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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. |
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Answer» Correct Answer - 7200 kg wt ; 0.28 cm 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` |
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