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 . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xdasuk3aid1u.png" /> (A) √(g sin θ (L2 - b2)/L) (B) (√2 g sin θ (L - b)/L) (C) √2 g sin θ (L - b) (D) √g sin θ (L - b)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-xdasuk3ax8ph.png" /> Given, u=0 Applying, v2-u2=2as a=gcos (90 - θ ) =gsin θ v2=2gsin 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 $B A$ is $L - b$ . Now, the chain is released and it slides down. Calculate the speed of the chain when its end reaches $B$ .

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

$\frac{2 g s in θ ( L - b )}{L}$

$2 g s in θ (L - b)$

$g s in θ (L - b)$

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

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(L−b)​​

2gsinθ(L−b)​

gsinθ(L−b)​