Multiple choice

A seconds pendulum clock has a steel wire. The clock shows correct time at $25^0C$. How much time does the clock loose or gain, in one week, when the temperature is increased to $35^0C$? ($\alpha{steel} = 1.2 \times 10^{-5} / ^0 C$)

  1. 321.5 s

  2. 3.828 s

  3. 82.35 s

  4. 36.28 s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Time lost/gained = (1/2) * alpha * deltaT * T_total. DeltaT = 10 degrees. T_total = 7 days * 24 * 3600 seconds = 604800 seconds. Change = 0.5 * 1.2 * 10^-5 * 10 * 604800 = 0.5 * 1.2 * 60.48 = 36.288 seconds.

AI explanation

Using the formula for time lost in a day, the loss is calculated as half the product of the coefficient of linear expansion, the temperature change, and the total seconds in a day. Substituting the given values gives a daily loss of (1/2) * (1.2 x 10^-5 /C) * (35C - 25C) * 86400 seconds, which equals 5.184 seconds. Multiplying this daily loss by 7 days for a week gives a total time loss of 36.28 seconds.

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