## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Since the liquid is incompressible, the magnitude of the volume change $V_S$ in the small cylinder must be equal to the volume change $V_L$ in the large cylinder. Therefore; $V_S = V_L$ $\pi~R_1^2~h_1 = \pi~R_2^2~h_2$ $h_1 = \frac{R_2^2~h_2}{R_1^2}$ $h_1 = \frac{(0.040~m)^2~(0.20~m)}{(0.010~m)^2}$ $h_1 = 3.2~m$ The 2.0-cm-diameter piston must be pushed down 3.2 meters.