A body of volume 0.046m³ is immersed in a liquid of density 980kgm‾³3 with 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

C. 112.70N

Explanation

m = D x V

= 980 x 0.046=45.08×10=450.80N

then ,3/4 of it vol immersed in water ; 3/4×0.046×980=33.81×10=338.10N

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