JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020
Answers
An open - ended U tube of uniform cross sectional area contain water (density 10³ kg/m³). initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water immiscible liquid) of density 800 kg/m³ is added to the left arm untill its length is 0.1 m, as shown in the schematic figure below.
To find : The ratio (h1/h2) of the heights of the liquid in the two arms is...
solution : here (h1 - 0.1 )+ h2 = 0.29 × 2
⇒h1 + h2 = 0.29 × 2 + 0.1 = 0.68......(1)
again, pressure due to 0.1 m kerosene oil, + pressure due to (h1 - 0.1) water = pressure due to h2 water
⇒ρ_k × g × 0.1 + ρ_w × g { (h1 - 0.1 ) = ρ_w × g × h2
⇒800 × 0.1 + 1000 × (h1 - 0.1) = 1000 × h2
⇒0.08 + h1 - 0.1 = h2
⇒h1 - h2 = 0.02 ........(2)
from equations (1) and (2) we get,
h1 = 0.35 and h2 = 0.33
so h1/h2 = 0.35/0.33 = 35/33
Therefore the correct option is (B)