A) 180°
B) 90°
C) 45°
D) 0°
Correct Answer: B
Solution :
\[[{{M}^{0}}{{L}^{2}}{{T}^{-2}}]\] Let A and B be two vectors, then from definition of scalar product, we have \[[ML{{T}^{-2}}]\] Given, \[[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[{{k}_{\alpha }}\] \[\overset{0}{\mathop{A}}\,\] \[{{k}_{\alpha }}\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[6\times {{10}^{-5}}\]
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