JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 1)
A heating element has a resistance of 100 $$\Omega $$ at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. what is the temperature coefficient of resistance of the heating element ?
0.5 $$ \times $$ 10$$-$$4 oC$$-$$1
5 $$ \times $$ 10$$-$$4 oC$$-$$1
1 $$ \times $$ 10$$-$$4 oC$$-$$1
2 $$ \times $$ 10$$-$$4 oC$$-$$1
Explanation
When temperature increased by 500oC then, nrw registance
Rt = $${{220} \over 2}$$ = 110 $$\Omega $$
We know,
Rt = R0 (1 + $$ \propto $$ $$\Delta $$ t)
$$ \Rightarrow $$ 110 = 100 (1 + $$ \propto $$ $$ \times $$ 500)
$$ \Rightarrow $$ $$ \propto $$ = $${{10} \over {100 \times 500}} = 2 \times {10^{ - 4}}$$ oC$$-$$1
Rt = $${{220} \over 2}$$ = 110 $$\Omega $$
We know,
Rt = R0 (1 + $$ \propto $$ $$\Delta $$ t)
$$ \Rightarrow $$ 110 = 100 (1 + $$ \propto $$ $$ \times $$ 500)
$$ \Rightarrow $$ $$ \propto $$ = $${{10} \over {100 \times 500}} = 2 \times {10^{ - 4}}$$ oC$$-$$1
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