Question

A second's pendulum clock has a steel wire. The clock is calibrated at 20^@ C. How much time does the clock lose or gain in one week when the temperature is increased to 30^@ C ? alpha_(steel = 1.2 xx 10^-5(^@ C)^-1.

Answer 1

time lost in a week would be 36.28 sec

given, The clock is calibrated at 20°C.

so, initial temperature, $T_i$ = 20°C

final temperature, $T_f$ = 30°C

coefficient of linear expansion of pendulum (steel), α = 1.2 × 10^-5/°C

using formula, ∆T = 1/2Tα($T_f-T_i$)

we know, time period of simple pendulum, T = 2sec

so, ∆T = 1/2 × 2sec × 1.2 × 10^-5/°C × (30°C - 20°C)

= 1.2 × 10^-5 × 10

= 1.2 × 10^-4 sec

so, T' = T + ∆T = 2sec + 1.2 × 10^-4 sec

= 2.00012 sec

so, time lost in a week, ∆t = (∆T/T') × t

= (1.2 × 10^-4/2.00012) × 1 week

= (1.2 × 10^-4/2.00012) × 7 × 24 × 3600

= 36.2858 ≈ 36.28 sec