Two pistons of hydraulic press have diameter of `30.0 cm` and `2.5 cm`, find the force exerted on the longer piston when 50.0 kg wt. is placed on smaller piston.

Answer

7200 kg wt ; 0.28 cm

Explanation

Here, `A_1 = (piD_1^2)/(4) = (22)/(7)xx(2.5)^2cm^2`
`A_2 = (pi D_2^2)/(4) = (22)/(7) xx(30)^2cm^2,`
`F_1 = 50 kg wt , l_1 = 4.0cm` Force on longer piston, `F_2 = (F_1 xxA_2)/(A_1) = 50 xx ((30)^2)/(2.5)^2`
=` 50 xx 144 = 7200 kg wt`
As, `F_1 l_1 =F_2 l_2, so `
`l_2 = (F_1l_1)/(F_2) = (50xx4)/(7200) = 0.028cm`
`:.` Distance covered in 10 strokes `= 10 xx 0.028 = 0.28 cm`