Question

answer the question in the attachment.

Answer 1

R = p × l/A

• R = Resistance
• p = Rho
• l = length
• A = Cross section area

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Let length of 1st wire be l and second wire be l'.

➜ R = p × l/A

➜ 0.6 = p × l/A ___(1)

The things given between two copper wires,

1. A' = 1/2A
2. p [rho] as material is same.

A' = 1A/2 because when mass is same ans length is increased by stretching then the area will automatically decrease.

➜ R' = p × l'/A'

Given : l' = 2l and A' = 1A/2

➜ R' = p × 2l/1A/2

➜ R' = p × 4l/A

➜ R' = 4[p × l/A]

By substituting (1) we get,

➜ R' = 4 × 0.6

➜ R' = 2.4 Ohms

Answer ➜ 2.4 Ohms