BELOW

We know that resistance is given by, R = ρxL/A

 

Lets take its inital diameter to be 'd' and initial lenght as 'l', therefore we get

A = pi (d/2)² = pixd²/4

R = ρxl/A

   = ρxl/(pixd²/4)

   = 4ρxl/pixd²

 

Now its given that the new diameter is half of the inital diameter,

ie, d' = d/2

Therefore new A' = pix(d'/2)² = pix(d/2x2)² = pix(d/4)² = pixd² /16 = A/4

Now to find out the new length of the wire we will apply mass conservation,

ie, Mass after elongation = Mass before elongation

 

Mass after elongation =σ A' l'

Mass after elongation =σ A l

where,

σ = Density of the wire

l' = new length of the wire after elongation

 

σ A' l' = σ A l

therefore, l' = Axl/A'

                 = pix(d²/4)xl/pix(d²/16)

                 = 4l

Now putting value of A' and l', for getting the value of new resistance R', we get

 

R' = ρxl'/A'

    = ρx(4l)/(A/4)

    = (4x4) [(ρxl/A)]

    = (16) (R)

hence, R' = 16R.

 

Hope this helps.

Team Topper