The circuit shown in figure is in the steady state with switch `S_(1)` closed.At `t=0`,`S_(1)` is opened and switch `S_(2)` is closed. The first instant `t` when energy in inductor becomes one third of that in capacitor is equal to `(pi xx 10^(-5))/x sec`. Then finout value of `x`.

Answer

C

Explanation

In a `LC` circuit, total energy remains always conserved.
`therefore` Electrical energy stored in capacitor +Magnetic stored in inductor=`(1 Q_(0)^(2))/(2 C_(2))`
`rArr U_(E)+U_(B)=(Q_(0)^(2))/(2C_(2))`
At the time `t=t_(1) U_(B)=1/3 U_(E) rArr U_(E)=3/4(1/2(Q_(0)^(2))/(C_(2)))`
`rArr 1/2 (Q^(2))/(C^(2))=3/4 (1/2(Q_(0)^(2))/(C_(2)))`
`rArr Q=sqrt3/2Q_(0) rArr Q_(0)cos omega t_(1)=sqrt3/2 Q_(0)`
`rArr omega t_(1)=pi/6 rArr t_(1)=pi/(6omega)=1.05xx10^(-5)=10.5 muS`