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  4. A wire has resistance of 10 Ω. If it is stretched by 1 / 10 th of its length, then its resistance is nearly

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Q. A wire has resistance of $10 \Omega$. If it is stretched by $1 / 10$ th of its length, then its resistance is nearly

Current Electricity

Solution:

$R=\rho l / A=10 \Omega$
New length, $ l_{1}=l+l / 10=11 l / 10$
$\therefore$ New area, $ A_{1}=A l / l_{1}=\frac{10 A}{11}$
$\therefore$ New resistance, $R_{1}=\rho l_{1} / A_{1}=\rho(11 l / 10) /(10 / 11) A$
$=\frac{121 \rho l}{100 A}=\frac{121}{100} \times 10=12.1 \Omega$