Question

A wooden cube just floats inside water with a $400\text{gm}$ mass placed on it. When the mass is removed, the cube floats with its top surface $4\text{cm}$ above the water level. What is the side of the cube?

• A. $6\text{cm}$
• B. $8\text{cm}$
• C. $10\text{cm}$
• D. $12\text{cm}$

Answer 1

The correct option is C $10\text{cm}$

Let $L$ and $d$ be the side of the cube and density of the cube respectively.
Given, weight of block, $W=400\text{gm}$
In first case,
weight of the cube+weight of mass = weight of water displaced
$(L^3\times d\times g)+W=L^3\times \rho \times g$
$(L^3\times d\times g)+400g=L^3\times 1\times g$
$L^3\times d=L^3-400$........(1)
$\because$ Density of water$,\rho =1{\text{g/cm}}^3$
In second case, $(L-4)$ part of cube remains in water
weight of the cube= weight of water displaced
$(L^3\times d\times g)=L^2\times (L-4)\times \rho \times g$
$L^3\times d=L^2\times (L-4)$.......(2)
From (1) and (2) equations,
$L^3-400=L^2\times (L-4)$
$-400=-4L^2$
$L=10\text{cm}$