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Question Number 17117 by tawa tawa last updated on 01/Jul/17
A clock has a pendulum made of iron rod of length 2.5m, if the clock keeps accurate time at 0°C. By how much time will it be late running at a temperature 30°C for 1 day. coefficient of linear expansivity of iron is 1.2 × 10^(−5) per k.
$$\mathrm{A}\:\mathrm{clock}\:\mathrm{has}\:\mathrm{a}\:\mathrm{pendulum}\:\mathrm{made}\:\mathrm{of}\:\mathrm{iron}\:\mathrm{rod}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2}.\mathrm{5m}, \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{clock}\:\mathrm{keeps}\:\mathrm{accurate}\:\mathrm{time}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}.\:\mathrm{By}\:\mathrm{how}\:\mathrm{much}\:\mathrm{time}\:\mathrm{will}\:\mathrm{it}\:\mathrm{be}\:\mathrm{late} \\ $$$$\mathrm{running}\:\mathrm{at}\:\mathrm{a}\:\mathrm{temperature}\:\mathrm{30}°\mathrm{C}\:\mathrm{for}\:\mathrm{1}\:\mathrm{day}.\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{expansivity}\:\mathrm{of} \\ $$$$\mathrm{iron}\:\mathrm{is}\:\:\mathrm{1}.\mathrm{2}\:×\:\mathrm{10}^{−\mathrm{5}} \mathrm{per}\:\mathrm{k}. \\ $$
Answered by ajfour last updated on 01/Jul/17
t=2π(√(L/g)) t+△t=2π(√((L(1+α△T))/g)) =2π(√(L/g))×(1+((α△T)/2)) t+△t =t(1+((α△T)/2)) ⇒ △t = t(((α△T)/2)) or ((△t)/t)=((α△T)/2) time lost = ((△t)/t)×86400 s =((α△T)/2)×86400 s =((1.2×10^(−5) ×30)/2)×86400 s =((1.2×30×0.864)/2)=18×0.864 s = 15.552 s .
$$\mathrm{t}=\mathrm{2}\pi\sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\ $$$$\mathrm{t}+\bigtriangleup\mathrm{t}=\mathrm{2}\pi\sqrt{\frac{\mathrm{L}\left(\mathrm{1}+\alpha\bigtriangleup\mathrm{T}\right)}{\mathrm{g}}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\pi\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}×\left(\mathrm{1}+\frac{\alpha\bigtriangleup\mathrm{T}}{\mathrm{2}}\right) \\ $$$$\:\:\mathrm{t}+\bigtriangleup\mathrm{t}\:=\mathrm{t}\left(\mathrm{1}+\frac{\alpha\bigtriangleup\mathrm{T}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\bigtriangleup\mathrm{t}\:=\:\mathrm{t}\left(\frac{\alpha\bigtriangleup\mathrm{T}}{\mathrm{2}}\right) \\ $$$$\mathrm{or}\:\:\:\frac{\bigtriangleup\mathrm{t}}{\mathrm{t}}=\frac{\alpha\bigtriangleup\mathrm{T}}{\mathrm{2}} \\ $$$$\mathrm{time}\:\mathrm{lost}\:=\:\frac{\bigtriangleup\mathrm{t}}{\mathrm{t}}×\mathrm{86400}\:\mathrm{s} \\ $$$$\:\:\:\:\:=\frac{\alpha\bigtriangleup\mathrm{T}}{\mathrm{2}}×\mathrm{86400}\:\mathrm{s} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}.\mathrm{2}×\mathrm{10}^{−\mathrm{5}} ×\mathrm{30}}{\mathrm{2}}×\mathrm{86400}\:\mathrm{s} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}.\mathrm{2}×\mathrm{30}×\mathrm{0}.\mathrm{864}}{\mathrm{2}}=\mathrm{18}×\mathrm{0}.\mathrm{864}\:\mathrm{s} \\ $$$$\:\:\:\:\:=\:\:\:\:\:\:\mathrm{15}.\mathrm{552}\:\mathrm{s}\:. \\ $$
Commented by tawa tawa last updated on 01/Jul/17
God bless you sir. i really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}.\: \\ $$

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