A U- tube with limbs of diameters 5, mm and 2 | vedclass

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Question

$P_{\mathrm{A}}+P_{\mathrm{o}}=$ atmosphere $\mathbf{P r}=\mathbf{P c}$

$P_{\mathrm{B}}=P-\frac{2 \mathrm{T}}{\mathrm{R}}, \mathrm{P}_{\mathrm{D}}=P-

Vedclass · Accepted Answer

$8.4\, mm$

Vedclass · Answer

$P_{\mathrm{A}}+P_{\mathrm{o}}=$ atmosphere $\mathbf{P r}=\mathbf{P c}$

$P_{\mathrm{B}}=P-\frac{2 \mathrm{T}}{\mathrm{R}}, \mathrm{P}_{\mathrm{D}}=P-\frac{2 \mathrm{T}}{\mathrm{r}}$

$P_{E}=P-\frac{2 T}{r}+\mathbf{h d g}$

$	ext { In equilibrium } \quad P_{\mathrm{B}}=P_{E}$

$P-\frac{2 T}{R}=P-\frac{2 T}{r}+h d g \Rightarrow h=\frac{2 T}{d g}\left[\frac{1}{r}-\frac{1}{R}ight]$

$=\frac{2 	imes 7 	imes 10^{-2}}{10^{3} 	imes 10 	imes 10^{-3}}\left[1-\frac{1}{2.5}ight] \mathrm{m}=8.4 \mathrm{mm}$

A $U-$ tube with limbs of diameters $5\, mm$ and $2\, mm$ contains water of surface tension $7 \times 10^{-2}$ newton per metre, angle of contact is zero and density $10^3\, kg/m^3$. If $g$ is $10 \,m/s^2$, then the difference in level of two limbs is :-