A uniform metal chain of mass m and length L | vedclass

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(D) Let $\lambda$ be the linear mass density of the chain,so $\lambda = \frac{m}{L}$.The mass of the chain on one side is $m_1 = \lambda l$ and on the

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$4$

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(D) Let $\lambda$ be the linear mass density of the chain,so $\lambda = \frac{m}{L}$.The mass of the chain on one side is $m_1 = \lambda l$ and on the other side is $m_2 = \lambda (L - l)$.The net force on the chain is $F_{net} = (m_2 - m_1)g = \lambda(L - l - l)g = \lambda(L - 2l)g$.The total mass of the chain is $m = \lambda L$.The acceleration $a$ of the chain is given by $a = \frac{F_{net}}{m} = \frac{\lambda(L - 2l)g}{\lambda L} = \frac{(L - 2l)g}{L}$.Given that $a = \frac{g}{2}$ when $l = \frac{L}{x}$,we substitute these values:$\frac{g}{2} = \frac{(L - 2(\frac{L}{x}))g}{L}$$\frac{1}{2} = 1 - \frac{2}{x}$$\frac{2}{x} = 1 - \frac{1}{2} = \frac{1}{2}$$x = 4$.

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