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Question #42127

Difficulty: 8/10 | 🟢 Free

The Question

A thermistor has a temperature-dependent resistance given by R(T)=R0eβ/TR(T) = R_0 e^{\beta/T}, where TT is the absolute temperature in Kelvin. It is given that R0=10 ΩR_0 = 10\ \Omega and β=300ln⁡(3) K\beta = 300 \ln(3)\ \text{K}. The thermistor is connected in series with an inductor of inductance L=0.4 HL = 0.4\ \text{H} and an AC voltage source V(t)=100sin⁡(100t) VV(t) = 100 \sin(100t)\ \text{V}. The temperature is measured using a faulty linear scale ∘X^\circ X, on which the freezing and boiling points of water are 10∘X10^\circ X and 110∘X110^\circ X, respectively. In steady state, the thermistor reaches a constant operating temperature measured as 37∘X37^\circ X, while the surrounding environment is at 7∘X7^\circ X. Assuming Newton's law of cooling holds, what is the thermal dissipation constant KK of the thermistor?

Options

  • A.

    2.0 W/K2.0\ \text{W/K}

  • B.

    4.0 W/K4.0\ \text{W/K}

  • C.

    103 W/K\frac{10}{3}\ \text{W/K}

  • D.

    1.5 W/K1.5\ \text{W/K}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Celsius-Fahrenheit Calibration Errors|Dynamic Resistance (AC Resistance)

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of thermal calibration, AC circuit impedance, and steady-state energy conservation.

It employs a 'Dynamic Resistance' trap where the resistance is coupled to the operating temperature, which in turn depends on the power dissipated.

A common JEE Advanced pitfall tested here is confusing apparent power (VrmsIrmsV_{rms}I_{rms}) with true active power (Irms2RI_{rms}^2 R) in circuits with reactive components, leading to Option C.

Another trap is using peak current instead of RMS current for power calculations, leading to Option B.

Finally, it forces students to correctly map a faulty linear scale to absolute Kelvin before evaluating the non-linear exponential resistance function.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptCelsius-Fahrenheit Calibration Errors
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree