A) 18
B) 13
C) 9
D) 7
Correct Answer: B
Solution :
Suppose, the initial intensity is \[\sqrt{{{l}_{1}}+{{l}_{2}}}\]and corresponding decibels levels is \[\frac{{{l}_{1}}{{T}_{2}}-{{l}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}\]. When intensity is increased to 20 I then corresponding increased in decibels level is let\[\frac{{{l}_{1}}{{T}_{2}}+{{l}_{2}}{{T}_{1}}}{{{T}_{1}}+{{T}_{2}}}\]Thus, \[\frac{v\,dv}{dX}=-{{\omega }^{2}}\text{x with the initial condition V}={{\text{V}}_{0}}\text{ at}\] and \[V=\sqrt{v_{0}^{2}+{{\omega }^{2}}{{X}^{2}}}\] This implies \[V=\sqrt{v_{0}^{2}-{{\omega }^{2}}{{X}^{2}}}\] \[V=\sqrt[3]{v_{0}^{3}+{{\omega }^{3}}{{X}^{3}}}\]You need to login to perform this action.
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