A cable breaks if stretched by more than 2mm. It is cut into two equal parts. By how much either part can be stretched without breaking 
 

Young's modulus connects the length and the stretch parameters.

$\mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A}∆\mathrm{L}}$

$$\begin{gathered} \mathrm{For} \mathrm{orginal} \mathrm{length}, \\ {\mathrm{Y}}_1=\frac{{\mathrm{F}}_1{\mathrm{L}}_1}{{\mathrm{A}}_1∆{\mathrm{L}}_1} \end{gathered}$$

$$\begin{gathered} \mathrm{When} \mathrm{the} \mathrm{cable} \mathrm{is} \mathrm{cut} \mathrm{into} 2 \mathrm{equal} \mathrm{parts} \\ {\mathrm{Y}}_2=\frac{{\mathrm{F}}_2{\mathrm{L}}_2}{{\mathrm{A}}_2∆{\mathrm{L}}_2} \end{gathered}$$

stretch per unit length remains constant

${\mathrm{Y}}_1={\mathrm{Y}}_2\Rightarrow \frac{{\mathrm{F}}_1{\mathrm{L}}_1}{{\mathrm{A}}_1∆{\mathrm{L}}_1}=\frac{{\mathrm{F}}_2{\mathrm{L}}_2}{{\mathrm{A}}_2∆{\mathrm{L}}_2}$

Since the cable's area and the force used to stretch it are both constants,

$\frac{{\mathrm{L}}_1}{∆{\mathrm{L}}_1}=\frac{{\mathrm{L}}_2}{∆{\mathrm{L}}_2}$

$\Rightarrow ∆{\mathrm{L}}_2=\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}∆{\mathrm{L}}_1=\frac{{\displaystyle \frac{\mathrm{L}}{2}}}{\mathrm{L}}\times ∆{\mathrm{L}}_1=\frac{1}{2}\times ∆{\mathrm{L}}_1=\frac{1}{2}\times 2=1\mathrm{mm}$

Hence the correct answer is 1mm.