Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.3k views
in Physics by (91.6k points)
closed by
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.

1 Answer

0 votes
by (91.1k points)
selected by
 
Best 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`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...