A) 3.92 N/m
B) 0.0392 N/m
C) 0.392 N/m
D) 0.00392 N/m
Correct Answer: B
Solution :
\[p=\frac{4T}{R}\] \[hpg=\frac{4T}{R}\] \[T=\frac{Rphg}{4}\] \[=\frac{1\times {{10}^{-2}}\times 2\times {{10}^{-3}}\times 0.8\times {{10}^{3}}\times 9.8}{4}\] \[=3.92\times {{10}^{-2}}N\text{/}m\] \[=0.0392\,N\text{/}m\]
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