Q25 / 75 · JEE Main 2026 (05 April Shift 2) · Chemistry
For a reaction $A \rightarrow P$ at $T\text{ K}$, the half life $(t_{1/2})$ is plotted as a function of initial concentration $[A]_0$ of $A$ as given below.

The value of $x$ in the given figure is _______ s (Nearest integer)

The value of $x$ in the given figure is _______ s (Nearest integer)
Correct answer: 90
Solution
From the given graph, the plot of half-life $t_{1/2}$ versus initial concentration $[A]_0$ is a straight line passing through the origin.
This indicates that $t_{1/2}$ is directly proportional to $[A]_0$, which means $t_{1/2} \propto [A]_0$.
Therefore, the ratio $\dfrac{t_{1/2}}{[A]_0}$ is constant.
Using the coordinates from the graph, we get:
$\dfrac{240}{4 \times 10^{-3}} = \dfrac{x}{1.5 \times 10^{-3}}$
Solving for $x$:
$x = \dfrac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}$
$x = 60 \times 1.5 = 90$
The value of $x$ is $90$.
Answer: $90$
This indicates that $t_{1/2}$ is directly proportional to $[A]_0$, which means $t_{1/2} \propto [A]_0$.
Therefore, the ratio $\dfrac{t_{1/2}}{[A]_0}$ is constant.
Using the coordinates from the graph, we get:
$\dfrac{240}{4 \times 10^{-3}} = \dfrac{x}{1.5 \times 10^{-3}}$
Solving for $x$:
$x = \dfrac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}$
$x = 60 \times 1.5 = 90$
The value of $x$ is $90$.
Answer: $90$
Hence, the correct answer is 90.