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Question

Two wires of the young's modulii Y and 2Y having lengths 2L,L and radii 2R,R respectively, are joined end to end as shown in the image. The elastic potential energy stored in the system in equilibrium, is


[Assume the wires are massless and w is the weight hung at the bottom]
  1. 4w2LπR2Y
  2. w2L2πR2Y
  3. w2L4πR2Y
  4. 2w2LπR2Y


Solution

The correct option is B w2L2πR2Y
Potential energy stored in the system is given by E=F2L2AY
where F= Force acting on wire
L= Length of wire
A= Area of cross section
Y= Young's Modulus

Total potential energy of system
E=w22L2AY+w2L2A(2Y)
[Here, w is the force acting on each wire]
E=w2×2L2(π(2R)2)Y+w2L2(πR2)×2Y
E=2w2L8πR2Y+w2L4πR2Y
E=w2L4πR2Y+w2L4πR2Y
E=2w2L4πR2Y
E=w2L2πR2Y
Hence, the correct option is (d).

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