Physics: Principles with Applications (7th Edition)

The pressure on the sample is $4.0\times 10^7~Pa$
The lever rotates about an axis of rotation. The sum of the torque about this axis equals zero. We can find the force on the small cylinder as: $(320~N)(2L)-F(L) = 0$ $F = 640~N$ We can find the pressure on the small cylinder as: $P = \frac{F}{A}$ $P = \frac{640~N}{\pi~(1.0\times 10^{-2}~m)^2}$ $P = 2.037\times 10^6~N/m^2$ By Pascal's Principle, the same pressure is exerted on the large cylinder. We can find the force on the large cylinder. $F = P~A$ $F = (2.037\times 10^6~N/m^2)(\pi)(5.0\times 10^{-2}~m)^2$ $F = 1.60\times 10^4~N$ We can find the pressure on the sample as: $P = \frac{F}{A}$ $P = \frac{1.60\times 10^4~N}{4.0\times 10^{-4}~m^2}$ $P = 4.0\times 10^7~Pa$ The pressure on the sample is $4.0\times 10^7~Pa$.