A wire stretched to increase its length by`5%`. Calculate percentage charge in its resistance.

A wire stretched to increase its length by`5%`. Calculate percentage c

1.	A wire stretched to increase its length by`5%`. Calculate percentage charge in its resistance.
Answer» Correct Answer - `10.25 %` When a wire is stretched, its volume remain constant hence `l_(1)A_(1) = l_(2)A_(2) = V`, the volume Now `R_(1) = (rhol_(1))/(A_(1)) = (rhol_(1) xx l_(1))/(l_(1)A_(1)) = (rhol_(1)^(2))/(V)` , i.e. `R_(1) prop l_(1)^(2)` Hence or `R_(2) = 1.1025 R_(1)` `:. %` change in resistance `= ((R_(2) - R_(1))/(R_(1))) xx 100= ((R_(2))/(R_(1)) - 1) xx 100` `= (1.1025 - 1) xx 100 = 10.25 %`

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A wire stretched to increase its length by`5%`. Calculate percentage charge in its resistance.

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