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Question Number 30866 by Tinkutara last updated on 27/Feb/18

Commented by Tinkutara last updated on 28/Feb/18

Is this the same reason for two times surface tension in Q 30094 also?

Commented by mrW2 last updated on 28/Feb/18

The water column has on both ends  contact to air. We have 2 times surface  tension to hold the water column.

$${The}\:{water}\:{column}\:{has}\:{on}\:{both}\:{ends} \\ $$$${contact}\:{to}\:{air}.\:{We}\:{have}\:\mathrm{2}\:{times}\:{surface} \\ $$$${tension}\:{to}\:{hold}\:{the}\:{water}\:{column}. \\ $$

Commented by mrW2 last updated on 28/Feb/18

14 mm?

$$\mathrm{14}\:{mm}? \\ $$

Commented by Tinkutara last updated on 28/Feb/18

Answer is 15 mm. Maybe you took g differently.

Commented by mrW2 last updated on 28/Feb/18

2×((2×73.5×10^(−3) )/(1000×9.81×0.002))=0.015 m=15 mm

$$\mathrm{2}×\frac{\mathrm{2}×\mathrm{73}.\mathrm{5}×\mathrm{10}^{−\mathrm{3}} }{\mathrm{1000}×\mathrm{9}.\mathrm{81}×\mathrm{0}.\mathrm{002}}=\mathrm{0}.\mathrm{015}\:{m}=\mathrm{15}\:{mm} \\ $$

Commented by Tinkutara last updated on 28/Feb/18

Yes this gives 15 mm but formula is  h=((2Scos θ)/(rρg)). I can′t understand when  we take it ((4Scos θ)/(rρg)).

$${Yes}\:{this}\:{gives}\:\mathrm{15}\:{mm}\:{but}\:{formula}\:{is} \\ $$$${h}=\frac{\mathrm{2}{S}\mathrm{cos}\:\theta}{{r}\rho{g}}.\:{I}\:{can}'{t}\:{understand}\:{when} \\ $$$${we}\:{take}\:{it}\:\frac{\mathrm{4}{S}\mathrm{cos}\:\theta}{{r}\rho{g}}. \\ $$

Commented by mrW2 last updated on 28/Feb/18

No. That is a different situation.

$${No}.\:{That}\:{is}\:{a}\:{different}\:{situation}. \\ $$

Commented by Tinkutara last updated on 01/Mar/18

Thanks! Got this question.

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