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Capillary tube with very thin walls is suspended vertically in a cup containing water. Before the capillary is suspended, the cup was balanced against weights on a beam balance. When capillary is dipped to a small distance in the water, water rises in the capillary. Water completely wets the capillary walls. To restore the balance, we have to decrease the load on the other side by $0.14 π g .$ If the surface tension of water is $0.07 N / m,$ determine the radius $r (i n m m)$ of the capillary.

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answer is 5.

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Detailed Solution

$F$ by capillary on water $= s \times 2 π r \times 2$
$4 π s r = m g$
$4 π \times 0.07 \times r = 0.14 π \times 10^{- 3} \times 10$
$r = \frac{20}{4} = 5 m m$

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