Question

A cubical block of side $0.5\text{m}$ 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 $={10}^3{\text{kg/m}}^3$]

• A. $30.1\text{kg}$
• B. $46.3\text{kg}$
• C. $87.5\text{kg}$
• D. $65.4\text{kg}$

Answer 1

The correct option is C $87.5\text{kg}$

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
$\Rightarrow V{\rho }_{b}g=\left(30\%V\right){\rho }_{w}g$
where, ${\rho }_b=$ density of block,
${\rho }_w=$ density of water
and $V=$ volume of block.
$\Rightarrow {\rho }_b=0.3{\rho }_w$
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 $m$


$\Rightarrow V{\rho }_{w}g=V{\rho }_{b}g+mg$
$\Rightarrow V({\rho }_w-{\rho }_b)=m\Rightarrow V\times 0.7{\rho }_w=m$
$[\because {\rho }_b=0.3{\rho }_w]$
Here, $V=(side{)}^3=(0.5{)}^3{\text{m}}^3$
${\rho }_w={10}^3{\text{kg m}}^{-3}$
Substituting these values in the above relation, we get
$m=(0.5{)}^3\times 0.7\times {10}^3=87.5\text{kg}$