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Properties of Matter question

2023 · 8 Apr · Shift 2 · Q42
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Properties of Matter question

2023 · 8 Apr · Shift 2 · Q42

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg5000 \mathrm{~kg}5000 kg. The area of cross section of the cylinder carrying the load is 250 cm2250 \mathrm{~cm}^{2}250 cm2. The maximum pressure the smaller piston would have to bear is [\left[\right.[ Assume g=10 m/s2]\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]g=10 m/s2]
  1. A
    20×10+6 Pa20 \times 10^{+6} \mathrm{~Pa}20×10+6 Pa
  2. B
    200×10+6 Pa200 \times 10^{+6} \mathrm{~Pa}200×10+6 Pa
  3. C
    2×10+5 Pa2 \times 10^{+5} \mathrm{~Pa}2×10+5 Pa
  4. D
    2×10+6 Pa2 \times 10^{+6} \mathrm{~Pa}2×10+6 Pa
View written solutionFree

Correct answer: D

  1. Use Pascal’s law

In a hydraulic lift, the pressure transmitted through the fluid is the same throughout. So the pressure needed to lift the vehicle is

P=FAP = \frac{F}{A}P=AF​

where:

  • F=mgF = mgF=mg
  • A=A =A= area of the larger piston carrying the load
  1. Calculate the load force

Given:

  • m=5000 kgm = 5000\,\text{kg}m=5000kg
  • g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2

So,

F=mg=5000×10=50000 N=5×104 NF = mg = 5000 \times 10 = 50000\,\text{N} = 5 \times 10^4\,\text{N}F=mg=5000×10=50000N=5×104N

  1. Convert area into SI units

Given area:

250 cm2250\,\text{cm}^2250cm2

Since,

1 cm2=10−4 m21\,\text{cm}^2 = 10^{-4}\,\text{m}^21cm2=10−4m2

therefore,

250 cm2=250×10−4=2.5×10−2 m2250\,\text{cm}^2 = 250 \times 10^{-4} = 2.5 \times 10^{-2}\,\text{m}^2250cm2=250×10−4=2.5×10−2m2

  1. Compute pressure

P=5×1042.5×10−2P = \frac{5 \times 10^4}{2.5 \times 10^{-2}}P=2.5×10−25×104​

P=2×106 PaP = 2 \times 10^6\,\text{Pa}P=2×106Pa

  1. Match with options

The required maximum pressure is

2×106 Pa\boxed{2 \times 10^6\,\text{Pa}}2×106Pa​

So the correct option is D.

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