The resistance of a bulb filament is 100Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200Ω at temperature of
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a
300°C
b
400°C
c
500°C
d
200°C
answer is B.
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Detailed Solution
α=R2−R1R1t2−R2t1 for t1>20∘C0.005=200−100100t2−200×100;100t2−200=1005×10−3t2−200=200⇒t2=400∘C