If both the number of turns and length of an inductor are tripled, keeping other factors constant, the self-inductance of the inductor will be

tripled

The correct answer is Option (1) → tripled

Self-inductance of a solenoid:

$L = \mu_0 \frac{N^2 A}{l}$, where N = number of turns, l = length of solenoid, A = cross-sectional area

Let original inductance be $L_0 = \mu_0 \frac{N^2 A}{l}$

If number of turns is tripled: $N' = 3N$

If length is tripled: $l' = 3l$

New inductance:

$L' = \mu_0 \frac{(3N)^2 A}{3l} = \mu_0 \frac{9 N^2 A}{3 l} = 3 \mu_0 \frac{N^2 A}{l} = 3 L_0$

∴ Self-inductance becomes three times the original value