Question

A wooden cylinder floats vertically in water with one fourth of its length immersed. What will be the density of wood?

• A. It equals to the density of water.
• B. It equals to half the density of water.
• C. It equals to one fourth the density of water.
• D. It equal to three fourth the density of water.

Answer 1

The correct option is C

It equals to one fourth the density of water.

Step 1: Given data

The length of the wooden cylinder immersed in water is one-fourth of the total length.

Step 2: To find the relation between the Density of the wooden cylinder and water.

From the Archimedes principle, we have,

Weight of the body = Weight of the fluid displaced

$$\begin{gathered} M_{wo}\times g=M_w\times g \\ {V}_{wo}\times d_{wo}=V_w\times d_w \end{gathered}$$

Where,$V_{wo}$ is the volume of the wooden cylinder, $d_{wo}$ is the density of the wooden cylinder,$V_w$ is the volume of the water displaced,$d_w$ is the density of the water.

Let r be the radius of the wooden cylinder and L be the length of the cylinder, then

$$\begin{gathered} \pi r^{2}L\times d_{wo}=\pi r^2\frac{L}{4}\times d_w \\ \pi r^{2}L=volumeofwoodencylinder \\ \pi r^2\frac{L}{4}=volumeofwaterdisplaced \end{gathered}$$

$d_{wo}=\frac{d_w}{4}$

Hence, Option C is correct.