The temperature coefficient of resistnace for a wire is $0.00125 /^{0} C$ . At 300K its resistance is 1 ohm. The temperature at which the resistance becomes 2 ohm is

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1100K

1400 K

1127 K

1154 K

answer is D.

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Detailed Solution

Let the resistance at 0^oC be $R_{0}$

$\frac{R_{1}}{R_{2}} = \frac{R_{0} (1 + {αt}_{1})}{R_{0} (1 + {αt}_{2})} \Rightarrow \frac{1}{2} = \frac{(1 + 0.00125 \times 27)}{(1 + 0.00125 \times t)}$

$\Rightarrow t = 854^{\circ} C \Rightarrow T = 1127 K$

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