The upper end of a wire of diameter $12\,mm$ and length $1\,m$ is clamped and its other end is twisted through an angle of $30^{\circ}$. The angle of shear is$........^{\circ}$
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(b)
$r \theta=\ell \phi \Rightarrow \phi=\frac{ r \theta}{\ell}=\frac{6 mm \times 30^{\circ}}{1 m }=0.18^{\circ}$
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