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प्रश्न
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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उत्तर
Given:
Radius of cylindrical vessel, r = 6.0 cm = 0.06 m
Surface tension of water, T = 0.075 J/m2
Area, A = πr2 = π × (0.06)2
Surface energy = T × A
= (0.075) × (3.14) × (0.06)2
= 8.5 × 10−4 J
Therefore, the surface energy of water kept in a cylindrical vessel is 8.5 × 10−4 J.
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संबंधित प्रश्न
'n' droplets of equal size of radius r coalesce to form a bigger drop of radius R. The energy liberated is equal to...................
(T =Surface tension of water)
`(a) 4piR^2T[n^(1/3)-1]`
`(b) 4pir^2T[n^(1/3)-1]`
`(c) 4piR^2T[n^(2/3)-1]`
`(d)4 pir^2T[n^(2/3)-1]`
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