In a capillary rise, find the heat developed taking all standard notations as described in the foregoing section.

Options

A. `Q=(2piTcos^(2)theta)/(rhog)`
B. `Q=(2pir^(2)Tcos^(2)theta)/(rhog)`
C. `Q=(2piT^(2)sin^(2)theta)/(rhog)`
D. `Q=(2piT^(2)cos^(2)theta)/(rhog)`

Answer

D

Explanation

As the liquid rises, positive work is done by surface tension in pulling the liquid up by a distance h which is given as
`W_(ST)=F_(y)h` where `F_(y)=(Tcostheta)2pir`
`W_(gr)=-DeltaU=mgy_(CM)`,
where `Y_(CM)=(h)/(2)` and `m=(pir^(2)h)rho`
Then, `DeltaU=(pir^(2)rhogh^(2))/(2)` where `h=(2Tcostheta)/(rhogr)` This gives applying first law of thermodynamics, the heat dissipated Q can be gives as `W_(ST)=Q+DeltaU`
From the above equations,
we have `Q=(2piT^(2)cos^(2)theta)/(rhog)`