A) \[7.5\times {{10}^{-4}}N-m\]
B) \[3.0\times {{10}^{-4}}N-m\]
C) \[1.5\times {{10}^{-4}}N-m\]
D) \[6.0\times {{10}^{-4}}N-m\]
Correct Answer: C
Solution :
\[T=MB\text{ }sin\,\theta \] \[=m\times (2l)\times B\text{ }sin\,\theta \] \[={{10}^{-4}}\times (0.1)\times 30\text{ }sin\text{ }30{}^\circ \] \[=1.5\times {{10}^{-4}}N-m\]
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