A clock is correct at 10°C and loses 17.28 s per day at another temperature. If α=20×10⁻⁶ °C⁻¹, and the latter temperature is N×10°C, find N.
Physics · Mixed Chapter
JEE Main Practice

Question

A clock is correct at 10°C and loses 17.28 s per day at another temperature. If α=20×10⁻⁶ °C⁻¹, and the latter temperature is N×10°C, find N.

G1 Physics · Accepted Answer

Step 1: For a pendulum, ΔT_period/T_period≈½αΔθ. The fractional rate of clock loss is 17.28/86400=2×10⁻⁴. Step 2: Hence (1/2)(20×10⁻⁶)Δθ=2×10⁻⁴ ⇒Δθ=20°C. Step 3: New temperature =10+20=30°C, giving N=3 for N×10°C. Step 4: The source prints N=2, which corresponds to temperature rise (20°C), not absolute temperature (30°C); original wording is internally inconsistent.