Multiple choice

A clock, with a brass pendulum, keeps correct time at $20^\circ C$, but losses $8.212 s$ per day, when the temperature rises to $30^\circ C$. The coefficient of linear expansion of brass is:-

  1. $25 \times 10^{-6}/^\circ C$
  2. $19 \times 10^{-6}/^\circ C$
  3. $20 \times 20^{-6}/^\circ C$
  4. $11 \times 10^{-6}/^\circ C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time lost is delta_t = (1/2) * alpha * delta_T * T, where T is the total time in seconds per day (86400). 8.212 = 0.5 * alpha * (30-20) * 86400. Solving for alpha: alpha = 8.212 / (0.5 * 10 * 86400) = 8.212 / 432000 = 1.9 * 10^-5 = 19 * 10^-6.

AI explanation

The fractional change in the time period of the pendulum is given by the formula for time loss per day, which is (1/2) times alpha times the change in temperature. Substituting the given values into the equation gives 8.212 = (1/2) * alpha * (30 - 20) * 86400. Solving for alpha yields alpha = 16.424 / 86400 = 19 x 10^-6 /C.

Keep practicing — related questions