A) \[\widehat{i}-\widehat{j}\]
B) \[\frac{1}{2}(\widehat{i}+\widehat{j})\]
C) \[(\widehat{i}-\widehat{j})\]
D) \[\frac{1}{2}(\widehat{i}-\widehat{j})\]
Correct Answer: B
Solution :
We can show the situation as : The centre of mass of square is at point \[D(x,y)\] The position co-ordinate of point D \[(x,y)\equiv \left( \frac{0+1}{2},\frac{1+0}{2} \right)\] \[=\left( \frac{1}{2},\frac{1}{2} \right)\] Hence, position vector or centre of mass D is \[=x\hat{i}+y\hat{j}\] \[=\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}\] \[=\frac{1}{2}(\hat{i}+\hat{j})\]You need to login to perform this action.
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