A wire is pulled at a rate of 3 cm/hr. What is the change in its resistance?

It is given that the wire is being pulled at the rate of 3
cm/hr. So, we can take the mass of the substance making up the wire and its volume as
constant.

Due to the pull, the length increases and there
is a corresponding decrease in the area of
cross-section.

Let’s denote the initial volume, length and
area of cross-section by V, L and A resp.

We have V =
L*A

dV/dt = d/dt
(L*A)

=> 0 = (dL/dt)*A + L
*(dA/dt)

=> 0 = 0.03A + L*(dA /
dt)

=> dA/dt =
-0.03A/L

Now resistance = R = rho*L/A, where rho is the
resistivity, L is the length of the wire and A is the area of
cross-section.

The rate of change in resistance is given
by

dR/dt = d/dt
(rho*L/A)

=> dR / dt = rho*[(A*dL/dt –
L*dA/dt)/A^2]

=> rho*[(0.03A –
L*dA/dt)/A^2]

substitute dA/dt =
-0.03A/L

=> rho*[(0.03A –
L*(-0.03A/L))/A^2]

=> rho*[(0.03A +
0.03A)/A^2]

=> rho*0.06/A
ohm/hr

Therefore the rate of change of
resistance is rho*0.06/A ohm/hr