1. Use Pascal’s law

In a hydraulic press, pressure applied to a confined liquid is transmitted equally throughout the liquid.

So, for equilibrium:

$P_1 = P_2$

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

where:

• $F_1 = 10\,\text{N}$ is the force on the thinner arm,
• $A_1$ and $A_2$ are cross-sectional areas of the thinner and thicker arms.

2. Relate area to diameter

For a circular cross-section,

$A = \frac{\pi d^2}{4}$

Hence,

$\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2$

From the figure, the diameter of the thicker arm is 10 times the diameter of the thinner arm. Therefore,

$\frac{d_2}{d_1} = 10$

So,

$\frac{A_2}{A_1} = 10^2 = 100$

3. Find the force on the thicker arm

Using

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

we get

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

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

4. Final answer

$\boxed{1000}$

So, the force required on the thicker arm is 1000 N.