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

Answer 1

Answer: (C)

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