A vertical U-tube contains a liquid of density` rho` and surface tension T. if the radius of the meniscus of liquid in the limbs of the U-tube are `R_(1)` and `R_(2)` find the difference in the liquid column in the limbs.

Options

A. `Deltah=(T(R_(1)-R_(2)))/(rhogR_(1)R_(2))`
B. `Deltah=(2T(R_(1)-R_(2)))/(rhogR_(1)R_(2))`
C. `Deltah=(2T(R_(1)+R_(2)))/(rhogR_(1)R_(2))`
D. `Deltah=(4T(R_(1)-R_(2)))/(rhogR_(1)R_(2))`

Answer

B

Explanation

Let the heights of liquid column in the limbs are `h_(1)` and `h_(2)` using the formula `Deltap=(2T)/(R)` for the meniscus in the limbs, we have the pressures at the points A and B gives as `P_(A)=P_(0)-(2T)/(R_(1))` and `P_(B)=P_(0)-(2T)/(R_(2))`
Using the formula `Deltap=rhogh`, we have the pressures at A and C `P_(A)=(P_(C))-P_(B)=rhog(h_(2)-h_(1))`, substituting `P_(A)` from eq. (i) from eq (ii) in eq (iii), we have
`[p_(0)-(2T)/(R_(1))]-[p_(0)-(2T)/(R_(2))]=rhogDeltah`
atmospheric pressure. Suppose a and b are the radio of the this gives `Deltah=(2T(R_(1)-R_(2)))/(rhogR_(1)R_(2))`