Question

The lengths of two copper wires A and B are 180 cm and 270 cm respectively. If the mass of wire A is twice the mass of wire B and the electrical resistance of wire A is \(200\Omega\), then the electrical resistance of wire B is:

• A. \(900\Omega\)
• B. \(400\Omega\)
• C. \(600\Omega\)
• D. \(300\Omega\)

Hint:

For wires of the same material: \[ R\propto \frac{L^2}{m}. \] This shortcut saves a lot of calculation.

Answer 1

The Correct Option is A

Concept: For wires of the same material, \[ R=\rho\frac{L}{A}. \] Also, \[ m=\rho_m AL. \] Hence, \[ A\propto \frac{m}{L}. \] Therefore, \[ R\propto \frac{L^2}{m}. \]

Step 1: Write resistance ratio.
\[ \frac{R_A}{R_B} = \frac{L_A^2/m_A}{L_B^2/m_B}. \] Given, \[ L_A=180cm, \qquad L_B=270cm, \] \[ m_A=2m_B. \] Thus, \[ \frac{R_A}{R_B} = \frac{180^2}{270^2} \times \frac{m_B}{2m_B}. \] \[ = \frac{4}{9}\times\frac12. \] \[ = \frac{2}{9}. \]

Step 2: Find \(R_B\).
\[ \frac{200}{R_B} = \frac{2}{9}. \] \[ R_B = 200\times\frac92. \] \[ R_B=900\Omega. \] Hence, \[ \boxed{900\Omega}. \]