A wire of resistance 4Ω is stretched to double its original length. The resistance of the stretched wire would be?
4Ω
2Ω
8Ω
16Ω
Solution:
The volume of the wire remains the same. A*l = constant. So when length is doubled, area becomes half. Using the formula R = ρl/A we get the new resistance as 4 times the initial value. Thus D is correct.