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

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) √(g sin θ (L2 - b2)/L) (B) √( 2 g sin θ (L2 - b2)/L) (C) √2 g sin θ (L2 - b2) (D) √ g sin θ (L2 - b2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/28fd326b950172916fe5b6708160447d-.png" /> Given, u=0 Applying, v2-u2=2as a=g cos(90-θ) = g sin θ v2=2g sin θ(L-b) v=√2g sin θ(L-b)

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

$\frac{g s i n θ ( L ^{2} - b ^{2} )}{L}$

$\frac{2 g s i n θ ( L ^{2} - b ^{2} )}{L}$

$2 g sin θ (L^{2} - b^{2})$

$g sin θ (L^{2} - b^{2})$

Given, $u = 0$
Applying, $v^{2} - u^{2} = 2 a s$
$a = g cos (90 - θ)$
$= g sin θ$
$v^{2} = 2 g sin θ (L - b)$
$v = 2 g sin θ (L - b)$

Q. 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

Lgsinθ(L2−b2)​​

L2gsinθ(L2−b2)​​

2gsinθ(L2−b2)​

gsinθ(L2−b2)​