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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Dec 2007 18:48:04 IST
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If a liquid rises to the same height in two capillaries of the same material at the same temperature (A) the hydrostatic pressure at the base of capillaries must be zero (B) the radius of meniscus must be zero (C) the weight of the liquid in both capillaries must be equal (D) the capillaries must be cylindrical and vertical
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19 Dec 2007 18:54:58 IST
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if given that radius of both capillaries are same then ans is C. but its not given.
A and D definitely not possible.
and i have never heard the phrase "radius of meniscus" before.
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19 Dec 2007 18:55:44 IST
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Is it multiple correct sort of question ?Otherwise I agree with pannaguma !
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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19 Dec 2007 18:56:30 IST
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anything...even i don no the answer...plz tell me yur method....
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19 Dec 2007 18:59:27 IST
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why is D not possible. almost all capillaries r cylindrical n vertical
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
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19 Dec 2007 18:59:41 IST
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If its multiple type, then it should be (A) and (C) .For single type , it should be (A) for sure !
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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19 Dec 2007 19:00:12 IST
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can you explain??
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19 Dec 2007 19:03:00 IST
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rhd, there is a concept which gives that downward hydrostatic pressure depends only on height of liquid column. thus even if we keep the capillary aada/theda it will rise to same height i think.
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19 Dec 2007 19:07:34 IST
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vo to thik hai, but it doesn't mean that its impossible, if we kip it aada/theda, quantity of liquid in capillary will change
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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19 Dec 2007 19:09:00 IST
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n it think cos(theta) in 2Tcos(theta)/rpg will also change
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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19 Dec 2007 19:09:40 IST
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^^yeah man. so option C is also out of consideration.
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19 Dec 2007 19:17:33 IST
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C is the correct option yaar because if both have same heighhs then they have same pressure at the bottom
Pressure is force by area
area being same the force is also equal therefore i think C
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19 Dec 2007 19:45:17 IST
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Pressure is definitely same but nothing is said about area
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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