A water drop of radius ' $r$ ' and volume ' $\mathrm{V}$ ' is kept in between the two identical glass plates…
- $\frac{F V}{2 A^2}$
- $\frac{\mathrm{A}^2}{\mathrm{FV}}$
- $\frac{\mathrm{AV}}{\mathrm{F}^2}$
- $\frac{\mathrm{FV}}{4 \mathrm{~A}^2}$
Solution
The Laplace pressure jump is given by:
On plugging into equation (1), $\frac{\mathrm{T}}{\left(\frac{\mathrm{V}}{2 \mathrm{~A}}\right)}=\frac{\mathrm{F}}{\mathrm{A}}$
$\therefore \mathrm{T}=\frac{\mathrm{FV}}{2 \mathrm{~A}^2}$Asked in: MHT CET 2022 (07 Aug Shift 1)
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