Question

The temperature coefficient of resistivity of a material is 0.0004/K. When the temperature of the material is increased by $ 50{}^\circ C $ , its resistivity increases by $ 2\times {{10}^{-8}} $ ohm- metre. The initial resistivity of the material in ohm- metre is:

• A. $ 50\times {{10}^{-8}} $
• B. $ 90\times {{10}^{-8}} $
• C. $ 100\,\times \,{{10}^{-8}} $
• D. $ 200\times {{10}^{-8}} $

Answer 1

Answer: (C)

Given: Temperature difference $ {{t}_{2}}-{{t}_{1}}=50{{\,}^{o}}C $ And resistivity difference $ {{\rho }_{t}}-{{\rho }_{0}}=2\times {{10}^{-8}}\Omega -m $ and $ \alpha =0.0004/K $ $ {{\rho }_{t}}={{\rho }_{0}}[1+\alpha ({{t}_{2}}-{{t}_{1}})] $ $ {{\rho }_{t}}-{{\rho }_{0}}={{\rho }_{0}}\alpha ({{t}_{2}}-{{t}_{1}}) $ or $ {{\rho }_{0}}=\frac{{{\rho }_{t}}-{{\rho }_{0}}}{\alpha ({{t}_{2}}-{{t}_{1}})} $ ?(i) Now, putting the given values in Eq. (i) we get $ {{\rho }_{0}}=\frac{2\times {{10}^{-8}}}{4\times {{10}^{-4}}\times 50} $ $ =100\times {{10}^{-8}}\Omega m $