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 gets

depressed by an amount h in a capillary tube of radius r_{1} , while water rises

by 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}

 

Answers (1)

 

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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