A) \[{{Y}^{2}}=XZ\]
B) \[{{Z}^{2}}=XY\]
C) \[{{Y}^{2}}=4XZ\]
D) \[{{X}^{2}}=YZ\]
Correct Answer: C
Solution :
[c] At resonance v will be maximum, tending to infinity. So term in the denominator, i.e. \[\sqrt{X{{\omega }^{2}}+Y\omega +Z}\] will tend to zero (minimum). \[v=\frac{1}{\sqrt{X{{\omega }^{2}}-Y\omega +Z}}\] \[\therefore \,\,\,\frac{d}{d\omega }\left[ \sqrt{X{{\omega }^{2}}-Y\omega +Z} \right]=0\] \[\Rightarrow \,\,\,\omega =\frac{Y}{2X}\] \[\therefore \,\] At, \[\omega =\frac{Y}{2X}\] \[X{{\omega }^{2}}-Y\omega +Z=0\] \[\Rightarrow \,\,\,X{{\left[ \frac{Y}{2X} \right]}^{2}}-Y\left[ \frac{Y}{2X} \right]+Z=0\,\,\,\,\Rightarrow \,\,\,{{Y}^{2}}=4XZ\]
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