A capillary tube of radius $r$ dipped in a liquid of surface tension $T$ and density $\rho$, contact angle $\theta$, gives rise (or fall) to height $h$ where:
A$h = 2T \cos\theta / (r \rho g)$
B$h = T \cos\theta / (r \rho g)$
C$h = r \rho g / (2T \cos\theta)$
D$h = 2 T r / (\rho g \cos\theta)$
Answer & Solution
Correct answer: A. $h = 2T \cos\theta / (r \rho g)$
Force balance: $\pi r^2 h \rho g = 2\pi r T \cos\theta$ ⇒ $h = 2T \cos\theta / (r\rho g)$. Narrower tube ⇒ greater rise. If $\theta > 90°$ (non-wetting like mercury–glass), $\cos\theta < 0$ ⇒ capillary depression.
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