A barometer contains two uniform capillaries of radii `1.4xx10^(-3)m` and `7.2xx10^(-4)m`. If the height of liquid in narrow tube is `0.2m` more than that in wide tube, calculate the true pressure difference. Density of liquid `=10^(3)kg//m^(3)`, surface tension `=72xx10^(-3)N//m` and `g=9.8ms^(-12)`.

Options

A. `1360(N)/(m^(2))`
B. `1260mm`
C. `860(N)/(m^(2))`
D. `1860(N)/(m^(2))`

Answer

D

Explanation

let the pressures in wide and narrow limbs `P_(1)` and `P_(2)` respectively. If `R_(1)` and `R_(2)` be the radii of wide and narrow tube. The pressure just below the meniscus of wide tube `=P_(1)-(2T)/(R_(1))` and pressure just below the meniscus of narrow limb `=P_(2)-(2T)/(R_(2))`
Therefore , difference of these pressures
`(P_(1)-(2T)/(R_(1)))-(P_(2)-(2T)/(R_(2)))=hrhog`
Therefore, true pressure difference,
`P_(1)-P_(2)=hrhog-2T((1)/(R_(2))-(1)/(R_(1)))`
For the water and glass surface, taking the angle of contact `theta` to be zero, we have
`R_(1)=(r_(1))/(costheta)approxr_(1),R_(2)=(r_(2))/(costheta)approxr_(2)`
Where `r_(1)` and `r_(2)` are wide and narrow limbs respectively
`P_(1)-P_(2)=hrhog-2T((1)/(r_(2))-(1)/(r_(1)))=0.2xx10^(3)xx9.8xx-2xx72xx10^(-3)xx((1)/(7.2xx10^(-4))-(1)/(1.44xx10^(-3)))`
`=1.96xx10^(3)-0.10xx10^(3)`
`=1.86xx10^(3)(N)/(m^(2))=1860(N)/(m^(2))`