A) 0.0125 \[N{{m}^{-1}}\]
B) 0.1 \[N{{m}^{-1}}\]
C) 0.05 \[N{{m}^{-1}}\]
D) 0.025 \[N{{m}^{-1}}\]
Correct Answer: D
Solution :
\[2TL=mg\] \[T=\frac{mg}{2L}=\frac{1.5\times {{10}^{-2}}}{2\times 30\times {{10}^{-2}}}=\frac{1.5}{600}=0.025\] N/mYou need to login to perform this action.
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