Question

The approximate depth of an ocean is 2700m. The compressibility of water is $45.4 \times 10^{-11} Pa^{-1}$ and density of water is $10^3 kg/m^3$. What fractional compression of water will be obtained at the bottom of the ocean?
A. 0.012
B. 0.014
C. 0.008
D. 0.01

Answer 1

Hint: Compressibility of a material is a quantity which defines how much a material will undergo compression when subjected to a particular pressure. More the value of compressibility, lesser will be the decrease in volume of the material.

Formula used:
fractional compression = $\dfrac{dV}{V} = \kappa \times P$

Complete step-by-step answer:
When a material is subjected to an external pressure, it undergoes decrease in volume. The amount of this volume decrease will depend upon the compressibility factor and amount of pressure applied. The expression regarding these values is given by:
$\dfrac{dV}{V} = \kappa \times P$
Where, $\kappa$ is the compressibility factor and P is the pressure on water under consideration.
Now, we know that pressure of water at a depth of ‘h’ is given by $P = \rho g h$
Hence $P = 10^3 \times 10 \times 2700 = 2.7\times 10^7\ pa$
Putting the values in the expression$\dfrac{dV}{V} = \kappa \times P$, we get;
$\dfrac{dV}{V} = 45.4\times 10^{-11} \times (2.7 \times 10^7) = 0.012$

So, the correct answer is “Option A”.

Additional Information: Compressibility can be defined as the property of being reduced to a smaller space by pressure. The compressibility of fluid is basically a measure of the change in density that will be produced in the fluid by a specified change in pressure. Gases are, in general, highly compressible whereas most liquids have very low compressibility.
Bulk modulus of a substance is defined as the ratio of infinitesimal pressure increase to a decrease of the volume. Bulk modulus is meaningful only for a fluid. It is denoted as B. Also $B = \dfrac1\kappa$.

Note: One might think that there must be a defined object that will lie inside water so that its change in volume can be calculated. But this is not true. Even water at huge depths undergo compression significantly, which is important to be studied and hence this method is important to calculate the compression of water.