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
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 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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