A chain of length L and of mass m is placed upon a smooth surface. The length of BA is L-b. Now, the chain is released and it slides down. Calculate the speed 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))`