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(B) The shear strain $\phi$ in a twisted rod is given by the relation $r 	heta = L \phi$,where $r$ is the radius of the rod,$	heta$ is the angle of

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$0.004$

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(B) The shear strain $\phi$ in a twisted rod is given by the relation $r 	heta = L \phi$,where $r$ is the radius of the rod,$	heta$ is the angle of twist,$L$ is the length of the rod,and $\phi$ is the shear strain.Given:Radius $r = 1 \,cm = 10^{-2} \,m$Length $L = 2 \,m$Angle of twist $	heta = 0.8 \,radians$Substituting the values into the formula:$10^{-2} 	imes 0.8 = 2 	imes \phi$$0.008 = 2 	imes \phi$$\phi = \frac{0.008}{2} = 0.004$Thus,the shear strain developed is $0.004$.

$A$ $2 \,m$ long rod of radius $1 \,cm$ which is fixed from one end is given a twist of $0.8$ radians. The shear strain developed will be

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