A thread of liquid is in a uniform capillary tube of length L, as measured by a ruler. The temperature of the tube and thread of liquid is raised by ΔT. If γ be the coefficient of volume expansion of the liquid and α be the coefficient of linear expansion of the material of the tube, then the increase ΔL, in the length of the thread, again measured by the ruler will be

Since a ruler is used, the scale used does not expand with the tube. If the radius of the capillary be r, the increase due to thermal expansion is given by dr=rαdT for a temperature rise of dT. Since area of cross section is A=πr2, we see thatdA/A=2dr/r or dA=A(2α)dT. Thus if the temperature is increased from T to T + dT, the cross-sectional area changes from A to A(1+2αdT). The volume expansion of the liquid gives V′=V+dV=V(1+γdT) , where γ is the coefficient of the volume expansion of the liquid. This causes change in length of thread and final length becomes L' = L + dL. The mass of liquid is constant; hence, L′A′=V′=V(1+γdT)=LA(1+γdT).But A′=A(1+2αdT)Hence, L′=L1+γdT1+2αdT         =L1+(γ−2α)dT−2αγ(dT)2The last term is negligible. Hence,  L′=L[1+(γ−2α)dT]                 ΔL=L(γ−2α)ΔT