Question

A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density $800kg{m}^{-3}$ and of surface tension $3\times {10}^{-2}N{m}^{-2}$.The angle of contact of liquid-glass is given by $cos\theta =0.3$. If $g=10m{s}^{-2}$, the rise of liquid in the capillary tube is______ $\left(Cm\right)$:

• A. $9$
• B. $0.9$
• C. $9x{10}^{-3}$
• D. $0.09$

Answer 1

Answer: (B)

$0.9$

From the formula $h=\frac{2Tcos\theta }{rg\rho }$we get

$h=\frac{2\times 3\times {10}^{-2}\times \left(0.3\right)}{.25\times {10}^{-3}\times 10\times 800}$

$=.009m$

$=0.9cm$