Question

A cube of wood supporting 200gm mass just floats in water (ρ=1g/cc). When the mass is removed, the cube is raised by 2cm. The size of the cube is _____.

Answer 1

Hint: To solve this question, we need to use the basic theory related to the density of the material. First in initial condition we consider the sum of Weight of the cube and weight of mass placed above (200 gm) is equal to the weight of the water displaced. And then after the mass is removed, Weight of the block is equal to the upthrust. Now, using these two equations which we get from above mentioned two conditions and solve these equations.

Complete step-by-step solution -
Let ρ be the density of the cube and l be the side of a cube.
Now, in initial condition
Weight of the cube + weight of mass placed above (200 gm) = weight of the water displaced
${l^3}{\rho _{wood}}g + 200g = {l^3}{\rho _w}g$
${l^3}{\rho _{wood}}$ $= {l^3}{\rho _w} - 200$……. .(I)
When the mass is removed
Then,
${l^3}{\rho _{wood}}g = {l^2}(l - 2)g$……. (II)
Weight of the block = upthrust
Now, from equation (I) and (II)
${l^3}{\rho _w} - 200 = {l^2}(l - 2)$
We know that,
${\rho _w} = 1$
So,
${l^3} - 200 = {l^3} - 2{l^2}$
${l^2} = 100$
$l = 10cm$
Hence the sides of the cube are 10 cm.

Note: As we know density of any material is defined as the ratio of mass of the substance to the volume of the material. That mean density of any material depends upon these two parameters i.e. mass and volume of the material.
${\text{density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}}$