A) \[-40\,\widehat{k}\,Nm\]
B) \[-50\,\widehat{k}\,Nm\]
C) \[-60\,\widehat{k}\,Nm\]
D) \[-70\,\widehat{k}\,Nm\]
Correct Answer: C
Solution :
(c) \[-60\,\widehat{k}\,Nm\] Here, \[\overrightarrow{M}=30\,\widehat{i}\,A{{m}^{2}}\] and \[\overrightarrow{B}=(2\,\widehat{i}\,+5\,\widehat{j})T\]. Since \[\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}\] \[=30\,\widehat{j}\times (2\widehat{i}\,+5\,\widehat{j})=60\,\widehat{j}\times \widehat{i}+150\,\widehat{i}\times \widehat{j}\] \[=60(-\widehat{k})+150\times 0\] \[=60\,\widehat{k}\,Nm[\because \,\widehat{j}\times \widehat{i}=-\widehat{k}and\,\widehat{j}\times \widehat{j}=0]\]
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