The rod AB has a cross sectional area 5 mm². Young's modulus of its material is $2\times {10}^{10}\mathrm{N}/{\mathrm{m}}^2$. If mass of the rod is 10 Kg, strain produced at its mid point is 

Acceleration $\mathrm{a}=\frac{100-50}{10}\mathrm{m}/{\mathrm{s}}^2=5\mathrm{m}/{\mathrm{s}}^2$

Tension produced at its mid section is

given by, $\mathrm{T}-50=5\times 5\Rightarrow \mathrm{T}=75\mathrm{N}$

$\therefore$ strain $=\frac{\text{stress}}{\text{young's modulus}}=\frac{75/5\times {10}^{-6}}{2\times {10}^{10}}=\frac{75}{{10}^5}\times 100\%$

$\Rightarrow \mathrm{strain}=0.075\mathrm{\%}$