Question

A cork ball when moved from the surface of a $250m$deep lake to its bottom. The volume of ball decreases by $0.15$ the density of lake water is $1.0\times {{10}^{3}}\dfrac{kg}{{{m}^{3}}}$. Find the bulk modulus of the cork.

Answer 1

When attempting questions like these, remember the concepts about the various elastic properties of solid or fluid, and keep in mind the different effects that pressure and volume and other factors play in changing the values of such properties. We will apply the formula for bulk modulus and you will get the required result.

Complete step-by-step solution:
When we attempt this question keep in mind the various mechanical properties of solids. These properties elaborate characteristics like resistance to deformation and the strength to bear stress and strain. Resistance to deformation is called how resistant that object is to the change in shape. Some mechanical properties of solids include elasticity, plasticity et cetera. One such property of solids or fluids is known as the bulk modulus.
Bulk modulus describes the elastic properties of solid or fluid – fluid in our case – when it is under pressure from all surfaces. The applied pressure reduces the volume of the mentioned material which returns back to its original volume when the pressure is removed. Sometimes referred to as the incompressibility, the bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.
A substance that is difficult to compress has a large bulk modulus but has a small compressibility whereas a substance that is easy to compress has a small bulk modulus and high compressibility. So from here we can say that compressibility is defined as a reciprocal to bulk modulus.
From question given above, we know that;
Density of water, assuming it to be $d$
d$$$$={{10}^{3}}\dfrac{Kg}{{{m}^{3}}}
The depth of lake assuming it to be $h$
h$$$$=250m
The decrease in volume is $0.15$
So the Bulk Modulus will be
$B=V\dfrac{P}{dV}$
$\Rightarrow \dfrac{dgh}{\dfrac{dV}{V}}$
$\Rightarrow \dfrac{{{10}^{3}}\times 9.8\times 250}{{{10}^{-3}}}$
$\Rightarrow 2450\times {{10}^{6}}$
$\therefore 2.45\times {{10}^{8}}\dfrac{N}{{{m}^{2}}}$
Hence from the above calculations the bulk modulus comes out to be $2.45\times {{10}^{8}}\dfrac{N}{{{m}^{2}}}$

Note: Another property similar to Bulk modulus is the Young’s modulus which describes the elastic properties of a solid undergoing tension or compression in one direction. It is the measure of the ability of a material to withstand the changes in length when under pressure.