Two wires of the same material (Young's modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy stored in the system is
A
3w2L4πR2Y
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B
3w2L8πR2Y
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C
5w2L8πR2Y
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D
w2LπR2Y
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Solution
The correct option is C5w2L8πR2Y As we know, using spring-rod analogy, k=YAL where, k= Spring constant Y= Young's modulus So, k1=Yπ(2R)2L,k2=Yπ(R)2L Equivalent spring constant =1k1+1k2=L4YπR2+LYπR2=5L4YπR2
Since block is in equilibrium, hence k1x1=k2x2=w [for springs in series. force acting is the same] Elastic potential energy of the system U=12k1x21+12k2x22 U=12k1(wk1)2+12k2(wk2)2 U=12w2{1k1+1k2}=12w2(5L4YπR2) ∴U=5w2L8πYR2