# 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 : l1

Length of the wire after it increases by 0.1% : l2 = l1 + (1/1000 )l1

= l1 + (1/1000)l1

= (1001/1000)l1

Area of  the wire initially  : A1

Area of the wire after it is stretched becomes : A2

Resistance initially of the wire : R1

Resistance after it is stretched : R2

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

Hence, l1A1 = l2A2

A1= l2A2/l1

= 1001 l1A2/1000l1

A1 = 1001A2/1000

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

R = p(l/A)

Where ‘p’ 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: R1/R2 = p(l1/A1)/p(l2/A2)

R1/R2 = (l1A2)/(A1l2)

Substituting the values of l2 and A1 in the above equation we get:

R1/R2 = (l1A2*1000*1000)/(1001*1001*A2l1)

R1/R2 = 1000000/1002001

R2 = (0.998)R1

R2 = (99.8/100)R1

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

= 0.2% decrease

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