Question

A rubber ball is taken to depth 1 km inside water so that its volume reduces by 0.05%. What is the bulk modulus of the rubber?
(A) $2 \times 10^{10} \,N/m^2$
(B) $2 \times 10^{9} \,N/m^2$
(C) $2 \times 10^{7} \,N/m^2$
(D) $2 \times 10^{11} \,N/m^2$

Answer 1

Hint
Bulk modulus is a ratio between applied pressure and the corresponding
relative decrease in the volume of the material.
$B = \Delta P/\left( {\Delta V/V} \right)$Where, B: Bulk modulus
ΔP: change of the pressure or force applied per unit area on the material
ΔV: change of the volume of the material due to the compression
V: Initial volume of the material in the units of in the English system and N/$m^2$ in the metric system

Complete step by step answer
Pressure at 1 km depth inside water,
$\Rightarrow P = \rho gh$
$\Rightarrow P = 1000 \times 10 \times 1000 = {10^7}N{m^2}$............[assuming that density of water 1000]
Now say, V= volume of the ball
The volume reduced by 0.05%
So the volume of the ball after reducing is,
$\Rightarrow \dfrac{{0.05}}{{100}} \times V = \Delta V$.........[here,$\Delta = \dfrac{{0.05}}{{100}}$]
Now Bulk Modulus,
$\Rightarrow B = \dfrac{{ - PV}}{{\Delta V}} = \dfrac{{ - {{10}^7} \times V}}{{\dfrac{{ - 0.05V}}{{100}}}}$
$\Rightarrow 2 \times {10^{10}}N{m^{ - 2}}$.
Option (A) is correct.

Note
Young’s Modulus is the ability of any material to resist the change along its length. Bulk Modulus is the ability of any material to resist the change in its volume.
$K = \dfrac{Y}{{3(1 - \dfrac{2}{\mu })}}$
Where, K is the Bulk modulus.
Y is Young’s modulus.
μ is the Poisson’s ratio.