The upper end of a wire 1 metre long and 2 mm in radius is clamped. The lower end is twisted through an angle of 45° . The angle of shear is

The correct option is 0.09°.

To find the angle of shear, we use the relationship between the angle of twist and the angle of shear for a cylindrical wire or rod.
Let:
- $L$ be the length of the wire,
- $r$ be the radius of the wire,
- $\theta$ be the angle of twist at the lower end,
- $\varphi$ be the angle of shear.

For a wire of radius $r$ and length $L$, the displacement of the outermost layer of the wire at the lower end is related to both the angle of twist and the angle of shear by the arc length formula:
$r\theta =L\varphi$

From this relationship, we can solve for the angle of shear ($\varphi$):
$\varphi =\frac{r\theta }{L}$

Given data in the problem:
- Length of the wire, $L=1\text{m}$
- Radius of the wire, $r=2\text{mm}=2\times {10}^{-3}\text{m}$
- Angle of twist, $\theta ={45}^{\circ }$

Substitute these values into the formula to find $\varphi$:
$\varphi =\frac{(2\times {10}^{-3}\text{m})\times {45}^{\circ }}{1\text{m}}$
$\varphi =90\times {10}^{-3}{}^{\circ }$
$\varphi ={0.09}^{\circ }$

Therefore, the angle of shear is 0.09°.