Chapter 10, Problem 32P

To determine

To Find: The SG of the wood on the basis of given information.

Answer to Problem 32P

The value specific gravity for the wood approximately is $0.88$ .

Explanation of Solution

Given:

Mass of the wood, $\mathrm{m}=0.48\mathrm{kg}$

For alcohol, $\mathrm{S}\mathrm{G}=0.79$

Apparent mass, ${\mathrm{m}}_{\mathrm{a}\mathrm{p}\mathrm{p}}=0.047\mathrm{kg}$

Formula used:

When an object is submerged into the fluid, then a force is provided by the fluid to the object, this force is known as Buoyancy force.

This force is written as,

  ${\mathrm{F}}_{\mathrm{B}}={\mathrm{\rho }}_{\mathrm{F}}\mathrm{V}\mathrm{g}$

Where, ${\mathrm{\rho }}_{\mathrm{F}}$ is the fluid density, ${\mathrm{F}}_{\mathrm{B}}$ is the buoyancy force, $\mathrm{V}$ is the volume of the fluid and $\mathrm{g}$ is the acceleration due to gravity.

And when the buoyancy force is equal to the weight force then

  ${\mathrm{F}}_{\mathrm{B}}={\mathrm{F}}_{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}$

Calculation:

The apparent weight of wood in alcohol,

  ${\mathrm{W}}_{\mathrm{a}}={\mathrm{m}}_{\mathrm{a}}\mathrm{g}=\mathrm{m}\mathrm{g}-{\mathrm{F}}_{\mathrm{b}}\mathrm{....}\left(1\right)$

Here, ${\mathrm{m}}_{\mathrm{a}}$ is the apparent mass,

  $\mathrm{m}$ is the mass of wood,

  ${\mathrm{F}}_{\mathrm{b}}$ is the buoyant force.

  ${\mathrm{W}}_{\mathrm{a}}={\mathrm{m}}_{\mathrm{a}}\mathrm{g}=\mathrm{m}\mathrm{g}-{\mathrm{F}}_{\mathrm{b}}\mathrm{....}\left(1\right)$

The buoyant force is given as,

  ${\mathrm{F}}_{\mathrm{b}}=\mathrm{m}\mathrm{g}-{\mathrm{m}}_{\mathrm{a}}\mathrm{g}\mathrm{....}\left(2\right)$

According to Archimedes principle,

  ${\mathrm{F}}_{\mathrm{b}}={\mathrm{\rho }}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{d}\mathrm{f}}\mathrm{g}\mathrm{.....}\left(3\right)$

Here, ${\mathrm{V}}_{\mathrm{d}\mathrm{f}}$ is the displaced volume.

By using the equation 2 and 3, then the result is

  $\begin{array}{l}{\mathrm{\rho }}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{d}\mathrm{f}}\mathrm{g}=\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)\mathrm{g}\\ {\mathrm{V}}_{\mathrm{d}\mathrm{f}}=\frac{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}{{\mathrm{\rho }}_{\mathrm{a}}}\end{array}$

According o the definition of the specific gravity is the density of the substance divided by the water’s density. So,

  $\begin{array}{l}\mathrm{S}{\mathrm{G}}_{\mathrm{a}}=\frac{{\mathrm{\rho }}_{\mathrm{a}}}{{\mathrm{\rho }}_{\mathrm{w}}}\\ {\mathrm{\rho }}_{\mathrm{a}}=\mathrm{S}{\mathrm{G}}_{\mathrm{a}}{\mathrm{\rho }}_{\mathrm{w}}\end{array}$

So, use this value in the above equation,

  ${\mathrm{V}}_{\mathrm{d}\mathrm{f}}=\frac{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}{\mathrm{S}{\mathrm{G}}_{\mathrm{a}}{\mathrm{\rho }}_{\mathrm{w}}}$ -

Now, since wood’s block sink in alcohol then it totally immersed in liquid, so that the volume of displaced fluid must be equal to the wood block’s volume. This follows that,

  ${\mathrm{V}}_{\mathrm{d}\mathrm{f}}={\mathrm{V}}_{\mathrm{w}}$

And,

  $\mathrm{M}={\mathrm{V}}_{\mathrm{W}}\times {\mathrm{\rho }}_{\mathrm{b}\mathrm{w}}={\mathrm{V}}_{\mathrm{d}\mathrm{f}}\times {\mathrm{\rho }}_{\mathrm{b}\mathrm{w}}$

Substitute the values of ${\mathrm{V}}_{\mathrm{W}}=\frac{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}{\mathrm{S}{\mathrm{G}}_{\mathrm{a}}}$

  ${\mathrm{V}}_{\mathrm{d}\mathrm{f}}\times {\mathrm{\rho }}_{\mathrm{b}\mathrm{w}}={\mathrm{\rho }}_{\mathrm{b}\mathrm{w}}\times \frac{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}{\mathrm{S}{\mathrm{G}}_{\mathrm{a}}}$

Thus,

  $\begin{array}{l}\mathrm{m}={\mathrm{\rho }}_{\mathrm{b}\mathrm{w}}\times \frac{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}{\mathrm{S}{\mathrm{G}}_{\mathrm{a}}{\mathrm{\rho }}_{\mathrm{w}}}\\ \mathrm{m}=\mathrm{S}{\mathrm{G}}_{\mathrm{w}}\times \frac{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}{\mathrm{S}{\mathrm{G}}_{\mathrm{a}}}\left(\because \mathrm{S}{\mathrm{G}}_{\mathrm{w}}=\frac{{\mathrm{\rho }}_{\mathrm{b}\mathrm{w}}}{{\mathrm{\rho }}_{\mathrm{w}}}\right)\end{array}$

So, rearrange the above equation, then the outcomes is

  $\mathrm{S}{\mathrm{G}}_{\mathrm{w}}=\frac{\mathrm{m}\times \mathrm{S}{\mathrm{G}}_{\mathrm{a}}}{\left(\mathrm{m}-{\mathrm{m}}_{\mathrm{a}}\right)}$

  $\mathrm{S}{\mathrm{G}}_{\mathrm{w}}=\frac{\mathrm{M}\times \mathrm{S}{\mathrm{G}}_{\mathrm{a}}}{\left(\mathrm{M}-{\mathrm{M}}_{\mathrm{a}}\right)}=\frac{0.48\mathrm{kg}\times 0.79}{\left(0.48-0.047\right)\mathrm{kg}}=\frac{0.3792}{0.433}=0.8757=0.88$ Approximately.

Conclusion:

The specific gravity of the wood is $0.88$ .