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The lower end of a capillary tube of rad...

The lower end of a capillary tube of radius r is placed vertically in water of density `rho` , surface tension S. The rice of water in the capillary tube is upto height h, then heat evolved is

A

`+(pi^(2)r^(2)h^(2))/(J)dg`

B

`+(pir^(2)h^(2)dg)/(2J)`

C

`-(pir^(2)h^(2)dg)/(2J)`

D

`-(pir^(2)h^(2)dg)/(J)`

Text Solution

Verified by Experts

The correct Answer is:
B
When the tube is placed vertically in water, water rises through height `h` given by `h=(2Tcos theta)/(rdg)`
Upward force `=2pir xx Tcos theta`
work done by this force in raising water column through height `h` is given by
`DeltaW = (2pirTcos theta)h = (2pir h cos theta)T = (2pirh cos theta)((rhdg)/(2cos theta)) = pir^(2)h^(2) dg`
However, the increase in potential energy `DeltaE_(P)` of the raised water column `=mg.(h)/(2)`
So, `DeltaE_(P) = (pir^(2)hd)((hg)/(2)) = (pir^(2)h^(2)dg)/(2)`
Further, `DeltaW - DeltaE_(P) = (pir^(2)h^(2)dg)/(2)`
The part `(DeltaW - DeltaE_(P))` is used in doing work against visous forces and frictional forces between water and glass surface and appears as heat. So heat released `=(DeltaW - DeltaE_(P))/(J) = (pir^(2)h^(2)dg)/(2J)`
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