Q136 / 160 · AP EAMCET 2021 (23 Aug Shift 1) · Physics
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
- Correct answerA$0.36^{\circ}$
- B$0.12^{\circ}$
- C$3.6^{\circ}$
- D$1.2^{\circ}$
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
$$
\begin{aligned}
\tan \theta & =\frac{r \phi}{l} \\
\theta & =\frac{r \phi}{l} \quad(\because \tan \theta \simeq \theta) \\
& =\frac{8 \times 10^{-3} \times 45}{1}=0.36^{\circ}
\end{aligned}
$$
$$
\therefore \quad \theta=\frac{r \phi}{l}
$$
Hence, the correct answer is option: 'A'.