Chapter 9, Problem 69QAP

To determine

(a)

The volume of marble shrinks in the cylinder.

Answer to Problem 69QAP

The volume of marble shrinks in the cylinder is $0.133\times {10}^{-8}{m}^3$.

Explanation of Solution

Given:

Radius of glass marble $r=0.500\times {10}^{-2}m$

Level of mercury in cylinder $h=20.0\times {10}^{-2}m$

Atmospheric Pressure at the surface $P_0=1.0\times {10}^{5}N/m^2$

Acting Pressure at the bottom (volume stress) $P=\frac{F}{A}=2.7\times {10}^{4}N/m^2+1.0\times {10}^{5}N/m^2=1.27\times {10}^{5}N/m^2$

Bulk modulus of glass $B=50\times {10}^{6}N/m^2$

Let $\Delta V$ be the shrink volume

Original volume of glass marble $V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(0.500\times {10}^{-2}m\right)}^3=0.524\times {10}^{-6}{m}^3$

Formula used:

Bulk Modulus $B=\frac{volumestress}{volumestrain}=\frac{F/A}{\Delta V/V}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}B=\frac{F/A}{\Delta V/V}\\ or,\Delta V=\frac{F/A}{B/V}\end{array}$

  $\begin{array}{l}or,\Delta V=\frac{1.27\times {10}^{5}N/m^2}{\left(50\times {10}^{6}N/m^2\right)}\times \left(0.524\times {10}^{-6}{m}^3\right)\\ or,\Delta V=0.133\times {10}^{-8}{m}^3\end{array}$

Hence, the volume of marble shrinks in the cylinder is $0.133\times {10}^{-8}{m}^3$.

Conclusion:

Thus, the volume of marble shrinks in the cylinder is $0.133\times {10}^{-8}{m}^3$.

To determine

(b)

The change in the radius of glass marble

Answer to Problem 69QAP

The change in the radius of glass marble is $0.68mm$.

Explanation of Solution

Given:

The volume of marble shrinks in the cylinder is $\Delta V=0.133\times {10}^{-8}{m}^3$.

Let $\Delta r$be the change in the radius of glass marble.

Formula used:

  $shrinkvolumeofglass-marble\left(\Delta V\right)=\frac{4}{3}\pi {\left(\Delta r\right)}^3$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}\left(\Delta V\right)=\frac{4}{3}\pi {\left(\Delta r\right)}^3\\ or,\Delta r=\sqrt[3]{\frac{3\left(\Delta V\right)}{4\times \pi }}\end{array}$

  $\begin{array}{l}or,\Delta r=\sqrt[3]{\frac{3\times \left(0.133\times {10}^{-8}{m}^3\right)}{4\times \pi }}\\ or,\Delta r=\sqrt[3]{0.32\times {10}^{-9}m}\\ or,\Delta r=0.68\times {10}^{-3}m=0.68mm\end{array}$

Hence, the change in the radius of glass marble is $0.68mm$.

Conclusion:

Thus, the change in the radius of glass marble is $0.68mm$.