A piece of conducting putty is shaped into a cylinder of uniform cross-sectional area, as shown - Edexcel - A-Level Physics - Question 4 - 2023 - Paper 1

When the putty is rolled out to a length of 2l, its volume remains constant. The initial resistance, R, is given by:

$R = \rho \frac{l}{A}$

where (\rho) is the resistivity and (A) is the cross-sectional area.

The volume of the putty can be expressed as:

$V = A \cdot l$

After rolling out to a new length of 2l, the new cross-sectional area, (A'), can be determined by setting the volumes equal:

$A \cdot l = A' \cdot 2l$

This simplifies to:

$A' = \frac{A}{2}$

Now, we can calculate the new resistance, R', as:

$R' = \rho \frac{2l}{A'} = \rho \frac{2l}{\frac{A}{2}} = 4 \times \rho \frac{l}{A} = 4R$

Since the original resistance R is 8.0Ω, we substitute to find R':

$R' = 4 \times 8.0Ω = 32.0Ω$

Thus, the new resistance is 32.0Ω.