- Home
- Standard 11
- Physics
The following observations were taken for determining surface tensiton $T$ ofwater by capillary method :
Diameter of capilary, $D= 1.25 \times 10^{-2}$ $m$ rise of water, $h=1.45 \times 10^{-2}$ $m$ Using $g= 9.80$ $m/s^2$ and the simplified relation $T = \frac{{rhg}}{2}$ $X$ $10^3$ $N/m$ the possible error in surface tension is closest to ......... $\%$
$0.15$
$1.5$
$2.4 $
$10$
Solution
$Surface\,tension,T = \frac{{rhg}}{2} \times {10^3}$
${\mathop{\rm Re}\nolimits} lative\,error\,in\,surface\,tension,$
$\frac{{\Delta T}}{T} = \frac{{\Delta r}}{r} + \frac{{\Delta h}}{h} + 0$
$Percentage\,error$
$100 \times \frac{{\Delta T}}{T} = \left( {\frac{{{{10}^{ – 2}} \times 0.01}}{{1.25 \times {{10}^{ – 2}}}} + \frac{{{{10}^{ – 2}} \times 0.01}}{{1.45 \times {{10}^{ – 2}}}}} \right)100$
$ = \left( {0.8 + 0.689} \right)$
$ = \left( {1.489} \right) = 1.489\% \cong 1.5\% $
