Question

A metal wire of resistance 6 ohm is stretched so that its length become double of it original length.what would be the final resistance

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf New\: resistance (R_2)=24\:ohm }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Resistance $\sf{(R_1)=6\:ohm}$

Let its ,

• Lenght = l

• Area = A

$\large\underline\pink{\sf To\:Find: }$

• New resistance when old resistance stretched to double $\sf{(R_2)=?}$

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We know that ,

$\large{\boxed{\sf R=\rho\frac{l}{A}}}$

if length become double of it original length then its area will become half of original.

$\large\implies{\sf R_2=\rho\frac{2l}{A/2}}$

$\large\implies{\sf R_2=\rho\frac{l}{A}×4 }$

$\large\implies{\sf R_2=R_1×4 }$

${\sf R_1=6\:ohm }$ (Given)

$\large\implies{\sf R_2=6×4 }$

$\large\implies{\sf R_2=24\:ohm}$

$\red{\boxed{\sf New\: resistance (R_2)=24\:ohm }}$

Hence ,

New resistance on doubling the Lenght of old resistance we get 24 ohm