Question

the length of wire is increased by 0.1% of its initial length, how many percentage by its resistance will increase​

Answer 1

Answer:

Resistance will also increase by 0.1% as resistance is directly proportional to length.

Answer 2

Percentage increase in resistance is 0.1 % .

Given :

The length of wire is increased by 0.1% of its initial length.

Let , initial length is l and initial resistance is R .

Therefore , $R=\rho\dfrac{l}{A}$

Now , new length ,

$l'=l+\dfrac{0.1}{100}l\\\\l'=\dfrac{100.1}{100}l$

Therefore , new resistance ,

$R'=\rho \dfrac{l'}{A}\\\\R'=\rho \dfrac{l'}{A}\\\\R'=\rho \dfrac{\dfrac{100.1}{100}l}{A}\\\\R'=\dfrac{100.1}{100}\dfrac{\rho l}{A}\\\\R'=\dfrac{100.1}{100}R$

Therefore percent increase , $(\dfrac{\dfrac{100.1}{100}R-R}{R})\times 100=0.1 \ percentage$

Hence , this the required solution.

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