JEE Main 2025 Jan 28 Shift 2, Chemistry Q15: Reactions of Various Orders

If $N_0$ is the initial count and $N$ the count at time $t$, the analogue of radioactive decay for population growth is

$\frac{dN}{dt}=kN\Rightarrow N=N_0{e}^{kt}$,

where $k$ is the (positive) growth constant. Plotting $N/N_0$ against $t$ gives a monotonically increasing exponential curve that starts at 1 and rises with positive curvature.

$\therefore$ the correct graph is the exponentially rising curve (option 4).