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b/h < μ. Two thresholds, each a pure ratio. Sliding begins when the driving component beats friction, W sin θ > μ W cos θ, that is at tan θ = μ. Overturning begins when the weight vector through the centroid falls outside the lower corner, which is at tan θ = b/h. Whichever threshold is reached first as θ rises is the one that governs, so the block tips first when b/h < μ. Check it on numbers: with b = 0.6 m, h = 1.2 m and μ = 0.4, b/h = 0.5 > 0.4, so this block slides first, at 21.80°, and never gets to its tipping angle of 26.57°. Make it slender at h = 2.0 m and now b/h = 0.3 < 0.4, so the same block on the same surface tips first, at 16.70°. Option B is the same comparison backwards and describes sliding. Option D is the trap worth understanding rather than just avoiding: W cancels out of both criteria, so a heavier block of identical proportions fails at exactly the same angle in exactly the same way — mass changes the forces but not which threshold arrives first. This is the reason a retaining wall must be checked separately against sliding and against overturning, and why widening its base cures one while roughening its foundation cures the other.