Question 14: Properties of Matter

Question

Surface tension of two liquids (having same densities), $T_1$ and $T_2$, are measured using capillary rise method utilizing two tubes with inner radii of $r_1$ and $r_2$ where $r_1 > r_2$. The measured liquid heights in these tubes are $h_1$ and $h_2$ respectively. [Ignore the weight of the liquid above the lowest point of miniscus]. The heights $h_1$ and $h_2$ and surface tensions $T_1$ and $T_2$ satisfy the relation :

Accepted Answer

When a capillary tube is dipped in a liquid, the liquid rises due to surface tension until the upward force of surface tension is balanced by the weight of the liquid column. The surface tension T acts along the circumference of the meniscus. If $ heta$ is the contact angle, the vertical component of the force is : $$ \mathrm{F}_{\mathrm{up}}=\mathrm{T} \cos heta imes(2 \pi \mathrm{r}) $$ The weight of the liquid column of height h is : $$ W=\left(\pi r^2 h ight) ho g $$ At equilibrium : $$ \mathrm{T} \cos heta(2 \pi \mathrm{r})=\pi \mathrm{r}^2 \mathrm{~h} ho \mathrm{~g} $$ $$\Rightarrow $$ $$ \mathrm{h}=\frac{2 \mathrm{~T} \cos heta}{\mathrm{r} ho \mathrm{~g}} $$ If we assume the two liquids refers to the same substance, then: $\mathrm{T}_1=\mathrm{T}_2 \ldots$ (i) (Surface tension is a property of the material and temperature). $ ho$ and $g$ are constant. $ heta$ is constant. This results the relationship: $\mathrm{h} \propto \frac{1}{\mathrm{r}}$. We are given that $r_1>r_2 \Rightarrow \frac{r_1}{r_2}>1$. So, the ratio of heights of capillary rise, $$ \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_2}<1 \Rightarrow \mathrm{~h}_1<\mathrm{h}_2 \ldots (ii)$$ From both the relations, $\mathrm{h}_1<\mathrm{h}_2$ and $\mathrm{T}_1=\mathrm{T}_2$. Hence, the correct option is (B).

Answer

$h_1 > h_2$ and $T_1 = T_2$

Answer

$h_1 < h_2$ and $T_1 = T_2$

Answer

$h_1 > h_2$ and $T_1 < T_2$

Answer

$h_1 = h_2$ and $T_1 = T_2$

Properties of Matter

Choose the correct answer: