Question

Figure shows a hydraulic press with the larger piston of diameter 35 cm at a height of 1.5 m relative to the smaller piston of diameter 10 cm.The mass on the smaller piston is 20 kg. What is the force exerted on the load by the larger piston. (The density of oil in the press is 750 kg/m${}^3$, g $=9.8m/s^2$).

• A. $F=2.3\times {10}^3$ N
• B. $F=1.3\times {10}^3$ N
• C. $F=3.3\times {10}^3$ N
• D. $F=4.3\times {10}^3$ N

Answer 1

Answer: (B)

$F=1.3\times {10}^3$ N

$\frac{F_1}{A_1}=\frac{20\times 9.8}{\frac{\pi (.1{)}^2}{4}}=24968.15N$

$\frac{F_2}{A_2}=\frac{F_2}{\frac{\pi (.35{)}^2}{4}}=10.4F_{2}N$

As the fluid is still, thus pressure at point A must be equal to pressure at point B.

$P_A=\frac{F_1}{A_1}+\rho gh=24968.15+\rho gh$

$P_B=\frac{F_2}{A_2}+\rho g(h+1.5)=10.4F_2+\rho g(h+1.5)$

Now $P_A=P_B⟹24968.15=\rho g\left(1.5\right)+10.4F_2$

$24968.15=750\left(9.8\right)\left(1.5\right)+10.4F_2$

$⟹F_2\approx 1.3\times {10}^{3}N$