A) 12 Newton × metre
B) 18 Newton × metre
C) 0.9 Newton × metre
D) None of the above
Correct Answer: C
Solution :
\[\tau =MB\sin \theta =48\times 25\times {{10}^{-2}}\times 0.15\times \frac{1}{2}\]\[=0.9\ N\times m\]
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