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Question Number 17117 by tawa tawa last updated on 01/Jul/17

$$\mathrm{A}\mathrm{clock}\mathrm{has}\mathrm{a}\mathrm{pendulum}\mathrm{made}\mathrm{of}\mathrm{iron}\mathrm{rod}\mathrm{of}\mathrm{length}2.5\mathrm{m},$$$$\mathrm{if}\mathrm{the}\mathrm{clock}\mathrm{keeps}\mathrm{accurate}\mathrm{time}\mathrm{at}0°\mathrm{C}.\mathrm{By}\mathrm{how}\mathrm{much}\mathrm{time}\mathrm{will}\mathrm{it}\mathrm{be}\mathrm{late}$$$$\mathrm{running}\mathrm{at}\mathrm{a}\mathrm{temperature}30°\mathrm{C}\mathrm{for}1\mathrm{day}.\mathrm{coefficient}\mathrm{of}\mathrm{linear}\mathrm{expansivity}\mathrm{of}$$$$\mathrm{iron}\mathrm{is}1.2\times {10}^{-5}\mathrm{per}\mathrm{k}.$$

Answered by ajfour last updated on 01/Jul/17

$$\mathrm{t}=2\pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$$$$\mathrm{t}+△\mathrm{t}=2\pi \sqrt{\frac{\mathrm{L}(1+\alpha △\mathrm{T})}{\mathrm{g}}}$$$$=2\pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\times (1+\frac{\alpha △\mathrm{T}}{2})$$$$\mathrm{t}+△\mathrm{t}=\mathrm{t}(1+\frac{\alpha △\mathrm{T}}{2})$$$$\Rightarrow △\mathrm{t}=\mathrm{t}\left(\frac{\alpha △\mathrm{T}}{2}\right)$$$$\mathrm{or}\frac{△\mathrm{t}}{\mathrm{t}}=\frac{\alpha △\mathrm{T}}{2}$$$$\mathrm{time}\mathrm{lost}=\frac{△\mathrm{t}}{\mathrm{t}}\times 86400\mathrm{s}$$$$=\frac{\alpha △\mathrm{T}}{2}\times 86400\mathrm{s}$$$$=\frac{1.2\times {10}^{-5}\times 30}{2}\times 86400\mathrm{s}$$$$=\frac{1.2\times 30\times 0.864}{2}=18\times 0.864\mathrm{s}$$$$=15.552\mathrm{s}.$$

Commented by tawa tawa last updated on 01/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}.$$

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