Questions

# A clock with a metallic pendulum is 5 seconds fast each day at a temperature of  ${15}^{\circ }\text{c}$ and 10 seconds slow each day at a temperature of  ${30}^{\circ }\text{c}$, find the coefficient of linear expansion of the metal?

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a
2.31×10−5 ∘c-1
b
2.03×10−5 ∘c-1
c
1.03×10−6 C -1  0
d
1.03×10−5 C -1  0
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detailed solution

Correct option is A

Graduation temperature= Tochange in time due to coefficient of linear expansion of material of pendulum is∆time = 12α(temperature difference)x86400 here α is coefficient of linear expansion of metal of pendulum5=12.α.To-15×86400----(1)10=12.α.30-To×86400----(2)on dividing eqn (1) with eqn(2) 12=T0-1530-T0 ⇒T0=200C substitute T0 vallue in eqn(1) 5=12α(20-15)86400⇒α=2.31 x 10-5 C -1  0=coefficient of linear expansion of metal

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