A copper wire of resistance R is pulled so that its length is double (temperature remaining constant). Find the resistance of the wire in terms of its original resistance R
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Answer:
R - Resistance
L - length
A - cross sectional area
R- rho × l/A
when length is doubled then Cross sectional area is halved
let R ' to be new resistance
R' - rho × 2 l/ A/2
R'- rho 4L/A
R' - 4 R ( R- rho l/A)
hence
new resistance will be 4 times the resistance
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