A) 3.9 N/m
B) \[3.9\text{ }\times {{10}^{2}}N/m\]
C) \[3.9\text{ }\times {{10}^{3}}N/m\]
D) 3.9 dyne/m
Correct Answer: B
Solution :
\[\frac{4T}{R}=hdg\]\ \[\]\[T=\frac{Rhdg}{4}\] T \[=\frac{{{10}^{-2}}\times 2\times {{10}^{-3}}\times 0.8\times {{10}^{3}}\times 9.8}{4}\]\[=3.9\times {{10}^{-2}}\,N/m\]
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