91Ó°ÊÓ

1. Height$\text{h}=1.28\text{″¾}$
2. Initial height${\text{h}}_0=\text{h}/2=0.64\text{″¾}$
3. Temperature$\mathrm{Δ}\text{T}={\text{T}}_2−{\text{T}}_1=30°−20°=10°\text{C}$
4. Coefficient of thermal expansion${\mathrm{Î\pm }}_{\mathrm{glass}}=1×{10}^{−5}/\text{K}$
5. Coefficient of liquid${\mathrm{β}}_{\mathrm{liquid}}=4×{10}^{−5}/\text{K}$

Consider a vertical glass tube filled with liquid at a particular temperature. If we heat the glass tube, the liquid inside the glass tube will change its height. In this case, the cross-section area expands according to how the glass expands on heating. This can be analyzed by considering the area. Also, the height of the glass tube can be calculated by taking the ratio of the volume of the liquid to the total area of the glass.

Formula:

Base area of the tube,$\text{A}=\mathrm{Ï€}{\text{r}}^2$ ,,,(i)

The height of the tube,$\text{h}=\frac{\text{V}}{\text{A}}$ …(¾±¾±)

The linear expansion of a body,$\mathrm{Δ}\text{L}=\text{L}\mathrm{Î\pm }\mathrm{Δ}\text{T}$ …(¾±¾±¾±)

Where$\mathrm{Î\pm }$ is the coefficient of linear expansion of body.

The volume change in expansion of a body,$\mathrm{Δ}\text{V}={\text{V}}_0\mathrm{βΔ}\text{T}$ …(¾±±¹)

Where, $\mathrm{β}$is the volume expansion coefficient of body.

To calculate the height change in the glass tube, let the initial volume of the liquidbe${\text{V}}_0$ , the initial height be ${\text{h}}_0$, and the initial cross section area be ${\text{A}}_0$.

On differentiating the area equation (i), we get

$\begin{array}{c}\mathrm{dA}=2\mathrm{Ï€°ù}\mathrm{dr}\\ =2\mathrm{Ï€°ù}(\mathrm{°ùÎ\pm }\mathrm{dT})(âˆ\mu \mathrm{dr}=\mathrm{°ùÎ\pm }\mathrm{dT},\text{fromequation}(\mathrm{iii}\left)\right)\\ =2\mathrm{Î\pm }\left({\mathrm{Ï€°ù}}^2\right)\mathrm{dT}\\ =2\mathrm{Î\pm ´¡}\mathrm{dT}\text{(fromequation(i))}\end{array}$

Therefore, the height from equations (ii), (iii) and (iv) will be given as:

$\begin{array}{c}\mathrm{h}=\frac{{\mathrm{V}}_0+{\mathrm{V}}_0{\mathrm{β}}_{\mathrm{liquid}}\mathrm{Δ°Õ}}{{\mathrm{A}}_0+{\mathrm{A}}_{0}2{\mathrm{Î\pm }}_{\mathrm{glass}}\mathrm{Δ°Õ}}(\mathrm{V}={\mathrm{V}}_0+{\mathrm{V}}_0{\mathrm{β}}_{\mathrm{liquid}}\mathrm{Δ°Õ}\text{and}\mathrm{A}={\mathrm{A}}_0+{\mathrm{A}}_{0}2{\mathrm{Î\pm }}_{\mathrm{glass}}\mathrm{Δ°Õ})\\ =\frac{{\mathrm{V}}_0(1+{\mathrm{β}}_{\mathrm{liquid}}\mathrm{Δ°Õ})}{{\mathrm{A}}_0(1+2{\mathrm{Î\pm }}_{\mathrm{glass}}\mathrm{Δ°Õ})}\\ ={\mathrm{h}}_0\left(\frac{1+{\mathrm{β}}_{\mathrm{liquid}}\mathrm{Δ°Õ}}{1+2{\mathrm{Î\pm }}_{\mathrm{glass}}\mathrm{Δ°Õ}}−1\right)(âˆ\mu {\mathrm{V}}_0/{\mathrm{A}}_0={\mathrm{h}}_0)\\ =\left(0.64\text{″¾}\right)\left(\frac{1+(4×{10}^{−5}\text{ }/\text{K})(10°\text{C})}{1+2(1×{10}^{−5}\text{ }/\text{K})(10°\text{C})}−1\right)\left(\text{consideringthegivenquantities}\right)\\ =1.3×{10}^{−4}\text{m}\end{array}$

Hence, the value of the height of the liquid column is$1.3×{10}^{−4}\text{m}$