A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg. The area of cross section of the cylinder carrying the load is . The maximum pressure the smaller piston would have to bear is [Assume ] [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg. The area of cross section of the cylinder carrying the load is $250 {cm}^{2}$ . The maximum pressure the smaller piston would have to bear is [Assume $g = 10 {m/s}^{2}$ ] [2023]

1 $2 \times 10^{+ 5} Pa$

2 $200 \times 10^{+ 6} Pa$

3 $20 \times 10^{+ 6} Pa$

4 $2 \times 10^{+ 6} Pa$

Ans.
(4)

$Force = m g = 5000 g$

$Area of cross section = 250 {cm}^{2} = 250 \times 10^{- 4} m^{2}$

$Maximum pressure = \frac{Force}{Area of cross section}$

$= \frac{5000 g}{250 \times 10^{- 4}} = \frac{20 \times g}{10^{- 4}} = 2 \times 10^{6} Pa$

Want Us to Email you About Special Offers & Updates?