Please
Solve This

Dear Student
V.R. of single movable pulley = No. of supporting segments of string = 2
$$\begin{gathered} M.A. of \sin gle movable pulley = \frac{L}{E} = \frac{250}{150} = \frac{5}{3} \\ \% efficency = \frac{M.A.}{V.R.} \times 100 \\ = \frac{5}{3\times 2}\times 100 = 83.33\% \end{gathered}$$
Regards
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