Question

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-shaped tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is $7.3 \times 10^{-2}Nm^{−1}$ Take the angle of contact to be zero and density of water to be $1.0 \times 10^3 kg m^{-3}$.$(g=9.8ms^{−2})$

Answer 1

Hint: At zero angle of contact, tangent made between two surfaces will be along the surface and When angle of contact is zero degree radius of meniscus and the radius of bores are equal.

Complete step by step solution:
$\eqalign{
  & {\text{As given }} r_1 = \dfrac{3}{2} \cr
  & \Rightarrow {{r_1 = 1}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{m}} \cr
  & {\text{Also }} r_2 = \dfrac{6}{2} \cr
  & \Rightarrow {\text{3}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{m}} \cr
  & {\text{Also surface tension = 7}}{\text{.3}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}} \cr
  & {{\rho = 1}}{\text{.0}} \times {\text{1}}{{\text{0}}^3}{\text{kg}}{{\text{m}}^{ - 1}}{\text{ and g = 9}}{\text{.8m/se}}{{\text{c}}^2} \cr
  & {\text{When angle of contact is zero degree; radius of meniscus and the radius of bores are equal}} \cr
  & {{P_1 = }}\dfrac{{2{\text{T}}}}{{{{r_2}}}} \cr
  & {\text{Excess pressure in first bore}} \cr
  & \Rightarrow {{P_1 = }}\dfrac{{2 \times 7.3 \times {{10}^{ - 2}}}}{{1.5 \times {{10}^{ - 3}}}} \cr
  & \Rightarrow {{P_1 = 97}}{\text{.3 Pa}} \cr
  & {\text{Excess pressure in second bore}} \cr
  & {{P_2 = }}\dfrac{{{\text{2T}}}}{{{{r_2}}}} \Rightarrow {{P_2 = }}\dfrac{{2 \times 7.3 \times {{10}^{ - 2}}}}{{3 \times {{10}^{ - 3}}}} \cr
  & \therefore {{P_2 = 48}}{\text{.7 Pa}} \cr
  & {\text{Hence, pressure difference in the two limbs of the tube}} \cr
  & \Rightarrow \Delta {{P = P_1 - P_2}} \cr
  & \Rightarrow \Delta {{P = h}}\rho {\text{g}} \cr
  & \Rightarrow {\text{h = }}\dfrac{{{{P_1 - P_2}}}}{{\rho g}} \cr
  & \Rightarrow {\text{h = }}\dfrac{{97.3 - 48.7}}{{1 \times {{10}^3} \times 9.8}} \cr
  & \therefore h = 5.0mm \cr} $

Note: Surface tension is the cohesive force of molecules on the surface of an element being attracted to each other to take the least possible surface area. In even simpler terms, it measures how much force it takes to hold a liquid together. The contact angle is measured as the angle where a liquid or vapor (but most often a liquid) makes contact with a solid surface. A high contact angle (picture correct) indicates that there is less wetting in the surface - that is, the liquid droplet will not spread too much over the surface. A low contact angle (left, bottom) indicates that the surface is high wet, meaning that small drops of water circulate over the surface.