A) \[\frac{\rho g}{2hr}\]
B) \[\frac{2}{hr\rho g}\]
C) \[\frac{r\rho g}{2h}\]
D) \[\frac{hr\rho g}{2}\]
Correct Answer: D
Solution :
\[T=\frac{rhdg}{2\cos \theta }.\] For pure water\[\theta ={{0}^{o}}\] so \[T=\frac{rhdg}{2}\]
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