Question

The bulk modulus of rubber is $9.1 \times {10^8}\,N/{m^2}$. To what depth (approximately) a rubber ball be taken in a lake so that its volume is decreased by $0.1\% ?$

Answer 1

Instead of directly solving the problem let us first get some idea about elasticity. Elasticity refers to a body's ability to withstand a distorting impact and then return to its original size and shape until the influence or force has been removed.

Complete step by step answer:
Let us understand bulk modulus. As the pressure on a gas is changed, the volume of the gas varies. The ratio of volumetric stress to volumetric strain is the bulk modulus of a gas.
$\beta = - \dfrac{{\Delta p}}{{{\dfrac{\Delta V}{V}}}}$
where $\Delta p$denotes a pressure change and $\Delta V$ denotes a volume change. As applied pressure is raised, the gas compresses (volume decreases), as shown by the negative symbol. Pascal is the unit of bulk modulus.

Now let us solve the problem: given: volume decrease as a percentage $= 0.1\%$
$\dfrac{\Delta V}{V}\times 100 = 0.1\%$
$\Rightarrow \dfrac{\Delta V}{V}\times 100 = {10^{ - 3}}$
$\beta = - \dfrac{{\Delta p}}{{{\dfrac{\Delta V}{V}}}}$
9.1 \times {10^8} = \dfrac{{\Delta p}}{{{{10}^{ - 3}}}} \\\ \Rightarrow \Delta p = 9.1 \times {10^5}N/{m^2} \\\
We know, $\Delta P = \rho gh$
Where $\rho$ is the density of seawater, $g$ is gravity's acceleration, and $h$ is the rubber ball's depth.
$\rho = {10^3}$ $kg/{m^3}$
$\Rightarrow h = \dfrac{{\Delta P}}{{\rho g}}$
$\Rightarrow h = \dfrac{{9.1 \times {{10}^5}}}{{{{10}^3} \times 9.8}} \\\ \therefore h = 92.85\,m$

Hence, the depth to which a rubber ball be taken in a lake is 92.85 m.

Note: The higher the modulus of elasticity of the material, the greater the rigidity of the product; doubling the modulus of elasticity doubles the rigidity of the product in equivalent products. The greater a structure's rigidity, the more force is required to achieve a given deformation.