Question

A clock with an iron pendulum keeps correct time at ${15}^{\circ }C$. If the room temperature rises to ${20}^{\circ }C$, the error in seconds per day will be (coefficient of linear expansion for iron is $0.000012{/}^{\circ }C$):

• A. $2.5sec$
• B. $2.6sec$
• C. $2.4sec$
• D. $2.2sec$

Answer 1

The correct option is B $2.6sec$

The initial time period of a pendulum is $T_i=2\pi \sqrt{\frac{l_i}{g}}$ and final time period is $T_f=2\pi \sqrt{\frac{l_f}{g}}$

If $\alpha$ be the coefficient of linear expansion, then $l_f=l_i(1+\alpha \Delta \theta )$ where $\Delta \theta =$ change in temperature.

So, $T_f=2\pi \sqrt{\frac{l_i(1+\alpha \Delta \theta )}{g}}=T_i(1+\alpha \Delta \theta {)}^{1/2}=T_i(1+\frac{\alpha \Delta \theta }{2})$ [for small value of $\alpha \Delta$, we can neglect the higher order term]

Or, $\frac{T_f-T_i}{T_i}=\frac{1}{2}\alpha \Delta \theta$

Thus, the error in seconds per day $=\frac{T_f-T_i}{T_i}\times 86400=\frac{1}{2}\alpha \Delta \theta \times 86400=0.5\left(0.000012\right)(20-15)\left(86400\right)=2.6sec$