Question

a copper wire is in the form a cylinder and has a resistance R . It is stretched till its thickness reduces by half of its initial size.Find its new resistance in terms of R

Answer 1

Solution:

Let R₀ be initial and R₁ be final resistance.

R₀ = ρL₀/π r₀² ..... equation 1

R₁ = ρL₁/ π r₁² ... equation 2

Divide 1 by 2 , We get,

R₁ /R₀  = L₁/L₀  r₀²/r₁²  

⇒ R₁ = R₀ [L₁/L₀  r₀²/r₁² ] .... equation 3

From Question, its given that r₁ =  r₀/2

⇒ r₁² =  r₀²/4 .... equation 4

Substitute equation 4 in 3 , We get,

R₁ = R₀ ( L₁/L₀ . 4 r₀²/ r₀² )

R₁ = 4 R₀ (L₁/L₀) .... equation 5

Now, volume of wire remain unchanged.

Initial volume = Final volume

πr₀²L₀ = πr₁²L₁

⇒ r₀²/ r₁² = L₁/L₀

From equation 4

⇒  L₁/L₀ = 4

substituting in equation 5 , we get,

R₁= 16 R₀