Respuesta :
Explanation:
Let us assume that the final resistance after stretching is [tex]R_{f}[/tex]. Since, the volume of rod does not change and the initial and final volume will be the same as follows.
V = [tex]V_{f}[/tex]
[tex]LA = L_{f}A_{f}[/tex]
It is given that, [tex]L_{f} = 3L[/tex].
Therefore, [tex]LA = L_{f}A_{f}[/tex]
[tex]LA = 3LA_{f}[/tex]
[tex]A_{f} = \frac{A}{3}[/tex]
We know that relation between resistance and area is as follows.
R = [tex]\frac{\rho L}{A}[/tex]
[tex]R_{f} = \frac{\rho L_{f}}{A_{f}}[/tex]
= [tex]\frac{\rho (3L)}{\frac{A}{3}}[/tex]
= 9 [tex](\frac{\rho L}{A})[/tex]
= 9R
Thus, we can conclude that the resistance of the stretched wire is 9R.
Answer:
Explanation:
initial length of the wire = l
initial area of crossetion of wire = A
initial resistance of the wire = R
Let the final resistance is r' area is A' and length is l' . Let be the resistivity of the material of the wire.
l' = 3l
So, volume of the wire is constant.
l x A = l' x A'
l x A = 3l x A'
A' = A/3
The resistance of the wire is given by
[tex]R=\rho \frac{l}{A}[/tex] ... (1)
now the new resistance is R'
[tex]R'=\rho \frac{l'}{A'}[/tex]
[tex]R'=\rho \frac{3\times 3l}{A}[/tex] .... (2)
divide equation (2) by equation (1)
R' = 9 R
Thus, the resistance of wire becomes 9 times.
