Answer:
a)
600 N
b)
1.2 x 10⁵ Pa
Explanation:
(a)
d₁ = diameter of small piston = 8 cm = 0.08 m
d₂ = diameter of large piston = 40 cm = 0.40 m
F₂ = force applied to large piston = 15000 N
F₁ = force applied to small piston = ?
Using pascal's law
[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]
Inserting the values
[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]
F₁ = 600 N
b)
Pressure applied is given as
[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]
[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]
P = 1.2 x 10⁵ Pa
