A pendulum clock has an iron pendulum 1m (coeficient linear expansion of iron =10^-5/°c).If the temperature rises by 10°c the clock ?
ashishsingh189:
complete ques. how many seconds the clock will gain or lose per day ?
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Hey dear,
You didn't complete the question. But I assume you wanna find out time error per unit time.
● Answer -
∆T/T = 5×10^-5 s
● Explaination -
# Given -
α = 10^-5 /°C
∆θ = 10 °C
# Solution -
Time period of pendulum is given by -
T = 2π √(l/g)
When temprature increases -
T' = 2π √(l'/g)
T' = 2π √[l(1+α∆θ)/g]
T' = T (1 + α∆θ/2)
Change in time period on heating,
∆T = T' - T
∆T/T = α∆θ/2
∆T/T = 10^-5 × 10 / 2
∆T/T = 5×10^-5 s
Therefore, clock will gain 5×10^-5 s each second.
Hope this helps you...
You didn't complete the question. But I assume you wanna find out time error per unit time.
● Answer -
∆T/T = 5×10^-5 s
● Explaination -
# Given -
α = 10^-5 /°C
∆θ = 10 °C
# Solution -
Time period of pendulum is given by -
T = 2π √(l/g)
When temprature increases -
T' = 2π √(l'/g)
T' = 2π √[l(1+α∆θ)/g]
T' = T (1 + α∆θ/2)
Change in time period on heating,
∆T = T' - T
∆T/T = α∆θ/2
∆T/T = 10^-5 × 10 / 2
∆T/T = 5×10^-5 s
Therefore, clock will gain 5×10^-5 s each second.
Hope this helps you...
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