A bar of elastic material is subjected to a direct compressive stress σ₁ in the longitudinal direction. Suitable lateral compressive stress σ₂ is applied along each of the other two lateral directions to limit the net strain in each of the lateral directions to half the magnitude of what it could be under σ₁ acting alone. If μ is the Poisson's ratio of the material, then the magnitude of σ₂ is

ANS:C -

Consider f1=longitudinal stress. f2 & f3 = lateral stress, m = poisson's ratio.
General expression for lateral strain, e2 = f2/E - m*f1/E - m*f3/E.

In first case, same stress is applied along each lateral direction. Therefore, f2 = f3.
Therefore, e2 = (1-m)*f2/E - m*f1/E.

In second case, f1 acting alone.
therefore, e2 = - m*f1/E.

Condition given is, lateral strain in first case = half of lateral strain in second case.
(1-m)*f2/E - m*f1/E = - 0.5*m*f1/E.
(1-m)*f2/E = 0.5*m*f1/E.
f2 = 0.5* [m/(1-m)] *f1.

Hence C is correct option.