A) \[9.6\times {{10}^{4}}m\]
B) \[9.6\times {{10}^{3}}m\]
C) \[9.6\times {{10}^{2}}m\]
D) \[9.6\times {{10}^{1}}m\]
Correct Answer: A
Solution :
\[\ h=\frac{2T\cos \theta }{rdg}\], for water \[\theta =0{}^\circ \] Þ \[r=\frac{2T}{hdg}=\frac{2\times 7.2\times {{10}^{-2}}}{3\times {{10}^{-2}}\times {{10}^{3}}\times 10}=4.8\times {{10}^{-4}}\] \ \[d=2r=9.6\times {{10}^{-4}}\,m\]
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