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Properties of Solids and Liquids: Physics | JEE Main

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

\text{g} = 10\text{ m/s}^2

] :

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A

200 \times 10^{+6}\text{ Pa}

B

20 \times 10^{+6}\text{ Pa}

C

2 \times 10^{+6}\text{ Pa}

D

4 \times 2 \times 10^{+5}\text{ Pa}

Step-by-step solutionView

Correct answer

According to Pascal's principle, the pressure on the smaller piston equals the pressure exerted by the load on the larger piston, which is $2 imes 10^6\text{ Pa}$ .

Option analysis

Why each option works or fails

A ·

200 \times 10^{+6}\text{ Pa}

Believing that $1\text{ cm}^2 = 10^{-2}\text{ m}^2$ or $10^{-6}\text{ m}^2$ leads to an extra factor of $100$ in the calculated pressure. Convert area using $1\text{ cm}^2 = (10^{-2}\text{ m})^2 = 10^{-4}\text{ m}^2$ to obtain the correct SI unit value.

B ·

20 \times 10^{+6}\text{ Pa}

Making an arithmetic order-of-magnitude slip by a factor of 10 when simplifying $\frac{50000}{250 \times 10^{-4}}$ . Carefully cancel common factors: $\frac{50000}{0.025} = 2{,}000{,}000\text{ Pa} = 2 \times 10^6\text{ Pa}$ .

C ·

2 \times 10^{+6}\text{ Pa}

None. This is the correct calculation using Pascal's principle. Correct: $P = \frac{F}{A} = \frac{5000 \times 10}{250 \times 10^{-4}} = 2 \times 10^6\text{ Pa}$ .

D ·

4 \times 2 \times 10^{+5}\text{ Pa}

Confusing the values of area and mass during the calculation, or introducing an extraneous numerical factor. Directly apply $P = \frac{m g}{A}$ without introducing extra scaling factors.

Reviewed route

Solution

StepWorking

01given

Mass $m = 5000\text{ kg}$ , area of load cylinder $A = 250\text{ cm}^2 = 250 \times 10^{-4}\text{ m}^2$ , $g = 10\text{ m/s}^2$ .

02find

Maximum pressure $P$ the smaller piston must bear.

03strategise

By Pascal's principle, pressure is transmitted undiminished throughout an enclosed fluid. Hence, the pressure on the smaller piston equals the pressure supporting the load: $P = \frac{F}{A} = \frac{mg}{A}$ .

04execute

$P = \frac{5000 \times 10}{250 \times 10^{-4}} = \frac{50000}{0.025} = 2 \times 10^6\text{ Pa}$ .

✓verify

Units are $\text{N/m}^2 = \text{Pa}$ . For a 5-ton vehicle spread over $250\text{ cm}^2$ , $P = 2\text{ MPa} \approx 20\text{ atm}$ , which is realistic for hydraulic systems.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is the pressure on the smaller piston equal to the pressure under the load cylinder?

Pascal's principle states that in an enclosed incompressible static fluid, any applied pressure is transmitted equally in all directions throughout the fluid.

Answer

According to Pascal's principle, the pressure on the smaller piston equals the pressure exerted by the load on the larger piston, which is $2 imes 10^6\text{ Pa}$ .

Why each option works or fails

A: $200 \times 10^{+6}\text{ Pa}$ - Believing that $1\text{ cm}^2 = 10^{-2}\text{ m}^2$ or $10^{-6}\text{ m}^2$ leads to an extra factor of $100$ in the calculated pressure. Convert area using $1\text{ cm}^2 = (10^{-2}\text{ m})^2 = 10^{-4}\text{ m}^2$ to obtain the correct SI unit value.
B: $20 \times 10^{+6}\text{ Pa}$ - Making an arithmetic order-of-magnitude slip by a factor of 10 when simplifying $\frac{50000}{250 \times 10^{-4}}$ . Carefully cancel common factors: $\frac{50000}{0.025} = 2{,}000{,}000\text{ Pa} = 2 \times 10^6\text{ Pa}$ .
C · correct: $2 \times 10^{+6}\text{ Pa}$ - None. This is the correct calculation using Pascal's principle. Correct: $P = \frac{F}{A} = \frac{5000 \times 10}{250 \times 10^{-4}} = 2 \times 10^6\text{ Pa}$ .
D: $4 \times 2 \times 10^{+5}\text{ Pa}$ - Confusing the values of area and mass during the calculation, or introducing an extraneous numerical factor. Directly apply $P = \frac{m g}{A}$ without introducing extra scaling factors.

Step-by-step solution

given: Mass $m = 5000\text{ kg}$ , area of load cylinder $A = 250\text{ cm}^2 = 250 \times 10^{-4}\text{ m}^2$ , $g = 10\text{ m/s}^2$ .
find: Maximum pressure $P$ the smaller piston must bear.
strategise: By Pascal's principle, pressure is transmitted undiminished throughout an enclosed fluid. Hence, the pressure on the smaller piston equals the pressure supporting the load: $P = \frac{F}{A} = \frac{mg}{A}$ .
execute: $P = \frac{5000 \times 10}{250 \times 10^{-4}} = \frac{50000}{0.025} = 2 \times 10^6\text{ Pa}$ .
verify: Units are $\text{N/m}^2 = \text{Pa}$ . For a 5-ton vehicle spread over $250\text{ cm}^2$ , $P = 2\text{ MPa} \approx 20\text{ atm}$ , which is realistic for hydraulic systems.