A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg The area of cross-sections of the piston carrying the load is 425 cm 2. What maximum pressure would the smaller piston have to bear?

Question

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg The area of cross-sections of the piston carrying the load is 425 cm 2. What maximum pressure would the smaller piston have to bear? (A) 3.16 × 105 Pa (B) 4.72 × 105 Pa (C) 5.84 × 105 Pa (D) 6.92 × 105 Pa

Tardigrade Admin · Accepted Answer

Pressure is equally transmitted in the fluid, so the pressure is same for both the pistons. P=(F/A)=((3000 kg )(9.8 ms -2)/(425 × 10-4 m 2)) =6.92 × 105 Nm -2

Q. A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000 k g$ The area of cross-sections of the piston carrying the load is $425 c m^{2}$ . What maximum pressure would the smaller piston have to bear?

$3.16 \times 1 0^{5} P a$

$4.72 \times 1 0^{5} P a$

$5.84 \times 1 0^{5} P a$

$6.92 \times 1 0^{5} P a$

Pressure is equally transmitted in the fluid, so the pressure is same for both the pistons.
$P = \frac{F}{A} = \frac{( 3000 k g ) ( 9.8 m s ^{- 2} )}{( 425 \times 1 0 ^{- 4} m ^{2} )}$
$= 6.92 \times 1 0^{5} N m^{- 2}$

Q. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg The area of cross-sections of the piston carrying the load is 425cm2. What maximum pressure would the smaller piston have to bear?

3.16×105Pa

4.72×105Pa

5.84×105Pa

6.92×105Pa