Changing force without mass or acceleration alteration A water tank put on scale measuring $50 kg$ , inside the tank a fully submerged balloon tied with a thread to the tank bottom. 

If the thread was cut, will there be a different reading on the scale (momentarily until the balloon reach the surface) ?

I believe that the forces downward are not only the mass but the water pressure at the tank bottom multiplied by the area. A balloon being submerged will cause displacement and raise water level thus total pressure of tank bottom, that can be observed during the process of submerging the balloon and until you tie it to the tank bottom, only then the buoyancy force will cancel the added pressure until that thread is cut .
 A: The water tank has forces on it only of gravity and the scale (ignoring air pressure/buoyancy affects in the atmosphere).  Since no mass leaves the tank, the total gravitational pull on it must be constant.
That leaves only the scale.  Any changes in the normal force from it would accelerate the tank.
At the moment the string is cut the balloon starts to accelerate up.  Presumably it would then reach some nearly-constant speed as drag increases until it reached the surface.  But the space occupied by the balloon is replaced with water.  The net result of these changes is that the center of mass of the tank moves downward while the balloon moves upward.
So momentarily after the string is cut, the scale will be slightly lighter as the COM accelerates downward.  It will then return to the initial static weight as the balloon ascends at constant speed.  When the balloon reaches the surface, the scale will show slightly higher as the COM decelerates to a stop.

I believe that the forces downward are not only the mass but the water pressure at the tank bottom multiplied by the area.

(Again, assuming we are ignoring air pressure affects here) the mass of the water is what generates the pressure.  As long as we account for all the masses, we don't have to worry about pressure separately.
A: This is not how it works. You don't sum up pressure to weight because the pressure water apply on the floor of the container IS its weight, and indeed if you calculate the total force ie you multiply the pressure $\rho gh$ for the surface of the base, suppose rectangular cuboid container you obtain $\rho hAg$ $h\cdot A$ is the volume $V$ and $\rho V$ is the mass so $mg$ with is indeed the weight.
The right way to think about it is using dynamic principle and so impose that the sum of the force on a resting body is zero. In this case the only force the water is subjected is its weight, so the force the scale apply must even be equal to the weight.
When you cut the thread the balloon accelerate up, so there is a force the water applies on it and the opposite force that the balloon apply on the water directed down (third law), so the weight measured on the scale will increase. The amount of the increase depends on the volume of the body : $\rho Vg$, where $\rho$ is the density of water and $V$ the volume of the balloon.
A: This answer will overlap with previous ones. When you cut the thread a mass of water equal to the volume will fall down, so the indicated weight will drop by this amount. If the balloon can escape to the surface without friction, the scale wil continue showing this weight until the water hits the remaining water. It will have accelerated so the scale will indicate a higher weight (overshoot) until this kinetic energy has dissipated and converted into heat. Since a balloon senses quite a bit of friction when escaping this overshoot is probably absent and the weight will gradually recover to its initial value.
Variation of gravity with height and increase of temperature of the water adding to its mass are ignored ... 
