question_answer

The upper end of a wire 1 m long and 2 mm radius is clamped. The lower end is twisted through an angle of\[45{}^\circ \]. The angle of shear is

A) \[0.09{}^\circ \]                                

B) \[0.9{}^\circ \]

C)  \[9{}^\circ \]                                     

D) \[90{}^\circ \]

Correct Answer: A

Solution :

                 In this figure,\[\theta \]is the angle of twist and\[\phi \]is the angle of shear. \[BB'=O'B\times \theta =AB\times \phi \] Or           \[\phi =\frac{O'B\times \theta }{AB}\]                 \[=r\times \frac{\theta }{l}\]                 \[=\frac{2\times {{10}^{-3}}\times {{45}^{o}}}{1}\]                 \[={{0.09}^{o}}\]