Question

recasting of a cylindrical conductor of resistance 20 ohm is done having its new radius half than the earlier and keeping its Length unchanged then the new resistance of the conductor will be

1) 80 ohm
2) 10 ohm
3) 20 ohm
4)5 ohm​

Answer 1

Answer:

60 ohm

Explanation:

R=( p*l) /A

P(rho)= resistivity of material

L= length

A= area of cross section

Now volume of the conductor will not change when you elongate it.

A(1)* l(1)= A(2) * l(2). Now we have l(2)= 2l(1). So we get A(2)= A(1)/2. So when you elongate the length the cross section of the conductor decreases .

R(2) = p * l(2)/ A(2) rho is independent of geometry of conductor.

R(2) = ( p* 2l(1)) / ((A(1)/2) )

R(2)= 4 ( p * l(1))/( A(1) )

= 4 * 20= 80 ohms

So change is 80–20= 60ohms.

Answer 2

Explanation:

Also volume of wire V=AL

⟹ R=ρ

V

L

2

We get R∝L

2

(as volume of the wire remains the same when it is recasted)

Now the wire is melted and recasted to half of its length i.e. L

′

=

2

L

20 ohm