AIIMS2019Physics-Fluid Mechanics

AIIMS 2019 Physics Hydraulic Lift MCQ Question

Type: MCQ-numerical-Medium-Class 11

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

A

15.82×10⁵ Pa

B

6.92×10⁵ Pa

C

2.63×10⁵ Pa

D

1.12×10⁵ Pa

Correct Answer

Option A

Detailed Explanation

To find the maximum pressure that the smaller piston must bear, we use the formula for pressure, P=FAP = \frac{F}{A}, where FF is the force (weight of the car) and AA is the area of the piston. The weight of the car is calculated as F=mg=3000kg9.81m/s2=29430NF = m \cdot g = 3000 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 29430 \, \text{N}. Converting the area from cm² to m² gives A=425cm2=0.0425m2A = 425 \, \text{cm}^2 = 0.0425 \, \text{m}^2. Thus, the pressure is P=29430N0.0425m26.92×105PaP = \frac{29430 \, \text{N}}{0.0425 \, \text{m}^2} \approx 6.92 \times 10^5 \, \text{Pa}, which corresponds to option B, not A.

Option A is incorrect because it miscalculates the pressure based on the given values. Options C and D are also incorrect as they do not match the calculated pressure.

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