question_answer

A capacitor of \[4\,\,Ml\] charged upto \[2\,\,Ml\] is connected in parallel with another capacitor of \[Ml\] charged upto 100V. The common potential is:                                                                   [BHU PMT-2002]

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,  \[\Rightarrow \]                 \[T=Ml{{\left( 2\pi \times \frac{2}{\pi } \right)}^{2}}\]       \[\Rightarrow \]                 \[T=16\,Ml\]      \[y=A\,\sin \left( \frac{2\pi }{\lambda }vt-\frac{2\pi }{\lambda }x+\phi  \right)\]