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प्रश्न
The surface tension of water at 0ºc is 75·5 dyne/cm. Find surface tension of water at 25°C. [ α for water = 0·0021/°C ]
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उत्तर
Thus, T = T0 ( 1 - αθ )
= 75.5 [ 1 - (0.0027)(25) ]
= 75.5 [ 1 - 0.0675 ]
= 75.5( 0.9325 )
= 70.40 dyne/cm
Thus, the surface tension of water at 25°C is 70.40 dyne/cm.
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