Answer
$a. \quad 13.2 \mathrm{N}$
$b. \quad 13.2 \mathrm{N}$
Work Step by Step
a.
Before the string breaks, the reading on the scales will be the total weight of the water and ball,
$(m_{w}+m_{b})g=(1.20+0.150 \mathrm{k}\mathrm{g} )(9.81\mathrm{m}/\mathrm{s}^{2})=13.2 \mathrm{N}$.
b.
When the string breaks, the center of mass of the water-ball system will move down,
but will not accelerate down (as the text states).
So, if the center of mass is not accelerating, the scale is not acting with a net force on the system. Therefore, there is no reaction force of the system on the scale. The scale reads as it did before the string broke.
