If $\mathrm{S}_1, \mathrm{~S}_2$ and $\mathrm{S}_3$ are the tensions at liquid-air, solid-air and…
- $\mathrm{S}_1, \cos \theta+\mathrm{S}_2 \sin \theta=\mathrm{S}_3$
- $S_1 \cos \theta+S_3=S_2$
- $S_2 \cos \theta+S_3=S_1$
- $S_3 \cos \theta+S_1=S_2$
Solution

At equilibrium of liquid drop, $S_2=S_1 \cos \theta+S_3$
Asked in: AP EAMCET 2024 (19 May Shift 2)
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