## Related questions with answers

A trash compactor can reduce the volume of its contents to 0.350 their original value. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased?

Solution

VerifiedHere a trash compactor reduces the volume of its contents to 0.350 their original value.

Let the original volume of the rubbish is $V_{\text{i}}$, then its volume after compression is

$\begin{align*} V_{\text{f}} = 0.350\,V_{\text{i}} \end{align*}$

Now as we know the density is the mass per unit volume of a substance or object. In equation form, density is defined as

$\begin{align*} \rho = \frac{m}{V}\tag{1} \end{align*}$

Where $m$ is the mass and $V$ is the volume of the substance or the object.

Hence for a constant mass

$\begin{align*} \rho \propto \frac{1}{V} \end{align*}$

Therefore the original density of the rubbish is

$\begin{align*} \rho_{\text{i}} \propto \frac{1}{V_{\text{i}}}\tag{2} \end{align*}$

And the final density of the rubbish is

$\begin{align*} \rho_{\text{f}} \propto \frac{1}{V_{\text{f}}}\tag{3} \end{align*}$

So from equations (2) and (3), we get

$\begin{align*} \frac{\rho_{\text{f}}}{\rho_{\text{i}}} & = \frac{V_{\text{i}}}{V_{\text{f}}}\\ & = \frac{V_{\text{i}}}{0.350\,V_{\text{i}}}\\ & = 2.86 \end{align*}$

Therefore, the density of the rubbish is

$\begin{align*} \rho_{\text{f}} = 2.86\,\rho_{\text{i}} \end{align*}$

Therefore the factor by which the density is increased

$\begin{align*} F = 2.86 \end{align*}$

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