What is the force needed to apply a piston of 2 cm radius in order to result a force of 6000N at working piston of radius 6 cm?

i) The hydraulic lift is an application of Pascal's law. Pascal's law states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.

The correct option is c) Pascal's law.

ii) When pressure is applied to an enclosed fluid, it is transmitted unchanged to every portion of the fluid and the walls of the container. This is a fundamental principle of fluid mechanics, known as Pascal's principle.

The correct option is d) transmitted unchanged to every portion of the fluid and the walls of the container.

iii) For a hydraulic lift, the work done by the input piston is equal to the work done on the output piston (assuming an ideal system).
Let $A_1$ and $d_1$ be the area and distance moved by piston P1, and $A_2$ and $d_2$ be the area and distance moved by piston P2.
Given: $A_1 = A$, $A_2 = 2A$, $d_1 = x$.
The volume of fluid displaced by piston P1 must be equal to the volume of fluid displaced by piston P2.

$$V_1 = V_2$$ $$A_1 d_1 = A_2 d_2$$

Substitute the given values:

$$A \cdot x = (2A) \cdot d_2$$

Divide both sides by $A$:

$$x = 2 d_2$$

Solve for $d_2$:

$$d_2 = \frac{x}{2}$$

The distance moved by piston P2 is $\boxed{\frac{x}{2}}$.

iv) We use Pascal's principle, which states that the pressure exerted on the fluid by the smaller piston is equal to the pressure exerted by the fluid on the larger piston.
Let $F_1$ be the force applied on the input piston and $F_2$ be the force on the working piston.
Let $r_1$ be the radius of the input piston and $r_2$ be the radius of the working piston.

Given:
$r_1 = 2 \text{ cm}$
$r_2 = 6 \text{ cm}$
$F_2 = 6000 \text{ N}$

The area of a circular piston is $A = \pi r^2$.
So, $A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2$.

According to Pascal's principle:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

Substitute the area formulas:

$$\frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2}$$

Cancel $\pi$ from both sides:

$$\frac{F_1}{r_1^2} = \frac{F_2}{r_2^2}$$

Solve for $F_1$:

$$F_1 = F_2 \left(\frac{r_1}{r_2}\right)^2$$

Substitute the given values:

$$F_1 = 6000 \text{ N} \times \left(\frac{2 \text{ cm}}{6 \text{ cm}}\right)^2$$ $$F_1 = 6000 \text{ N} \times \left(\frac{1}{3}\right)^2$$ $$F_1 = 6000 \text{ N} \times \frac{1}{9}$$ $$F_1 = \frac{6000}{9} \text{ N}$$ $$F_1 = \frac{2000}{3} \text{ N}$$ $$F_1 \approx 666.67 \text{ N}$$

Rounding to the nearest whole number, $F_1 \approx 667 \text{ N}$.

The force needed is $\boxed{667 \text{ N}}$.