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(B) Let the mass of the chain be $m$ and its length be $L$.The mass per unit length of the chain is $\lambda = m/L$.When a part of the chain of length

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$t_a < t_b$

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(B) Let the mass of the chain be $m$ and its length be $L$.The mass per unit length of the chain is $\lambda = m/L$.When a part of the chain of length $x$ is hanging,the acceleration of the chain in situation $(a)$ is given by $a = \frac{(\lambda x) g}{m} = \frac{gx}{L}$.Initially,when $x = L/2$,the acceleration is $g/2$,and it increases as $x$ increases.In situation $(b)$,when two identical masses $M$ are attached to the ends,the total mass of the system is $m + 2M$. The force causing acceleration is the weight of the hanging part of the chain plus the weight of the mass $M$ at the hanging end. The mass $M$ on the table does not contribute to the driving force but increases the inertia.The acceleration $a'$ in situation $(b)$ is $a' = \frac{(\lambda x) g + Mg}{m + 2M} = \frac{g(mx/L + M)}{m + 2M}$.Comparing $a$ and $a'$,we find that $a > a'$ for the range of motion. Since the acceleration in situation $(a)$ is consistently higher than in situation $(b)$ throughout the motion,the time taken to slip off the table will be less in situation $(a)$.Therefore,$t_a < t_b$.

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