Given : Force applied F = 10⁵ N; side of cube x = 10 m = 10 $$\times$$ 10^$$-$$2 m; shift dx = 0.5 cm = 0.5 $$\times$$ = 10^$$-$$2 m

On cube of side x = 10 cm;

Stress = $${F \over {{x^2}}} = {{{{10}^5}} \over {{{(10 \times {{10}^{ - 2}})}^2}}} = {10^7}$$ N/m²

Strain = $${{dx} \over x} = {{0.5 \times {{10}^{ - 2}}} \over {10 \times {{10}^{ - 2}}}} = 0.5$$

On cube of side x' = 20 cm;

Stress' = $${F \over {x{'^2}}} = {{{{10}^5}} \over {{{(20 \times {{10}^{ - 2}})}^2}}}$$

Strain' = $${{dx'} \over {x'}} = {{dx'} \over {20 \times {{10}^{ - 2}}}}$$

Now, it is given that the material is same for both cubes, therefore,

$${{Stress} \over {Strain}} = {{Stress'} \over {Strain'}} \Rightarrow {{{{10}^7}} \over {0.05}} = {{{{10}^5}} \over {(20 \times {{10}^{ - 2}})}} \times {{20 \times {{10}^{ - 2}}} \over {dx'}}$$

$$ \Rightarrow {{{{10}^7} \times {{10}^2}} \over 5} = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}}{1 \over {dx'}}$$

$$ \Rightarrow dx' = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}} \times {5 \over {{{10}^7} \times {{10}^2}}} = {{5 \times {{10}^{ - 2}}} \over {20}} = 0.25 \times {10^{ - 2}}$$ m

$$ \Rightarrow dx' = 0.25$$ cm