The lower end of capillary tube is immresed in mercury. The level...

Given data: Level of mercury in the tube is, (\mathrm{h}^{\prime}=2 \mathrm{~cm}). Level of mercury in the tube is given as: (\mathrm{h}^{\prime}=\frac{2 \mathrm{~T}^{\prime} \cos \theta^{\prime}}{\mathrm{r} \rho^{\prime} \mathrm{g}} . \quad \ldots (1)) where (\rho^{\prime}) is the density of mercury. Level of water in the tube is given as: (\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \quad \ldots (2)) where (\rho) is the density of water. Here, the same capillary is immersed in both the liquids hence, the radius of the capillary is same i.e. $r$ Using the standard values for variables and taking (\theta^{\prime}=135^{\circ}), in equation (1) gives: ($$\begin{aligned} & 2=\frac{2(484.5) \cos 135^{\circ}}{r(13.6)(980)} \\ & r=\frac{685.083}{(26656)} \\ & r=0.025 \mathrm{~cm} \end{aligned}$$) Using value of $r$ and (\theta=0^{\circ}) in equation (2), gives: ($$\begin{aligned} & \mathrm{h}=\frac{2(72.7) \cos ^{\circ}}{(0.025)(1)(980)} \\ & \mathrm{h}=\frac{145.4}{(24.5)} \\ & \mathrm{h}=5.9 \mathrm{~cm} \end{aligned}$$)