A) \[\pi rT\cos \theta \]
B) \[\pi rT\sin \theta \]
C) \[\pi rT\cos \left[ \theta -{{\tan }^{-1}}\left( \frac{r}{h} \right) \right]\]
D) \[\pi rT\cos \left[ \theta -{{\tan }^{-1}}\left( \frac{h}{r} \right) \right]\]
Correct Answer: C
Solution :
[c] \[dF=Tdl\] Horizontal components will cancel out. \[F=\int{dF\,\cos \,\,\left( \theta -\alpha \right)}=\int{Td\ell \,\cos \,\left( \theta -\alpha \right)}\] \[=T\cos \,\,\left( \theta -\alpha \right)\int{d\ell }=T\cos \left( \theta -\alpha \right)2\pi r'\] Since only \[\frac{h}{2}\] portion is inside the fluid, \[\therefore \,\,\,r'=\frac{r}{2}\] or \[F=\pi rT\cos \left[ \theta -{{\tan }^{-1}}\left( \frac{r}{h} \right) \right]\]You need to login to perform this action.
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