Saved Bookmarks
| 1. |
A wire of length L is strech such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 Omega, its new resistanc would be |
|
Answer» `40 Omega` SUPPOSE diameter of wire of length `l_(1) ` is `d_(1) ` and wire of length `l_(2) ` is `d_(2) ` . `therefore PI d_(1)^(2) = pi d_(2)^(2) l_(2)` `therefore ((d_(1))/(d_(2)))^(2) = (l_(2))/(l_(1))` `therefore (2)^(2) = (l_(2))/(l_(1))"" therefore 4 = (l_(2))/(l_(1))` Now R. = `((l_(2))/(l_(1)))^(2) .R = (4)^(2) XX 10 = 160 Omega ` |
|