A pendulum clock, comprising a light steel rod attached to a heavy ball, indicates the correct time at 20^ ^ ^ C in a location where g = 9.800 m s^ -2 . If the clock is moved to a new location where g = 9.788 m s^ -2

Question

Quantrex Academy · Accepted Answer

Correct answer: D. -82^ ^ ^ C

The time period of a pendulum is given by

T = 2\pi \sqrt{\dfrac{L}{g}}

.

For the clock to indicate the correct time at the new location, its time period must remain constant, which implies:

\dfrac{L_1}{g_1} = \dfrac{L_2}{g_2}

\dfrac{L_2}{L_1} = \dfrac{g_2}{g_1}

Using the formula for linear expansion,

L_2 = L_1(1 + \alpha \Delta \theta)

, where

\Delta \theta = \theta_2 - \theta_1

:

1 + \alpha \Delta \theta = \dfrac{g_2}{g_1}

\alpha \Delta \theta = \dfrac{g_2 - g_1}{g_1}

Substituting the given values

g_1 = 9.800 \text{ m s}^{-2}

,

g_2 = 9.788 \text{ m s}^{-2}

, and

\alpha = 12 \times 10^{-6} ^{}^{{}^{\circ}}\textC^{-1}

:

12 \times 10^{-6} \times \Delta \theta = \dfrac{9.788 - 9.800}{9.800}

12 \times 10^{-6} \times \Delta \theta = \dfrac{-0.012}{9.800}

\Delta \theta = \dfrac{-0.012}{9.800 \times 12 \times 10^{-6}}

\Delta \theta = \dfrac{-1000}{9.8} \approx -102^{}^{{}^{\circ}} \text{C}

Since

\Delta \theta = \theta_2 - \theta_1

and

\theta_1 = 20^{}^{{}^{\circ}} \text{C}

:

\theta_2 - 20 = -102

\theta_2 = -82^{}^{{}^{\circ}} \text{C}

A pendulum clock, comprising a light steel rod attached to a heavy...

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