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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

At room temperature (27.0°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the

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प्रश्न

At room temperature (27.0°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10−4 °C−1.

संख्यात्मक
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उत्तर

Given: Initial temperature, T1 = 27°C

Initial resistance, R1 = 100 Ω

Let T2 is the increased temperature of the filament.

Final resistance, R2 = 117 Ω

The temperature coefficient of the material of the filament,

α = 1.70 × 10−4 °C−1

α is given by the relation,

α = `(R_2 - R_1)/(R_1(T_2 - T_1))`

`T_2 - T_1 = (R_2 - R_1)/(R_1α)`

`T_2 - 27 = (117 - 100)/(100 xx 1.7 xx 10^-4)`

`T_2 - 27 = 17/(170 xx 10^-4)`

`T_2 - 27 = 17/(0.017)`

T2 − 27 = 1000

T2 = 1000 + 27

∴ T2 = 1027°C

Therefore, at 1027°C, the resistance of the element is 117 Ω.

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पाठ 3: Current Electricity - EXERCISES [पृष्ठ १०५]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 3 Current Electricity
EXERCISES | Q 3.3 | पृष्ठ १०५
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 3 Current Electricity
Exercise | Q 3.5 | पृष्ठ १२७

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