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 Question:
We are given a spherical jellyfish of radius $R$ and bulk modulus $B$.
When it dives to a depth $d$, the hydrostatic pressure increases, causing its volume to decrease, which in turn reduces its radius to $R - \Delta R$.
We need to find the fractional change in radius $\frac{\Delta R}{R}$.

Step 2: Key Formula or Approach:
The bulk modulus $B$ is defined as:
\[ B = -\frac{\Delta P}{\Delta V / V} \implies \frac{\Delta V}{V} = \frac{\Delta P}{B} \] where $\Delta P$ is the change in hydrostatic pressure at depth $d$, given by $\Delta P = \rho g d$.
The volume of a sphere is $V = \frac{4}{3}\pi R^3$.

Step 3: Detailed Explanation:

• Let the initial volume at the surface be $V_i = \frac{4}{3}\pi R^3$.

• Let the final volume at depth $d$ be $V_f = \frac{4}{3}\pi (R - \Delta R)^3$.

• The change in volume is $\Delta V = V_i - V_f$.

• The fractional change in volume is:
\[ \frac{\Delta V}{V_i} = \frac{V_i - V_f}{V_i} = 1 - \frac{V_f}{V_i} \]
• Substituting the formulas for volume:
\[ \frac{\Delta V}{V_i} = 1 - \left( \frac{R - \Delta R}{R} \right)^3 = 1 - \left( 1 - \frac{\Delta R}{R} \right)^3 \]
• From the definition of bulk modulus:
\[ \frac{\Delta V}{V_i} = \frac{\Delta P}{B} = \frac{\rho g d}{B} \]
• Equating the two expressions:
\[ 1 - \left( 1 - \frac{\Delta R}{R} \right)^3 = \frac{\rho g d}{B} \] \[ \left( 1 - \frac{\Delta R}{R} \right)^3 = 1 - \frac{\rho g d}{B} \]
• Taking the cube root on both sides:
\[ 1 - \frac{\Delta R}{R} = \left( 1 - \frac{\rho g d}{B} \right)^{1/3} \] \[ \frac{\Delta R}{R} = 1 - \left( 1 - \frac{\rho g d}{B} \right)^{1/3} \]

Step 4: Final Answer:
The fractional change in radius $\frac{\Delta R}{R}$ is $1 - \left( 1 - \frac{\rho g d}{B} \right)^{1/3}$.