Question

A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a diameter of 0.15 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the pistons.
A. 5600 kg
B. 1400 kg
C. 2800 kg
D. 250 kg
E. 700 kg

Answer 1

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,

    $\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}$

    $\dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}$

    $\dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}$

    $F_2= \dfrac{250}{0.01^2}\times {0.075^2}$

               F₂ = 14062.5 N

F = m g

$m = \dfrac{F}{g}$

$m = \dfrac{14062.5}{9.8}$

m = 1435 Kg

hence, the correct answer is option B