A) 1 :1
B) 1 : 2
C) 1 : 4
D) 2 : 1
Correct Answer: C
Solution :
Given, \[{{y}_{1}}=0.03\sin 2\pi (60t+{{\phi }_{1}})m\] and \[{{y}_{2}}=0.06\sin 2\pi (30t+{{\phi }_{2}})m\] \[\therefore \] \[{{a}_{1}}=0.03\] and \[{{a}_{2}}=0.06\] \[\therefore \] \[\frac{{{l}_{1}}}{{{l}_{2}}}={{\left( \frac{{{a}_{1}}}{a} \right)}^{2}}\] \[={{\left( \frac{0.03}{0.06} \right)}^{2}}\] \[=\frac{9}{36}=\frac{1}{4}\]You need to login to perform this action.
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