The following observations were taken for determining surface tension of water by capillary tube method. Diameter of capillary, D=1.25×10−2m and rise of water in capillary, h=1.45×10−2 m. Taking g=9.80 ms−2 and using the relation T=(rgh/2)×103 Nm−1, what is the possible error in surface tension T ?

h-2Trdg T=hrdg2Percentage error in T=0.011.25×100+0.019.80×100+0.011.45×100=0.8+0.1+0.7=1.6%

The following observations were taken for determining surface tension of water by capillary tube method. Diameter of capillary, D=1.25×10−2m and rise of water in capillary, h=1.45×10−2 m. Taking g=9.80 ms−2 and using the relation T=(rgh/2)×103 Nm−1, what is the possible error in surface tension T ?

h-2Trdg T=hrdg2Percentage error in T=0.011.25×100+0.019.80×100+0.011.45×100=0.8+0.1+0.7=1.6%

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The following observations were taken for determining surface tension of water by capillary tube method. Diameter of capillary, $D = 1.25 \times 10^{- 2} m$ and rise of water in capillary, $h = 1.45 \times 10^{- 2} m .$ Taking $g = 9.80 m s^{- 2}$ and using the relation $T = (r g h / 2) \times 10^{3} N m^{- 1},$ what is the possible error in surface tension T ?

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2.4%

15 %

1.6%

0.15%

answer is C.

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Detailed Solution

$h - \frac{2 T}{r d g} T = \frac{h r d g}{2}$

Percentage error in $T = \frac{0.01}{1.25} \times 100 + \frac{0.01}{9.80} \times 100 + \frac{0.01}{1.45} \times 100$

$= 0.8 + 0.1 + 0.7 = 1.6$ %

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