Refer to the circuit diagram given below. The heat generated across the 6Ω resistance in 100 second is α/100 J . The value of α is (Nearest integer)

Physics · Current Electricity

JEE Main 2026 — 6 April, Morning Shift — Question 16

Refer to the circuit diagram given below. The heat generated across the $6\Omega$ resistance in 100 second is $\frac{\alpha}{100}\mathrm{J}$ . The value of $\alpha$ is (Nearest integer)

Answer: 3477

Numerical answer — enter this value.

Step-by-step solution

From the circuit (not fully shown), solution calculates heat = 34.77 J = 3477/100 J, so $\alpha = 3477$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Current Electricity
Topic: Heating Effects of Current and Thermal Powe