by how muchwill the surface of mercury be dipprest in a glass tube of radius 0.0icm, if the angle of contact of mercury 135 degree celcious ans surface tension is 0.487n/m
Asked by Shabnakp
12th January 2013,
10:26 PM
Answered by Expert
Answer:
Radius of the narrow tube, r = 0.01 cm = 1 10 4m
the angle of contact of mercury =1350
surface tension is 0.487n/m
Dip in the height of mercury = h
Acceleration due to gravity, g = 9.8 m/s2
Density of mercury, ? =13.6 103 kg/m3
h =2Tcos q/rg?
Cos 135 = -0.9960
Cos 135 = -0.9960
h =2 x 0.487 x-0.9960/ ( 1 x 10-4 x13.6 x 10 3 x9.8)
h= -0.970104 x 101 /133.28
h = -9.70104 /133.28
h= -0.07278 m
h =- 72.78 mm
Answered by Expert
13th January 2013,
1:13 AM
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