If the length of a clock pendulum increases by $0.2 \%$ due to atmospheric temperature rise, then the loss in time of clock per day is ........... $s$
Medium
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(a)
Time period $=2 \pi \sqrt{\frac{1}{g}}$
$T \propto \sqrt{I}$
$\frac{T^{\prime}}{T} \propto \sqrt{\frac{I^{\prime}}{I}}$
$T=T \sqrt{\frac{1+l \propto \Delta \theta}{I}}$
$T=T\left(1+\frac{1}{2} \propto \Delta \theta\right)[\alpha \Delta \theta=0.002]$
$\Delta T=T-T=\frac{1}{2} T \propto \Delta \theta=T \times 0.001$
Time lost in time $t$ is
$\Delta T=\frac{1}{2} \quad t=1 \text { day }=24 \times 3600 \,s =86400 \,s$
$\Delta T=\left(\frac{\Delta T}{T}\right) \times t$
$\Delta T=0.001 \times 86400$
$\Delta T=86.4 \,s$
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