A uniform chain of mass m and length L is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming the chain does not form a heap on the floor. Calculate the force exerted by it on the floor when the whole chain is on the floor.

\frac{3 m}{L}

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\frac{3 m g x}{L}

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\frac{g x}{L}

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Solution

The correct option is B $\frac{3 m g x}{L}$
$\begin{matrix} L e t u s t a k e a s m a l l e l e m e n t a t a d i s t a n c e x f r o m t h e f l o o r o f l e n g t h d y . N o w, d m = \frac{M d x}{L} T h e v e l o c i t y w i t h w h i c h e l e m e n t w i l l s t r i k e t h e f l o o r, v = \sqrt{2 g x} S o, M o m e n t u m t r a n s f e r r e d t o f l o o r i s M = (d m) v = \frac{[M \times d x \times \sqrt{2 g x}]}{L} N o w, t h e f o r c e e x e r t e d o n f l o o r c h a n g e i n m o m e n t u m i s g i v e n b y F_{1} = \frac{d M}{d t} = \frac{[M \times d x \times \sqrt{2 g x}]}{L \times d t} a n d v = \frac{d x}{d t} = \sqrt{2 g x} F_{1} = \frac{[M \times \sqrt{2 g x} \times \sqrt{2 g x}]}{L} F_{1} = \frac{M (2 g x)}{L} F_{1} = \frac{M g x}{L} N o w, a g a i n t h e f o r e c e e x e r t e d d u e t o x l e n g t h o f c h a i n o n f l o o r d u e t o i t s o w n w e i g h t i s, w = \frac{M g x}{L} T h e r e f o r e T o t a l f o r e c e e x e r t e d D = F_{1} + W = \frac{2 M g x}{L} + \frac{m g x}{L} = \frac{3 M g x}{L} H e n c e, o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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