Liquid drops are falling slowly one by one from vertical glass tube. The relation between the weight of a…

Liquid drops are falling slowly one by one from vertical glass tube. The relation between the weight of a drop ' $w$ ', the surface tension ' $T$ ' and the radius ' $r$ ' of the bore of the tube is (Angle of contact is zero)
  1. $\mathrm{W}=\pi \mathrm{r}^2 \mathrm{~T}$
  2. $\mathrm{W}=2 \pi^2 \mathrm{r} T$
  3. $\quad \mathrm{W}=\left(\frac{4}{2}\right) \pi^2 \mathrm{rT}$
  4. $\quad \mathrm{W}=2 \pi \mathrm{rT}$

Solution

$\begin{array}{ll} & \text { Weight of liquid drop }=\text { Force due to surface } \\ & \text { tension } \\ \therefore \quad & \mathrm{W}=\text { Circumference of the tube } \times \mathrm{T} . \\ \therefore \quad & \mathrm{W}=2 \pi \mathrm{rT}\end{array}$

Asked in: MHT CET 2024 (02 May Shift 2)

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