Question

If a copper wire is stretched to make it 0.1% longer, then what is the approximate percentage change in its resistance? Plz answer fast.

Answer 1

Let

X,  be the initial length, and r

Y be the cross-sectional area

R be the resistance of wire.

Let X', Y' and R' be the new length, cross- sectional area and resistance of wire after the length of wire is increased.

Length is increased by 0.1%, therefore,

L' = L+(0.1/100)L

L' = 1.001 L

Since the volume of wire remains same, thus,

V' = V

A'*L' = A*L

A' * 1.001 L = A*L

A' = A/1.001

New Resistance,

R' = ρL'/A'

= ρ*1.001 L/(A/1.001)

= (1.001)²(ρL/A)

R' = 1.002*R

% change in resistance = (Final resistance - Initial resistance)/Initial resistance * 100%

= (R' - R)/R*100%

= (1.002R - R)/R*100%

= 0.002 * 100%

= 0.2%

Thus, resistance of wire is increased by 0.2%.

Answer 2

here the initial length is given by 0.1% has a length increases the so area of cross section will decreases .

because the length is inversely proportional to the area of cross section

here the volume is constant

A=V/L

so here we can negelect the V

remaining answer I am given in the picture

I hope this answer will help you ❣