Discussion Forum

The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, d=1.25 × 10^-2 m rise of water, h=1.45 × 10^-2 m. Using g=9.80 m/s² and the simplified relation  $T= \frac{rhg}{2}\times 10^{3} N/m.$  The possible error in surface tension is closest to

Answer:

Explanation:

By ascent formula, we have surface tension.

                               $T= \frac{rhg}{2}\times 10^{3} N/m.$

                                 $= \frac{dhg}{4}\times 10^{3} N/m.$                 ( $\because r=\frac{d}{2}$ )

                           $\Rightarrow \frac{\triangle T}{T}=\frac{\triangle d}{d}+\frac{\triangle h}{h}$  [ given g is constant]

   So ,percentage

               $= \frac{\triangle T}{T}\times 100=(\frac{\triangle d}{d}+\frac{\triangle h}{h})\times 100$

  $=  (\frac{0.01 \times 10^{-2}}{1.25 \times 10^{-2}}+\frac{0.01 \times 10^{-2}}{1.45 \times 10^{-2}})\times 100$=1.5%

$\therefore \frac{\triangle T}{T}\times 100$= 1.5%