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]

Nedextra

9 years ago

m = D x V

= 980 x 0.046=45.08×10=450.80N

then ,3/4 of it vol immersed in water ; 3/4×0.046x980=33.81×10=338.10N



so,the upthrust= weight in Air- weight in water = 450.80N-338.10N=112.70N

so the correct answer is option C

Nedextra

9 years ago

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