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The flexible bicycle type chain of lengt...

The flexible bicycle type chain of length `(pir)/2` and mass per unit length `rho` is released from rest with `theta =0^@` In the smooth circular channel and falls through the hole in the supporting surface, Determine the velocity v of the chain as the last link leaves the slot.
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Text Solution

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The correct Answer is:
B, D

Initial PE,
`U_(i) =int_(theta=0^@)^(theta=pi//2) (rd theta)(rho)(g)(r cos theta)`
`=(rhogr^(2))[sin theta]_(0)^(pi//2) =rhogr^(2)`
Find `PE`,
`U_(f) =((pir)/2xxrho)(g)(-(pir//2)/2)=-(pi^(2)r^(2)rhog)/8`
`DeltaU =r^(2)rho g(1+pi^(1)/8)`
`DehtaU=KE`
`DeltaU=KE`
or `r^(2)gh(1+(pi^(2))/(8))=(1)/(2)((pir)/(2))(rho)v^(2)`
`rArr v=sqrt(4rg(1/(pi)+(pi)/8))`
or `v=sqrt(rg((pi)/2+4/(pi)))`.
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