or

A cylindrical wire of resistance 'R' is reformed in a new wire of double length then resistance will be 

A. 2R

B. 4R

C. 6R

D. 8R

As the length of the wire is changed, it's cross sectional area will also change but volume remains same.

       Radius  decreases by factor of 2^(1/2).

       Resistance=(rho× L)/A

       Using above relation , new resistance will be 4R