A pendulum clock loses 12 s a day if the temperature is 40∘C and gains 4 s a day if the temperature is 20∘C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively :
(A)60∘C;α=1.85×10−4/∘C
(B)30∘C;α=1.85×10−3/∘C
(C)55∘C;α=1.85×10−2/∘C
(D)25∘C;α=1.85×10−5/∘C✓ Correct
Answer: (D)
Solution
A pendulum clock runs fast when cool and slow when warm, and the daily error is proportional to the temperature difference from the correct-time temperature θ0. Losing 12 s at 40∘C and gaining 4 s at 20∘C means the error is zero somewhere between, in the ratio 40−θ012=θ0−204; cross-multiplying, 3(θ0−20)=40−θ0, so θ0=25∘C. The fractional daily error is N0ΔN=21αΔθ with N0=86400 s; using the 12 s loss over Δθ=15∘C, α=86400×152×12≈1.85×10−5∘C−1.
JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option