A) \[{{f}_{0}}=200cm\]
B) \[{{u}_{0}}=-2km=-2\times {{10}^{5}}cm\]
C) \[O=50m=5\times {{10}^{3}}cm\]
D) \[\therefore \]
Correct Answer: C
Solution :
Key Idea: When resistors are connected in parallel potential difference across them is same. In the given circuit the resistors \[\Rightarrow \] and \[{{K}_{f}}=500-25=475J\] are connected in parallel hence potential difference (V) across them is same. In order that they undergo same energy loss, \[\sim 1\,eV\] \[h{{v}_{0}}=\frac{GMh}{r{{c}^{2}}}{{v}_{0}}\]must be equal to \[r=\frac{GM}{{{c}^{2}}}\]. i.e., \[{{v}_{escape}}=\sqrt{\frac{2GM}{r}}\ge c\] Now resistor \[B=\frac{normal\,stress}{volume\,strain}\] is in series with \[B=-\frac{\Delta P}{\Delta V/V}\], hence energy through them is \[B\propto \frac{1}{\frac{\Delta V}{V}}\] where \[\Rightarrow \] is current across \[\frac{\Delta V}{V}\propto \frac{1}{B}\]. Since \[y=\sin \omega t\,-\cos \,\omega t\], therefore current through them is \[\frac{dy}{dt}=\omega \cos \omega t+\omega \sin \omega t\] \[\frac{{{d}^{2}}y}{d{{t}^{2}}}=-{{\omega }^{2}}\sin \omega t+{{\omega }^{2}}\cos \omega t\] \[a=-{{\omega }^{2}}(\sin \,\omega t-\cos \omega t)\] \[a=-{{\omega }^{2}}y\] \[\Rightarrow \]
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