A) 0.5656 ´ 10?4 N-m
B) 0.5656 ´ 10?3 N-m
C) 0.656 ´ 10?4 N-m
D) 0.656 ´ 10?5 N-m
Correct Answer: B
Solution :
\[\tau =MB\sin \theta =\left( mL \right)B\sin \theta \] \[=(40\times 10\times {{10}^{-2}})\times 2\times {{10}^{-4}}\times \sin 45{}^\circ \] \[=0.565\times {{10}^{-3}}N-m\]
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