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Grade: 12th pass
        
Can anybody plz solve the Q in the given attachment asap.. plz
Ans is opn b)
3 months ago

Answers : (1)

Arun
18654 Points
							

Initial temperature, T1 = 27°C
Length of the brass wire at T1l = 1.8 m
Final temperature, T2 = –39°C
Diameter of the wire, d = 2.0 mm = 2 × 10–3 m
Tension developed in the wire = F
Coefficient of linear expansion of brass, α = 2.0 × 10–5 K–1
Young’s modulus of brass, = 0.91 × 1011 Pa
Young’s modulus is given by the relation:
γ = Stress / Strain  =  (F/A) / (∆L/L)
L = F X L / (A X Y)      ……(i)

 

Where,
= 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) = FL / [ π(d/2)2X Y ]
F = α(T2 – T1)πY(d/2)2
F = 2 × 10-5 × (-39-27) × 3.14 × 0.91 × 1011 × (2 × 10-3 / 2 )2
= -3.8 × 102 N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×102 N.

3 months ago
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