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 × 10-2m Using g = 9.80 m/s2 and the simplified relation T = rhg/2 × 10³ N/m, the possible error in surface tension is closest to: (a) 10 % (b) 0.15 % (c) 1.5 % (d) 2.4 %

Answer: (c). T = rhg/2 with r = D/2, so (Δ T)/(T) = (Δ D)/(D) + (Δ h)/(h) + (Δ g)/(g) Errors from last significant digit: (0.01)/(1.25) = 0.8%, (0.01)/(1.45) = 0.69%, (0.01)/(9.80) = 0.10% Total ≈ 1.59%, closest to 1.5% So the answer is (c) 1.5%.

The following observations were taken for determining surface tension… | JEE Main 2017

Q102JEE Main 2017Difficult

Errors in Measurement

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 $\times$ 10-2m

Using g = 9.80 m/s2 and the simplified relation T = $\frac{rhg}{2} \times 10^{3}\text{ N/m}$ , the possible error in surface tension is closest to:

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Answer: (c)

$T = \frac{rhg}{2}$ with $r = D/2$ , so $\frac{\Delta T}{T} = \frac{\Delta D}{D} + \frac{\Delta h}{h} + \frac{\Delta g}{g}$
Errors from last significant digit: $\frac{0.01}{1.25} = 0.8\%$ , $\frac{0.01}{1.45} = 0.69\%$ , $\frac{0.01}{9.80} = 0.10\%$
Total $\approx 1.59\%$ , closest to $1.5\%$
So the answer is (c) $1.5\%$ .

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