Question

A 4 ohm resistance wire is stretched to double itself find the new resistance of the wire

Answer 1

Double (2X)
More than double (>2X)
Quadruple (4X)

I vote for quadruple.

You start with a wire of length L and cross section area A.
Your wire has a fixed "amount" of copper.
If you stretch it to 2L, the cross sectional area must halve to A/2.

Think of a rectangle length L by width A. It has area M.
If you change your rectangle to 2L, with area M, the width must now be A/2.

Anyway, resistance of a wire is linearly proportional to length, and inverse linear proportional to cross sectional area.

R = x * L/A

The length of the wire is double - this doubles resistance.
The cross sectional area is half, which doubles the resistance (again).

Start with R = x * L/A
Stretch
R = x * 2L/(1/2A) = x * 4 * L/A

This gives us quadruple...

Answer 2

Answer 1 ohm

∴ Let the resistance of the wire originally 'R' of length 'L' and area of cross-section 'A' with resistivity of material is 'p',

Then

R = p l/A = 4 ohms

Now, for new arrangement,

p' = p

l' = l / 2

A' = 2A

Thus,

R' = p' l'/A' = p (l /2) ÷ 2A = !/4 (p l / A) = 1/4R = 1/4 × 4 = 1 ohms