A wire of resistance 5Ω is drawn out so that its length is increased by thrice its original length. The new resistance would be:
Answer & explanation
Correct answer: option 4
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, $A'=\frac{A}{3}$
$∴R'=ρ\frac{3L}{(A/3)}=9\frac{ρL}{A}$
$=9×5=45Ω$