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Ber [7]
1 year ago
12

To measure a base lone, a steel stape 30m lang, Standardised at 15°C with a pull of 80N was used. find the correction per tape l

ength, if the tempreture at the time of measurements is 25°c and the pull exerted is 150N. Take young's modulus E= 2X10^5 N/mm2 and coefficient of thermal expansion 211.2x10^6c°Cross-sectional area of tape is 8mm2​
Geography
1 answer:
iren2701 [21]1 year ago
8 0

The correction of thermal expansion per tape length, if the temperature at the time of measurements is 25°c and the pull exerted is 150N is 4.67 ×10 ⁻³m.

Thermal expansion is the tendency of remembering to change in form, volume, and region in response to an alternate in temperature. Temperature is a monotonic characteristic of the common molecular kinetic power of a substance.

Given,

Length of steel tape = 30m

Temperature at standardization = 15°C

The temperature at the time of measurement = 25°C

Standardized pull (Po) = 80N

Pull exerted (Pm) = 150N

Young modulus of elasticity = E =  E= 2X10^5 N/mm2

Coefficient of thermal expansion = 211.2x10^6c°

Cross-sectional area of tape = 8 mm²

For temperature correction  Cm = (Tm - T₀)αL₀

Putting α = ( 25 - 15 ) × 11.2 × 10⁻⁶ × 30

            Cm = 3.36 ×  10⁻³ m

For full correction ;

         C pull = (Pm - Po ) × Lo

Putting value'

    C pull =  (150 - 80) × 30 ×  1000 /8 ×2 × 10⁵

    C pull = 1.31 × 10 ⁻³

Total correction = 3.36 × 10⁻³ m + 1.31 × 10 ⁻³

Total correction = 4.67 ×10 ⁻³

Learn more about thermal expansion here:-brainly.com/question/1166774

#SPJ9

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