In a U-shaped tube the radius of one limb is 2 mm and that of other limb is 4 mm. A liquid of surface tension $0.03 N m^{- 1}$ , density $1500 {kgm}^{- 3}$ and angle of contact zero is taken in the tube. The difference in the heights of the levels of the liquid in the two limbs is (Acceleration due to gravity = $10 m s^{- 2}$ )

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3 m

2.5 mm

1 mm

1.5 mm

answer is C.

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Detailed Solution

Height of liquid in a capillary $= \frac{2 T \cos θ}{r ρ g}$

$h_{1} - h_{2} = \frac{2 T \cos θ}{ρ g} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$

$= \frac{2 \times 0.03 \cos 0^{o}}{1500 \times 10} \times (\frac{1}{2 \times 10^{- 3}} - \frac{1}{4 \times 10^{- 3}})$

$= \frac{0.06}{15 \times 10^{3}} \times (500 - 250) = 10^{- 3} m = 1 m m$

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