Multiple choice

A thermostated chamber at a height h above earth's surface and maintained at $30^{\mathrm{o}}\mathrm{C}$ has a clock fitted with an uncompensated pendulum. The maker of the clock has wrongly designed it to maintain correct time at $20^{\mathrm{o}}\mathrm{C}$. It is found that if the chamber is brought to earth's surface the clock clicks correct time. $\mathrm{R}$ is the radius of earth. What is the linear compressibility of the material of the pendulum?

  1. $\displaystyle \frac{\mathrm{h}}{10\mathrm{R}}$
  2. $\displaystyle \frac{5\mathrm{R}}{h}$
  3. $\displaystyle \frac{\mathrm{h}}{5\mathrm{R}}$
  4. $\displaystyle \frac{\mathrm{h}}{20\mathrm{R}}$
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C Correct answer
Explanation

The time period of a pendulum is T = 2*pi*sqrt(L/g). The change in length due to temperature is dL = L*alpha*dT, and the change in g due to height is dg = -2*g*h/R. For the clock to keep correct time, the fractional change in length must balance the fractional change in gravity, leading to alpha*dT = h/R, where dT = 10 degrees. Thus, alpha = h/(10R).

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