The 4/5th part of volume of a piece of wood is swiming as
well as sunk in a fluid. if the relative dencity of wood is
0.8 then what will be the dencity of fluid in KG/M3.
PLEASE ANSWER WITH SOLUTION.
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Answer / mechatronics89
let p1 = density of wood and p2 be dat of fluid.
applying force balance
p1*V*g= p2*(4/5)V*g (gravity = buoyant force)
800=.2*p2
p2=4000 kg/m^3
Is This Answer Correct ? | 3 Yes | 2 No |
Answer / arul - mz
Kindly be clear in the words used in the question...
Taking the question in the following way I solve it...
The 4/5th part of volume of a piece of wood foating is sunk
in a fluid. if the relative dencity of wood is
0.8 then what will be the dencity of fluid in KG/M3.
PLEASE ANSWER WITH SOLUTION.
Assume wolume of wood = 1000 cu.m
As per the data avl...
qty immersed = 800 cu.m (4/5 th sunk)
by Archimedes principle it must have displaced 800 cu.m of
liquid.
RD=0.8 implies that
density = 800 kg/cu.m
so the mass of wood = vol of wood floating*density
= 1000*800
Den of liq. = mass of liq displaced / vol of liq
As per archimedes second law for floating bodies,
mass of liq displaced = mass of floating object
so, Den of liq = 1000*800/800=1000 kg/sq.cm
Is This Answer Correct ? | 1 Yes | 1 No |
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