Answer:
[tex]x=0.730*10^{-3}m[/tex]
Explanation:
From the question we are told that:
Tensile stress [tex]\sigma_c = 14 MPa=>14*10^6[/tex]
Modulus of elasticity [tex]E=225GPa=>225*10^9[/tex]
Surface energy of MgO [tex]\gamma=1N/m[/tex]
Generally the equation for maximum allowable surface crack length is mathematically given by
[tex]x=\frac{2E \gamma}{\pi\sigma_c^2}[/tex]
[tex]x=\frac{2(225*10^9)(1)}{3.142*(14*10^6)^2}[/tex]
[tex]x=0.730*10^{-3}m[/tex]
