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In a measurement of surface tension by the falling-drop method, 5 drops of a liquid of density 0.797 g/mL weighed 0.220 g. Calculate the surface tension of the liquid. |
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Answer» Solution :Mass of the average drop = `0.22/5 = 0.044 g` VOLUME of the drop `=("mass")/("DENSITY")` `=(0.044)/(0.797) = 0.0552` mL Assuming the drop to be spherical in SHAPE Volume of the drop `=4/3 pir^(3) = 0.0552` `r = root(3)((3 xx 0.0052)/(4 xx 3.14)) = 0.236 cm` `lambda = (mg)/(2pir) = (0.044 xx 981)/(2 xx 3.14 xx 0.236)` `=29.12 "dyne" cm^(-1)` `=0.02912 N m^(-1)` |
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