Water rises to a height of 10, cm in capillar | vedclass

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Question

$\mathrm{h}=\frac{2 \sigma \cos 	heta}{\mathrm{r} ho \mathrm{g}} \Rightarrow \sigma \propto \frac{\mathrm{h} ho}{\cos 	heta}$

$\Rightarrow \f

Vedclass · Accepted Answer

$1 : 6$

Vedclass · Answer

$\mathrm{h}=\frac{2 \sigma \cos 	heta}{\mathrm{r} ho \mathrm{g}} \Rightarrow \sigma \propto \frac{\mathrm{h} ho}{\cos 	heta}$

$\Rightarrow \frac{\sigma_{\mathrm{w}}}{\sigma_{\mathrm{m}}}=\frac{\mathrm{h}_{\mathrm{w}} ho_{\mathrm{w}}}{\cos 	heta_{\mathrm{w}}} 	imes \frac{\cos 	heta_{\mathrm{m}}}{\mathrm{h}_{\mathrm{m}} ho_{\mathrm{m}}}$

$=\frac{10 	imes 1}{\cos 0^{\circ}} 	imes \frac{\cos 135^{\circ}}{-3.1 	imes 13.6}$

$=\frac{10 	imes(-0.707)}{-3.1 	imes 13.6} \approx \frac{1}{6}$

Water rises to a height of $10\, cm$ in capillary tube and mercury falls to a depth of $3.1\,cm$ in the same capillary tube. If the density of mercury is $13.6$ and the angle of contact for mercury is $135^o$, the approximate ratio of surface tensions of water and mercury is

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