A ring is cut from a platinum yube

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06/04/2008 23:27:21

Title: A ring is cut from a platinum yube

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vaishali gupta

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A ring is cut from a platinum yube of 8.5 mm internal & 8.7 mm external supported horizontally from the par of a balance so that it comes in contact with water in a glass vessel. it is found that an extra weight of 3.97 gm is required to pull the ring out of water. Find surface tension of water.

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07/04/2008 13:24:03

Title: Re:A ring is cut from a platinum yu...

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Indranil Chakraborty

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Is the answer 0.36N/m (i am not very sure, coz ques is bit unclear).

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08/04/2008 13:09:00

Title: A ring is cut from a platinum yube

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Indranil Chakraborty

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vaishali gupta wrote:A ring is cut from a platinum yube of 8.5 mm internal & 8.7 mm external supported horizontally from the par of a balance so that it comes in contact with water in a glass vessel. it is found that an extra weight of 3.97 gm is required to pull the ring out of water. Find surface tension of water.

I think I have made some assumptions which need to be made clear. Firstly I have assumed that the values 8.5mm and 8.7 mm are the values of the radii of the ring (actually I believe that they should be in cm since according to standard value of Surface Tension of water which I know, my answer is bit different). Then I have proceeded to solve the question as $T=mg/(2\pi({r}_{1}+{r}_{2})$ where ${r}_{1}$ and ${r}_{2}$ are the external and internal radii of the ring since surface tension acts on both the inner and outer edges of the ring. Substituting the values we get 0.36N/m which is quite high. If the values given are for the diameter then the value is 0.72 N/m and if the values are in cm instead of mm, then the answer should be .072N/m. Please tell me if I am correct in my assumptions and if not then please check my alternate answers and tell me whether they are correct or not.

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10/04/2008 21:43:49

Title: Re:A ring is cut from a platinum yu...

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Akshat Anand

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surface tension ( $\sigma$ ) will act along the length

L = 2 $\pi$ (r1 + r2)where

r1 : internal radius $\\$ &
r2 : external radius $\\$

(force which is balancing surface tension is mg where m= 3.97 gms) $\\$

now $\\$

$\sigma$ L = mg $\\$
by

substituting values we get $\\$
$\sigma$ = 0.72 N/m $\\$

This message was edited 1 time. Last update was at 10/04/2008 21:47:28

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