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 `DeltaT`. If `gamma` be the coefficient of volume expansion of the liquid and `alpha` be the coefficient of linear expansion of the material of the tube, then the increase `DeltaL` in the length of the thread, again measured by the ruler will be

Options

A. `DeltaL=L(gamma-alpha)DeltaT`
B. `DeltaL=L(gamma-2alpha)DeltaT`
C. `DeltaL=L(gamma-3alpha)DeltaT`
D. `DeltaL=LgammaDeltaT`

Answer

B

Explanation

Since a ruler is used the scale used dows not expand with the tube. If the radius of the capillary be r, the increase due to thermal expansion is given by `dr=ralphadT` for a temparature rise of dT. Since area of cross section is `A=pir^2`, we see that `(dA)/(A)=(2dr)/(r)` or `dA=A(2alpha)dT`. Thus if the temperature is increased from T to `T+dT`, the cross sectional area changes from A to `A(1+2alphadT)`. The volume expansion of the liquid gives `V'=V+dV=V(1+gammadT)`, where `gamma` 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+gammadT)=LA(1+gammadT)`.
`A'=A(1+2alphadT)`
Hence, `L'=L((1+gammadT)/(1+2alphadT))`
`=L[1+(gamma-2alpha)dT-2alphagamma(dT)^2]`
the last term is negligible
Hence, `L'=L[1+(gamma-2alpha)dT]`
`DeltaL=L(gamma-2alpha)dT`