a steel scale is to be prepared such that the millimeter intervals are to be accurate with in 6×10^-5mm. The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (a =12×10^-6k^-1)
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Use formula of thermal expansion ,
∆L = L₀α∆T
where ∆L is the change in length of object after changing temperature
L₀ is the initial length of object
∆T is the temperature change
and α is the coefficient of linear expansion
Here, ∆L = 6 × 10⁻⁵ mm = 6 × 10⁻⁸ m
L₀ = 1mm = 10⁻³ m
α = 12 × 10⁻⁶ /K
Now, 6 × 10⁻⁸ = 10⁻³ × 12 × 10⁻⁶ ∆T
60 = 12∆T ⇒ ∆T = 5K or 5 /°C
∆L = L₀α∆T
where ∆L is the change in length of object after changing temperature
L₀ is the initial length of object
∆T is the temperature change
and α is the coefficient of linear expansion
Here, ∆L = 6 × 10⁻⁵ mm = 6 × 10⁻⁸ m
L₀ = 1mm = 10⁻³ m
α = 12 × 10⁻⁶ /K
Now, 6 × 10⁻⁸ = 10⁻³ × 12 × 10⁻⁶ ∆T
60 = 12∆T ⇒ ∆T = 5K or 5 /°C
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