Question

A seconds pendulum clock having steel wire is calibrated at $20{}^{\circ }C$ . When temperature is increased to $30{}^{\circ }C$ , then calculate how much time does the clock lose or gain in one week $\left[{\alpha }_{\text{steel}}=1\text{.}2\times 10^{-5}{}^{{}^{\circ }}{\text{C}}^{-1}\right]$

• A. $\text{0.3628}s$
• B. $\text{3.626}s$
• C. $\text{362.8}s$
• D. $\text{36.28}s$

Answer 1

Answer: (D)

The time period of the second's pendulum = $2s$
Change in the time period, $\Delta \text{T}=\frac{1}{2}\text{T}\alpha \Delta \theta$
$\Delta \text{T}=\left(\frac{1}{2}\right)\left(2\right)\left(1\text{.}2\times 10^{-5}\right)\left(30^{{}^{\circ }}-20^{{}^{\circ }}\right)$
$=1\text{.}2\times 10^{-4}\text{ s}$
New time period, ${\text{T}}^{{}^{\prime }}=\text{T}+\Delta \text{T}$
${\text{T}}^{{}^{\prime }}=2\text{.}00012\text{ s}$
Time lost in a week,
$\Delta \text{t}=\frac{\Delta \text{T}}{{\text{T}}^{{}^{\prime }}}\times \text{t}$
$=\frac{1\text{.}2\times 10^{-4}}{2\text{.}00012}\times \left(7\times 24\times 3600\right)$
$=36\text{.}28\text{ s}$