Question ID - 172782 | SaraNextGen Top Answer
A cylindrical capillary tube of $0.2$ mm radius is made by joining two capillaries T1 and T2 of different materials having water contact angles of $0^{\circ}$ and $60^{\circ}$, respectively. The capillary tube is dipped vertically in water in two different configurations, case I and II as shown in figure. Which of the following option(s) is(are) correct?
(Surface tension of water $=0.075 \mathrm{~N} / \mathrm{m}$, density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$, take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )
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(1) The correction in the height of water column raised in the tube, due to weight of water contained in the meniscus, will be different for both cases.
(2) For case I, if the capillary joint is $5 \mathrm{~cm}$ above the water surface, the height of water column raised in the tube will be more than $8.75 \mathrm{~cm}$. (Neglect the weight of the water in the meniscus)
(3) For case I, if the joint is kept at $8 \mathrm{~cm}$ above the water surface, the height of water column in the tube will be $7.5 \mathrm{~cm}$. (Neglect the weight of the water in the meniscus)
(4) For case II, if the capillary joint is $5 \mathrm{~cm}$ above the water surface, the height of water column raised in the tube will be $3.75 \mathrm{~cm}$. (Neglect the weight of the water in the meniscus)
$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gR}} \quad ; \mathrm{h}_{1}=\frac{2 \times 0.075 \times \cos 0^{\circ}}{1000 \times 10 \times 0.2 \times 10^{-3}}$
$\Rightarrow \mathrm{h}_{1}=75 \mathrm{~mm}$ (in T1) [If we assume entire tube of T1]
$\Rightarrow \mathrm{h}_{2}=\frac{2 \times 0.075 \times \cos 60^{\circ}}{1000 \times 10 \times 0.2 \times 10^{-3}}=37.5 \mathrm{~mm}$ (in T2) [If we assume entire tube of T2]
Option (1): Since contact angles are different so correction in the height of water column raised in the
tube will be different in both the cases, so option (1) is correct
Option (2) : If joint is $5 \mathrm{~cm}$ is above water surface, then lets say water crosses the joint by height $\mathrm{h}$, then:
$\Rightarrow P_{0}-\frac{2 T}{r}+\rho g h+\rho g \times 5 \times 10^{-2}$
$=\mathrm{P}_{0}$
$\Rightarrow \cos \theta=\frac{\mathrm{R}}{\mathrm{r}}, \mathrm{r}=\frac{\mathrm{R}}{\cos \theta}$
$\Rightarrow \rho g\left(\mathrm{~h}+5 \times 10^{-2}\right)=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{R}}$
$\Rightarrow \mathrm{h}=\frac{2 \times 0.075 \times \cos 60}{0.2 \times 10^{-3} \times 1000 \times 10}-5 \times 10^{-2}$
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$\Rightarrow \mathrm{h}=-\mathrm{ve}$, not possible, so liquid will not cross the interface, but angle of contact at the interface will change, to balance the pressure,
So option (2) is wrong.
Option (3): If interface is $8 \mathrm{~cm}$ above water then water will not even reach the interface, and water will
rise till $7.5 \mathrm{~cm}$ only in $\mathrm{T} 1$, so option (3) is right.
Option (4): If interface is $5 \mathrm{~cm}$ above the water in vessel, then water in capillary will not even reach the
interface. Water will reach only till $3.75 \mathrm{~cm}$, so option $(4)$ is right.
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