  Question

A clock with an iron pendulum keeps correct time at $$15^{\circ}C$$. If the room temperature rises to $$20^{\circ}C$$, the error in seconds per day will be (coefficient of linear expansion for iron is $$0.000012/^{\circ}C$$):

A
2.5sec  B
2.6sec  C
2.4sec  D
2.2sec  Solution

The correct option is B $$2.6sec$$The initial time period of a pendulum is $$T_i=2\pi \sqrt{\dfrac{l_i}{g}}$$ and final time period is $$T_f=2\pi \sqrt{\dfrac{l_f}{g}}$$ If $$\alpha$$ be the coefficient of linear expansion, then $$l_f=l_i(1+\alpha\Delta \theta)$$  where $$\Delta \theta=$$ change in temperature. So, $$T_f=2\pi \sqrt{\dfrac{l_i(1+\alpha\Delta\theta)}{g}}=T_i(1+\alpha\Delta\theta)^{1/2}=T_i(1+\dfrac{\alpha\Delta\theta}{2})$$   [for small value of $$\alpha\Delta$$, we can neglect the higher order term]Or, $$\dfrac{T_f-T_i}{T_i}=\dfrac{1}{2}\alpha\Delta\theta$$Thus, the error in seconds per day $$=\dfrac{T_f-T_i}{T_i}\times 86400=\dfrac{1}{2}\alpha\Delta\theta \times 86400=0.5(0.000012)(20-15)(86400)=2.6 sec$$Physics

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