Capillary tubes with very thin walls are suspended vertically in a cup containing water. Before the capillary was 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 scales by 0.14g.If the surface tension of water is $0.07N/m$,determine the radius r(in mm) of the capillary.

Hint: Load is decreased on the other side of the beam which should be balanced by the surface tension force and liquid water completely wets the capillary walls which implies that the angle of contact of water with capillary walls is zero.

Complete step by step answer:
Surface tension force must be balanced by the load decreased
And Surface tension force(F) $= 2 \times T.2\pi r = 2 \times T \times 2\pi r\cos \theta $ where T is surface tension
Putting values we have,
$F = 2 \times 0.07 \times 2\pi r\cos 0 = 2 \times 0.07 \times 2 \times \dfrac{{22}}{7} \times r \times 1 = 0.88r$
Now load is $0.14g$ so load force $F' = \dfrac{{0.14}}{{1000}} \times 10 = 0.0014$
Both these forces must be balanced so,
$F = F' \Rightarrow 0.88r = 0.0014 \Rightarrow r = \dfrac{{0.0014}}{{0.88}} = 0.00159m = 1.59\;mm$

Hence final answer is $1.59\;mm$.

Note: Here load is decreased to balance the beam it means force by surface tension of water must be upward this is indeed true because if force wouldn’t act upwards there will be no rise in capillary tube. Or else you can also solve this by calculating the height of the capillary rise and then equating the mass of those risen water with the decreased load.