Respuesta :
Answer:
option B
Explanation:
given,
Force exerted by the hydraulic jack piston = F₁ = 250 N
diameter of piston, d₁ = 0.02 m
r₁ = 0.01 m
diameter of second piston, d₂ = 0.15 m
r₂ = 0.075 m
mass of the jack to lift = ?
now,
[tex]\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}[/tex]
[tex]\dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}[/tex]
[tex]\dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}[/tex]
[tex]F_2= \dfrac{250}{0.01^2}\times {0.075^2}[/tex]
F₂ = 14062.5 N
F = m g
[tex]m = \dfrac{F}{g}[/tex]
[tex]m = \dfrac{14062.5}{9.8}[/tex]
m = 1435 Kg
hence, the correct answer is option B
