Question

The length of a conductor is doubled and its radius is halved, its resistance is
(A) unchanged (B) doubled
(C) quadrupled (D) eight times its value
Plz Give One Example Too....

Answer 1

l is length at first and A is original area of cross section.henceR~l/AA=pi*r^2A’=pi*(r/2)^2A’=pi*r^2/4=A/4l’=2lhence, new resistance, R’R’=l’/A’R’=2l/(A/4)R’=8length/AR’=8RNew resistance is 8 times the original resistance

Answer 2

HEY!!
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✴Resistance of a wire is given as:-

✴R=ρ(l/A)=ρ(l/πr2)

✴According to the question:-

✴l'→2l; r'→r/2

✴⇒R'=ρl'/πr'2

✴=ρ(2l×4/πr2)

✴=8ρl/A=8R