Question

A cylinder of wood floats vertically in water with one-fourth of its length out of water. The density of wood is :

• A. $0.5\mathrm{g}{\mathrm{cm}}^{-3}$
• B. $0.5\mathrm{g}{\mathrm{cm}}^3$
• C. $0.75\mathrm{g}{\mathrm{cm}}^{-3}$
• D. $1\mathrm{g}{\mathrm{cm}}^{-3}$

Answer 1

The correct option is C

$0.75\mathrm{g}{\mathrm{cm}}^{-3}$

Step 1: Given Data

Firstly we need to find the Immersed Volume of the body.

$\mathrm{Total}\mathrm{Volume}=\mathrm{V}$

$\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{outside}\mathrm{the}\mathrm{Water}=\frac{\mathrm{V}}{4}$

$\mathrm{Immersed}\mathrm{Volume}=\mathrm{Total}\mathrm{Volume}-\mathrm{Volume}\mathrm{outside}$

${\mathrm{V}}_2=\mathrm{V}-{\mathrm{V}}_1$

$$\begin{gathered} {\text{V}}_2=\mathrm{Immersed}\mathrm{Volume} \\ {\mathrm{V}}_1=\mathrm{Volume}\mathrm{outside} \end{gathered}$$

Step 2: Substituting the values,

${\mathrm{V}}_2=\mathrm{V}-\mathrm{V}/4$

${\mathrm{V}}_2=4\mathrm{V}-\mathrm{V}/4$

${\mathrm{V}}_2=3\mathrm{V}/4$

As this is a case where,

Density of Wood < Density of Water.

Step 3: Finding ratio of immersed volume and total volume

We need to Apply,

$\frac{\sigma }{\rho }=\frac{Immersedvolume}{Totalvolume}$

Where,

$\rho =$Density of water$=1g/c{m}^3$

$\sigma =$Density of wood (cylinder)

$\frac{\sigma }{\rho }=\frac{v_2}{v}$

Step 4: Finding density of wood

Substituting the values

$$\begin{gathered} \frac{\sigma }{\rho }=\frac{{\displaystyle \frac{3v}{4}}}{v} \\ \frac{\sigma }{\rho }=\frac{3v}{4v} \\ \frac{\sigma }{\rho }=\frac{3}{4} \\ \rho =1g/c{m}^3 \\ \frac{\sigma }{1}=\frac{3}{4} \\ \sigma =\frac{3}{4}\times 1 \\ \sigma =\frac{3}{4} \\ \sigma =0.75g/m^3 \end{gathered}$$

Hence, Density of Wood (Cylinder) is $0.75g/cm³.$

Hence, option C is correct.