A clock with an iron pendulum keeps correct time at 15-C- If the room temperature rises to 20-C- the error in seconds per day will be coefficient of linear expansion for iron is 0-000012-C-

The correct option is B 2.6sec The initial time period of a pendulum is T_i=2π√(l_ig) and final time period is #160; T_f=2π√(l_fg) #160;If ...

A clock with ...

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 $0.000012 /^{\circ} C$ ):

2.5 s e c

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2.6 s e c

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2.4 s e c

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2.2 s e c

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Solution

The correct option is B $2.6 s e c$
The initial time period of a pendulum is $T_{i} = 2 π \sqrt{\frac{l_{i}}{g}}$ and final time period is $T_{f} = 2 π \sqrt{\frac{l_{f}}{g}}$
If $α$ be the coefficient of linear expansion, then $l_{f} = l_{i} (1 + α Δ θ)$ where $Δ θ =$ change in temperature.
So, $T_{f} = 2 π \sqrt{\frac{l_{i} (1 + α Δ θ)}{g}} = T_{i} (1 + α Δ θ)^{1 / 2} = T_{i} (1 + \frac{α Δ θ}{2})$ [for small value of $α Δ$ , we can neglect the higher order term]
Or, $\frac{T_{f} - T_{i}}{T_{i}} = \frac{1}{2} α Δ θ$
Thus, the error in seconds per day $= \frac{T_{f} - T_{i}}{T_{i}} \times 86400 = \frac{1}{2} α Δ θ \times 86400 = 0.5 (0.000012) (20 - 15) (86400) = 2.6 s e c$

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