The strain in the upper wire, when the whole system moves upwards with an acceleration of 2ms^–2, will be (Given:area of cross -section of each string = 0.5mm² and Y = 2×10¹¹Nm^–2 ,
g = 10ms^–2 ) $\_\_\_\_\_\times {10}^{-4}.$

 

From the figure for 3 kg,

${\mathrm{T}}_2-({\mathrm{T}}_1+3\mathrm{g})=3\mathrm{a}$

$\Rightarrow {\mathrm{T}}_2-{\mathrm{T}}_1=3(\mathrm{a}+\mathrm{g})\Rightarrow {\mathrm{T}}_2-{\mathrm{T}}_1=3(10+2)=36\mathrm{\_\_\_\_\_\_\_\_}\left(1\right)$

for 2 kg,

${\mathrm{T}}_1-2\mathrm{g}=2\mathrm{a}\Rightarrow {\mathrm{T}}_1=2(\mathrm{g}+\mathrm{a})=24$

${\mathrm{T}}_2-24=36\Rightarrow {\mathrm{T}}_2=60\mathrm{N}$

Now tension in the upper string =${\mathrm{T}}_2=60\mathrm{N}$

$\mathrm{A}=0.5{\mathrm{mm}}^2=(0.5\times {10}^{-6}){\mathrm{m}}^2, \mathrm{Y}=2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2$

$\mathrm{stress}=\frac{{\mathrm{T}}_2}{\mathrm{A}}=\frac{600}{0.5\times {10}^{-6}}=120\times {10}^{-6}$

$\mathrm{Strain}=\frac{\mathrm{Stress}}{\mathrm{Y}}=\frac{120\times {10}^6}{2\times {10}^{11}}=6\times {10}^{-4}$

Hence the correct answer is $6\times {10}^{-4}.$