Question

There is a hydraulic pressure is loaded up with water (ρwater = 1 g/cm3) and has the pistons of 264 cm2 and 31 cm2 regions each. At first the pistons are offset, at that point a 60‑kg load is set on the bigger piston so it descends until new equilibrium is reached. To what height relative to it's initial level the small piston will rise?

h = Unit in cm

What mass ought to be set on the little piston so it moves back to the initial level

m = Unit in kg

Answer 1

Isoln 60kg. Load А, #264cm² 31cm2 (Let instal level gnitial pressures and fren on Left at ' and 'B' are equal Let Big piston descends through & height Small piston will rige height the there is (hth's between gap the Jenets of liquid and right sides. pg (hth) = mg [m= 60 kg AT → looo (hth) - CO 264 Xloy Also, Ah,="Agh' (volumed liquid 264 movement would be Constant) from @ 1000 hith' = 60 12648104 → 1000 X 295h' = 9.8 264 269x10ry h = 0.3322 m h = 33.22.com AL > h=31 hl 31 264 00

we on in Let place small piston so that it Comes to original position Again a pressure difference of pg(hth") hully fy by small (hthy mg gglhth) Az from equation pg (hth) = mg 0 Gold AI > ng Ar Mg AL m = 60x Az Al m= 60x31 264 Im= 7.05 kg A.