Physics
WAEC 2009
A body of volume 0.046m
3 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]
-
A.
11.27N
-
B.
33.81N
-
C.
112.70N
-
D.
338.10N
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
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