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ICSE Class 10 Physics Question 13 of 19

Machines — Question 9

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Question

Question 9

A boy uses a single fixed pulley to lift a load of 50 kgf to some height. Another boy uses a single movable pulley to lift the same load to the same height. The ratio of the two effort is:

  1. 1:2
  2. 2:1
  3. 1:1
  4. none of these
Answer

2:1

Reason — We know that in case of a single fixed pulley, the effort applied is equal to the load.

Hence,

Ef=LE_f = L \\[0.5em]

In the case of a single movable pulley the effort needed to lift a load is equal to half the load.

Em=L2Em=50 kgf2⇒Em=25 kgfE_m = \dfrac{L}{2} \\[1em] E_m = \dfrac{50 \text{ kgf}}{2} \\[1em] \Rightarrow E_m = 25\text{ kgf} \\[1em]

Hence, the ratio of efforts applied by the respective pulley is

EfEm=21\dfrac{E_f}{E_m} = \dfrac{2}{1} \\[0.5em]

Therefore, Ef : Em = 2 : 1