A length-L conducting wire is heated by ΔT over time t using two identical ideal cells in series. If N such cells in series heat a wire of the same material and cross-section but length 2L by the same ΔT in the same time, find N. Ignore heat loss.
Physics · Mixed Chapter
JEE Mains

Question

A length-L conducting wire is heated by ΔT over time t using two identical ideal cells in series. If N such cells in series heat a wire of the same material and cross-section but length 2L by the same ΔT in the same time, find N. Ignore heat loss.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Resistance and mass scale with wire length: For identical material and cross-section, R∝L and m∝L. Hence R₂=2R₁, m₂=2m₁. Step 2 — First configuration: Voltage V₁=2ℰ (each ideal cell has emf ℰ). Heating energy Q₁ = (V₁²/R₁)t = (4ℰ²/R₁)t. Step 3 — Second configuration: Voltage V₂=Nℰ and resistance 2R₁. Q₂=(N²ℰ²/(2R₁))t. Step 4 — Same temperature rise requires double heat in double mass: Q₂=2Q₁ ⇒ N²ℰ²/(2R₁) = 8ℰ²/R₁. Step 5 — Solve for N: N²=16 ⇒ N=4. Answer: 4.