Potential energy of the mass at a height above the surface is given by
$ = \frac{ 75 }{ 100 } \times 12 = 9 $ J $ \hspace20mm$ ..(i)
Now, KE of the mass at the end of fall
KE = $ \frac{1}{2} mv^2 $ $ \hspace20mm$ ..(ii)
Applying law of conservation of energy
$ \frac{1}{2} mv^2 = 9 $
v = $ \sqrt{ \frac{ 2 \times 9 }{ m }} $
$ \sqrt{ \frac{ 18 }{ 1}} = \sqrt{18 } \, ms^{ - 1}$