Question

A wire is stretched, show that it's diameter reduced by 50% . Find resistance.​

Answer 1

Answer:

Answer is

Let initial length of the wire be l.

Resistance of the wire R=

V

ρl

2

New length of the wire l

′

=l+0.5l=1.5l

New resistance of the wire R

′

=

V

ρ(1.5l)

2

=2.25R

Percentage change in resistance

R

R

′

−R

×100=

R

2.25R−R

×100=125 %

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