The following observations were taken for determining surface tension T of water by capillary method- Diameter of capillary- D -1-25 - 10-2 m rise of water- h -1-45 - 10-2 m-Using g -9-80 m - s 2 and the simplified relation T - rgh 2 - 103 N - m- the possible error in surface tension is closest toA- 10 -B- 0-15 -C- 1-5 -D- 2-4 -

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Standard XII

Physics

Dimensional Analysis

The following...

Question

The following observations were taken for determining surface tension $T$ of water by capillary method. Diameter of capillary, $D = 1.25 \times 10^{- 2} m$ rise of water, $h = 1.45 \times 10^{- 2} m.$
Using $g = 9.80 m / s^{2}$ and the simplified relation $T = \frac{r g h}{2} \times 10^{3} N / m$ , the possible error in surface tension is closest to

10 %

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0.15 %

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1.5 %

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2.4 %

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Solution

The correct option is C 1.5 %
$T = \frac{r g h}{2} \times 10^{3} N / m$
$Δ r = 0.01 m Δ h = 0.01 m \frac{Δ T}{T} = (\frac{Δ r}{r} ∣ ∣ + (\frac{Δ h}{h} ∣ ∣ = \frac{0.01}{1.25} + \frac{0.01}{1 . 45$}$

$\frac{Δ T}{T} \times 100 = (\frac{1}{125} + \frac{1}{145}) \times 100 = 0.8 + 0.69 = 1.5 %$

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The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, D=1.25×10−2 m rise of water, h=1.45×10−2 m. Using g=9.80m/s2and the simplified relation T=rgh2×103 N/m , the possible error in surface tension is closest to

The correct option is C 1.5 %T=rgh2×103 N/m Δr=0.01 mΔh=0.01 mΔTT=(Δrr∣∣+(Δhh∣∣=0.011.25+0.011.45$ ΔTT×100 = (1125+1145)×100 = 0.8+0.69 =1.5 %