When current $I$ passes through an inductor having self-inductance of $4 \ H$. If the current is made double,what will be the new self-inductance of the inductor?

Two solenoids $A$ and $B$ of equal number of turns have their lengths and radii in the same ratio $1 : 3$. The ratio of the self-inductance of solenoid $A$ to that of $B$ will be

$A$ coil of wire of a certain radius has $600$ turns and a self-inductance of $108 \, mH$. The self-inductance of a $2^{nd}$ similar coil of $500$ turns will be.........$mH$.

Two coils have self-inductance $L_1 = 4 \, mH$ and $L_2 = 1 \, mH$ respectively. The currents in the two coils are increased at the same rate. At a certain instant of time,both coils are given the same power. If $I_1$ and $I_2$ are the currents in the two coils at that instant of time respectively,then the value of $\frac{I_1}{I_2}$ is:

An average induced emf of $0.5 \,V$ appears in a coil when the current in it is changed from $10 \,A$ in one direction to $10 \,A$ in the opposite direction in $0.8 \,s$. The self-inductance of the coil is: (in $\,mH$)

In an inductor of self-inductance $L = 2 \, mH$,current changes with time according to the relation $i = t^2 e^{-t}$. At what time is the $emf$ zero? (in seconds)