The increase in length of a wire on stretching is 0.025%. If its poisson ratio is 0.4, then the percentage decrease in the diameter is : 
 

Assuming that D is the wire's diameter and the Poissons ratio,

$$\begin{gathered} \sigma =\frac{lateral strain}{longitudinal strain}; \\ \sigma =\frac{{\displaystyle \frac{∆D}{D}}}{{\displaystyle \frac{∆L}{L}}} \end{gathered}$$

$$\begin{gathered} \frac{∆L}{L}=0.025 \\ \sigma =0.004 \\ \sigma =\frac{{\displaystyle \frac{∆D}{D}}}{{\displaystyle \frac{1}{40}}} \end{gathered}$$

$$\begin{gathered} \frac{∆D}{D}=\frac{1}{40}\times 0.4 \\ \frac{∆D}{D}=0.01 \end{gathered}$$

Hence the correct answer is 0.01.