# The resistance if a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:

It is given that there is a wire of radius r, length l1 having a resistance R.

Now this wire is said to be extended to a new length l2 , which is n times l1 .

Since , there is no change in radius of the wire, let us assume that the volume before and after stretching are constant. V1=V2 ,

where

• V1 is volume before stretching
• V2 is volume after stretching.

Now we know that volume is a product of area and height.

In our case, height is the length of the wire A1 × l= A2 × l2 We know that , l= n×l1 .

Substituting this , we get ⇒A1 × l=A2 × n × l⇒ A1/n = A2

We know that resistance R of a material is calculated by using the formula R=ρ l/A , where

• ρ is represented as resistivity of the material
• A is area and l is length of the wire.

Now , before extension , resistivity is given as , R1 = ρl1/A1

After extension, resistivity is given as, R=ρl/A2

Substituting for A2 in the equation for R2 , we get

⇒ R2 = ρl2 × n/A1

Now substituting l2= n ×l1 ,

we get ⇒ R2 = ρl× n2/A1 (The term ρl1A1 is equal to R1 and R1 value given in the question is R )

⇒ R2 = R × n2

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