Question

Resistance is 10 ohm, l is increased by 10% by stretching, what is the new resistance

Answer 1

Answer:

Check the attachment..

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Answer 2

$\Large\underline{\underline{\sf \orange{Given}:}}$

• Resistance of wire $\sf{(R_1)}$ = 10Ω

• Length is increased by = 10%

$\Large\underline{\underline{\sf \orange{To\:Find}:}}$

• New resistance (R') = ?

$\Large\underline{\underline{\sf \orange{Solution}:}}$

Let ,

Original Lenght = L

A = VL

Resistance (R) $\implies{\rho \dfrac{L}{A}\:\:\:\:\implies \:\:10Ω}$

New Lenght (L')

$\implies{\sf L'=L+\dfrac{10}{100}L }$

$\implies{\sf L'=1.1L }$

New Area (A')

$\implies{\sf A'=\dfrac{AL}{1.1L}\:\:\:\implies\:\:\: \dfrac{A}{1.1} }$

New Resistance (R')

$\large\boxed{\boxed{\sf R'=\rho\dfrac{L'}{A'} }}$

$\implies{\sf R'=\rho \dfrac{1.1L}{\dfrac{A}{1.1}}}$

$\implies{\sf R'=1.21×\rho\dfrac{L}{A}}$

$\implies{\sf R'=1.21×10}$

$\implies{\sf R'=12.1Ω }$

$\Large\underline{\underline{\sf \orange{Answer}:}}$

⛬ New Resistance is 12.1Ω