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A particle is moving uniformly in a circular path of radius $r$. When it moves through an angular displacement $\theta$, then the magnitude of the corresponding linear displacement will be
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Verified Answer
The correct answer is:
$2 r \sin \left(\frac{\theta}{2}\right)$
$\ln \Delta A O B \quad \sin \frac{\theta}{2}=\frac{A B}{A O}$
$\Leftrightarrow A O=r)$
$\begin{aligned}
A B &=A O \sin \frac{\theta}{2} \Rightarrow A B=r \sin \frac{\theta}{2} \\
A C &=A B+B C \quad(\because A B=B C) \\
&=r \sin \frac{\theta}{2}+r \sin \frac{\theta}{2} \\
A C &=2 r \sin \frac{\theta}{2}
\end{aligned}$
So, the magnitude of the corresponding linear displacement will be $2 r \sin \frac{\theta}{2}$
$\Leftrightarrow A O=r)$
$\begin{aligned}
A B &=A O \sin \frac{\theta}{2} \Rightarrow A B=r \sin \frac{\theta}{2} \\
A C &=A B+B C \quad(\because A B=B C) \\
&=r \sin \frac{\theta}{2}+r \sin \frac{\theta}{2} \\
A C &=2 r \sin \frac{\theta}{2}
\end{aligned}$
So, the magnitude of the corresponding linear displacement will be $2 r \sin \frac{\theta}{2}$
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