Question

The following observations were taken for determining surface tension $T$ of water by capillary method:
Diameter of capillary, $D=1.25\times {10}^{-2}m$
rise of water, $h=1.45\times {10}^{-2}m$
Using $g=9.80m/s^2$ and simplified relation $T=\frac{rhg}{2}\times {10}^{3}N/m$, the possible error in surface tension is closest to :

• A. $10\%$
• B. $0.15\%$
• C. $1.5\%$
• D. $2.4\%$

Answer 1

The correct option is C $1.5\%$

$T=\frac{rhg}{2}\times {10}^3$

Taking $\log$ on both sides

$\log x=\log r+\log h+\log \left(g\right)+\log \left({10}^3\right)-\log 2$

Differentiating

$\frac{\Delta T}{T}=\frac{\Delta r}{r}+\frac{\Delta h}{h}+0$ ($g=$ constant $\therefore \log \left(g\right)=$ constant $\therefore$ differentiation $=0$)

For $/$

$100\times \frac{\Delta T}{T}=(\frac{{10}^{-2}\times 0.01}{1.25\times {10}^{-2}}+\frac{{10}^{-2}\times 0.1}{1.45\times {10}^{-2}})\times 100$

$=0.8+0.689$

$=1.489$

$\therefore 100\frac{\Delta T}{T}=1.5\%$