Question

a wire of resistance X ohm is drawn out so that its length is increased to twice its original length and its new resistance become 20 ohm then X will be?? ​

Answer 1

suppose the initial length of wire is l and becomes l' after stretching. the initial cross section of wire is A and after stretching it becomes A'

so Al must be equal to A'l'

A/A'=l'/l if we supposed it equal to n as wire is only stretches

so R=p l/A---(1)
and after stretching
R'=pl'/A'----(2)

so R'/R=[l'/l][A/A']
=n×n
=n²
so R'=n²R

so if R=X
and wire streaches twice of its length then
n=2

and R'=20
so
R=R'/n²
=20/4
=5.

hope dis help

Answer 2

Here, let L be the length of wire

A be area of cross section

Resistivity = ϱ

Resistance = R

R = ϱ l/A

X = ϱ l/A -------(CASE1)

Now,

l = 2l, A = A/2 (When wire is doubled, area is halved)

R = ϱ l/A

20 = ϱ 2l*2/A

20 = ϱ 4l/A -------(CASE2)

Divide 2 by 1

X/20 = 1/4

X = 5 ohm