The following observations were tak...
Question
The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, d=1.25×10^−2m rise of water, h=145×10^−2
Fullscreen

The following observations were taken for determining surface tension ( T ) of water by capillary method. Diameter of capillary, ( d=1.25 times 10^{-2} mathrm{m} ) rise of water, ( h=145 times 10^{-2} mathrm{m} )
Using ( g=9.80 quad mathrm{m} / mathrm{s}^{2} quad ) and ( quad ) the ( quad ) simplified relation ( T=frac{r h g}{2} times 10^{3} mathrm{N} / mathrm{m}, ) the possible error in surface tension is

NEET/Medical Exams
Physics
Solution
160
Rating
1.0 (2 ratings)
The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, d=1.25×10^−2m rise of water, h=145×10^−2m Using g=9.80m/s^2 and  the  simplified relation T=rhg/2×10^3N/m, the possible error in surface tension is
Fullscreen

By ascent formula, we have surface tension, ( begin{array}{rlr}T & =frac{r h g}{2} times 10^{3} frac{mathrm{N}}{mathrm{m}} & =frac{d h g}{4} times 10^{3} frac{mathrm{N}}{mathrm{m}} & Rightarrow quad frac{Delta T}{T} & =frac{Delta d}{d}+frac{Delta h}{h} & {[text { given, } g text { is constant }]}end{array} )
So, percentage ( =frac{Delta T}{T} times 100=left(frac{Delta d}{d}+frac{Delta h}{h}right) times 100 )
( begin{aligned} &=left(frac{0.01 times 10^{-2}}{1.25 times 10^{-2}}+frac{0.01 times 10^{-2}}{1.45 times 10^{-2}}right) times 100 &=1.5 % therefore quad frac{Delta T}{T} times 100 &=1.5 % end{aligned} )

Quick and Stepwise Solutions Just click and Send Download App OVER 20 LAKH QUESTIONS ANSWERED Download App for Free