Login/Sign Up
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 $\sqrt{2}=1.414$ ) (A) 4646 approx1mnately (B) 5056 approximnately (C) 5656 approximately (D) $4OO\sqrt{2 }$ approxınnate]y
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 $\sqrt{2}=1.414$ ) (A) 4646 approx1mnately (B) 5056 approximnately (C) 5656 approximately (D) $4OO\sqrt{2 }$ approxınnate]y
Maths
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 $\sqrt{2}=1.414$ ) (A) 4646 approx1mnately (B) 5056 approximnately (C) 5656 approximately (D) $4OO\sqrt{2 }$ approxınnate]y

The correct option is option (C) 5656 approximately

The rate of growth is proportional to the number present.
$\begin{array}{l}
\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right) \propto \mathrm{N}\\
\Rightarrow\left(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right)=\mathrm{kN}\\
\Rightarrow\left(\frac{\mathrm{dN}}{\mathrm{N}}\right)=\mathrm{kdt}\\
\Rightarrow \log \mathrm{N}=\mathrm{kt}+\log \mathrm{C}\\
\Rightarrow \mathrm{N}=\mathrm{ce}^{\mathrm{kt}}\\
\mathrm{At} \mathrm{t}=0\\
\mathrm{N}=\mathrm{C}=1000\\
\therefore N=1000 \mathrm{e}^{\mathrm{kt}}\\
\mathrm{At} \mathrm{t}=1 \mathrm{hr}\\
\mathrm{N}=2000\\
\therefore 2000=1000 e^{k}\\
\mathrm{N}=1000 \mathrm{e}^{0.693 \mathrm{t}}\\
\mathrm{At} \mathrm{t}=2.5 \mathrm{hr}\\
\mathrm{N}=1000 \mathrm{e}^{(0.693 \times 2.5)}\\
\mathrm{N}=5654.77
\end{array}$

i

More Material