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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 $1500 k g m^{- 3},$ and the angle of contact zero is taken in the tube. The difference in heights of the levels of the liquid in the two limbs is $(g = 10 m s^{- 2})$

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$3 m m$

$2.5 m m$

$1 m m$

$1.5 m m$

answer is C.

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Detailed Solution

$h = \frac{2 T \cos θ}{r ρ g} h_{1} - h_{2} = \frac{2 T \cos θ}{ρ g} [\frac{1}{r_{1}} - \frac{1}{r_{2}}] = \frac{2 \times 0.03 \times \cos 0 °}{1500 \times 10} [\frac{1}{2 \times 10^{- 3}} - \frac{1}{4 \times 10^{- 3}}] ∴ 1 m m$

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