Back to Directory
NEET PHYSICSMedium

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 m−3650 \text{ kg m}^{-3}

B

425 kg m−3425 \text{ kg m}^{-3}

C

800 kg m−3800 \text{ kg m}^{-3}

D

928 kg m−3928 \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 m−3\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=1000×1314≈928.57 kg m−3\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 m−3928 \text{ kg m}^{-3}.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started