A) \[x\]
B) \[y=7\,\sin \left[ 7\,\,\pi t-0.4\,\pi x+\frac{\pi }{3} \right]\]
C) \[\frac{2\pi }{49}m/s\]
D) \[\frac{49}{2\pi }m/s\]
Correct Answer: A
Solution :
Key Idea: The equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances. Let the change on the capacitors be \[F=Mr{{\omega }^{2}}\]. Then total charge is \[r=l\,\,\sin \,\,\theta \] \[\therefore \] Since, capacitors are connected in parallel equivalent capacitance is \[T\,\,\sin \,\,\theta =M\,\,l\,\,\sin \,\,\theta \,{{\omega }^{2}}\] \[\Rightarrow \] \[T=Ml{{\omega }^{2}}=Ml{{\left( 2\pi f \right)}^{2}}\] Given, \[\Rightarrow \] \[T=Ml{{\left( 2\pi \times \frac{2}{\pi } \right)}^{2}}\] \[\Rightarrow \] \[T=16\,Ml\] \[y=A\,\sin \left( \frac{2\pi }{\lambda }vt-\frac{2\pi }{\lambda }x+\phi \right)\]You need to login to perform this action.
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