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Question

An effort of 5 units is applied on a 10-unit load. The distance travelled by the effort and the load are 50 and 20 units, respectively. Find the efficiency of this machine.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
80%

Calculating Machine Efficiency

The efficiency of a machine represents the ratio of useful work output (work done on the load) to the total work input (work done by the effort). It is typically expressed as a percentage.

Given Data:

  • Effort Applied = 5 units
  • Load = 10 units
  • Distance travelled by effort (\(d_e\)) = 50 units
  • Distance travelled by load (\(d_l\)) = 20 units

Calculations:

1. Work Done by Effort (\(W_e\)):

The work done by the effort is calculated as the product of the effort applied and the distance it travels.

\( W_e = \text{Effort} \times d_e \)

\( W_e = 5 \text{ units} \times 50 \text{ units} = 250 \text{ units}^2 \)

2. Work Done on Load (\(W_l\)):

The work done on the load is calculated as the product of the load and the distance it moves.

\( W_l = \text{Load} \times d_l \)

\( W_l = 10 \text{ units} \times 20 \text{ units} = 200 \text{ units}^2 \)

3. Efficiency (\(\eta\)):

Efficiency is the ratio of work done on the load to the work done by the effort, multiplied by 100%.

\( \eta = \left( \frac{W_l}{W_e} \right) \times 100\% \)

\( \eta = \left( \frac{200 \text{ units}^2}{250 \text{ units}^2} \right) \times 100\% \)

\( \eta = \left( \frac{4}{5} \right) \times 100\% \)

\( \eta = 0.8 \times 100\% = 80\% \)

Conclusion:

The efficiency of the machine is 80%.

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