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Question

A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water?
[Take density of water =103 kg/m3]

A
30.1 kg
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B
46.3 kg
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C
87.5 kg
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D
65.4 kg
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Solution

The correct option is C 87.5 kg
When only block is floating on water, 30% of its volume is in water as shown below



By Archimede's principle,
weight of block = weight of displaced water
Vρbg=(30%V)ρwg
where, ρb= density of block,
ρw= density of water
and V= volume of block.
ρb=0.3ρw
Now, let a mass m is placed over block to just keep the cube fully immersed in water.
Now, by Archimedes principle,
weight of water displaced = weight of block + weight of mass m


Vρwg=Vρbg+mg
V(ρwρb)=mV×0.7ρw=m
[ρb=0.3ρw]
Here, V=(side)3=(0.5)3 m3
ρw=103 kg m3
Substituting these values in the above relation, we get
m=(0.5)3×0.7×103=87.5 kg

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