Question

The following observations were taken for determining surface tension $T$ of water by capillary method: Diameter of capillary $(D)$ = $1.50\times 10^{-2}m$, Rise of water $(h)=3.00\times 10^{-2}m$. Using $g=9.80m/s^2$ and the simplified relation $T=\frac{\text{ rhg }}{2}\times 10^{3}N/m$, what is the possible error (in %) in surface tension?

Answer 1

Answer: 1

$D=1.50\times 10^{-2}m;h=3.00\times 10^{-2}m$
The maximum permissible error in $D$
$=\Delta D=0.01\times 10^{-2}m$
The maximum permissible error in $h$
$=\Delta h=0.01\times 10^{-2}m$
$T=\frac{rhg}{2}\times 10^{3}N/m=\frac{dhg}{4}\times 10^{3}N/m$
$\therefore$ Relative error $\frac{\Delta T}{T}=\frac{\Delta d}{d}+\frac{\Delta h}{h}+\frac{\Delta g}{g}$
Since $g$ is given as a constant and is errorless
$\therefore$ Percentage error,
$\frac{\Delta T}{T}\times 100=\frac{\Delta d}{d}\times 100+\frac{\Delta h}{h}\times 100$
$=\left[\left(\frac{0.01\times 10^{-2}}{1.50\times 10^{-2}}+\frac{0.01\times 10^{-2}}{3.00\times 10^{-2}}\right)\times 100\right]\%$
$=1\%$