A resistor of wire 24 ~cm length and resistance 8 is stretched in to a uniform wire of 48 ~cm length, then the new resistance will be :

Step-by-step solution

The resistance R of a wire is given by R = l A , where is the resistivity, l is the length, and A is the cross-sectional area. Since the volume V = A l of the wire remains constant during stretching, we have A₁ l₁ = A₂ l₂ . Given l₁ = 24 cm and l₂ = 48 cm , the ratio of lengths is l₂ l₁ = 48 24 = 2 . From the volume conservation, A₂ = A₁ l₁ l₂ = A₁ 2 . The new resistance R₂ is given by R₂ = l₂ A₂ = 2l₁ A₁ / 2 = 4 l₁ A₁ = 4 R₁ . Given R₁ = 8 , the new resistance is R₂ = 4 8 = 32 . Answer: 32