The two resistors shown are both uniform cylinders of equal length made from the same conducting putty - AQA - A-Level Physics - Question 27 - 2018 - Paper 1

Given that the diameter of Y is twice that of X, the radius of Y will be two times that of X.

The area of the cross-section of a cylinder is given by:

A = rac{ heta^2}{4}

For resistor X, the area is ( A_X = \pi r_X^2 ).

For resistor Y, since the diameter is double, the radius will be: ( r_Y = 2r_X ), thus:

$A_Y = \pi (2r_X)^2 = 4\pi r_X^2 = 4A_X$

Since both resistors have the same length and are made from the same material, the resistances can be expressed as ratios of their areas;

Using the formula for resistance: $R = \frac{L}{\sigma A}$, we find:

• Resistance of Y ( R_Y = R )
• Resistance of X ( R_X = \frac{L}{\sigma (A_X)} + k = \frac{L}{\sigma (\frac{1}{4} A_Y)} = \frac{R}{4} ). Therefore, ( R_X = \frac{R}{4} ).