The following observations were taken for determining surface tension T of water by capillary method -diameter of capillary- D-1-25- 10-2mrise of water- h-1-45- 10-2m-Using g-9-80 m-s2 and the simplified relation- T-rhg2- 103 N-m- the possible error in surface tension is closest to -

The correct option is A 1.5% T=rhg2× 10^3 N/m = Dhg4× 10^3 N/m (Since D = 2r) ⇒log T = log D + log h + log g 4+3 ⇒ TT= DD+ hh =0.011 ...

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Standard XII

Physics

Friction and Vertical Circular Motion Problems

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. $T = \frac{r h g}{2} \times 10^{3} N / m$ , 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 A $1.5 %$
$T = \frac{r h g}{2} \times 10^{3} N / m = \frac{D h g}{4} \times 10^{3} N / m$ (Since D = 2r)
$\Rightarrow log T = log D + log h + log g - 4 + 3$
$\Rightarrow \frac{△ T}{T} = \frac{△ D}{D} + \frac{△ h}{h}$
$= \frac{0.01}{1.25} + \frac{0.01}{1.45} = 0.015 = 1.5 %$
Hence correct answer is option (3).

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The following observations were taken for determining surface tension T of water by capillary method :diameter of capillary, D=1.25×10−2mrise of water, h=1.45×10−2m.Using g=9.80m/s2 and the simplified relation. T=rhg2×103N/m, the possible error in surface tension is closest to :

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