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 $g=9.80\, m / s ^{2}$ 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

$D =1.50 \times 10^{-2} m ; h =3.00 \times 10^{-2} m$
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,
$\frac{\Delta T }{ T } \times 100 =\frac{\Delta d }{ d } \times 100+\frac{\Delta h }{ h } \times 100$
$=\left[\left(\frac{0.01 \times 10^{-2}}{1.50 \times 10^{-2}}+\frac{0.01 \times 10^{-2}}{3.00 \times 10^{-2}}\right) \times 100\right] \%$
$=1 \%$