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For a system of n pulleys, the weight of the lower block along with pulleys is w, then its efficiency is given by:

  1. η=1−wnE\eta = 1 - \dfrac{w}{nE}
  2. η=1+wnE\eta = 1 + \dfrac{w}{nE}
  3. η=wnE\eta = \dfrac{w}{nE}
  4. η=n−wE\eta = n - \dfrac{w}{E}

Machines

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Answer

η=1−wnE\eta = 1 - \dfrac{w}{nE}

Reason — For a system of n pulleys, let w be the total weight of the lower block along with the pulleys in it. In the balanced position,

E = T and L + w = nT

or L = nT - w = nE - w

M.A. = LE=nE−wE=n−wE\dfrac{L}{E} = \dfrac{nE-w}{E} = n - \dfrac{w}{E}

Thus, the mechanical advantage is less than the ideal value n.

The velocity ratio does not change, it remains n, i.e., V.R. = n

Hence, efficiency η=M.A.V.R.=n−wEn=1−wnE\eta = \dfrac{M.A.}{V.R.} = \dfrac{n - \dfrac{w}{E} }{n} = 1 - \dfrac{w}{nE}.

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