The end of a wire of length 0.5m and radius 10⁻³m is twisted through 0.80 radian. The shearing strain at the surface of wire will be

The correct answer is 1.6 x 10⁻³.

Step-by-Step Explanation:

When a wire of length $L$ and radius $r$ is twisted at one end through an angle of twist $\theta$ while the other end is kept fixed, the shearing strain at the surface of the wire is given by the relation:
$\varphi =\frac{r\theta }{L}$

Here, the given parameters are:
Length of the wire, $L=0.5\text{m}$
Radius of the wire, $r={10}^{-3}\text{m}$
Angle of twist, $\theta =0.80\text{radian}$

Substituting the given values into the shearing strain formula:
$\varphi =\frac{{10}^{-3}\times 0.80}{0.5}$

Simplifying the calculation:
$\varphi =\frac{0.80}{0.5}\times {10}^{-3}$

Since $0.80/0.5=1.6$, we get:
$\varphi =1.6\times {10}^{-3}$

Therefore, the shearing strain at the surface of the wire is $1.6\times {10}^{-3}$.