A wire of length L m has 2 ohms resistance. It is stretched twice its original length and its area becomes half of its original area. What is its new resistance?

A wire's resistance is proportional to its length and inversely proportional to its area.

So if the original resistance was R=2Ω, the new resistance is then:
R1=2Ω * (2L/L)* (A/(A/2))
= 2Ω * (2)*(2)
= 8Ω

To find the new resistance of the wire after it is stretched and its area changes, we can use the formula for the resistance of a wire:

\[R = (ρ * L) / A\]

Where:
- R is the resistance of the wire
- ρ (rho) is the resistivity of the material the wire is made of
- L is the length of the wire
- A is the cross-sectional area of the wire

Given:
Original wire length, \(L\): L m
Original wire resistance, \(R_1\): 2 ohms
New length of the wire, \(L_{\text{new}}\): 2L m
New area of the wire, \(A_{\text{new}}\): (1/2)A

To find the resistivity, we need more information. If we assume that the wire is made of a material with a constant resistivity, then we can solve the equation without it. Let's proceed with this assumption.

First, let's find the original cross-sectional area, \(A\), using the given resistance and length:

\[R_1 = (ρ * L) / A\]
\[2 = (ρ * L) / A\]
\[A = (ρ * L) / 2\]

Now, let's substitute the given new length and area values into the formula and solve for the new resistance, \(R_{\text{new}}\):

\[R_{\text{new}} = (ρ * L_{\text{new}}) / A_{\text{new}}\]
\[R_{\text{new}} = (ρ * 2L) / [(1/2)A]\]
\[R_{\text{new}} = (4ρ * L) / A\]

Since the original area, \(A\), can be expressed as \((ρ * L) / 2\), we can substitute this expression for \(A\) in the above equation:

\[R_{\text{new}} = (4 * ρ * L) / [(ρ * L) / 2]\]
\[R_{\text{new}} = (4 * ρ * L) * (2 / (ρ * L))\]
\[R_{\text{new}} = 8\]

Therefore, the new resistance of the wire after it is stretched and its area changes is 8 ohms.