The length of a metal wire is `l_(1)` when the tension in it is `T_(1)` and is `l_(2)` when the tension is `T_(2)`. Then natural length of the wire is

Options

A. `(l_(1) + l_(2))/(2)`
B. `sqrt(l_(1)l_(2))`
C. `(l_(1)T_(2)- l_(2)T_(1))/(T_(2)-T_(1))`
D. `(l_(1)T_(2) + l_(2)T_(1))/(T_(1)+ T_(1))`

Answer

C

Explanation

Let the natural length of the wire be `l_(0)`. The elongation `x_(1)= (l_(1)-l_(0))` and the elongation `x_(2)= (l_(2)-l_(0))`
As `Y= (F.L)/(A.x) rArr (F)/(x)= (YA)/(L)`= constant for same wire.
For different tension of the same wire,
`(F_(1))/(x_(1)) = (F_(2))/(x_(2)) rArr (T_(1))/((l_(1)-l_(0)))= (T_(2))/((l_(2)-l_(0)))`
`T_(1) (l_(2)-l_(0)) = T_(2) (l_(1)-l_(0))`
`T_(1) l_(2)- T_(1) l_(0) = T_(2) l_(1)- T_(2) l_(0)`
`(T_(2)-T_(1)) l_(0) = T_(2) l_(1) - T_(1) l_(2) rArr l_(0) = (l_(1) T_(2) - l_(2) T_(1))/(T_(2)- T_(1))`
`rArr` ( c) is correct