Question

a piece of wire of resistance 20 Ohm is drawn out so that its length is increased to twice its original length. calculate the resistance of the wire in the new situation.

Answer 1

resistance = 20

original length be l

new length = 2l

we know length is directly proportional to resistance

R/20 = 2l/l

R = 40

the new resistance is 40 ohm

Answer 2

Given resistance , R = 20Ω

And we have to find the new resistance with the help of the original length .

Solution :

We know very well that ,

R = (pl/A)

We can write it as : R = (pl²/v)

Then , It is given that l' = 2l

So , R' = (4pl²)/V = 4R ----(eq.1)

R = 20 ohm = 20 Ω

So , putting R = 20 ohm in equation 1 ,

We have : R' = 4R = 4(20) = 80 Ω

The required new resistance is of 80 ohm = 80Ω