The minimum acceleration with which a fireman can slide down a rope of breaking strength two-third of his weight is

We have $\mathrm{Mg}-\mathrm{T}=\mathrm{Ma}$ clearly, the minimum acceleration corresponds to maximum T. Now, the maximum possible T is equal to the breaking strength of the string. So, $\mathrm{Mg}-\frac{2}{3}\mathrm{Mg}=\mathrm{Ma}$

$\Rightarrow \mathrm{a}=\mathrm{g}/3$