Answer :

Correct answer is : 0.1

P = 250 N/mm2, a = 200 mm, Elastic modulus (E) = 2 × 105 N/mm2, Poisson's ratio (μ) = 0.3

σx = σy = σz = -P

$ϵ=\frac{{\sigma }_x}{E}-\mu \frac{{\sigma }_y}{E}-\mu \frac{{\sigma }_z}{E}$

$ϵ=\frac{\delta a}{a}=\frac{1}{E}\left[P_x-\mu \left(P_y+P_z\right)\right]$

$\frac{\delta a}{a}=\left(\frac{-P}{E}\right)\left(1-2\mu \right)$

$\delta a=-\frac{250\times 200\times \left\{1-(2\times 0.3)\right\}}{2\times {10}^5}$ = -0.10 mmThe reduction in the side of a cube is 0.10 mm.