A clock with an iron pendulum keeps the correct time at 14^ ^ C . If the room temperature is 20^ ^ C , the error in a second per day will be nearly (coefficient of linear expansion of iron is 1.2 × 10^ -5 / ^ ^ C )

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 3.11 s

The time period of a simple pendulum is given by

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

Fractional change in time period due to temperature change is

\dfrac{\Delta T}{T} = \dfrac{1}{2} \alpha \Delta \theta

Error in time per day is

\Delta t = \dfrac{1}{2} \alpha \Delta \theta \times t

where

t = 24 \times 60 \times 60 = 86400 \text{ s}

Given

\alpha = 1.2 \times 10^{-5} / ^{}^{\circ}\text{C}

and

\Delta \theta = 20^{}^{\circ}\text{C} - 14^{}^{\circ}\text{C} = 6^{}^{\circ}\text{C}

Substituting the values:

\Delta t = \dfrac{1}{2} \times 1.2 \times 10^{-5} \times 6 \times 86400

\Delta t = 3.6 \times 10^{-5} \times 86400 = 3.1104 \text{ s}

The error per day is approximately

3.11 \text{ s}

.

Answer:

3.11 \text{ s}

A clock with an iron pendulum keeps the correct time at 14^ ^ C ....

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