The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are 27 gms of certain substance and 3 hours later it is found that \(8 \mathrm{gms}\) are left, then the amount left after one more hour is

(C)
Let \(x\) gms be the amount of the substance left at time \(t\). Then the rate of decay be \(\frac{d x}{d t}\), which is proportional to \(x\).
\(\begin{array}{l}
\therefore \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{k} \cdot \mathrm{x}, \text { where } \mathrm{k}>0 \\
\therefore \int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}=\int-\mathrm{k} \cdot \mathrm{dt} \\
\therefore \log \mathrm{x}=-\mathrm{kt}+\mathrm{c} \\
\text { Initially i.e. when } \mathrm{t}=0, \mathrm{x}=27 . \\
\quad \log 27=-\mathrm{k} \times 0+\mathrm{C} \Rightarrow \mathrm{C}=\log 27 \\
\therefore \log \mathrm{x}=-\mathrm{kt}+\log 27 \\
\therefore \log \mathrm{x}-\log 27=-\mathrm{kt} \\
\text { Now, when } \mathrm{t}=3, \mathrm{x}=8
\end{array}\)
\(\begin{array}{l}
\therefore \quad \log \left(\frac{8}{27}\right)=-3 \mathrm{k} \Rightarrow-3 \mathrm{k}=\log \left(\frac{2}{3}\right)^{3}=3 \log \left(\frac{2}{3}\right) \\
\quad \mathrm{k}=-\log \left(\frac{2}{3}\right) \\
\therefore \quad \log \left(\frac{\mathrm{x}}{27}\right)=\mathrm{t} \cdot \log \left(\frac{2}{3}\right) \\
\text { When } \mathrm{t}=4, \\
\quad \log \left(\frac{\mathrm{x}}{27}\right)=4 \log \left(\frac{2}{3}\right)=\log \left(\frac{2}{3}\right)^{4} \\
\frac{\mathrm{x}}{27}=\frac{16}{81} \Rightarrow \mathrm{x}=\frac{16}{3} \mathrm{gms}
\end{array}\)