NCERT Solutions
Class 11 Physics
Thermal Properties of Matter
Q. 10.9
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 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.
Initial Temperature = 27°C
Length of the brass wire at T1, l = 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, Y = 0.91 × 1011 Pa
Young’s modulus is given by the relation:
Y = (Stress)/ (Strain)
= (F/A)/ (ΔL/L)
ΔL = (F x L)/ (A x Y)
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) = (FL)/( π(d/2)2 x Y ) 
F = α (T2 – T1) Y π (d/2)2
F= 2x10-5x (-39x-27) x3.14x0.91x1011x ((2 x 103)/ (2)) 2
=3.8 x 102 N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×102 N.