Question

When a capillary tube of diameter $0.8 \,mm$ is dipped in a liquid having density $ 800\,kgm^{-3}$ then the height of liquid in the capillary tube rises to $4 \,cm$. The surface tension of liquid is $ (g=9.8\,m/s^{2}): $

• A. $ 4.3\times {{10}^{-2}}N{{m}^{-1}} $
• B. $ 5.6\times {{10}^{-2}}N{{m}^{-1}} $
• C. $ 6.3\times {{10}^{-2}}N{{m}^{-1}} $
• D. $ 7.3\times {{10}^{-2}}N{{m}^{-1}} $

Answer 1

Answer: (C)

Given that
Density of liquid $(D)=800\, kgm ^{-3}$
Height of liquid $(h)=4 \,cm =0.04\, m$
Acceleration due to gravity $(g)=9.8\, m / s ^{2}$
Diameter of tube $(d)=0.8\, mm$
Radius of tube $(r)=0.4\, mm =4 \times 10^{-4} m$
Surface tension $(T)=?$
By using
$T =\frac{r h D g}{2} $
$=\frac{\left(4 \times 10^{-4}\right) \times(.04) \times 800 \times 9.8}{2}$
$=4 \times 10^{-4} \times .04 \times 400 \times 9.8 $
$=4 \times 4 \times 4 \times 98 \times 10^{-5} $
Hence, $ T=6.272 \times 10^{-2}$
$=6.3 \times 10^{-2} Nm ^{-1} $