
Answer & explanation
Correct answer: option 3
For solenoid, if N is the number of turns, A is cross section area of solenoid, l is the length of wire then, Self Inductance is $L = \frac{\mu_0 N^2 A}{l_0}$
$ A = \pi r^2 , \text{Length of the wire used} l'= 2\pi r \times N$
$\Rightarrow L = \frac{\mu_0 N^2 \pi r^2}{l_0} = \frac{\mu_0 (2\pi rN)^2}{4\pi l} = \frac{\mu_0 l'^2}{4\pi l_0} $
$4\pi L l_0 = \mu_0 l'^2$
$\Rightarrow l'= \sqrt {\frac{4\pi Ll_0}{\mu_0}}$