Chapter 9, Problem 52QAP

To determine

(a)

The change in volume of the sphere

Answer to Problem 52QAP

The change in volume of the sphere is $2.38\times {10}^{-5}{m}^3$.

Explanation of Solution

Given:

Original radius of sphere $r_0=5.00cm=5.00\times {10}^{-2}m$

Original volume of sphere $V_0=\frac{4}{3}\pi r_0{}^3=\frac{4}{3}\pi {\left(5.00\times {10}^{-2}m\right)}^3=5.24\times {10}^{-4}{m}^3$

Bulk Modulus of copper $B=140\times {10}^{9}N{m}^{-2}$

Surface Area of sphere $A=4\pi r_0{}^2=4\pi \times {\left(5.00\times {10}^{-2}m\right)}^2=3.14\times {10}^{-2}{m}^2$

Let the change in volume be $\Delta V$

Applied force $F=2.00\times {10}^{8}N$

Formula used:

Bulk Modulus, $B=\frac{F/A}{\Delta V/V_0}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}\Delta V=\frac{F}{B}\times \frac{V_0}{A}\\ \Delta V=\frac{2.00\times {10}^{8}N}{140\times {10}^{9}N{m}^{-2}}\times \frac{5.24\times {10}^{-4}{m}^3}{3.14\times {10}^{-2}{m}^2}\\ \Delta V=2.38\times {10}^{-5}{m}^3\end{array}$

Hence, the change in volume of the sphere is $2.38\times {10}^{-5}{m}^3$.

Conclusion:

Thus, the change in volume of the sphere is $2.38\times {10}^{-5}{m}^3$.

To determine

(b)

The radius of the sphere.

Answer to Problem 52QAP

The final radius of the sphere is $5.08cm$.

Explanation of Solution

Given:

Original radius of sphere $r_0=5.00cm=5.00\times {10}^{-2}m$

Original volume of sphere $V_0=\frac{4}{3}\pi r_0{}^3=\frac{4}{3}\pi {\left(5.00\times {10}^{-2}m\right)}^3=5.24\times {10}^{-4}{m}^3$

Let the final radius of the sphere be $r_f$

Let the final volume of the sphere be $V_f$

The change in volume of the sphere is $\Delta V=2.38\times {10}^{-5}{m}^3$.

Formula used:

Final volume of the sphere $V_f=\Delta V+V_0$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}{V}_f=\Delta V+V_0\\ V_f=2.38\times {10}^{-5}{m}^3+5.24\times {10}^{-4}{m}^3\\ V_f=0.238\times {10}^{-4}{m}^3+5.24\times {10}^{-4}{m}^3\\ V_f=5.48\times {10}^{-4}{m}^3\end{array}$

But final volume of sphere $V_f=\frac{4}{3}\pi r_f{}^3=5.48\times {10}^{-4}{m}^3$

So, final radius of the sphere

  $\begin{array}{l}{r}_f=\sqrt[3]{\frac{3}{4\pi }\times 5.48\times {10}^{-4}{m}^3}\\ r_f=\sqrt[3]{\frac{3}{4\pi }\times 5.48\times {10}^{-4}{m}^3}\\ r_f=\sqrt[3]{130.8\times {10}^{-6}{m}^3}\\ r_f=5.08\times {10}^{-2}m\\ r_f=5.08cm\end{array}$

Hence, the final radius of the sphere is $5.08cm$.

Conclusion:

Thus, the final radius of the sphere is $5.08cm$.