Physics - Fluid Question with Solution | TestHub

Given data: Level of mercury in the tube is, ${\mathrm{h}}^{\prime }=2\mathrm{cm}$. Level of mercury in the tube is given as: ${\mathrm{h}}^{\prime }=\frac{2{\mathrm{T}}^{\prime }\cos {\theta }^{\prime }}{\mathrm{r}{\rho }^{\prime }\mathrm{g}}.\dots (1)$ where ${\rho }^{\prime }$ is the density of mercury. Level of water in the tube is given as: $\mathrm{h}=\frac{2\mathrm{T}\cos \theta }{\mathrm{r}\rho \mathrm{g}}\dots (2)$ 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: $\begin{array}{rl}& 2=\frac{2(484.5)\cos 135^{\circ }}{r(13.6)(980)}\\ & r=\frac{685.083}{(26656)}\\ & r=0.025\mathrm{cm}\end{array}$ Using value of $r$ and $\theta =0^{\circ }$ in equation (2), gives: $\begin{array}{rl}& \mathrm{h}=\frac{2(72.7){\cos }^{\circ }}{(0.025)(1)(980)}\\ & \mathrm{h}=\frac{145.4}{(24.5)}\\ & \mathrm{h}=5.9\mathrm{cm}\end{array}$