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Standard XII

Physics

Surface Tension

Two capillary...

Question

Two capillary tubes of same diameter are put vertically one each in two liquids whose relating densities are 0.8 and 0.6 and surface tensions are $60 d y n e / c m$ and $50 d y n e / c m$ respectively. Ratio of heights of liquids in the two tubes $\frac{h_{1}}{h_{2}}$ is :-

\frac{10}{9}

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\frac{3}{10}

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\frac{10}{3}

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\frac{9}{10}

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Solution

The correct option is C $\frac{9}{10}$
The capillary rise in a tube is given by the expression,

h = $\frac{2 T c o s θ}{ρ g r}$

Therefore, $h \propto \frac{T}{ρ}$

So, $\frac{h_{1}}{h_{2}} = \frac{T_{1} ρ_{2}}{T_{2} ρ_{1}}$ = $\frac{60 \times 0.6}{50 \times 0.8}$ = $\frac{9}{10}$

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Two capillary tubes of same diameter are put vertically one each in two liquids whose relating densities are 0.8 and 0.6 and surface tensions are 60dyne/cm and 50dyne/cm respectively. Ratio of heights of liquids in the two tubes h1h2 is :-

The correct option is C 910The capillary rise in a tube is given by the expression,h = 2TcosθρgrTherefore, h∝TρSo, h1h2=T1ρ2T2ρ1= 60×0.650×0.8= 910

Two capillary tubes of same diameter are put vertically one each in two liquids whose relating densities are 0.8 and 0.6 and surface tensions are $60 d y n e / c m$ and $50 d y n e / c m$ respectively. Ratio of heights of liquids in the two tubes $\frac{h_{1}}{h_{2}}$ is :-

The correct option is C $\frac{9}{10}$
The capillary rise in a tube is given by the expression,

h = $\frac{2 T c o s θ}{ρ g r}$

Therefore, $h \propto \frac{T}{ρ}$

So, $\frac{h_{1}}{h_{2}} = \frac{T_{1} ρ_{2}}{T_{2} ρ_{1}}$ = $\frac{60 \times 0.6}{50 \times 0.8}$ = $\frac{9}{10}$