Multiple choice

A second pendulum clock having steel wire is calibrated at ${ 20 }^{ 0 }C$. When temperature is increased to ${ 30 }^{ 0 }C$, then how much time does the clock lose or gain in one week ? $[{ \alpha }_{ steel }=1.2\times { 10 }^{ -5 }{ (^{ 0 }C) }^{ -1 }]$ :

  1. $0.3628 s$
  2. $3.626 s$
  3. $362.8 s$
  4. $36.28 S$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A pendulum's period increases by approximately half the fractional increase in wire length. The fractional period increase is 0.5 x 1.2 x 10^-5 x 10 = 6 x 10^-5, so the clock loses 604800 x 6 x 10^-5 = 36.28 seconds in one week.

AI explanation

The daily loss of time is calculated using the formula (1/2) times the coefficient of linear expansion times the temperature change times the total seconds in a day. Substituting the given values gives a daily loss of (1/2) * (1.2 x 10^-5 /C) * (30C - 20C) * 86400 seconds, which equals 5.184 seconds. Since one week has 7 days, the total time lost is 5.184 * 7 = 36.28 seconds.

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