Question

the wire of length L and cross sectional area a is stretched to double its length find its new resistance

Answer 1

given old area of crossection = a
old length= l
now wire stretched length become 2l and rho be resistivity of material
since
New volume = old volume
new area of crossection * new length = old area of crossection * old length
=> new area of crossection = a*l/2l
= a/2
old resistance (r) = rho l/a
new resistance = rho 2l/(a/2)
= 4 rho *l/a
= 4r
thus new resistance become 4 times of old resistance

Answer 2

$Hi \: \: there \: \\ \\ Let \: the \: cross \: \: section \: area \: of \: the \: wire \: be - \: A \: \\ \\ Length \: of \: the \: wire \: be - \: L \\ \\ \: Resistance \: of \: the \: wire \: be - \: R \\ \\ \: and \: resistivity \: of \: \: the \: wire \: be - \: p$
Let me tell you an relation among these -

R ∝ L

and R ∝ 1/A

Thus we get

R = ( p × L ) / A

or ( RA ) / L = p ...... ( 1 )


where p is resistivity and is constant [ that doesn't change ]

Now coming to our Question -

it says when

New Cross section area - A/2

Length of the wire - 2L

Then ,

New resistance

substituting the value of p in equation ( 2 )
$R' \: \: = p\: \frac{2L \: }{ \frac{A}{2} } ......... \: (2) \\ \\ \: \: \: \: \: \: \: \: = \frac{RA \: }{L \: } \times \frac{4L \: }{A \: } \\ \\ \: \: \: \: \: \: \: \: = 4R \:$

Hence New Resistance is 4th of Resistance


Hope this help ya ☺