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]
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Correct Answer: Option D
Explanation:
M = D x V
= 980 x 0.046 x \(\frac{3}{4}\) = 33.81kg
upthrust = mg = 33.81 x 10
= 338.1N
M = D x V
= 980 x 0.046 x \(\frac{3}{4}\) = 33.81kg
upthrust = mg = 33.81 x 10
= 338.1N