Answer
${\Delta V\over \Delta t}=1.66\times 10^{-11}\,\mathrm{m^3/s}$
Work Step by Step
From the graph, it is seen that a maximum occurs at 6 seconds. And the subsequent minimum at 12 seconds. The corresponding conditions are:
$2d_1=(m+{1\over 2})\lambda_n$
And $2d_2=m\lambda_n$
Then the change in the thickness can be computed by subtracting these two equations:
$\Delta d=d_2-d_1=-{1\over 2}{\lambda_n\over 2}=-{\lambda_n\over 4}$
The corresponding time interval is $\Delta t=12-6=6\,\mathrm{s}$
The rate of change of the film thickness is therefore,
${\Delta d\over\Delta t}=-{\lambda_n/4\over 6}=-{\lambda\over 24n}$
The volume of the film at thickness $d$ is $V=\pi r^2d$
$\implies \Delta V=\pi r^2\Delta d$
The rate of change of volume is
\begin{align*}
{\Delta V\over \Delta t}&={\pi r^2\Delta d\over\Delta t}\\
&=-\pi (1.8\times 10^{-2})^2{\lambda\over 24n}\\
&=-\pi (1.8\times 10^{-2})^2{550\times10^{-9}\over 24\times 1.4}\\
&=-1.66\times 10^{-11}\,\mathrm{m^3/s}
\end{align*}
The minus sign indicates a decrease in volume with time.
