Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = $0.075 J m^{- 2}$ .

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Solution

T = 0.075 $J / m^{2}$ , r = 6.0 cm = 0.06 m,

A = $π r^{2} = π \times (0.06)^{2}$

So, surface energy = $(0.075) \times (3.14) \times (0.06)^{2}$

= $8.5 \times 10^{- 4} J$

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Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water =0.075Jm−2.

T = 0.075 J/m2, r = 6.0 cm = 0.06 m, A = πr2=π×(0.06)2 So, surface energy = (0.075)×(3.14)×(0.06)2 =8.5×10−4J

Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = $0.075 J m^{- 2}$ .

T = 0.075 $J / m^{2}$ , r = 6.0 cm = 0.06 m,

A = $π r^{2} = π \times (0.06)^{2}$

So, surface energy = $(0.075) \times (3.14) \times (0.06)^{2}$

= $8.5 \times 10^{- 4} J$