume of water equal to half its own volume, or 200cm3. Since the density of water is 1gcm-3, the weight of this displaced water is: 200cm3 x 1gcm-3 = 200g According to the principle of buoyancy, the object experiences an upward force equal to the weight of the displaced water, which is 200g. Therefore, the reading on the spring balance will be the weight of the object minus the upward buoyant force: 1000g - 200g = 800g So the reading on the spring balance is 800g.
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