A metal wire of resistance `3 omega` is elongated to make a unform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle `60^(@)` at the centre, the equivalent resistance between these two points will be :
A metal wire of resistance 3Omega is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60^(@) at the center, the equivalent resistance between these two points will be :
A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire.
A wire of resistance 4 Omega is stretched to four times of its original length resistance of wire now becomes
A uniform wire of resistance =3Omega and length l is stretched to double its length. Now it is bent to form a circular loop and two point P & Q lies on the loop such that they subtend 60^(@) angle at centre. The equivalent resistance between two point P & Q is:
A metal wire has a resistance of 35Omega . If its length is increased to double by drawing it, then its new resistance will be
A wire has a resistance of 32Omega . It is melted and drawn into a wire of half of its original length. What is the resistance of the new wire?
A wire is bent in the form of a triangle now the equivalent resistance between its one end and the mid point of the side is
A wire of resistance 1 Omega is stretched to double its length. What is the new resistance ?
