A) \[r=\sqrt{\frac{3T}{(2d-\rho )g}}\]
B) \[r=\sqrt{\frac{T}{(d+\rho )g}}\]
C) \[r=\sqrt{\frac{T}{(d-\rho )g}}\]
D) \[r=\sqrt{\frac{2T}{3(d+\rho )g}}\]
Correct Answer: A
Solution :
\[T.2\,\,\pi r+\frac{2}{3}\pi {{r}^{3|}}\rho g=\frac{4}{3}\pi {{r}^{3}}dg\] \[T=\frac{{{r}^{2}}}{3}\,(2d-\rho )g\] \[r=\sqrt{\frac{3T}{(2d-\rho )g}}\]
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