Class 12 NEET Physics Current Electricity Questions Part 1

Here, Length of each rod, $l = 1 \text{ m}$ Area of cross-section of each rod, $A = 0.01 \text{ cm}^{2} = 0.01 \times 10^{-4} \text{ m}^{2}$ Resistivity of copper rod, $\rho_{\text{cu}} = 1.7 \times 10^{-6} \Omega \text{cm}$ $= 1.7 \times 10^{-6} \times 10^{-2} \Omega \text{m}$ $= 1.7 \times 10^{-8} \Omega \text{m}$ Resistivity of iron rod, $\rho_{\text{Fe}} = 10^{-5} \Omega \text{cm}$ $= 10^{-5} \times 10^{-2} \Omega \text{m} = 10^{-7} \Omega \text{m}$ $\therefore$ Resistance of copper rod, $R_{\text{Cu}} = \rho_{\text{Cu}} \frac{l}{A}$ and resistance of iron rod, $R_{\text{Fe}} = \rho_{\text{Fe}} \frac{l}{A}$As copper and iron rods are connected in series, therefore equivalent resistance is $R = R_{\text{Cu}} + R_{\text{Fe}}$ $= \rho_{\text{Cu}} \frac{l}{A} + \rho_{\text{Fe}} \frac{l}{A}$ $= (\rho_{\text{Cu}} + \rho_{\text{Fe}}) \frac{l}{A}$ Voltage required to produce 1 A current in the rods is $V = I R = (1) (R_{\text{Cu}} + R_{\text{Fe}})$ $= (\rho_{\text{Cu}} + \rho_{\text{Fe}}) \left(\frac{1}{0.01 \times 10^{-4}}\right) \text{V}$ $= 10^{-7} (0.17 + 1) (10^{6}) \text{V}$ $= 1.17 \times 10^{-1} \text{ V}$ $= 0.117 \text{ V}$ $0.00145 \text{ V}$ $0.0145 \text{ V}$