General equation learned for Normal stress on inclined plane σn=‌
σx+σy
2
+‌
σx−σy
2
‌cos‌2‌θ+τxy‌sin‌2‌θ And shear stress for inclined plane τn=−‌
σx−σy
2
‌sin‌2‌θ+τxy‌cos‌2‌θ when θ is measured clockwise from left face. Hence θ must be replaced by (−θ) in above equations and σy=0,τxy=0 σn=‌
σx
2
+‌
σx
2
‌cos‌2(−30)+0 σn=‌
4
2
+‌
4
2
‌cos(−60) σn=3MPa τn=−‌
σx
2
‌sin‌2(−θ) τn=‌
σx
2
‌sin(2θ) τn=‌
4
2
‌sin(60) τn=1.73 Since both the stress exceeds the given limits, answer is option (C).