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Question

A cubical body floats on mercury with 4 % of its volume below the mercury surface. What fraction of the volume of the body will be immersed in the mercury if a layer of water poured on top of the mercury covers the body completely?

[Assume, ρm=13.6 g/cm3 ; ρw=1 g/cm3 ]

A
0.25
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B
0.35
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C
0.45
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D
0.50
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Solution

The correct option is B 0.35
Case A: When cubical body is only immersed in mercury.
Let the weight of the body be W.
Weight of the body = upthrust provided by mercury
W=0.4Vρmg ........(1)
Here, V= Total volume of the cubical body and ρm= Density of mercury.



Case B: When water is filled on the mercury surface.
Let x be the volume of block extra immersed in mercury.
Then, in equilibrium,
W=xρmg+(Vx)ρwg
Using (1) in the above equation we get ,
0.4Vρmg=xρmg+(Vx)ρwg
x=(0.4ρmρw)Vρmρw
x=(0.4×13.61)V13.61=0.35V
So, the fraction of volume of the cubical body immersed in the mercury will be xV=0.35
Thus, option (b) is the correct answer.

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