A wire of length $5 m$ is twisted through $30^{\circ}$ at free end. If the radius of wire is $1\ mm$, the shearing strain in the wire is $..........$
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$\theta=\frac{r}{L} \phi$
$\theta=\frac{1 \times 10^{-3} \times 30^{\circ}}{5}$ $\left\{\begin{array}{l}\text { Where } \\ \theta=\text { Angle of shear } \\ \phi=\text { Angle of twist } \\ r=\text { Radius of rod } \\ I=\text { length or rod }\end{array}\right.$
$\theta=6 \times 10^{-3}$
$=0.36^{\prime}$
$\theta=\frac{1 \times 10^{-3} \times 30^{\circ}}{5}$ $\left\{\begin{array}{l}\text { Where } \\ \theta=\text { Angle of shear } \\ \phi=\text { Angle of twist } \\ r=\text { Radius of rod } \\ I=\text { length or rod }\end{array}\right.$
$\theta=6 \times 10^{-3}$
$=0.36^{\prime}$
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