Question

A cubical ball is taken to a depth of 200 m in a sea. The decrease in volume is observed to be 0.1%. The bulk modulus of the ball is $\left( {g = 10m{s^{ - 2}}} \right)$
A. $2 \times {10^7}Pa$
B. $2 \times {10^6}Pa$
C. $2 \times {10^9}Pa$
D. $1.2 \times {10^9}Pa$

Answer 1

This problem can be solved by direct application of the definition of Bulk’s Modulus.
The Bulk’s modulus is defined as the ratio of pressure acting on the body to the fractional change in the volume.
$K = - \dfrac{P}{{\dfrac{{dV}}{V}}}$
where P = pressure applied, $\dfrac{{dV}}{V}$ is the fractional change in the volume.

Complete step by step answer:
In a fluid like air and liquid, the pressure applied on a unit element of the substance is uniform in all the directions.
This pressure applied on all the directions attempts to decrease the volume of the material. However, the resistance is offered by the fluid.
The Bulk modulus of a fluid represents the resistance offered by the fluid for the change in the volume when the compressive pressure is uniformly applied on all ends. This is one of three moduli of elasticity, other two being the Young’s Modulus for linear stress and Shear Modulus for shear stress.
Mathematically, the Bulk modulus is given by –
$K = - \dfrac{P}{{\dfrac{{dV}}{V}}}$
where P = pressure applied, $\dfrac{{dV}}{V}$ is the fractional change in the volume.
In this problem, the decrease in the volume is given as, $dV = - 0.1\% \times V$
Hence,
The fractional change in the volume,
$\Rightarrow \dfrac{{dV}}{V} = - 0.1\% = \dfrac{{0.1}}{{100}} = - 0.001$
Given the depth of the sea, the pressure is calculated by the formula,
$P = \rho gh$
where $\rho$= density of water, $1000kg{m^{ - 3}}$
Acceleration due to gravity, $g = 10m{s^{ - 2}}$
Depth of the sea, $h = 200 m$
Therefore, pressure –
$P = \rho gh$
$\Rightarrow P = 1000 \times 10 \times 200$
$\Rightarrow P = 2 \times {10^6}Pa$
Substituting the pressure in the equation for Bulk modulus, we get,
$K = - \dfrac{P}{{\dfrac{{dV}}{V}}}$
$\Rightarrow K= - \dfrac{{2 \times {{10}^6}}}{{ - 0.001}}$
$\Rightarrow K = \dfrac{{2 \times {{10}^6}}}{{{{10}^{ - 3}}}} = 2 \times {10^9}Pa$

Therefore, the correct option is Option C.

Note:
The speed of sound in a medium depends on the Bulk modulus as shown –
Speed of sound in a medium, $c = \sqrt {\dfrac{K}{\rho }}$
where $\rho$= density of the medium and K is the bulk modulus.
The bulk modulus of liquid is higher than gas since the liquid can resist compression to a greater extent than a gas.
Hence, the speed of sound is higher in liquids than in air mediums.