A chain of length `L` and mass `m` is placed upon a smooth surface. The length of `BA` is `(L-b)`. Calculate the velocity of the chain when its end reaches `B`

Options

A. `sqrt((2gsintheta)/(L))(L^(2)-b^(2))`
B. `2sqrt((gsintheta)/(L))(L^(2)-b^(2))`
C. `sqrt((gsintheta)/(L))(L^(2)-b^(2))`
D. `sqrt((gsintheta)/(2L))(L^(2)-b^(2))`

Answer

C

Explanation

Loss in `PE=` gain in `KE`
`rArrmg((1)/(2)sintheta)-((m)/(L)b)g((b)/(2)sintheta)=(1)/(2)mv^(2)`
`rArrv=sqrt((gsintheta)/(L)(L^(2)-b^(2))`