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Q9.8: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.8 Solution: A hydraulic automobile lift is designed to lift cars with a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often convert the piston area by using $1\text{ cm}=0.01\text{ m}$ directly, forgetting that area units must be squared; the correct conversion is $1\text{ cm}^2=10^{-4}\text{ m}^2$. Using the wrong factor inflates or deflates the pressure by a factor of $10^4$.

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

Step-by-Step Verified Solution

**Answer:** 6.92×105 Pa6.92 \times 10^5 \text{ Pa} **Step-by-Step Explanation:** Force on large piston: F=mg=3000×9.8=29400F = mg = 3000 \times 9.8 = 29400 N. Area of large piston: A=425A = 425 cm_ =425×10−4=0.0425= 425 \times 10^{-4} = 0.0425 m_. Pressure:
P=FA=294000.0425=6.92×105 PaP = \frac{F}{A} = \frac{29400}{0.0425} = 6.92 \times 10^5 \text{ Pa}
By Pascal's law, this pressure is transmitted to the smaller piston.

All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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