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A pendulum clock keeps correct time at $0°C$. Its mean coefficient of linear expansions is $\alpha /^\circ C$, then the loss in seconds per day by the clock if the temperature rises by $t°C$ is
$\frac{{\frac{1}{2}\alpha \,t \times 864000}}{{1 - \frac{{\alpha \,t}}{2}}}$
$\frac{1}{2}\alpha \,t \times \,86400$
$\frac{{\frac{1}{2}\alpha \,t \times 86400}}{{{{\left( {1 - \,\frac{{\alpha \,t}}{2}} \right)}^2}}}$
$\frac{{\frac{1}{2}\alpha \,t \times 86400}}{{1 + \frac{{\alpha \,t}}{2}}}$
Solution
(b) Loss in time per second $\frac{{\Delta T}}{T} = \frac{1}{2}\alpha \Delta \theta = \frac{1}{2}\alpha \,(t – 0)$
$\Rightarrow$ loss in time per day
$\Delta t = \left( {\frac{1}{2}\alpha t} \right)\,t = \frac{1}{2}\alpha t \times (24 \times 60 \times 60) = \frac{1}{2}\alpha t \times 86400$
