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