A liquid rises to a height of 50cm in a capillary tube of diameter 0.04mm. If density of liquid is 0.8 ×103 kg m3 and angle of contact is 20°, the surface tension of the liquid is (Cos 2 0° = 0.94)

h=50cm=50×10−2mr=0.02×10−3,ρ=0.8×103cos⁡20∘=0.94;Tcos⁡θ=ρghr2T(0.94)=0.8×103×10×50×10−2×0.02×10−32T=4.17×10−2N/m

A liquid rises to a height of 50cm in a capillary tube of diameter 0.04mm. If density of liquid is 0.8 ×103 kg m3 and angle of contact is 20°, the surface tension of the liquid is (Cos 2 0° = 0.94)

h=50cm=50×10−2mr=0.02×10−3,ρ=0.8×103cos⁡20∘=0.94;Tcos⁡θ=ρghr2T(0.94)=0.8×103×10×50×10−2×0.02×10−32T=4.17×10−2N/m

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A liquid rises to a height of 50cm in a capillary tube of diameter 0.04mm. If density of liquid is 0.8 ×10³ kg m³ and angle of contact is 20°, the surface tension of the liquid is (Cos 2 0° = 0.94)

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$0.417 \frac{N}{m}$

$4.17 x 10^{- 2} \frac{N}{m}$

$4.17 \frac{N}{m}$

$4.17 x 10^{- 3} \frac{N}{m}$

answer is C.

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Detailed Solution

$\begin{array}{l} h = 50 cm = 50 \times 10^{- 2} m \\ r = 0 .02 \times 10^{- 3}, ρ = 0 .8 \times 10^{3} \\ \cos 20^{\circ} = 0 .94; Tcos θ = \frac{ρghr}{2} \\ T (0 .94) = \frac{0 .8 \times 10^{3} \times 10 \times 50 \times 10^{- 2} \times 0 .02 \times 10^{- 3}}{2} \\ T = 4 .17 \times 10^{- 2} N / m \end{array}$