The upper end of a wire of diameter 12,mm and | vedclass

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(B) Given: Diameter $d = 12\,mm$,so radius $r = 6\,mm = 6 	imes 10^{-3}\,m$.Length of the wire $\ell = 1\,m$.Angle of twist $	heta = 30^{\circ}$.The

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$0.18$

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(B) Given: Diameter $d = 12\,mm$,so radius $r = 6\,mm = 6 	imes 10^{-3}\,m$.Length of the wire $\ell = 1\,m$.Angle of twist $	heta = 30^{\circ}$.The relationship between the angle of shear $\phi$ and the angle of twist $	heta$ for a wire is given by $r 	heta = \ell \phi$.Therefore,$\phi = \frac{r 	heta}{\ell}$.Substituting the values: $\phi = \frac{6 	imes 10^{-3}\,m 	imes 30^{\circ}}{1\,m} = 180 	imes 10^{-3}\,^{\circ} = 0.18^{\circ}$.Thus,the angle of shear is $0.18^{\circ}$.

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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