Question

Review. A solid sphere of brass (bulk modulus of$\mathbf{14}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\raisebox{1ex}{$\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 $\mathbf{1}\mathbf{\text{\hspace{0.17em}km}}$?

Short answer

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

Step-by-step solution

Given Data:

The magnetic field,$B=14.0\times 10{}^{10}\raisebox{1ex}{$\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,

$\begin{array}{rcl}r& =& \frac{D}{2}\\ & =& \frac{3.0\text{m}}{2}\\ & =& 1.5\text{\hspace{0.17em}m}\end{array}$

A concept:

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

Find the decrease in diameter:

Applying the formula of Bulk modulus here, you get

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

While pressure is,

$P=\rho gh$

Combining both, you get

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

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

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

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 $\Delta D=0.124\text{m}$.