Question

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Answer 1

Answer:

Initial temperature, T1=27oC

Length of the brass wire at T1,l=1.8m

Final temperature, T2=39C 

Diameter of the wire, d=2.0mm=2×10−3m

Tension developed in the wire =F

Coefficient of linear expansion of brass,  =2.0×10−5K−1

Youngs modulus of brass, Y=0.91×1011 Pa

Youngs modulus is given by the relation:

Y= Stress / Strain 

Y=ΔL/LF/A

ΔL=F×L/(A×Y)   ......(i)

Where,

F= Tension developed in the wire 

A= Area of cross-section of the wire. 

ΔL= Change in the length, given by the relation:

ΔL= αL(T2 -T1)  .....(ii)

Equating equations (i) and (ii), we get:

αL(T2−T1)=π(d/2)2YFL

F=α(T2−T1)Yπ(d/2)2

F=2×10−5×(−39−27)×3.14×0.91×1011×(

Explanation:

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