Question

A ball falling in a lake of depth 200 m shows a decrease of 0.1% in its volume at the bottom. The bulk modulus of the elasticity of the material of the ball is (take $g = 10m/{s^2}$)
${\text{A}}{\text{. 1}}{{\text{0}}^9}N/{m^2}$
${\text{B}}{\text{. 2}} \times {\text{1}}{{\text{0}}^9}N/{m^2}$
${\text{C}}{\text{. 3}} \times {\text{1}}{{\text{0}}^9}N/{m^2}$
${\text{D}}{\text{. 4}} \times {\text{1}}{{\text{0}}^9}N/{m^2}$

Answer 1

Hint – Use the formula that bulk modulus is given by $B = - V\dfrac{{dP}}{{dV}}$ and then solve the question and find the answer. Bulk modulus is the relative change in the volume of a body produced by a unit compressive or tensile stress acting uniformly over its surface.
Formula used:- $B = - V\dfrac{{dP}}{{dV}}$ , $dP = \rho gh$

Complete step-by-step solution -
We have been given the depth of the lake is 200 m
Decrease in volume 0.1%
Now using the formula of Bulk Modulus,
$B = - V\dfrac{{dP}}{{dV}}$
Also, the change in pressure is, $dP = \rho gh$
Where $\rho $ is the density, g is acceleration due to gravity and h is the depth
Now, putting the value of these we get-
$dP = 1000 \times 10 \times 200 = 2 \times {10^6}$
It is given that change in volume is by 0.1 %.
So, $\dfrac{{dV}}{V} = - 0.001$
Therefore, $B = - V\dfrac{{dP}}{{dV}}$ will be-
$ B = - V\dfrac{{2 \times {{10}^6}}}{{ - 0.001 \times V}}\left\{ {\because dV = - 0.001 \times V} \right\} \\
   = 2 \times {10^9} \\ $
Hence, the correct option is B.

Note – Whenever such types of questions appear then always write down the things given in question and then solve the question. As mentioned in the solution, first we found out the change in pressure and then the change in volume, substituted them in the formula of Bulk Modulus and found the answer.