The temp at which the resistance of a conductor becomes 30% more than that of its resistance at 47°C will be:
(given the value of temperature coefficient of resistance of the conductor is \( 2\times10^{-4}\,\mathrm{K^{-1}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1820 K
The resistance of a conductor rises with temperature as \( R_2 = R_1\left(1 + \alpha\,\Delta T\right) \), where \( \alpha \) is the temperature coefficient of resistance. Taking the resistance at 47°C as \( R_1 \), the new resistance is 30% more:
\( \displaystyle R_2 = 1.3\,R_1 \;\Rightarrow\; 1 + \alpha\,\Delta T = 1.3 \)
\( \displaystyle \Delta T = \frac{0.3}{2\times10^{-4}\,\mathrm{K^{-1}}} = 1500\,\mathrm{K} \)
A temperature difference is the same in kelvin and in degrees Celsius, so
\( \displaystyle T = 47^\circ\mathrm{C} + 1500^\circ\mathrm{C} = 1547^\circ\mathrm{C} \)
The options are in kelvin, so convert:
\( \displaystyle T = 1547 + 273 = 1820\,\mathrm{K} \)