Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are $2.2 c m$ and $6.6 c m$ . The ratio of radii of the tubes will be

1 : 9

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1 : 3

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9 : 1

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3 : 1

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Solution

The correct option is D $3 : 1$
Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,
height $\propto \frac{1}{r a d i u s}$

$\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{2.2 c m}{6.6 c m} = \frac{r_{2}}{r_{1}}$

So, $\frac{r_{1}}{r_{2}} = 3$

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Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are 2.2cm and 6.6cm. The ratio of radii of the tubes will be

The correct option is D 3:1Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,height ∝1radiush1h2=r2r1⇒2.2cm6.6cm=r2r1So, r1r2=3

Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are $2.2 c m$ and $6.6 c m$ . The ratio of radii of the tubes will be

The correct option is D $3 : 1$
Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,
height $\propto \frac{1}{r a d i u s}$

$\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{2.2 c m}{6.6 c m} = \frac{r_{2}}{r_{1}}$

So, $\frac{r_{1}}{r_{2}} = 3$