a copper wire is streched to make it 0.1%lonnger.what is the percentage change in resistance?

Hi,

Given: length of the wire initially : l_{1}

Length of the wire after it increases by 0.1% : l_{2 }= l_{1} + (1/1000 )l_{1}

_{ = }l_{1 }+ (1/1000)l_{1}

= (1001/1000)l_{1}

Area of_{ } the wire initially : A_{1}

Area of the wire after it is stretched becomes : A_{2}

Resistance initially of the wire : R_{1}

Resistance after it is stretched : R_{2}

Now, the most important thing to be kept in mind while solving this problem is that volume of the wire remains constant.

Hence, l_{1}A_{1} = l_{2}A_{2}

A_{1}= l_{2}A_{2}/l_{1}

_{ }= 1001 l_{1}A_{2}/1000l_{1}

A_{1} = 1001A_{2}/1000

We know that resistance( R) is given by the following equation:

R = p(l/A)

Where āpā is the resistivity of the wire which will remain the same throughout as it is the property of the material of which the wire is made up.

Taking the ratio of the resistances: R_{1}/R_{2} = p(l_{1}/A_{1})/p(l_{2}/A_{2})

R_{1}/R_{2} = (l_{1}A_{2})/(A_{1}l_{2})

Substituting the values of l_{2} and A_{1} in the above equation we get:

R_{1}/R_{2} = (l_{1}A_{2}*1000*1000)/(1001*1001*A_{2}l_{1})

R_{1}/R_{2} = 1000000/1002001

R_{2} = (0.998)R_{1}

R_{2} = (99.8/100)R_{1}

Or the percentage decrease in the resistance of the wire is = 0.2%

= 0.2% decrease

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