When the number of turns and the length of the solenoid are doubled keeping the area of cross-section same, the inductance

When a current of $2 \ A$ is passed through a coil of $100$ turns,the magnetic flux associated with it is $5 \times 10^{-5} \ Wb$. Find the self-inductance of the coil.

$A$ long solenoid has $500$ turns. When a current of $2 \, A$ is passed through it,the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \, Wb$. The self-inductance of the solenoid is ....... $H$.

The self-inductance of a coil having $500$ turns is $50\,mH$. The magnetic flux through the cross-sectional area of the coil when the current through it is $8\,mA$ is found to be:

The self-inductance of an air-cored solenoid of length $40 \ cm$,diameter $7 \ cm$ having $200$ turns is nearly: (in $\mu H$)

The equivalent quantity of mass in electricity is