The rate of growth of bacteria is proportional to number present. If initially there
were 1000 bacteria and the number doubles in 1 hour then the number of bacteria
after \(2 \frac{1}{2}\) hours are (Given \(\sqrt{2}=1.414\) )

The rate of growth is proportional to the number present
\(\therefore \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \propto \mathrm{N} \Rightarrow \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{kN} \Rightarrow \frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}=\mathrm{k} \mathrm{dt}\)
\(\therefore\) On integrating we get
\(\int \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{k} \int \mathrm{dt}\)
\(\therefore \log \mathrm{N}=\mathrm{kt}+\mathrm{C}\)
When \(\mathrm{t}=0, \mathrm{~N}=1000 \Rightarrow \mathrm{C}=\log 1000\)
\(\therefore \log \mathrm{N}=\mathrm{kt}+\log 1000\)
\(\therefore \quad \log \left(\frac{\mathrm{N}}{1000}\right)=\mathrm{kt}\)
\(\mathrm{N}=1000 \mathrm{e}^{\mathrm{kt}}\) ...(1)
When \(\mathrm{t}=1, \mathrm{~N}=2000\)
\(\therefore \mathrm{e}^{\mathrm{k}}=2 \quad \Rightarrow \mathrm{N}=1000 \times 2^{1} \quad \ldots[\) from (1) \(]\)
When \(t=2 \frac{1}{2}\), we get
\(\mathrm{N}=1000 \times 2^{\frac{5}{2}}=1000 \times 4 \sqrt{2}\)
\(=1000 \times 4 \times 1.414=5656\)