A uniform rope of length `'L` and linear density `'mu'` is on a smooth horizontal table with a length `'I'` lying on the table. The wrok done in pulling the hanging part on to the table is

Options

A. `(mu g(L-l)^(2))/(2)`
B. `(mu g(L-l)^(2))/(2l^(2))`
C. `(mu g(L-l)^(2))/(2L^(2))`
D. `(mu g L)/(2(L-l))`

Answer

A

Explanation

`W = m^(1)gh = mu(L-l)g"((L-l))/(2)`
Where `m^(1)` is mass of hanging part