Question

An exotic spherical jellyfish has a bulk modulus $B$. Close to the surface of the sea (depth $d=0$), its radius is $R$. When it dives to a depth $d$ ($d \gg R$), its radius is reduced by $\Delta R > 0$. Given the density of the incompressible sea water $\rho$, and the uniform acceleration due to gravity $g$ such that $\rho g d \ll B$, what is $\frac{\Delta R}{R}$?

• A. $1 - \left( 1 - \frac{\rho g d}{B} \right)^{1/3}$
• B. $1 - \left( 1 - \frac{\rho g d}{B} \right)^{2/3}$
• C. $\left( 1 + \frac{\rho g d}{B} \right)^{2/3} - 1$
• D. $\left( 1 + \frac{\rho g d}{B} \right)^{1/3} - 1$

Hint:

For any small fractional changes, the volumetric strain is approximately three times the linear strain ($\frac{\Delta V}{V} \approx 3 \frac{\Delta R}{R}$).
Since we are given exact expressions in the options, working with exact volume equations yields the precise formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Bulk Modulus relates the change in pressure to the fractional change in volume of an object.
Key Formula or Approach:
Bulk Modulus \( B = -\frac{\Delta P}{\Delta V / V} \implies \frac{\Delta V}{V} = \frac{\rho gd}{B} \).

Step 2: Detailed Explanation:
1. Volume relation: Let \( V_{0} \) be initial volume and \( V \) be final volume.
\( \frac{V_{0} - V}{V_{0}} = \frac{\rho gd}{B} \implies \frac{V}{V_{0}} = 1 - \frac{\rho gd}{B} \).
2. Radius relation: For a sphere, \( V = \frac{4}{3}\pi r^{3} \).
\( \frac{\frac{4}{3}\pi(R - \Delta R)^{3}}{\frac{4}{3}\pi R^{3}} = 1 - \frac{\rho gd}{B} \)
\[ \left( \frac{R - \Delta R}{R} \right)^{3} = 1 - \frac{\rho gd}{B} \]
3. Solve for fractional change:
\[ 1 - \frac{\Delta R}{R} = \left( 1 - \frac{\rho gd}{B} \right)^{1/3} \implies \frac{\Delta R}{R} = 1 - \left( 1 - \frac{\rho gd}{B} \right)^{1/3} \]

Step 3: Final Answer:
The ratio is \( 1 - (1 - \rho gd/B)^{1/3} \).