Question

A wire of 1m length and 4mm radius is clamped at upper end. The lower end is twisted by an angle of 30°.

• A. 0.120
• B. 1.20
• C. 120
• D. 0.0120

Answer 1

Answer: (A)

0.120°

Answer 2

The angle of shear ($\phi$) is related to the angle of twist ($\theta$) by the formula $\phi = \frac{r\theta}{l}$, where $r$ is the radius of the wire, $l$ is the length of the wire, and $\theta$ is the angle of twist in radians. If the angle of twist is given in degrees ($\theta_{deg}$), and we want the angle of shear in degrees ($\phi_{deg}$), the relation becomes: $\phi_{deg} = \frac{r}{l} \times \theta_{deg}$ Given: Length of wire, $l = 1$ m Radius of wire, $r = 4$ mm $= 4 \times 10^{-3}$ m Angle of twist, $\theta_{deg} = 30^\circ$

Substituting the values into the formula: $\phi_{deg} = \frac{4 \times 10^{-3} \text{ m}}{1 \text{ m}} \times 30^\circ$ $\phi_{deg} = 0.004 \times 30^\circ$ $\phi_{deg} = 0.12^\circ$ The value $0.12^\circ$ can be written as $0.120^\circ$.