Question

A wooden cylinder is floating in water. The length of the cylinder is 40 cm and diameter is 14 cm. The density of wood is 0.38 gm/cc. At the equilibrium position the cylinder makes 30 degree with the surface of water. Find the length of cylinder submerged in water (both the longer and shorter side - when you imagine a 2D picture of the submerged cylinder, there is a shorter length and a longer length submerged in water).

Answer 1

As the cylinder is in equilibrium, Volume of the cylinder inside the water is given by (Using Archemdes Principle)

$\rho _{c}Vg - \rho _{w}V_{inside}g = 0$

$0.38g/cc V = 1g /ccV_{inside}$

$V_{inside} = 0.38 V$ Where V is the volume of the cylinder.

$V_{inside} = 0.38 \times \left ( \pi r^{2}h \right ) = 0.38\times \left ( \frac{22}{7}\times 7^{2}\times 40 \right ) cm^{3}= 2340.8 cm^{3}$

From the diagram Now Volume of submerged part is given by

$V_{inside} = \frac{1}{2}\frac{22}{7}\times 7^{2}\times (h_{2}+h_{1})= 2340.8 cm^{3}$

$h_{2}+h_{1}=30.4 cm$.............................(1)

From the geometry

$h_{2}-h_{1}=d\tan \theta = (14 cm )\times \tan 30^{0}= 8.08 cm$

$h_{2}-h_{1}= 8.08 cm$.........................(2)

Adding eq1 and eq2 we get

$2h_{2}= 30.4 cm + 8.08 cm= 38.48 cm$

$h_{2}= 19.24 cm$

$h_{1}=30.4 cm-h_{2} = 30.4 cm - 19.24 cm = 11.16 cm$