A wire of resistance 5Ω is drawn out so that its length is increased by thrice its original length. The new resistance would be:

45 Ω

The correct answer is Option (4) → 45 Ω

When a wire is stretched, its resistance changes. The resistance is given by the formula -

$R=ρ\frac{L}{A}$

where,

$R$ - Resistance

$ρ$ - Resistivity

$L$ - Length of wire

$A$ - Cross-sectional area

Length of wire is increased three times, $L'=3L$

Area of wire, $A'=\frac{A}{3}$

$∴R'=ρ\frac{3L}{(A/3)}=9\frac{ρL}{A}$

$=9×5=45Ω$