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 $250 \mathrm{~cm}^{2}$. The maximum pressure the smaller piston would have to bear is [Assume $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$]:

(C) $2 \times 10^{+6} \mathrm{~Pa}$

(A) $200 \times 10^{+6} \mathrm{~Pa}$

(B) $20 \times 10^{+6} \mathrm{~Pa}$

(D) $2 \times 10^{+5} \mathrm{~Pa}$

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Fluid Mechanics

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}

(A) $200 \times 1 0^{+ 6} Pa$
(B) $20 \times 1 0^{+ 6} Pa$
(C) $2 \times 1 0^{+ 6} Pa$ ✓ Correct
(D) $2 \times 1 0^{+ 5} Pa$

Answer: (C)

Solution

The force exerted by the vehicle is

F = m g = 5000 \times 10 = 50000

N. The area of cross-section is

A = 250 cm^{2} = 250 \times 1 0^{- 4} m^{2}

. The maximum pressure is

P = \frac{F}{A} = \frac{50000}{250 \times 1 0 ^{- 4}} = \frac{50000}{0.025} = 2 \times 1 0^{6}

Pa.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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