Question

A chain of length $L$ and of mass $m$ is placed upon a smooth surface. The length of $BA$ is $L - b$. Now the Chain is released and it slides down. Calculate the speed of the chain when its end reaches $B$

• A. $\sqrt{\frac{g \, \sin \, \theta (L^2 - b^2)}{L}}$
• B. $\sqrt{\frac{ 2 g \, \sin \, \theta (L^2 - b^2)}{L}}$
• C. $\sqrt{2 g \, \sin \, \theta (L^2 - b^2)}$
• D. $\sqrt{ g \, \sin \, \theta (L^2 - b^2)}$

Answer 1

Answer: (B)

Given, $u=0$
Applying, $v^{2}-u^{2}=2as$
$a=g\,cos\left(90-\theta\right)$
$= g\,\sin\,\theta$
$v^{2}=2g\,\sin\,\theta\left(L-b\right)$
$v=\sqrt{2g\,\sin\,\theta\left(L-b\right)}$