Explanation

(a) Let length of chain be $l$ and mass $m$. Let a part $x$ of chain can hang over one edge of table having coefficient of friction
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$\therefore$ Pulling force, $F = \frac{mx}{l}g$
and friction force, $f = \mu N = \mu\frac{m}{l}(l - x)g$
For equilibrium, $F = f,$ hence
$\frac{mx}{l}.g = \mu\frac{m}{l}(l - x)g = 0.25\frac{m}{l}(l - x)g$
$\Rightarrow \ x = \frac{l}{5}$ or $\frac{x}{l} = \frac{1}{5} = 20\%$