First slide

Thermal expansion

Question

A clock with an iron pendulum keeps correct time at $15^{0} C$ . If the coefficient of linear expansion of iron is $0.000012 /^{0} C$ and the room temperature is $20^{0} C$ it loses time in a day in seconds is

Moderate

A

1.35
B

2.6
C

3.25
D

3.85

Solution

$l o s s o r g a i n o f t i m e b y c l o c k p e r d a y i s Δ T = \frac{1}{2} α Δ t \times 86400 s e c o n d h e r e α i s c o e f f i c i e n t o f l i n e a r e x p a n s i o n = 12 x 10^{- 6} / {}^{0}C; Δ t i s c h a n g e i n t e m p e r a t u r e = 5 {}^{0}C Δ T = \frac{1}{2} \times 12 \times 10^{- 6} \times 5 \times 86400 Δ T = 3 \times 864 \times 10^{- 3} Δ T = 2592 \times 10^{- 3} = 2.592 ≃ 2.6 s e c o n d s = t i m e l o s t i n a d a y$

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