The resistance of a bulb filament is 100 Ω at a temperature of 100^ . If its temperature coefficient of resistance be 0.005 per ^ , its resistance wil - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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The resistance of a bulb filament is 100 Ω at a temperature of 10 0^{\circ} C . If its temperature coefficient of resistance be 0.005 per^{\circ} C, its resistance will become 200 Ω at a temperature of:

Solution & Step-by-step Explanation

Resistance at temperature T is R_{T} = R_{0} (1 + α T) .First, find R_{0} using the 10 0^{\circ} C data: 100 = R_{0} (1 + 0.005 \times 100) = R_{0} (1 + 0.5) = 1.5 R_{0} R_{0} = 100/1.5 = 200/3 Ω .Now find T for R_{T} = 200 Ω : 200 = \frac{200}{3} (1 + 0.005 T) 3 = 1 + 0.005 T 2 = 0.005 T ⟹ T = \frac{2}{0.005} = 40 0^{\circ} C .Note: FIITJEE solution uses R_{2} = R_{1} [1 + α Δ T], which is an approximation: 200 = 100 [1 + 0.005 (T - 100)] 2 = 1 + 0.005 (T - 100) ⟹ 1 = 0.005 (T - 100) T - 100 = 200 ⟹ T = 30 0^{\circ} C .Standard exam convention usually follows the R_{2} = R_{1} [1 + α Δ T] formula.

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The resistance of a bulb filament is 100 Ω at a temperature of 10 0^{\circ} C . If its temperature coefficient of resistance be 0.005 per^{\circ} C, its resistance will become 200 Ω at a temperature of:

20 0^{\circ} C

30 0^{\circ} C

40 0^{\circ} C

50 0^{\circ} C

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