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 cm². The maximum pressure the smaller piston would have to bear is [ Assume g = 10 m/s²] (a) 20 × 10⁺⁶ Pa (b) 200 × 10⁺⁶ Pa (c) 2 × 10⁺⁵ Pa (d) 2 × 10⁺⁶ Pa

Answer: (d). Pressure is the same throughout the fluid (Pascal's law), so the small piston bears the same pressure as the load cylinder. P = mg/A = 5000×10250×10⁻⁴ = 2×10⁶ Pa So the answer is (d).

A hydraulic automobile lift is designed to lift vehicles of mass 5000… | JEE Main 2023 (8 Apr, Shift 2)

Q13JEE Main 2023 · 8 APR · SHIFT 2Easy

Density, Compressibility & Pascal's Law

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

Show answer and solution

Answer: (d)

Pressure is the same throughout the fluid (Pascal's law), so the small piston bears the same pressure as the load cylinder.
$P = \dfrac{mg}{A} = \dfrac{5000\times10}{250\times10^{-4}} = 2\times10^6\text{ Pa}$
So the answer is (d).

Report a mistake

Density, Compressibility & Pascal's Law question from JEE Main 2023 (8 Apr, Shift 2)

More questions on Density, Compressibility & Pascal's Law