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MCQEasyJEE 2024Ohm's Law & Resistance

JEE Physics 2024 Question with Solution

An electric toaster has resistance of 60 Ω60 \, \Omega at room temperature (27∘C27^\circ \text{C}). The toaster is connected to a 220 V220 \, \text{V} supply. If the current flowing through it reaches 2.75 A2.75 \, \text{A}, the temperature attained by toaster is around: (if α=2×10−4 ∘C−1\alpha = 2\times 10^{-4} \, ^\circ \text{C}^{-1})

  • A

    694∘C694^\circ \text{C}

  • B

    1235∘C1235^\circ \text{C}

  • C

    1694∘C1694^\circ \text{C}

  • D

    1667∘C1667^\circ \text{C}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Initial resistance at room temperature is R0=60 ΩR_0 = 60 \, \Omega at T0=27∘CT_0 = 27^\circ \text{C}, supply voltage is V=220 VV = 220 \, \text{V}, current is I=2.75 AI = 2.75 \, \text{A}, and temperature coefficient is α=2×10−4 ∘C−1\alpha = 2 \times 10^{-4} \, ^\circ \text{C}^{-1}.

Find: The temperature attained by the toaster.

First find the resistance at the operating temperature using Ohm's law:

R=VI=2202.75=80 ΩR = \frac{V}{I} = \frac{220}{2.75} = 80 \, \Omega

Now use the resistance-temperature relation:

R=R0(1+α(T−T0))R = R_0\left(1 + \alpha (T - T_0)\right)

Substituting the known values:

80=60(1+2×10−4(T−27))80 = 60\left(1 + 2 \times 10^{-4}(T - 27)\right)

Divide by 6060 and solve step by step:

8060=1+2×10−4(T−27)\frac{80}{60} = 1 + 2 \times 10^{-4}(T - 27) 43−1=2×10−4(T−27)\frac{4}{3} - 1 = 2 \times 10^{-4}(T - 27) 13=2×10−4(T−27)\frac{1}{3} = 2 \times 10^{-4}(T - 27) T−27=13×2×10−4≈1667T - 27 = \frac{1}{3 \times 2 \times 10^{-4}} \approx 1667 T=1667+27=1694∘CT = 1667 + 27 = 1694^\circ \text{C}

Therefore, the temperature attained by the toaster is approximately 1694∘C1694^\circ \text{C}. The correct option is C.

Detailed Algebra

Given: R0=60 ΩR_0 = 60 \, \Omega, T0=27∘CT_0 = 27^\circ \text{C}, V=220 VV = 220 \, \text{V}, I=2.75 AI = 2.75 \, \text{A}, α=2×10−4 ∘C−1\alpha = 2 \times 10^{-4} \, ^\circ \text{C}^{-1}.

Find: Final temperature TT.

Resistance at elevated temperature:

R=2202.75=80 ΩR = \frac{220}{2.75} = 80 \, \Omega

Using

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

where

ΔT=T−27\Delta T = T - 27

we get

80=60(1+2×10−4(T−27))80 = 60\left(1 + 2 \times 10^{-4}(T - 27)\right)

Rearranging:

1.3333=1+2×10−4(T−27)1.3333 = 1 + 2 \times 10^{-4}(T - 27) 0.3333=2×10−4(T−27)0.3333 = 2 \times 10^{-4}(T - 27) T−27=0.33332×10−4=1666.5T - 27 = \frac{0.3333}{2 \times 10^{-4}} = 1666.5 T=1666.5+27=1693.5T = 1666.5 + 27 = 1693.5

Rounding to the nearest option, the temperature is 1694∘C1694^\circ \text{C}. Hence, the correct option is C. The solution stating option A conflicts with the actual working and final value, so the worked result is used.

Common mistakes

  • Using the initial resistance 60 Ω60 \, \Omega directly in Ohm's law is incorrect because the current given is at the operating temperature. First compute the hot resistance using R=VIR = \frac{V}{I}.

  • Forgetting to use T−27T - 27 as the temperature rise is wrong because the reference temperature is room temperature 27∘C27^\circ \text{C}, not 0∘C0^\circ \text{C}. Use R=R0(1+α(T−T0))R = R_0(1 + \alpha (T - T_0)).

  • Stopping at T−27≈1667T - 27 \approx 1667 and marking 1667∘C1667^\circ \text{C} as the final answer is incorrect. That value is only the increase in temperature; add 27∘C27^\circ \text{C} to get the final temperature.

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