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Q18P

Expert-verifiedFound in: Page 439

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Review. A solid sphere of brass (bulk modulus of **${\mathbf{14}}{\mathbf{.0}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{10}}}{\mathbf{\text{\hspace{0.17em}}}}\raisebox{1ex}{$\mathbf{\text{N}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{m}}}^{\mathbf{\text{2}}}$}\right.$**) with a diameter of **${\mathbf{3}}{\mathbf{.00}}{\mathbf{\text{\hspace{0.17em}m}}}$** is thrown into the ocean. By how much does the diameter of the sphere decrease as it sinks to a depth of** ${\mathbf{1}}{\mathbf{\text{\hspace{0.17em}km}}}$**?**

The decrease in diameter is $\Delta D=2.66\times {10}^{-3}\text{\hspace{0.17em}m}$.

The magnetic field, $B=14.0\times 10{}^{10}\text{\hspace{0.17em}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$

The depth of ocean, $h=1\text{km}=1000\text{m}$

The diameter of sphere, $D=3.0\text{\hspace{0.17em} m}$

The radius is,

$\begin{array}{rcl}r& =& \frac{D}{2}\\ & =& \frac{3.0\text{\hspace{0.17em}}\text{m}}{2}\\ & =& 1.5\text{\hspace{0.17em}m}\end{array}$The pressure *$P$* in a fluid is the force per unit area exerted by the fluid on a surface:

$P=\rho gh$

In the SI system, pressure has units of Newton’s per square meter $\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{${m}^{2}$}\right.$, and $1\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.=1\text{Pascal}\left(\text{Pa}\right)$.

Bulk modulus is given by,

role="math" localid="1663696968659" $B=\frac{P}{\raisebox{1ex}{$\u2206v$}\!\left/ \!\raisebox{-1ex}{$v$}\right.}$

Applying the formula of Bulk modulus here, you get

$B=\frac{P}{\raisebox{1ex}{$\u2206v$}\!\left/ \!\raisebox{-1ex}{$v$}\right.}$

While pressure is,

$P=\rho gh$

Combining both, you get

$\Delta V=\frac{\rho gh\times V}{B}\phantom{\rule{0ex}{0ex}}\frac{4}{3}\pi {\left(\Delta r\right)}^{3}=\frac{\rho gh\times \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi {r}^{3}}{B}$

$\begin{array}{rcl}{\left(\Delta r\right)}^{3}& =& \frac{\rho gh\times {r}^{3}}{B}\\ & =& \frac{1000\times 10\times 1000\times {\left(1.5\right)}^{3}}{14.0\times {10}^{10}}\\ & =& 2.4\times {10}^{-4}\end{array}$

$\begin{array}{rcl}\Delta r& =& \sqrt[3]{2.4\times {10}^{-4}}\\ & =& 0.062\text{m}\end{array}$

Also diameter, $D=2r$

Therefore decrease in diameter is,

$\begin{array}{rcl}\Delta D& =& 2\Delta r\\ & =& 2\times 0.062\text{m}\\ & =& 0.124\text{m}\end{array}$

Hence, the decrease in diameter is $\Delta D=0.124\text{m}$.** **

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