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Question

The system shown in the figure consists of three springs and two rods of the same material as shown in the figure. The springs are initially relaxed and there is no friction. The temperature of the rods is increased by ΔT. The coefficient of linear expansion of the material of the rods is α. Then, energy stored in the spring 1, when the system is in equilibrium, is:


A
75242kL2α2(ΔT)2
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B
27242kL2α2(ΔT)2
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C
81242kL2α2(ΔT)2
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D
49242kL2α2(ΔT)2
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Solution

The correct option is C 81242kL2α2(ΔT)2
Given:
Temperature change =ΔT
To find:
Energy stored in the spring 1, E=?
Let us assume:
Compression in spring 1, 2 and 3 is x1, x2 and x3 respectively due to expansion of the rods on heating.


So, total expansion of the rods =x1+x2+x3
LαΔT+(L2)αΔT=x1+x2+x3
[ from formula of linear expansion ]
32LαΔT=x1+x2+x2 ............(1)

The F.B.D of the rods is shown in the figure below.


Considering the equilibrium of rods,
we have, x1=2x2=3x3 ............(2)
[ Spring force, F=kx ]
From (1) and (2), we get
x1+x12+x13=32LαΔT
116x1=32LαΔT
x1=911LαΔT
So, energy stored in spring 1,
E=12kx21=12k[911LαΔT]2
=81242kL2α2ΔT2

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