If the initial concentration is reduced to $1/4^{\text{th }}$ of the initial value of a zero order reaction, the half life of the reaction

$$\begin{gathered} \text{For a zero order reaction,}{\mathrm{t}}_{1/2}=\frac{[\mathrm{A}{]}_0}{2\mathrm{k}} \\ \text{ Let }[\mathrm{A}{]}_0=\mathrm{a} \\ {\mathrm{t}}_{1/2}=\frac{\mathrm{a}}{2\mathrm{k}}....(1\left) \\ \text{ When the intial concentration is reduced to one fourth, }\right[\mathrm{A}{]}_0=\frac{1}{4}\times \mathrm{a}\text{. } \\ {\mathrm{t}}_{1/2}^{\mathrm{\prime }}=\frac{\frac{1}{4}\times \mathrm{a}}{2\mathrm{k}}\dots \dots \left(2\right) \\ \text{ By dividing equation (2) by (1), we get } \\ \frac{{\mathrm{t}}_{1/2}^{\mathrm{\prime }}}{{\mathrm{t}}_{1/2}}=\frac{\frac{1}{4}\times \mathrm{a}}{\frac{2\mathrm{k}}{\mathrm{a}}}=\frac{1}{4} \end{gathered}$$

Therefore, the correct option is (B).