Q103 / 120 · VITEEE 2008 · Physics

Equal amounts of a metal are converted into cylindrical wires of different lengths (L) and cross-sectional area (A). The wire with the maximum resistance is the one, which has

  • A
    length $=\mathrm{L}$ and area $=\mathrm{A}$
  • B
    length $=\frac{L}{2}$ and area $=2 \mathrm{~A}$
  • Correct answer
    C
    length $=2 \mathrm{~L}$ and area $=\frac{\mathrm{A}}{2}$
  • D
    all have the same resistance, as the amount of the metal is the same

Solution

We know that $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\therefore \text { For } l=\mathrm{L} \text { and } \mathrm{A}=\mathrm{A} ; \mathrm{R}_{1}=\rho \frac{\mathrm{L}}{\mathrm{A}}$ For $l=\frac{\mathrm{L}}{2}$ and $\mathrm{A}=2 \mathrm{~A}$; $R_{2}=\rho \frac{\left(\frac{L}{2}\right)}{2 A}=\frac{1}{4}\left(\rho \frac{L}{A}\right)=\frac{1}{4} R_{1}$ For $l=2 \mathrm{~L}$ and $\mathrm{A}=\frac{\mathrm{A}}{2}$; $\mathrm{R}_{3}=\rho \frac{2 \mathrm{~L}}{\left(\frac{\mathrm{A}}{2}\right)}=4\left(\rho \frac{\mathrm{L}}{\mathrm{A}}\right)=4 \mathrm{R}_{1}$ Thus $R_{3}>R_{1}>R_{2}$ Therefore, the wire having length $2 \mathrm{~L}$ and area $\frac{\text { A }}{2}$ has the maximum resistance.

Hence, the correct answer is option: 'C'.

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