Question 11. 9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of – 39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 x 10-5K-1; Young’s modulus of brass = 0.91 x 1011 Pa
Answers
stress ( F/A) = Young's modulus × strain
= Y∆L/L
= Ya∆T
Where a is coefficient of linear expansion of wire and ∆T temperature change
Length of wire = 1.8 m
Intial temperature= 27°C
final temperature = -39°C
Temperature change ( ∆T) = -39-27 = -66°C
Diameter of wire ( d) = 2mm = 2 × 10^-3 m
coefficient of linear expansion ( a) = 2 × 10^-5 /K
Young's modulus ( Y) = 0.91 × 10¹¹ N/m²
So, tension developed in wire ( F) = YAa∆T
= Y( πd²/4)a.∆T
= 0.91 × 10¹¹ × 3.14 × (2 × 10^-3)² × 2 ×10^-5 × (-66)/4
= 0.91 × 3.14 × (-66)
= -377N
Answer:
Concept :- when a wire or rod held taut between two rigid support is cooled then , stress is produced in it , called thermal stress.
Thermal stress ( F/A) = Young's modulus × strain
= Y∆L/L
= Ya∆T
Where a is coefficient of linear expansion of wire and ∆T temperature change .
Here,
Length of wire = 1.8 m
Intial temperature= 27°C
final temperature = -39°C
Temperature change ( ∆T) = -39-27 = -66°C
Diameter of wire ( d) = 2mm = 2 × 10^-3 m
coefficient of linear expansion ( a) = 2 × 10^-5 /K
Young's modulus ( Y) = 0.91 × 10¹¹ N/m²
So, tension developed in wire ( F) = YAa∆T
= Y( πd²/4)a.∆T
= 0.91 × 10¹¹ × 3.14 × (2 × 10^-3)² × 2 ×10^-5 × (-66)/4
= 0.91 × 3.14 × (-66)
= -377N
Here, negative sign shows that force is acting inwards due to contraction of the wire .