Question
A clock with an iron pendulum keeps the correct time at $20^{\circ} C$. How much will it lose or gain if the temperature changes to $40^{\circ} C$ ? Coefficient of cubical expansion of iron $=36 \times 10^{-6}{ }^{\circ} C ^{-1}$.
✓
Answer
Time period of simple pendulum,
$
T_{20}=2 s
$
Let $T _{40}$ be the time period at $40^{\circ} C$. If $l _0, l _{20}, l _{40}$ be the lengths of the pendulum at $0^{\circ} C , 20^{\circ} C$ and $40^{\circ} C$ respectively, then
$
\begin{aligned}
& l_{20}=l_0(1+20 \alpha) \\
& l_{40}=l_0(1+40 \alpha)
\end{aligned}
$
$
\begin{aligned}
& T_{20}=2 \pi \sqrt{\frac{l_{20}}{g}}=2 \pi \sqrt{\frac{l_0(1+20 \alpha)}{g}} \\
& T_{40}=2 \pi \sqrt{\frac{l_{40}}{g}}=2 \pi \sqrt{\frac{l_0(1+40 \alpha)}{g}} \\
& \therefore \frac{T_{40}}{T_{20}}=\sqrt{\frac{1+40 \alpha}{1+20 \alpha}}=(1+40 \alpha)^{1 / 2}(1+20 \alpha)^{-1 / 2}
\end{aligned}
$
$
\begin{aligned}
& =\left(1+\frac{1}{2} \times 40 \alpha\right)\left(1-\frac{1}{2} \times 20 \alpha\right) \text { [Using Binomial theorem] } \\
& =(1+20 \alpha)(1-10 \alpha)=1+10 \alpha
\end{aligned}
$
Fractional loss in time
$
\begin{aligned}
& =\frac{T_{40}-T_{20}}{T_{20}}=10 \alpha \\
& =10 \times 1.2 \times 10^{-5}=1.2 \times 10^{-4}
\end{aligned}
$
As the temperature increases, time period also increases. The clock runs slow.
Time lost in 24 hours
$
=1.2 \times 10^{-4} \times 24 \times 3600=10.368 s
$
$
T_{20}=2 s
$
Let $T _{40}$ be the time period at $40^{\circ} C$. If $l _0, l _{20}, l _{40}$ be the lengths of the pendulum at $0^{\circ} C , 20^{\circ} C$ and $40^{\circ} C$ respectively, then
$
\begin{aligned}
& l_{20}=l_0(1+20 \alpha) \\
& l_{40}=l_0(1+40 \alpha)
\end{aligned}
$
$
\begin{aligned}
& T_{20}=2 \pi \sqrt{\frac{l_{20}}{g}}=2 \pi \sqrt{\frac{l_0(1+20 \alpha)}{g}} \\
& T_{40}=2 \pi \sqrt{\frac{l_{40}}{g}}=2 \pi \sqrt{\frac{l_0(1+40 \alpha)}{g}} \\
& \therefore \frac{T_{40}}{T_{20}}=\sqrt{\frac{1+40 \alpha}{1+20 \alpha}}=(1+40 \alpha)^{1 / 2}(1+20 \alpha)^{-1 / 2}
\end{aligned}
$
$
\begin{aligned}
& =\left(1+\frac{1}{2} \times 40 \alpha\right)\left(1-\frac{1}{2} \times 20 \alpha\right) \text { [Using Binomial theorem] } \\
& =(1+20 \alpha)(1-10 \alpha)=1+10 \alpha
\end{aligned}
$
Fractional loss in time
$
\begin{aligned}
& =\frac{T_{40}-T_{20}}{T_{20}}=10 \alpha \\
& =10 \times 1.2 \times 10^{-5}=1.2 \times 10^{-4}
\end{aligned}
$
As the temperature increases, time period also increases. The clock runs slow.
Time lost in 24 hours
$
=1.2 \times 10^{-4} \times 24 \times 3600=10.368 s
$
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