A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 , its new resistance would be

Step-by-step solution

Let the original diameter of the wire be D . Therefore the new diameter is D / 2 . Original area of cross-section is D^2 4 and the final area of cross-section is D^2 16 . The new length of the wire is given by L D^2 4 =L^ D^2 16 L^ = 16 4 L=4 L Now, we know that the resistance is given by R= L A . aligned & R^ = L^ A^ = 4 L A / 4 =16 R . & [ A^ = D^2 16 = A 4 ] & R^ =16 10=160 . aligned