The approximate depth of an ocean is $2700 ext{ m}$. The compressibility of water is $45.4 imes 10^{-11} e

Question

The approximate depth of an ocean is $2700	ext{ m}$. The compressibility of water is $45.4 	imes 10^{-11}	ext{ Pa}^{-1}$ and density of water is $10^3	ext{ kg/m}^3$. What fractional compression of water will be obtained at the bottom of the ocean?

Accepted Answer

1. **Identify the Formula:** Fractional compression ($\frac{\Delta V}{V}$) is related to the change in pressure ($\Delta P$) and compressibility ($k$) by the definition of Bulk Modulus ($B$). Compressibility is the reciprocal of Bulk Modulus ($k = 1/B$).
   $$ k = \frac{1}{B} = \frac{\Delta V/V}{\Delta P} \implies \frac{\Delta V}{V} = k 	imes \Delta P $$
   
2. **Calculate Pressure Change:** The pressure at the bottom of the ocean due to the water column is given by hydrostatic pressure formula:
   $$ \Delta P = ho g h $$
   where $ho = 10^3	ext{ kg/m}^3$, $g \approx 10	ext{ m/s}^2$ (or $9.8	ext{ m/s}^2$), and $h = 2700	ext{ m}$.
   $$ \Delta P = 10^3 	imes 10 	imes 2700 = 2.7 	imes 10^7	ext{ Pa} $$
   
3. **Calculate Fractional Compression:**
   $$ \frac{\Delta V}{V} = (45.4 	imes 10^{-11}	ext{ Pa}^{-1}) 	imes (2.7 	imes 10^7	ext{ Pa}) $$
   $$ \frac{\Delta V}{V} = 122.58 	imes 10^{-4} $$
   $$ \frac{\Delta V}{V} = 1.2258 	imes 10^{-2} $$
   Rounding to one decimal place matches the option $1.2 	imes 10^{-2}$.

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