A 8Ω resistance wire is doubled on it. Calculate the new resistance of the wire.
When the wire is doubled the length will be halved and area will be double or twice
$$R = \frac{P\times L}{A}$$
$$R1 = \frac{P\times \frac{L}{2}}{2A}$$ , where R1 is the new resistance
$$\frac{R1}{R}=\frac{\frac{P\times \frac{L}{2}}{2A}}{\frac{P\times L}{A}}$$
$$\frac{R1}{R} = \frac{1}{4}$$
$$R1 = \frac{R}{4}$$
Putting the value of R=8 Ω,
$$R1=\frac{8}{4} = 2Ω$$
Option D is correct.
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