When the number of turns in a coil is doubled without any change in the length of the coil,its self-inductance

The self-inductance of a coil is $L$. Keeping the length and area the same,the number of turns in the coil is increased to four times. The self-inductance of the coil will now be

$A$ solenoid with an air core has length $48 \pi \ cm$,area of cross section $12 \ cm^2$ and $1200$ turns. The self-inductance of the solenoid is: (in $mH$)

$A$ long solenoid has $1500$ turns. When a current of $3.5 \, A$ flows through it, the magnetic flux linked with each turn of the solenoid is $2.8 \times 10^{-3} \, Wb$. The self-inductance of the solenoid is (in $ \, H$)

$A$ toroid is a long coil of wire ($N$ turns) wound over a circular core. The coefficient of self-induction of the toroid is [The magnetic field in it is uniform and $R >> r$,where $r=$ radius of wire,$R=$ radius of coil] ($\mu_0=$ permeability of free space).

Current in a circuit falls from $5.0 \ A$ to $0.00 \ A$ in $0.1 \ s$. If an average emf of $200 \ V$ is induced,then the self-inductance of the circuit is . . . . . . $H$.