Q15 / 75 · JEE Main 2026 (28 January Shift 1) · Chemistry
An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(\frac{1}{8}\right)^{\text {th }}$ and $\left(\frac{1}{10}\right)^{\text {th }}$ of its initial concentration are $\mathrm{t}_{1 / 8}$ and $\mathrm{t}_{1 / 10}$ respectively. What is the value of $\frac{\mathrm{t}_{1 / 8}}{\mathrm{t}_{1 / 10}} \times 10$ ? ($\log 2=0.3$)
- A3
- Correct answerB9
- C0.9
- D30
Solution
For first-order decomposition with rate equation $[A] = [A]_0 e^{-kt}$, the time for concentration to reduce to fraction f is: $t = \frac{1}{k}\ln\frac{1}{f} = \frac{\ln(1/f)}{k}$
For reduction to 1/8: $t_{1/8} = \frac{\ln 8}{k} = \frac{3\ln 2}{k}$
For reduction to 1/10: $t_{1/10} = \frac{\ln 10}{k}$
Therefore: $\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{3\log 2}{\log 10} = \frac{3 \times 0.3}{1} = 0.9$
Thus: $\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$
For reduction to 1/8: $t_{1/8} = \frac{\ln 8}{k} = \frac{3\ln 2}{k}$
For reduction to 1/10: $t_{1/10} = \frac{\ln 10}{k}$
Therefore: $\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{3\log 2}{\log 10} = \frac{3 \times 0.3}{1} = 0.9$
Thus: $\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$
Hence, the correct answer is option: 'B'.