Question

A wooden ball of density $0.8g/c{m}^3$ is placed in water.
the ratio of the volume above the water surface to that below the water surface is.

• A. 0.25
• B. 0.33
• C. 2
• D. 4

Answer 1

The correct option is B 0.33

Let the volume of the block of wood be $Vc{m}^3$ and its density be $d_{w}gc{m}^{-3}$

So the weight of the block$=V{d}_{w}g$ dyne, where g is the acceleration due to gravity$=980cm{s}^{-2}$

The block floats in liquid of density $0.8gc{m}^{-3}$ with $\frac{1}{4}$th of its volume

submerged. So the upward buoyant force acting on the block is the weight of displaced liquid $=\frac{1}{4}V\times 0.8\times$g dyne.

Hence by condition of floatation

$V\times d_w\times g=\frac{1}{4}\times V\times 0.8\times g$

$\Rightarrow d_w=0.2gc{m}^{-3}$

Now let the density of oil be $d_{o}gc{m}^{-3}$

The block floats in oil with $60\%$ of its volume submerged. So the buoyant force balancing the weight of the block is the weight of displaced oil$=60\%\times V\times d_o\times g$ dyne

Now applying the condition of floatation we get

$60\%\times V\times d_o\times g=V\times d_w\times g$

$\Rightarrow \frac{60}{100}\times V\times d_o\times g=V\times 0.2\times g$

$\Rightarrow d_o=0.2\times \frac{10}{6}=\frac{1}{3}=0.33gc{m}^{-3}$.