Question

A glass tube has inner diameter of 1 mm, outer diameter of 1.1 mm. It is kept vertical and partially dipped in water. Calculate the downward pull due to surface tension. (Surface tension of water = 75 dyne/cm) (Ans : 49.49 dyne)

Answer 1

Given :

d1 = 1mm  

∴ r 1 = d 1/2 = 1/ 2

= 0.5 mm = 0.05 cm

d2 = 1.1 mm

∴ r 2 = d2/ 2 = 1.1/ 2 = 0.55 mm = 0.55 cm

T = 75 dyne/cm

F = Tl  

l = 2π (r1 + r2 )

F = Tl  

= T × 2π (r1 + r2 )  

∴ F = 2πT (r1 + r2 )

= 2 × 3.142 × 75 (0.05 + 0.055)

= 2 × 3.142 × 75 × 0.105  

∴ F = 150 × 3.142 × 0.105

∴ F = 49.49 dynes

Answer 2

Answer:

Given :

d1 = 1mm  

∴ r 1 = d 1/2 = 1/ 2

= 0.5 mm = 0.05 cm

d2 = 1.1 mm

∴ r 2 = d2/ 2 = 1.1/ 2 = 0.55 mm = 0.55 cm

T = 75 dyne/cm

F = Tl  

l = 2π (r1 + r2 )

F = Tl  

= T × 2π (r1 + r2 )  

∴ F = 2πT (r1 + r2 )

= 2 × 3.142 × 75 (0.05 + 0.055)

= 2 × 3.142 × 75 × 0.105  

∴ F = 150 × 3.142 × 0.105

∴ F = 49.49 dynes

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