Question

A second's [pendulum clock having steel wire is calibrated at 20$^{\circ}$C . When temperature is, increased to 30$^{\circ}$C, then how much time does the clock lose or gain in one week ? $[\alpha_{steel}=1.2 \times 10^{-5}(^{\circ}\,C)^{-1}]$:

• A. 0.3628 s
• B. 3.628 s
• C. 362.8 s
• D. 36.28 s

Answer 1

Answer: (D)

36.28 s

Answer 2

Let $T _{0}$ be the initial time period $=2 \pi \sqrt{\frac{ l _{0}}{ g }}$ Hence, $T _{1}=2 \pi \sqrt{\frac{l_{0}(1+\alpha t )}{ g }}$ $\Longrightarrow T _{1}- T _{0}= T _{0}\left(\frac{\alpha t }{2}\right)$ This much time is lost in one second, since period $T _{0}$ is $1 s$. Hence total time lost $=\frac{\alpha t }{2} \times 7 \times 24 \times 60 \times 60=36.28 s$