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NEET PHYSICSMechanical Properties of FluidMedium

Question

A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm10 \text{ mm} above the water level on the other side. Meanwhile, the water rises by 65 mm65 \text{ mm} from its original level (see diagram). The density of the oil is:

A

650 kg m3650 \text{ kg m}^{-3}

B

425 kg m3425 \text{ kg m}^{-3}

C

800 kg m3800 \text{ kg m}^{-3}

D

928 kg m3928 \text{ kg m}^{-3}

Step-by-Step Solution

  1. Identify the Principle: In a U-tube containing static fluids, the pressure at the same horizontal level in a continuous fluid must be equal . Let's choose the interface between the oil and water in the left arm as our reference level.
  2. Determine Heights:
  • The water rises by 65 mm65 \text{ mm} from its original level in the right arm. Since the liquid is incompressible, the water level must have gone down by 65 mm65 \text{ mm} in the left arm. Therefore, the total height of the water column above the reference level (interface) in the right arm is hwater=65 mm+65 mm=130 mmh_{water} = 65 \text{ mm} + 65 \text{ mm} = 130 \text{ mm}.
  • The oil column in the left arm stands 10 mm10 \text{ mm} above the final water level in the right arm. So, the total height of the oil column is hoil=hwater+10 mm=130+10=140 mmh_{oil} = h_{water} + 10 \text{ mm} = 130 + 10 = 140 \text{ mm}.
  1. Set up the Pressure Equation: At the reference level (interface), the pressure due to the oil column equals the pressure due to the water column. Poil=PwaterP_{oil} = P_{water} ρoilghoil=ρwaterghwater\rho_{oil} g h_{oil} = \rho_{water} g h_{water} ρoilhoil=ρwaterhwater\rho_{oil} h_{oil} = \rho_{water} h_{water}
  2. Calculate Density:
  • ρwater=1000 kg m3\rho_{water} = 1000 \text{ kg m}^{-3}
  • hwater=130 mmh_{water} = 130 \text{ mm}
  • hoil=140 mmh_{oil} = 140 \text{ mm} ρoil=ρwater×hwaterhoil\rho_{oil} = \rho_{water} \times \frac{h_{water}}{h_{oil}} ρoil=1000×130140=1000×1314928.57 kg m3\rho_{oil} = 1000 \times \frac{130}{140} = 1000 \times \frac{13}{14} \approx 928.57 \text{ kg m}^{-3} Rounding to the nearest integer, we get 928 kg m3928 \text{ kg m}^{-3}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Mechanical Properties of Fluid. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMechanical Properties of Fluidatmospherepartiallyfilledimmisciblepoured

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