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The interanal radius of one limb of a ca...

The interanal radius of one limb of a capillary `U-`tube is `r_(1) = 1 mm` and the internal radius of the second limb is `r_(2) = 2 mm`, the tube is filled with same mercury, and one of the limbs is connected to a vecuum pump. The surface tension & density of mercury are `480 dyn//cm` & `13.6 gm//cm^(3)` respectively (assume contact angle to be `theta = 180^(@)`)`(g = 9.8 m//s^(2)`)
Which limb of the should be connected to the pump ?

A

Limb having radius `2 mm`

B

Limb having radius `1 mm`

C

Any of the limb

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B

Pressure difference `= (2T)/(r_(1))-(2T)/(r_(2))` (`:'` angle of contact is `180^(@)`)
`rArr rho_(Hg) h_(Hg) xx g = 2T((r_(2) - r_(1))/(r_(1)r_(2)))`
`rArr h_(Hg) = (2T)/(rhog)((r_(2) - r_(1))/(r_(1)r_(2))) = 3.53 mm` of `Hg`.
As `P_(A) gt P_(B)`, althorugh they are at same height, hence the air above the point `B` has been evacuated.
So the bigger limb of the tube should be connected to the pump.
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