A metal wire of resistance of 6Ω is stretched so that its length is increased to thrice its original length.

Calculate its new resistance.

Asked by om.chaudhari1673 | 23rd Jan, 2019, 04:43: PM

Expert Answer:

When the metal wire is streched so that its length is increased thrice, its diameter will decrease.
This fact has to be taken into account while calculating the change in resistance.
 
when the wire is stretched its volume remain constant. If L and D are initial length and diameter respectively,
and the modified length and diameter are 3L  and D' due to stretching,
 
then we have  L×D2 = 3L×(D')2
 
Resistance is directly proportional to length and inversley proportional to area or square of diameter.
 
Let R be the initial resistance and R' be the modified resistance due to stretching

Hence begin mathsize 12px style fraction numerator R over denominator R apostrophe end fraction space equals space fraction numerator L over denominator 3 L end fraction cross times fraction numerator D apostrophe squared over denominator D squared end fraction space equals space fraction numerator L over denominator 3 L end fraction cross times fraction numerator L over denominator begin display style 3 L end style end fraction space equals space 1 over 9 end style
hence new resistance will be 9 times of its initial value . New resistance is 9×6 = 54Ω

Answered by Thiyagarajan K | 23rd Jan, 2019, 05:41: PM

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