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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A nichrome heating element has a resistance of 100\ at room...

A nichrome heating element has a resistance of 100 Ω100\ \Omega at room temperature (20∘C20^\circ\text{C}). When it is connected to a 220 V220\ \text{V} supply, a steady current of 2.0 A2.0\ \text{A} flows through it. If the temperature coefficient of resistance of nichrome averaged over the temperature range involved is 2.0×10−4 ∘C−12.0 \times 10^{-4}\ ^\circ\text{C}^{-1}, what is the steady temperature of the heating element?

Options

  1. A500∘C500^\circ\text{C}
  2. B480∘C480^\circ\text{C}
  3. C520∘C520^\circ\text{C}
  4. D70∘C70^\circ\text{C}

Correct answer

C. 520∘C520^\circ\text{C}

Step-by-step solution

Resistance at room temperature T1=20∘CT_1 = 20^\circ\text{C} is R1=100 ΩR_1 = 100\ \Omega.

Resistance at steady temperature T2T_2 is R2=VI=2202.0=110 ΩR_2 = \dfrac{V}{I} = \dfrac{220}{2.0} = 110\ \Omega.

Using the relation R2=R1[1+α(T2−T1)]R_2 = R_1 [1 + \alpha (T_2 - T_1)], we get:

110=100[1+2.0×10−4(T2−20)]110 = 100 [1 + 2.0 \times 10^{-4} (T_2 - 20)]

1.1=1+2.0×10−4(T2−20)1.1 = 1 + 2.0 \times 10^{-4} (T_2 - 20)

0.1=2.0×10−4(T2−20)0.1 = 2.0 \times 10^{-4} (T_2 - 20)

T2−20=0.12.0×10−4=500T_2 - 20 = \dfrac{0.1}{2.0 \times 10^{-4}} = 500

T2=520∘CT_2 = 520^\circ\text{C}

Answer: 520∘C520^\circ\text{C}
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