A metal wire of circular cross-section has a resistance R. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes `R_2` then `R_2 : R_1` is

Options

A. `1:1`
B. `1 :2`
C. `4 :1`
D. `1:4`

Answer

C

Explanation

As `R= ( rho l )/( A ) = ( rho l xx l )/(A xx l) =( rho l^2)/(V)`
here ` rho ` and V are constant `therefore R prop l^2 therefore (R_1)/( R_2) =(l_(1)^(2))/( l_(2)^(2))`
given `l_2 = 2l_1 `. Hence ` (R_1)/( R_2) =(l_1^(2) )/( 4l_(1)^(2)) = 1/4 implies R_2 : R_1 = 4 :1`