The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass,are close to $135^{\circ}\: \: and\: \: 0^{\circ}$ , respectively. It is observed that mercury getsdepressed by an amount h in a capillary tube of radius $r_{1}$ , while water risesby the same amount h in a capillary tube of radius $r_{2}$. The ratio, $\frac{r_{1}}{r_{2}}$ , is then close to : Option 1) $\frac{4}{5}$ Option 2) $\frac{2}{5}$ Option 3) $\frac{3}{5}$ Option 4) $\frac{2}{3}$

Ascent Formula -

$\frac{2T\cos \Theta }{\rho gr}$

- wherein

$T- surface Tension$

$r- radius\: of\: capillary\: tube$

$\rho -liquid\: density$

$\theta - Angle\ of \ contact$

$h=\frac{2T cos\theta}{\rho rg}$

So, $h_{Hg}=h_w$

$\frac{2\times T_{Hg}\times cos\theta_{Hg}}{\rho _{Hg}R_{Hg}g}=\frac{2\times T_{w}\times cos\theta_{w}}{\rho _{w}R_{w}g}$

=> $\frac{R_{Hg}}{R_{w}}=\frac{r_1}{r_2}=(\frac{T_{Hg}}{T_w})(\frac{\rho _w}{\rho _{Hg}})(\frac{cos\theta_{Hg}}{cos\theta_w})$

=> $\frac{r_1}{r_2}=\frac{1}{13.6}\times 7.5\times \frac{1}{\sqrt2}$

=> $\frac{r_1}{r_2}\approx 0.4$

So, $\frac{r_1}{r_2}=\frac{2}{5}$

Option 1)

$\frac{4}{5}$

Option 2)

$\frac{2}{5}$

Option 3)

$\frac{3}{5}$

Option 4)

$\frac{2}{3}$

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