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

Answer

B

Explanation

Let initial length of rod be `l_(0)` and Area A
As, `(T)/(A) = Y (Delta l)/(l)`
So, `(T_(1))/(A) = (Y (l_(1)- l_(0)))/(l_(0)) , (T_(2))/(A) = (Y(l_(2)- l_(0)))/(l_(0))`
Dividing, `(T_(1))/(T_(2))= (l_(1)- l_(0))/(l_(2)- l_(0)) , T_(1) l_(2)- T_(1) l_(0)= T_(2) l_(1) - T_(2) l_(0), l_(0)= (l_(1)T_(2)- l_(2)T_(1))/(T_(2)- T_(1))`