A) \[{{\alpha }_{1}}+{{\alpha }_{2}}=0\]
B) \[{{\rho }_{1}}{{\alpha }_{1}}={{\rho }_{2}}{{\alpha }_{2}}\]
C) \[{{\rho }_{1}}{{\alpha }_{1}}+{{\rho }_{2}}{{\alpha }_{2}}=0\]
D) \[{{\rho }_{1}}{{\alpha }_{2}}+{{\rho }_{2}}{{\alpha }_{1}}=0\]
Correct Answer: C
Solution :
\[R={{R}_{1}}+{{R}_{2}}\]\[=\frac{1}{A}\left[ {{\rho }_{1}}\left( 1+{{\alpha }_{1}}t \right)+{{\rho }_{2}}\left( 1+{{\alpha }_{2}}t \right) \right]\] |
For, \[\frac{dR}{dt}=0\] |
Or \[{{\rho }_{1}}{{\alpha }_{1}}+{{\rho }_{2}}{{\alpha }_{2}}=0\] |
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