The magnetic energy in an inductor changes from maximum value to minimum value in $5 \ ms$. When connected to an $A.C.$ source,the frequency of the source is (in $Hz$)

$A$ capacitor of capacitance $C$ is charged to a potential $V_1$. It is then connected to an ideal inductor of inductance $L$. When the potential across the capacitor drops to $V_2$,what is the current flowing through the inductor?

An ideal capacitor of capacitance $0.2\, \mu F$ is charged to a potential difference of $10\, V$. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self-inductance $0.5\, mH$. The current at a time when the potential difference across the capacitor is $5\, V$,is.....$A$

The natural frequency of an $L-C$ circuit is $1,25,000 \text{ cycle/s}$. Then the capacitor $C$ is replaced by another capacitor with a dielectric medium of dielectric constant $K$. In this case,the frequency decreases by $25 \text{ kHz}$. The value of $K$ is

In an oscillating $LC$ circuit, the maximum charge on the capacitor is $Q$. When the energy is stored equally between the electric and magnetic fields, the charge on the capacitor becomes

Initially,the key was placed on $(1)$ until the capacitor got fully charged. Now,the key is placed on $(2)$ at $t = 0$. Find the minimum time when the energy in both the capacitor and the inductor will be the same.