A piece of wire of resistance `4Omega` is bent through `180^(@)` at its midpoint and the two halves are twisted together.Then the resistance is

A wire of resistance 4 Omega is bent through 180^@ at its mid point and the two halves are twisted together. Then the resistance is

A piece of wire of resistance 4 ohm s is bent through 180^(@) at its mid point and the two halves are twisted together, then the resistance is

A piece of wire of resistance 4 ohm s is bent through 180^(@) at its mid point and the two halves are twisted together, then the resistance is

A piece of wire of resistance 4 ohm is bent through 180 at its mid point and the two halves are twisted together. Then the resistance is

A 8 Omega resistance wire is bent through 180^(@) at its mid point and the two halves are twisted together. Then the resistance is

A wire of resistance 4 Omega is bent to form a circle. The resistance point A and B will be

(a) State four factors on which the resistance of a conductor depends. (b) A piece of wire of resistance 20 Omega is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

A 4 Omega non insulated resistance wire is bent in the middle by 180^@ and both the halves are twisted with each other. What will be its new resistance ?