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प्रश्न
The surface tension of water at 0°C is 75.5 dyne/cm. Calculate surface tension of water at 25°C.
(α for water = 2.7×10-3/°C)
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उत्तर
Given:
T0 = 75.5 dyne/cm
αwater = 2.7 × 10-3/°C
To find:
Surface tension of water at 25°C
Formula:
T1 = T0(1 - αΔt)
Solution:
T25 = T0(1 - αΔt)
T25 = T0(1 - α(25 - 0))
T25 = 75.5(1 - 2.7 × 10-3 × 25)
T25 = 75.5(1 - 0.0675)
T25 = 70.4 dyne/cm
The surface tension of water at 25°C is 70.4 dyne/cm.
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