1+sinα1−cosα
1+tanα1−tanα
1+sinα1−sinα
1+cosα1−cosα
Answer is A
Let the point be O of tube touching the ground.
Pressure due to liquid of density, d1, on the left side of O = Pressure due to liquid of density, d1 + Pressure due to liquid of density, d2, on the right side of O. Let P₀ be the pressure due to air above liquid, then
P₀ + d₁g(R − Rcosα) + d₂gRsin(π/2 − α) + d₂gRsinα = P₀ + d₁g(R − Rcos(π/2 − α))
Therefore, d₁(cosα − sinα) = d₂(cosα + sinα)
d₁/d₂ = (1 + tanα)/(1 − tanα).