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# Question and Answer

## QuestionPhysicsClass 12

At room temperature $${(27.0 {\degree}C)}$$  the resistance of a heating element is $$\text{100 \Omega .}$$  What is the temperature of the element if the resistance is found to be $$\text{17 \Omega ,}$$  given that the temperature coefficient of the material of the resistor is $${1.70 {\times} 10^{-4} {\degree}C}.$$

Let, Room temperature as $$T1$$
Resistance of heating element at room temperature as $$R1$$
Resistance of heating element at final temperature $$(T2 )$$ as $$R2$$
Here, we are given $$T1$$ as $$27^{\circ} {C}$$. Resistance $$R1$$ as $$100 \;\text{ohms}$$ ad Resistance $$R2$$ as $$17 \;\text{ohms}.$$
The $$\alpha$$ is $$1.70 \times 10^{-4}{^\circ} {C}$$
$$T 1=27^{\circ} C$$
$$R1=100\; \text{ohm}$$
$$R2=17\; \text{ohm}$$
$$\alpha=1.70 \times 10^{-4 \ \circ} C$$
We have the relation-
$$R2=R1 \times[1+\alpha \times(T 2-T 1)]$$
$$17=10\left [1+(1.7 \times 10^{-4})(T 2-27)\right]$$
On solving,
$$T 2=1027^{\circ} {C}$$