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)

Open in App

Solution

Given :
$d_{1} = 1 m m$
∴ $r_{1} = d \times \frac{1}{2} = \frac{1}{2}$ $= 0.5 m m = 0.05 c m$
$d_{2} = 1.1 m m$
∴ $r_{2} = \frac{d_{2}}{2} = \frac{1.1}{2} = 0.55 m m = 0.55 c m$
$T = 75 d y n e / c m$
$F = T \times l$
$l = 2 π (r_{1} + r_{2})$
$F = T l$
$= T \times 2 π (r_{1} + r_{2})$
∴ $F = 2 π T (r_{1} + r_{2})$
$= 2 \times 3.142 \times 75 (0.05 + 0.055)$
$= 2 \times 3.142 \times 75 \times 0.105$
∴ $F = 150 \times 3.142 \times 0.105$
∴ $F = 49.49 d y n e s$

Suggest Corrections

Similar questions

Q. If a 5 cm long capillary tube with 0.1 mm internal diameter, open at both ends, is slightly dipped in water having surface tension 75 dyne/cm, state whether: water will rise halfway in the capillary?

Q. If a

5 cm

long capillary tube with

0.2 mm

internal diameter open at both ends is slightly dipped in water having surface tension

75 dyne/cm

. What will be the rise of water in capillary

(

Take

g = 1000 cm / s^{2}

)

Water rises in a capillary tube when its one end is dipped vertically in it, is 3 cm. If the surface tension of water is $75 \times 10^{- 3}$ N/m, then the diameter of capillary will be

Q. If a 5 cm long capillary tube with 0.1 mm internal diameter, open at both ends, is slightly dipped in water having surface tension 75 dyne/cm, state whether:
water will overflow out of the upper end of capillary. Explain your answer.

Q. If a

10 cm

long capillary tube with

0.2 mm

internal diameter open at both ends is slightly dipped in water having surface tension

80 dyne/cm

, then:
[

g = 981 {g/cm}^{2}

]

Join BYJU'S Learning Program

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)

Given :d1=1mm∴ r1=d×12=12=0.5mm=0.05cm d2=1.1mm ∴ r2=d22=1.12=0.55mm=0.55cm T=75dyne/cmF=T×l 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=49.49dynes

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)