Thermal Properties of Matter
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Question : 9
Total: 22
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? Coefficient of linear expansion of brass = 2.0 × 10 – 5 K – 1 = 0.91 × 10 11 P a .
Solution:
Here l 1 = 1.8 m , t 1 = 27 ° C ,
t 2 = − 39 ° C
∴ t = t 2 − t 1
= 39 − 27 = − 66 ° C
l 2 = t 2 ° C
For brass,α = 2 × 10 − 5 K − 1
∴ Y = 0.91 × 10 11 P a
diameter of wire,d = 2.0 mm = 2.0 × 10 − 3 m
IfA A =
= π ( 10 − 3 ) 2
If F be the tension developed in the wire, then using the relation
Y =
, Δ l =
AlsoΔ l = l α Δ t ∴
= α l Δ T
orF = α Δ T A Y
= α ( T 2 − T 1 ) π r 2 Y
= 2 × 10 − 5 ( − 66 ) ×
( 10 − 3 ) 2 × 0.91 × 10 11 = − 3.8 × 10 2 N
Negative sign indicates that the force is inwards due to the contraction of the wire.
For brass,
diameter of wire,
If
If F be the tension developed in the wire, then using the relation
Also
or
Negative sign indicates that the force is inwards due to the contraction of the wire.
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