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 $\left[\alpha _{\text{steel}} = 1 \text{.} 2 \times 1 0^{- 5} \,{}^{^\circ } \text{C}^{-1}\right]$

• 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 = $2 \, s$
Change in the time period, $\Delta \text{T} = \frac{1}{2} \text{T} \alpha \Delta \theta $
$\Delta \text{T} = \left(\frac{1}{2}\right) \left(2\right) \left(1 \text{.} 2 \times 1 0^{- 5}\right) \left(3 0^{^\circ } - 2 0^{^\circ }\right)$
$= 1 \text{.} 2 \times 1 0^{- 4} \text{ s}$
New time period, $\text{T}^{'} = \text{T} + \Delta \text{T}$
$\text{T}^{'} = 2 \text{.} 0 0 0 1 2 \text{ s}$
Time lost in a week,
$\Delta \text{t} = \frac{\Delta \text{T}}{\text{T}^{'}} \times \text{t}$
$= \frac{1 \text{.} 2 \times 1 0^{- 4}}{2 \text{.} 0 0 0 1 2} \times \left(7 \times 2 4 \times 3 6 0 0\right)$
$= 3 6 \text{.} 2 8 \text{ s}$