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CBSE Class 12-science Answered

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Asked by www1028 | 07 Jul, 2010, 07:25: PM
Expert Answer
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
Answered by | 12 Jul, 2010, 04:02: PM
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