In a typical hydraulic press, a force of 20 N is exerted on a small piston with an area of 0.050$m^2$. What is the force exerted by a large piston on load if it has an area of 0.50$m^2$ ? 

Given that there are two pistons—one tiny and one huge—we must determine the force that the large piston will exert. For this, we are aware that the hydraulic pressure corresponds to the force-to-area ratio. Both the hydraulic pressure for small and big pistons is written. Equating both hydraulic pressures and substituting the aforementioned numbers will allow us to determine the big piston's needed force.

Because the force is applied to the small piston  $F_1=20N$
And the area where the force is applied by the small piston  $A_1=0.050m^2$
Water pressure is the ratio of force to the area where force is applied. 
Therefore, for small pistons, there is hydraulic pressure  $P_1={\displaystyle \frac{20}{0.050}}$ …. (1)
The force applied to the load by the large piston stand  $F_2$.
And the area where the force is applied by the big piston is  $A_2=0.50m^2$
Therefore, hydraulic pressure is for large pistons  $P_2={\displaystyle \frac{F_2}{0.50}}$ ……. (2)
Since both water pressures are equal, the balance between equations (1) and (2) is as follows.
$\Rightarrow {\displaystyle \frac{20}{0.05}}={\displaystyle \frac{F_2}{0.5}}$
$\Rightarrow F_2=200N$
Thus, the force exerted by the large piston is 200 N.
Hence, the correct answer is option A.