KCET · Physics · Electrostatics
A piece of copper is to be shaped into a conducting wire of maximum resistance. The suitable length and diameter are and respectively.
- A \( \mathrm{L} \) and \( \mathrm{d} \)
- B \( 2 \mathrm{~L} \) and \( d \)
- C \( \mathrm{L} / 2 \) and \( 2 \mathrm{~d} \)
- D \( 2 \mathrm{~L} \) and \( \mathrm{d} / 2 \)
Answer & Solution
Correct Answer
(D) \( 2 \mathrm{~L} \) and \( \mathrm{d} / 2 \)
Step-by-step Solution
Detailed explanation
Resistance, \( R=\frac{\rho L}{A} \)
where \( \rho \) is resistivity; \( L \) is length; \( A \) is area
Now, \( A=\Pi r^{2}=\Pi\left(\frac{d}{2}\right)^{2} \)
Therefore, \( R=\frac{\rho L}{\Pi(d / 2)^{2}} \)
\( \Rightarrow R \propto L \) and \( R \propto \frac{1}{d^{2}} \)
Thus, the length and diameter of copper wire should be \( 2 \mathrm{~L} \) and \( \mathrm{d} / 2 \), respectively.
where \( \rho \) is resistivity; \( L \) is length; \( A \) is area
Now, \( A=\Pi r^{2}=\Pi\left(\frac{d}{2}\right)^{2} \)
Therefore, \( R=\frac{\rho L}{\Pi(d / 2)^{2}} \)
\( \Rightarrow R \propto L \) and \( R \propto \frac{1}{d^{2}} \)
Thus, the length and diameter of copper wire should be \( 2 \mathrm{~L} \) and \( \mathrm{d} / 2 \), respectively.
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