question_answer

Two equal vectors have a resultant equal to either of them, then the angle between them will be:

A) \[110{}^\circ C\]

B) \[120{}^\circ C\]

C) \[60{}^\circ C\]

D) \[150{}^\circ C\]

Correct Answer: B

Solution :

                Key Idea: Two vectors (inclined at any angle) and their sum vector form a triangle. It is given that two vectors have a resultant equal to either of them, hence these three vectors form an equilateral triangle each angle of \[=\frac{Force}{Length}=\frac{[ML{{T}^{-2}}]}{[L]}=[M{{T}^{-2}}]\]. In the figure \[[ML{{T}^{-2}}]\] and \[=[M{{L}^{-1}}{{T}^{-2}}]\] are two vectors \[[M{{L}^{2}}{{T}^{-1}}]\] having their sum vectors \[\alpha =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{g}}}\] such that \[\Delta {{i}_{C}}\] Thus, the vectors\[\Delta {{i}_{g}}\] and \[\alpha =0.96,\Delta {{i}_{E}}=7.2,\] of same magnitude have the resultant vectors \[\vec{R}\] of the same magnitude. In this case angle between \[\Delta {{i}_{C}}=0.96\times 7.2=6.91mA\] and \[\Delta {{i}_{E}}=\Delta {{i}_{C}}+\Delta {{i}_{B}}\],    is \[\therefore \]. Alternate: Let there be two vectors \[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] and \[=7.2-6.91=0.29mA\] where, \[\frac{(2n+1)\lambda }{2}\]. Their sum is \[\frac{(n+1)\lambda }{2}\] Taking self product of both sides, we get \[n(\lambda +1)\] \[n\lambda \] \[[M{{L}^{3}}{{T}^{-3}}]\] where \[[M{{L}^{-1}}{{T}^{-1}}]\] is angle between \[[M{{L}^{2}}{{T}^{-2}}]\]and \[[M{{T}^{-2}}]\]. When \[\sqrt{\frac{1}{{{\varepsilon }_{0}}{{\mu }_{0}}}}\], then we have \[\sqrt{\frac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}\] \[\frac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}\]   \[{{\varepsilon }_{0}}{{\mu }_{0}}\] \[{{V}_{0}}\]    \[{{V}_{e}}\] \[{{V}_{e}}=2{{V}_{0}}\]   \[{{V}_{e}}=\sqrt{3{{V}_{0}}}\] In this condition angle between gives vectors should be \[{{V}_{e}}={{V}_{0}}\sqrt{2}\].