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MCQEasyJEE 2024Resistivity & Conductivity

JEE Physics 2024 Question with Solution

At room temperature (27∘C27^\circ\text{C}), the resistance of a heating element is 50 Ω50 \, \Omega. If the temperature coefficient of the material is 2.4×10−4 ∘C−12.4 \times 10^{-4} \, ^\circ\text{C}^{-1}, find the temperature of the element when its resistance is 62 Ω62 \, \Omega:

  • A

    927∘C927^\circ\text{C}

  • B

    1027∘C1027^\circ\text{C}

  • C

    1127∘C1127^\circ\text{C}

  • D

    1227∘C1227^\circ\text{C}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Initial resistance R0=50 ΩR_0 = 50 \, \Omega, final resistance R=62 ΩR = 62 \, \Omega, temperature coefficient α=2.4×10−4 ∘C−1\alpha = 2.4 \times 10^{-4} \, ^\circ\text{C}^{-1}, and initial temperature T0=27∘CT_0 = 27^\circ\text{C}.

Find: The final temperature TT of the heating element.

Use the relation between resistance and temperature:

R=R0(1+αΔT)R = R_0(1 + \alpha \Delta T)

Substituting the given values:

62=50(1+2.4×10−4ΔT)62 = 50(1 + 2.4 \times 10^{-4} \Delta T) 6250=1+2.4×10−4ΔT\frac{62}{50} = 1 + 2.4 \times 10^{-4} \Delta T 1.24−1=2.4×10−4ΔT1.24 - 1 = 2.4 \times 10^{-4} \Delta T 0.24=2.4×10−4ΔT0.24 = 2.4 \times 10^{-4} \Delta T ΔT=0.242.4×10−4=1000∘C\Delta T = \frac{0.24}{2.4 \times 10^{-4}} = 1000^\circ\text{C}

Now calculate the final temperature:

T=T0+ΔT=27+1000=1027∘CT = T_0 + \Delta T = 27 + 1000 = 1027^\circ\text{C}

Therefore, the temperature of the element is 1027∘C1027^\circ\text{C}. The correct option is B.

The solution explicitly concludes with 10271027, which matches option B.

Rearranged Formula Method

Given: R0=50 ΩR_0 = 50 \, \Omega, R=62 ΩR = 62 \, \Omega, α=2.4×10−4 ∘C−1\alpha = 2.4 \times 10^{-4} \, ^\circ\text{C}^{-1}, and T0=27∘CT_0 = 27^\circ\text{C}.

Find: Final temperature TT.

Starting from

R=R0(1+αΔT)R = R_0 \left(1 + \alpha \Delta T\right)

rearrange for ΔT\Delta T:

ΔT=R−R0αR0\Delta T = \frac{R - R_0}{\alpha R_0}

Substitute the values:

ΔT=62−50(2.4×10−4)⋅50\Delta T = \frac{62 - 50}{(2.4 \times 10^{-4}) \cdot 50} ΔT=120.012=1000∘C\Delta T = \frac{12}{0.012} = 1000^\circ\text{C}

Then

T=T0+ΔT=27+1000=1027∘CT = T_0 + \Delta T = 27 + 1000 = 1027^\circ\text{C}

Therefore, the final temperature is 1027∘C1027^\circ\text{C}, so the correct option is B.

Common mistakes

  • Using TT directly in the formula instead of the temperature rise ΔT\Delta T is incorrect because the relation here is written relative to the initial temperature. First find ΔT\Delta T, then add the initial temperature 27∘C27^\circ\text{C}.

  • Forgetting to add the initial temperature after computing ΔT=1000∘C\Delta T = 1000^\circ\text{C} gives the wrong final answer. The required temperature is T=27+1000=1027∘CT = 27 + 1000 = 1027^\circ\text{C}, not just 1000∘C1000^\circ\text{C}.

  • Substituting the denominator incorrectly in R−R0αR0\frac{R - R_0}{\alpha R_0} can cause major error. Compute αR0=2.4×10−4×50=0.012\alpha R_0 = 2.4 \times 10^{-4} \times 50 = 0.012 carefully before dividing.

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