A water drop is divided into 8 equal droplets. The pressure difference between the inner and outer side of…
A water drop is divided into 8 equal droplets. The pressure difference between the inner and outer side of the big drop will be
- same as the smaller droplet.
- half of that for smaller droplet
- $\left(\frac{1}{4}\right)^{\text {th }}$ of that for smaller droplet.
- twice that for smaller droplet
Solution
Volume of 8 smaller drops $=$ Volume of the bigger drop
$\begin{array}{ll}
& \Rightarrow 8 \times \frac{4}{3} \pi r^3=\frac{4 \pi}{3} R^3 \\
\therefore & 2 r=R \text { or } r=\frac{R}{2}
\end{array}$
$\begin{aligned} & \text { Excess pressure }\left(P_i-P_o\right)=\frac{2 T}{r} \\ & \Delta P_s=\frac{T}{2 r}, \Delta P_B=\frac{T}{2 R} \\ \therefore \quad & \frac{\Delta P_B}{\Delta P_S}=\frac{r}{R}=\frac{1}{2} \\ & \Rightarrow \Delta P_B=\frac{\Delta P_S}{2}\end{aligned}$
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Asked in: MHT CET 2024 (16 May Shift 2)
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