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A thread of liquid is in a uniform capil...

A thread of liquid is in a uniform capillary tube of length L. As measured by a ruler. The temperature of the tube and thread of liquid is raised by `DeltaT`. If `gamma` be the coefficient of volume expansion of the liquid and `alpha` be the coefficient of linear expansion of the material of the tube, then the increase `DeltaL` in the length of the thread, again measured by the ruler will be

A

`DeltaL=L(gamma-alpha)DeltaT`

B

`DeltaL=L(gamma-2alpha)DeltaT`

C

`DeltaL=L(gamma-3alpha)DeltaT`

D

`DeltaL=LgammaDeltaT`

Text Solution

Verified by Experts

The correct Answer is:
B
`L=(V)/(A)`
So `L_(2)-L_(1)=(V_(2))/(A_(2))-(V_(1))/(A_(1))=(V(1+gammaDeltaT))/(A(1+2alphaDeltaT))-(V)/(A)=(V)/(A)[(1+gammaDeltaT)/(1+2alphaDeltaT)-1]`
`DeltaL=(V)/(A)[(gammaDeltaT-2alphaDeltaT)/(1+2alphaDeltaT)]`
`DeltaLsquare(V)/(A)DeltaT(gamma-2alpha)`
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