The correct answer is D because:

Total length of wire L = Circumference = πD
Length of shorter sections = $\frac{L}{4} \times \frac{1}{2} = \frac{L}{8}$
Length of longer sections = $\frac{3 L}{4} \times \frac{1}{2} = \frac{3 L}{8}$
To find the total resistance of the circuit, an equivalent circuit diagram can be drawn as follows:

4-2-mcq-h-ma3-ocr-al-physics

Where 1 and 2 denote the shorter sections of wire and 3 and 4 denote the longer sections
Using the equation for resistivity: $R = \frac{ρ L}{A}$
- $R_{1} = R_{2} = \frac{ρ π D}{8 A} = \frac{ρ π D}{8 π {(\frac{d}{2})}^{2}} = \frac{ρ D}{2 d^{2}}$
- $R_{3} = R_{4} = \frac{3 ρ π D}{8 A} = \frac{3 ρ π D}{8 π {(\frac{d}{2})}^{2}} = \frac{3 ρ D}{2 d^{2}}$

Since $R_{1}$ and $R_{2}$ are parallel:
- $R_{1 & 2} = {(\frac{1}{R_{1}} + \frac{1}{R_{2}})}^{- 1} = {(\frac{2 d^{2}}{ρ D} + \frac{2 d^{2}}{ρ D})}^{- 1} = \frac{ρ D}{4 d^{2}}$

Since $R_{3}$ and $R_{4}$ are parallel:
- $R_{3 & 4} = {(\frac{1}{R_{3}} + \frac{1}{R_{4}})}^{- 1} = {(\frac{2 d^{2}}{3 ρ D} + \frac{2 d^{2}}{3 ρ D})}^{- 1} = \frac{3 ρ D}{4 d^{2}}$

Since the two smaller hoops are in series with each other:
- $R_{T} = \frac{ρ D}{4 d^{2}} + \frac{3 ρ D}{4 d^{2}} = \frac{ρ D}{d^{2}}$

Finally, rearrange e.m.f for I and substitute R:
- $ε = I (r + R)$
- $I = \frac{ε}{r + R} = \frac{ε}{r + \frac{ρ D}{d^{2}}}$

Questions like this can look scary and complicated, but once you decipher the situation and work out an equivalent circuit diagram, it can simplify the problem greatly. Always take a second to make sure you understand which parts contribute to the resistance (remember: even wires have resistance!) and which components are in series and parallel

Resistance & Resistivity (OCR A Level Physics)

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