A body of volume 0.046m3 is immersed in a liquid of density 980kgm-3 with \(\frac{3}{4}\) of its volume submerged. Calculate the upthrust on the body. [g = ...
Question 1 Report
A body of volume 0.046m3 is immersed in a liquid of density 980kgm-3 with \(\frac{3}{4}\) of its volume submerged. Calculate the upthrust on the body. [g = 10ms-2]
Answer Details
When an object is immersed in a fluid, it experiences an upward force called buoyant force or upthrust. This force is equal to the weight of the fluid displaced by the object. In this case, the volume of the object submerged in the fluid is \(\frac{3}{4}\) of its total volume. So, the volume of fluid displaced by the object is \(\frac{3}{4}\) x 0.046m3 = 0.0345m3 The weight of this displaced fluid can be calculated as: mass = density x volume = 980kgm-3 x 0.0345m3 = 33.81kg weight = mass x g = 33.81kg x 10ms-2 = 338.1N Therefore, the upthrust on the body is 338.1N. So, the correct answer is (C) 338.10N.
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