The upper end of a wire of radius $4\, mm$ and length $100\, cm$ is clamped and its other end is twisted through an angle of $60^o$. Then angle of shear is .......... $^o$
Medium
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Angle of shear $\phi=\frac{\mathrm{r} \theta}{\mathrm{L}}=\frac{4 \times 10^{-1}}{100} \times 60^{\circ}=0.24^{\circ}$
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