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Standard VIII

Chemistry

Surface Tension

The surface t...

Question

The surface tension of a liquid is 0.245Nm $^{- 1}$ . Its angle of contact is 60 $^{o}$ . Its density is 1000 kg/m $^{3}$ . When a capillary tube of diameter 1 mm is dipped vertically in it, the capillary rise is (g $=$ 10ms $^{- 2}$ )

49 cm

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4.9 cm

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98 cm

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9.8 cm

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Solution

The correct option is B 4.9 cm
$h = \frac{2 T c o s θ}{ρ r g}$
= $\frac{2 \times 0.245 \times c o s 60^{0}}{1000 \times \frac{1}{2} \times 10^{- 3} \times 10}$
= $\frac{4 \times 0.245 \times \frac{1}{2}}{1000 \times 10^{- 3} \times 10} (c o s 60^{0} = 1 / 2)$
= $4.9 \times 10^{- 2} m$
= 4.9 cm

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The surface tension of a liquid is 0.245Nm−1. Its angle of contact is 60o. Its density is 1000 kg/m3. When a capillary tube of diameter 1 mm is dipped vertically in it, the capillary rise is (g =10ms−2)

The correct option is B 4.9 cmh=2Tcosθρrg = 2×0.245×cos6001000×12×10−3×10 = 4×0.245×121000×10−3×10(cos600=1/2) = 4.9×10−2m = 4.9 cm

The surface tension of a liquid is 0.245Nm $^{- 1}$ . Its angle of contact is 60 $^{o}$ . Its density is 1000 kg/m $^{3}$ . When a capillary tube of diameter 1 mm is dipped vertically in it, the capillary rise is (g $=$ 10ms $^{- 2}$ )

The correct option is B 4.9 cm
$h = \frac{2 T c o s θ}{ρ r g}$
= $\frac{2 \times 0.245 \times c o s 60^{0}}{1000 \times \frac{1}{2} \times 10^{- 3} \times 10}$
= $\frac{4 \times 0.245 \times \frac{1}{2}}{1000 \times 10^{- 3} \times 10} (c o s 60^{0} = 1 / 2)$
= $4.9 \times 10^{- 2} m$
= 4.9 cm