A rod of 2m length and radius 1 cm is twisted at one end by 0.8 rad with respect to other end being clamped. The shear strain developed in its rod will be

To find the shear strain developed in the twisted rod, we can use the relationship between the angle of twist and the shear strain at the outer surface of the rod.

The shear strain ($\phi$) on the surface of a cylindrical rod of length $L$ and radius $r$ when twisted through an angle $\theta$ is given by the formula:
$\phi =\frac{r\theta }{L}$

Given values in the problem:
Length of the rod, $L=2\text{m}$
Radius of the rod, $r=1\text{cm}=0.01\text{m}$
Angle of twist, $\theta =0.8\text{rad}$

Substituting these values into the formula:
$\phi =\frac{0.01\text{m}\times 0.8\text{rad}}{2\text{m}}$

Simplify the expression in the numerator:
$0.01\times 0.8=0.008$

Now, divide by the length:
$\phi =\frac{0.008}{2}=0.004$

Thus, the shear strain developed in the rod is 0.004.