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A bar of square section of area $$a^2$$ is held such that one of its diameters is vertical. The maximum shear stress will develop at a depth h where h is
$$\frac{(2 \surd 3)}{4}$$
$$\frac{(3 \surd 2)}{4}$$
$$\frac{2}{\surd 3}$$
$$\frac{\surd 3}{4}$$
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