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 kgWhen 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(ρw−ρb)=m⇒V×0.7ρw=m [∵ρb=0.3ρw] Here, V=(side)3=(0.5)3 m3 ρw=103 kg m−3 Substituting these values in the above relation, we get m=(0.5)3×0.7×103=87.5 kg

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