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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Dec 2007 23:24:47 IST
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.......SURFACE OF THE LIQUID WHOSE SURFACETENSION IS 8*10NM^-1 .THE FOESE REQUIRED TO LIFT THE RING UPWARD FROM THE LIQUID SURFACE IS..?EXPLAIN.
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30 Dec 2007 23:44:05 IST
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F = 2pi R T ??
then F=0.8*0.06*pi
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30 Dec 2007 23:51:51 IST
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EDIT-TENSION=8*10^-2
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30 Dec 2007 23:55:20 IST
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i am not confident about the method....
by defintion Surface Tension os N/m
so Force = ST * length
that what i did. forget the values....
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30 Dec 2007 23:59:20 IST
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Depends on how exactly the ring is...
But in the normal case, it is 2 pi R x T
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Will nip in at times to solve problems :)
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31 Dec 2007 00:04:47 IST
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THANKS FOR TRYING.BUT I THING ITS NOT THE OPTION GIVEN BY PUTTING THE VALUES ON THE FORMULA GIVEN BY U.THE OPTIONS R NOT WITH ME NOW.THAT TIME I CHAKED BY THIS FORMULA ONLY.THAT I CAN SAY.
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31 Dec 2007 00:05:24 IST
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I had a hitch... It should be 2R x T.. check this
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Will nip in at times to solve problems :)
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31 Dec 2007 00:19:27 IST
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SORRY KARTHIK.THE ABOVE STATEMENT WAS @PANNAGUMA.
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31 Dec 2007 00:26:13 IST
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@KARTHIK
SURELY I WILL TRY BOTH.BUT NOW I DON'T HAVE OPTIONS.
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1 Jan 2008 10:18:17 IST
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the solution given is (2PiRT)*2
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1 Jan 2008 10:35:30 IST
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Yep. 4PiRT is the right answer. since there ST will act at two places..one along the circumference at radius R and again at R + t where t = thickness of ring
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