A uniform chain of 6 , m length is placed on | vedclass

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(D) Let the total length of the chain be $L = 6 \, m$ and the length of the chain hanging over the edge be $x$. The length of the chain on the table i

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$2$

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(D) Let the total length of the chain be $L = 6 \, m$ and the length of the chain hanging over the edge be $x$. The length of the chain on the table is $(L - x)$.Let $\lambda$ be the mass per unit length of the chain.The mass of the chain on the table is $M_{table} = \lambda(L - x)$,and the mass of the hanging part is $M_{hanging} = \lambda x$.The normal force $N$ exerted by the table on the chain is $N = M_{table} g = \lambda(L - x)g$.The maximum static frictional force is $f_{s,max} = \mu_s N = \mu_s \lambda(L - x)g$.For the system to be at rest,the hanging weight must be balanced by the maximum static friction:$f_{s,max} = M_{hanging} g$$\mu_s \lambda(L - x)g = \lambda x g$$\mu_s(L - x) = x$Given $\mu_s = 0.5$ and $L = 6 \, m$:$0.5(6 - x) = x$$3 - 0.5x = x$$3 = 1.5x$$x = \frac{3}{1.5} = 2 \, m$.Thus,the maximum length of the chain hanging from the table is $2 \, m$.

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