The resistance of a wire is $10 \Omega$. Its length is increased by 10% by stretching. The new resistance of the wire will now be:

$12.1 \Omega$

The correct answer is Option (1) → $12.1 \Omega$

The resistance of a wire is -

$R=ρ\frac{L}{A}$

where,

$ρ$ - Resistivity of Material

L - Length of the wire

A - Area of the cross-section.

When the wire is stretched, its length increases, and its cross-sectional area decreases.

$L_{new}=L+\frac{10}{100}×L=1.1L$

Since, the volume always remain constant.

$1.1L×A_{new}=AL$

$A_{new}=\frac{A}{1.1}$

$∴R_{new}=ρ\frac{1.1L×1.1}{A}=(1.1)^2×R$

$=1.21×R$

$=1.21×10$

$=12.1Ω$