Question

A wire is stretched so that its length becomes 6 by 5 times its original length if its original resistance is 25 find its new resistance and resistivity

Answer 1

R=p*(l/A)

here according to question the length becomes 6/5l

Now area also changes as volume remains same.

V=l*A

l*A=6/5l*new area

new area=5/6A

Now substituting these values we get R=p*(6/5l/5/6A)

R=p*(36/25(l/A))

So new resistance is 36/25 times old one.

36/25*25=36 ohm.

Hope this helps!

Answer 2

R = p L/A

R = 25 ohm

Volume of material in the conductor = LA

Vol of the conductor remains the same

Vol = AL

= A'L'

A' = AL/L'

After streching the wire

L' = 6L/5

A' = 5AL/6L

= 5A/6

Resistance of streched wire = R'

= pL'/A'

= p(6L/5) × {6 / (5A) }

R' = 36pL / 25A

R' = 36R / 25

Now Resistance ,

= 36×25 / 25ohm

= 36 Ohm