A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of $$10$$ N is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is ______ N.

For the hydraulic press to be in equilibrium, the pressure in both arms must be equal (Pascal's Law)

$$P_1 = P_2 \implies \frac{F_1}{A_1} = \frac{F_2}{A_2}$$

Since the cross-sectional area $$A$$ is proportional to the square of the diameter ($$A = \pi \frac{d^2}{4}$$), 

$$F_2 = F_1 \times \left( \frac{d_2}{d_1} \right)^2$$

$$F_2 = 10 \times \left( \frac{14}{1.4} \right)^2$$

$$F_2 = 10 \times (10)^2$$

$$F_2 = 10 \times 100 = 1000\text{ N}$$