One end of a wire of $8 \mathrm{~mm}$ radius and $100 \mathrm{~cm}$ length is fixed and the other end is…

One end of a wire of $8 \mathrm{~mm}$ radius and $100 \mathrm{~cm}$ length is fixed and the other end is twisted through an angle of $45^{\circ}$. The angle of shear is
  1. $0.36^{\circ}$
  2. $0.12^{\circ}$
  3. $3.6^{\circ}$
  4. $1.2^{\circ}$

Solution

Given, radius of wire $ \begin{aligned} r & =8 \mathrm{~mm} \\ & =8 \times 10^{-3} \mathrm{~m} \end{aligned} $ Length of wire, $l=100 \mathrm{~cm}=1 \mathrm{~m}$ Twist, $\phi=45^{\circ}$ Let shear angle be $\theta$. and $ \begin{aligned} \tan \theta & =\frac{r \phi}{l} \\ \theta & =\frac{r \phi}{l} \quad(\because \tan \theta \simeq \theta) \\ & =\frac{8 \times 10^{-3} \times 45}{1}=0.36^{\circ} \end{aligned} $ $ \therefore \quad \theta=\frac{r \phi}{l} $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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