Question

A piece of wood of mass $6\text{kg}$ is floating in water with its $\frac{1}{3}$rd part inside the water. On this floating piece of wood, what maximum mass can be put such that the whole of the piece of wood gets drowned in the water?

• A. $15\text{kg}$
• B. $14\text{kg}$
• C. $10\text{kg}$
• D. $12\text{kg}$

Answer 1

The correct option is D $12\text{kg}$

Weight of the submerged part of the block,
$W=\frac{1}{3}V{\rho }_{w}g$ ..........$\left(i\right)$
where ${\rho }_w$ is density of water.
Now,
Excess weight($W^{\prime }$) = weight of water having $\frac{2}{3}$rd volume of the block.
$W^{\prime }=\frac{2}{3}V{\rho }_{w}g$ ........$\left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$, $\frac{W^{\prime }}{W}=\frac{\frac{2}{3}}{\frac{1}{3}}$
$\therefore W^{\prime }=2W\Rightarrow W^{\prime }=2\times 6=12\text{kg}$